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Algebra Mark Steinberger The University at Albany State University of New York August 31, 2006 Preface The intent of this book is to introduce the student to algebra from a point of view that stresses examples and classiﬁcation. Thus, whenever possible, the main theorems are treated as tools that may be used to construct and analyze speciﬁc types of groups, rings, ﬁelds, modules, etc. Sample constructions and classiﬁcations are given in both text and exercises. It is also important to note that many beginning graduate students have not taken a deep, senior-level undergraduate algebra course. For this reason, I have not assumed a great deal of sophistication on the part of the reader in the introductory portions of the book. Indeed, it is hoped that the reader may acquire a sophistication in algebraic techniques by reading the presentations and working the exercises. In this spirit, I have attempted to avoid any semblance of proofs by intimidation. The intent is that the exercises should provide suﬃcient opportunity for the student to rise to mathematical challenges. The ﬁrst chapter gives a summary of the basic set theory we shall use throughout the text. Other prerequisites for the main body of the text are trigonometry and the diﬀerential calculus of one variable. We also presume that the student has seen matrix multiplication at least once, though the requisite deﬁnitions are provided. Chapter 2 introduces groups and homomorphisms, and gives some examples that will be used as illustrations throughout the material on group theory. Chapter 3 develops symmetric groups and the theory of G-sets, giving some useful counting arguments for classifying low order groups. We emphasize actions of one group on another via automorphisms, with conjugation as the initial example. Chapter 4 studies consequences of normality, beginning with the Noether Isomorphism Theorems. We study simple groups and composition series, and then classify all ﬁnite abelian groups via the Fundamental Theorem of Finite Abelian Groups. The automorphism group of a cyclic group is calculated. Then, semidirect products are introduced, using the calculations of automorphism groups to construct numerous examples that will be essential in classifying low order groups. We then study extensions of groups, developing some useful classiﬁcation tools. In Chapter 5, we develop some additional tools, such as the Sylow Theorems, and apply the theory we’ve developed to the classiﬁcation of groups of many of the orders ≤ 63, in both text and exercises. The methods developed are suﬃcient to classify the rest of the orders in that range. We also study solvable and nilpotent groups. Chapter 6 is an introduction to basic category theoretic notions, with examples drawn from the earlier material. The concepts developed here are useful in understanding rings and modules, though they are used sparingly in the text on that material. Pushouts of groups are constructed and the technique of generators and relations is given. Chapter 7 provides a general introduction to the theory of rings and modules. Exam p , the cyclotomic ples are introduced, including the quaternions, H, the p-adic integers, Z ix Preface x integers and rational numbers, Z[ζn ] and Q[ζn ], polynomials, group rings, and more. Free modules and chain conditions are studied, and the elementary theory of vector spaces and matrices is developed. The chapter closes with the study of rings and modules of fractions, which we shall apply to the study of P.I.D.s and ﬁelds. Chapter 8 develops the theory of P.I.D.s and covers applications to ﬁeld theory. The exercises treat prime factorization in some particular Euclidean number rings. The basic theory of algebraic and transcendental ﬁeld extensions is given, including many of the basic tools to be used in Galois theory. The cyclotomic polynomials over Q are calculated, and the chapter closes with a presentation of the Fundamental Theorem of Finitely Generated Modules over a P.I.D. Chapter 9 gives some of the basic tools of ring and module theory: Nakayama’s Lemma, primary decomposition, tensor products, extension of rings, projectives, and the exactness properties of tensor and hom. In Section 9.3, Hilbert’s Nullstellensatz is proven, using many of the tools so far developed, and is used to deﬁne algebraic varieties. In Section 9.5, extension of rings is used to extend some of the standard results about injections and surjections between ﬁnite dimensional vector spaces to the study of maps between ﬁnitely generated free modules over more general rings. Also, algebraic K-theory is introduced with the study of K0 . Chapter 10 gives a formal development of linear algebra, culminating in the classiﬁcation of matrices via canonical forms. This is followed by some more general material about general linear groups, followed by an introduction to K1 . Chapter 11 is devoted to Galois theory. The usual applications are given, e.g. the Primitive Element Theorem, the Fundamental Theorem of Algebra, and the classiﬁcation of ﬁnite ﬁelds. Sample calculations of Galois groups are given in text and exercises, particularly for splitting ﬁelds of polynomials of the form X n − a. The insolvability of polynomials of degree ≥ 5 is treated. Chapter 12 gives the theory of hereditary and semisimple rings with an emphasis on Dedekind domains and group algebras. Stable and unstable classiﬁcations of projective modules are given for Dedekind domains and semisimple rings. The major dependencies between chapters are as follows, where A → B means that B depends signiﬁcantly on material in A. 1−4 / 7 5 ? 8 ?? ?? ?? ?? ? 6 11 / 9 / 10 / 12 Additionally there are some minor dependencies. For instance, Chapter 9 assumes an understanding of the basic deﬁnitions of categories and functors, as given in the ﬁrst two sections in Chapter 6. Some categorical notions would also be useful in Chapter 7, but are optional there. Also, the last section in Chapter 5 assumes an understanding of some of the material on linear algebra from Chapter 10. Some material on linear algebra is also used in the closing section of Chapter 11. And optional references are made in Chapter 10 to the material on exterior algebras from Chapter 9. Other than the references to linear algebra in Chapter 5, the chapters may be read in order. Preface xi A solid box symbol ( ) is used to indicate either the end of a proof or that the proof is left to the reader. Those exercises that are mentioned explicitly in the body of the text are labeled with a dagger symbol (†) in the margin. This does not necessarily indicate that the reference is essential to the main ﬂow of discussion. Similarly, a double dagger symbol (‡) indicates that the exercise is explicitly mentioned in another exercise. Of course, many more exercises will be used implicitly in other exercises. Acknowledgments Let me ﬁrst acknowledge a debt to Dock Rim, who ﬁrst introduced me to much of this material. His love for the subject was obvious and catching. I also owe thanks to Hara Charalambous, SUNY at Albany, and David Webb, Dartmouth College, who have used preliminary versions of the book in the classroom and made some valuable suggestions. I am also indebted to Bill Hammond, SUNY at Albany, for his insights. I’ve used this material in the classroom myself, and would like to thank the students for many useful suggestions regarding the style of presentation and for their assistance in catching typos. Mark Steinberger Ballston Lake, NY Contents 1 A Little Set Theory 1.1 Properties of Functions . . . . . . . 1.2 Factorizations of Functions . . . . . 1.3 Relations . . . . . . . . . . . . . . . 1.4 Equivalence Relations . . . . . . . . 1.5 Generating an Equivalence Relation 1.6 Cartesian Products . . . . . . . . . . 1.7 Formalities about Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 3 6 7 10 11 13 2 Groups: Basic Deﬁnitions and Examples 2.1 Groups and Monoids . . . . . . . . . . . . 2.2 Subgroups . . . . . . . . . . . . . . . . . . 2.3 The Subgroups of the Integers . . . . . . . 2.4 Finite Cyclic Groups: Modular Arithmetic 2.5 Homomorphisms and Isomorphisms . . . . 2.6 The Classiﬁcation Problem . . . . . . . . 2.7 The Group of Rotations of the Plane . . . 2.8 The Dihedral Groups . . . . . . . . . . . . 2.9 Quaternions . . . . . . . . . . . . . . . . . 2.10 Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 15 19 24 27 29 35 37 38 40 42 3 G-sets and Counting 3.1 Symmetric Groups: Cayley’s Theorem . 3.2 Cosets and Index: Lagrange’s Theorem 3.3 G-sets and Orbits . . . . . . . . . . . . . 3.4 Supports of Permutations . . . . . . . . 3.5 Cycle Structure . . . . . . . . . . . . . . 3.6 Conjugation and Other Automorphisms 3.7 Conjugating Subgroups: Normality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 47 51 55 64 66 71 77 4 Normality and Factor Groups 4.1 The Noether Isomorphism Theorems . . . . 4.2 Simple Groups . . . . . . . . . . . . . . . . 4.3 The Jordan–Hölder Theorem . . . . . . . . 4.4 Abelian Groups: the Fundamental Theorem 4.5 The Automorphisms of a Cyclic Group . . . 4.6 Semidirect Products . . . . . . . . . . . . . 4.7 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 . 83 . 90 . 96 . 98 . 105 . 111 . 119 xii . . . . . . . . . . . . . . . . . . . . . CONTENTS xiii 5 Sylow Theory, Solvability, and Classiﬁcation 5.1 Cauchy’s Theorem . . . . . . . . . . . . . . . 5.2 p-Groups . . . . . . . . . . . . . . . . . . . . 5.3 Sylow Subgroups . . . . . . . . . . . . . . . . 5.4 Commutator Subgroups and Abelianization . 5.5 Solvable Groups . . . . . . . . . . . . . . . . 5.6 Hall’s Theorem . . . . . . . . . . . . . . . . . 5.7 Nilpotent Groups . . . . . . . . . . . . . . . . 5.8 Matrix Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 136 137 141 149 150 153 156 158 6 Categories in Group Theory 6.1 Categories . . . . . . . . . . . . . . . . . 6.2 Functors . . . . . . . . . . . . . . . . . . 6.3 Universal Mapping Properties: Products 6.4 Pushouts and Pullbacks . . . . . . . . . 6.5 Inﬁnite Products and Coproducts . . . . 6.6 Free Functors . . . . . . . . . . . . . . . 6.7 Generators and Relations . . . . . . . . 6.8 Direct and Inverse Limits . . . . . . . . 6.9 Natural Transformations and Adjoints . 6.10 General Limits and Colimits . . . . . . . . . . . . . . . . . . . . . . . . . . . and Coproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 164 167 171 176 183 186 189 191 195 198 7 Rings and Modules 7.1 Rings . . . . . . . . . . . . . . 7.2 Ideals . . . . . . . . . . . . . . 7.3 Polynomials . . . . . . . . . . . 7.4 Symmetry of Polynomials . . . 7.5 Group Rings and Monoid Rings 7.6 Ideals in Commutative Rings . 7.7 Modules . . . . . . . . . . . . . 7.8 Chain Conditions . . . . . . . . 7.9 Vector Spaces . . . . . . . . . . 7.10 Matrices and Transformations . 7.11 Rings of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 202 215 222 234 239 245 250 268 273 277 284 8 P.I.D.s and Field Extensions 8.1 Euclidean Rings, P.I.D.s, and U.F.D.s 8.2 Algebraic Extensions . . . . . . . . . . 8.3 Transcendence Degree . . . . . . . . . 8.4 Algebraic Closures . . . . . . . . . . . 8.5 Criteria for Irreducibility . . . . . . . . 8.6 The Frobenius . . . . . . . . . . . . . 8.7 Repeated Roots . . . . . . . . . . . . . 8.8 Cyclotomic Polynomials . . . . . . . . 8.9 Modules over P.I.D.s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 296 308 314 317 320 324 325 327 332 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS xiv 9 Radicals, Tensor Products, and Exactness 9.1 Radicals . . . . . . . . . . . . . . . . . . . . . 9.2 Primary Decomposition . . . . . . . . . . . . 9.3 The Nullstellensatz and the Prime Spectrum 9.4 Tensor Products . . . . . . . . . . . . . . . . 9.5 Tensor Products and Exactness . . . . . . . . 9.6 Tensor Products of Algebras . . . . . . . . . . 9.7 The Hom Functors . . . . . . . . . . . . . . . 9.8 Projective Modules . . . . . . . . . . . . . . . 9.9 The Grothendieck Construction: K0 . . . . . 9.10 Tensor Algebras and Their Relatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 342 346 351 365 377 383 386 392 398 406 10 Linear algebra 10.1 Traces . . . . . . . . . . . . . 10.2 Multilinear alternating forms 10.3 Properties of determinants . . 10.4 The characteristic polynomial 10.5 Eigenvalues and eigenvectors 10.6 The classiﬁcation of matrices 10.7 Jordan canonical form . . . . 10.8 Generators for matrix groups 10.9 K1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 416 418 424 429 432 434 440 443 447 11 Galois Theory 11.1 Embeddings of Fields . . . . 11.2 Normal Extensions . . . . . . 11.3 Finite Fields . . . . . . . . . 11.4 Separable Extensions . . . . . 11.5 Galois Theory . . . . . . . . . 11.6 The Fundamental Theorem of 11.7 Cyclotomic Extensions . . . . 11.8 n-th Roots . . . . . . . . . . 11.9 Cyclic Extensions . . . . . . . 11.10Kummer Theory . . . . . . . 11.11Solvable Extensions . . . . . . 11.12The General Equation . . . . 11.13Normal Bases . . . . . . . . . 11.14Norms and Traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450 451 454 458 460 465 474 476 480 485 488 493 498 500 502 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506 507 512 521 526 529 531 539 . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Hereditary and Semisimple Rings 12.1 Maschke’s Theorem and Projectives 12.2 Semisimple Rings . . . . . . . . . . . 12.3 Jacobson Semisimplicity . . . . . . . 12.4 Homological Dimension . . . . . . . 12.5 Hereditary Rings . . . . . . . . . . . 12.6 Dedekind Domains . . . . . . . . . . 12.7 Integral Dependence . . . . . . . . . Bibliography . . . . . . . 550 Chapter 1 A Little Set Theory Here, we discuss the basic properties of functions, relations and cartesian products. The ﬁrst section discusses injective and surjective functions and their behavior under composition. This is followed in Section 1.2 by a discussion of commutative diagrams and factorization. Here, duality is introduced, and injective and surjective maps are characterized in terms of their properties with respect to factorization. In the latter case, the role of the Axiom of Choice is discussed. Some introductory remarks about the axiom are given, to be continued in Chapter 6, where inﬁnite products are introduced. In the next three sections, we discuss relations, including equivalence relations and partial orderings. Particular attention is given to equivalence relations. For an equivalence relation ∼, the set of equivalence classes, X/∼, is studied, along with the canonical map π : X → X/∼. In Section 1.5, we analyze the process by which an arbitrary relation may be used to generate an equivalence relation. We shall make use of this in constructing free products of groups. We then describe the basic properties of cartesian products, showing that they satisfy the universal mapping property for the categorical product. We close with some applications to the theory of functions and to the basic structure of the category of sets. With the exception of the discussion of the Axiom of Choice, we shall not delve into axiomatics here. We shall assume a familiarity with basic naive set theory, and shall dig deeper only occasionally, to elucidate particular points. 1.1 Properties of Functions We use the notation f : X → Y to denote that f is a function from X to Y , and say that X is its domain, and Y its codomain. We shall assume an intuitive understanding of functions at this point in the discussion, deferring a more formal discussion to Section 1.7. We shall use the word “map” as a synonym for function. Deﬁnitions 1.1.1. Let X be a set. The identity map 1X : X → X is the function deﬁned by 1X (x) = x for all x ∈ X. If Y is a subset of X, then the inclusion map i : Y → X is obtained by setting i(y) = y for all y ∈ Y . We give names for some important properties of functions: Deﬁnitions 1.1.2. A function f : X → Y is one-to-one, or injective, if f (x) = f (x ) for x = x . An injective function is called an injection. 1 CHAPTER 1. A LITTLE SET THEORY 2 A function f : X → Y is onto, or surjective, if for each y ∈ Y , there is at least one x ∈ X with f (x) = y. A surjective function is called a surjection.1 A function is bijective if it is both one-to-one and onto. Such a function may also be called a one-to-one correspondence, or a bijection. Composition of functions is useful in studying the relationships between sets. The reader should verify the following properties of composition. Lemma 1.1.3. Suppose given functions f : X → Y , g : Y → Z, and h : Z → W . Then 1. (h ◦ g) ◦ f = h ◦ (g ◦ f ). 2. f ◦ 1X = f . 3. 1Y ◦ f = f . Here are some more interesting facts to verify regarding composition. Lemma 1.1.4. Suppose given functions f : X → Y and g : Y → Z. Then the following relationships hold. 1. If f and g are injective, so is g ◦ f . 2. If f and g are surjective, so is g ◦ f . 3. If g ◦ f is injective, so is f . 4. If g ◦ f is surjective, so is g. Two sets which are in one-to-one correspondence may be thought of as being identical as far as set theory is concerned. One way of seeing this is in terms of inverse functions. Deﬁnition 1.1.5. Let f : X → Y . We say that a function g : Y → X is an inverse function for f if the composite f ◦ g is the identity map of Y and the composite g ◦ f is the identity map of X. Proposition 1.1.6. A function f : X → Y has an inverse if and only if it is bijective. If f does have an inverse, the inverse is unique, and is deﬁned by setting f −1 (y) to be the unique element x ∈ X such that f (x) = y. Proof If f is bijective, then for each y ∈ Y , there is a unique x ∈ X such that f (x) = y. Thus, there is a function f −1 : Y → X deﬁned by setting f −1 (y) to be equal to this unique x. The reader may now easily verify that f −1 is an inverse function for f . Conversely, suppose that f has an inverse function g. Note that identity maps are always bijections. Thus, f ◦g = 1Y is surjective, and hence f is surjective by Lemma 1.1.4. Also, g ◦ f = 1X is injective, so that f is injective as well. Finally, since f (g(y)) = y for all y, we see that g(y) is indeed the unique x ∈ X such that f (x) = y. Deﬁnition 1.1.7. Let f : X → Y be a function. Then the image of f , written im f , is the following subset of Y : im f = {y ∈ Y | y = f (x) for some x ∈ X}. 1 “One-to-one” and “onto” are the terms native to the development of mathematical language in English, while “injection” and “surjection” come from French mathematics. CHAPTER 1. A LITTLE SET THEORY 3 Thus, the image of f is the set that classically would be called the range of f . Since the values f (x) of the function f : X → Y all lie in im f , f deﬁnes a function, which we shall call by the same name, f : X → im f . Notice that by the deﬁnition of the image, f : X → im f is surjective. Notice also that f factors as the composite f X / im f ⊂ Y, where the second map is the inclusion of im f in Y . Since inclusion maps are always injections, the next lemma is immediate. Lemma 1.1.8. Every function may be written as a composite g ◦ f , where g is injective and f is surjective. Exercises 1.1.9. 1. Give an example of a pair of functions f : X → Y and g : Y → X, where X and Y are ﬁnite sets, such that g ◦ f = 1X , but f is not surjective, and g is not injective. 2. Show that if X and Y are ﬁnite sets with the same number of elements, then any injection f : X → Y is a bijection. Show also that any surjection f : X → Y is a bijection. 3. Give an example of a function f : X → X which is injective but not bijective. 4. Give an example of a function f : X → X which is surjective but not bijective. 1.2 Factorizations of Functions Commutative diagrams are a useful way to summarize the relationships between diﬀerent sets and functions. Deﬁnition 1.2.1. Suppose given a diagram of sets and functions. [email protected] @@ @@ f @@ h Y / Z. > } }} } }g }} We say that the diagram commutes if h = g ◦ f . Similarly, given a diagram X f g g Z / Y f / W we say the diagram commutes if g ◦ f = f ◦ g. More generally, given any diagram of maps, we say the diagram commutes if, when we follow two diﬀerent paths of arrows which start and end at the same points (following the directions of the arrows, of course) and compose the maps along these paths, then the compositions from the two paths agree. Commutative diagrams are frequently used to model the factorization of functions. CHAPTER 1. A LITTLE SET THEORY 4 Deﬁnitions 1.2.2. Let f : X → Y be a function. We say that a function g : X → Z factors through f if there is a function g : Y → Z such that the following diagram commutes. [email protected] @@ @@ f @@ g Y /Z ? ~ ~ ~ ~~ ~~ g We say that g factors uniquely through f if there is exactly one function g : Y → Z that makes the diagram commute. The language of factorization can be confusing, as there is a totally diﬀerent meaning to the words “factors through,” which is also used frequently: Deﬁnitions 1.2.3. Let f : X → Y be a function. We say that a function g : Z → Y factors through f if there is a function g : Z → X such that the following diagram commutes. X? ??f ? ?? / Y Z g g We say that g factors uniquely through f if there is exactly one function g : Z → X that makes the diagram commute. Notice that despite the fact that there are two possible meanings to the statement that a function factors through f : X → Y , the meaning in any given context is almost always unique. Thus, if we say that a map g : X → Z factors through f : X → Y , then the only possible interpretation is that g = h ◦ f for some h : Y → Z. Similarly, if we say that a map g : Z → Y factors through f : X → Y , then the only possible interpretation is that g = f ◦ h for some h : Z → X. Indeed, the only way that ambiguity can enter is if we ask whether a function g : X → X factors through another function f : X → X. In this last case, one must be precise as to which type of factorization is being discussed. Factorization properties may be used to characterize injective and surjective maps. Proposition 1.2.4. Let f : X → Y . Then the following two statements are equivalent. 1. The map f is injective. 2. Any function g : Z → Y with the property that im g ⊂ im f factors uniquely through f. Proof Suppose that f is injective. Let g : Z → Y , and suppose that im g ⊂ im f . Then for each z ∈ Z there is an x ∈ X such that g(z) = f (x). Since f is injective, this x is unique, and hence there is a function g : Z → X deﬁned by setting g (z) equal to the unique x ∈ X such that g(z) = f (x). But then f (g (z)) = g(z), and hence f ◦ g = g, so that g factors through f . The factorization is unique, since if f ◦ h = g, we have f (h(z)) = g(z), and hence f (h(z)) = f (g (z)) for all z ∈ Z. Since f is injective, this says that h(z) = g (z) for all z ∈ Z, and hence h = g . CHAPTER 1. A LITTLE SET THEORY 5 Conversely, suppose that any function g : Z → Y with im g ⊂ im f factors uniquely through f . Let Z be the set with one element: Z = {z}. Then if f (x) = f (x ), deﬁne g : Z → Y by g(z) = f (x). Then if we deﬁne h, h : Z → X by h(z) = x and h (z) = x , respectively, we have f ◦ h = f ◦ h = g. By uniqueness of factorization, h = h , and hence x = x . Thus, f is injective. The uniqueness of the factorization is the key in the above result. There is an entirely diﬀerent characterization of injective functions in terms of factorizations, where the factorization is not unique. Proposition 1.2.5. Let f : X → Y with X = ∅. Then the following two statements are equivalent. 1. The map f is injective. 2. Any function g : X → Z factors through f . Proof Suppose that f is injective and that g : X → Z. Choose any z ∈ Z,2 and deﬁne g : Y → Z by g(x) if y = f (x) g (y) = z if y ∈ im f . Since f is injective, g is a well deﬁned function. By construction, g ◦ f = g, and hence g provides the desired factorization of g. (Note that if Z has more than one element, then g is not unique. For instance, we could choose a diﬀerent element z ∈ Z.) Conversely, suppose that every map g : X → Z factors through f . We apply this by setting g equal to the identity map of X. Thus, there is a function g : Y → X such that g ◦ f = 1X . But then f is injective by Lemma 1.1.4. There is a notion in mathematics called dualization. The dual of a statement is the statement obtained by turning around all of the arrows in it. It is often interesting to see what happens when you dualize a statement, but one should bear in mind that the dual of a true statement is not always true. The dual of Proposition 1.2.4 is true, but its proof is best deferred to our discussion of equivalence relations below. The dual of Proposition 1.2.5 is also true. In fact, it turns out to be equivalent to the famous Axiom of Choice. Axiom of Choice (ﬁrst form) Let f : X → Y be a surjective function. Then there is a function s : Y → X such that f ◦ s = 1Y . In this form, the Axiom of Choice seems to be obviously true. In order to deﬁne s, all we have to do is choose, for each y ∈ Y , an element x ∈ X with f (x) = y. Since such an x is known to exist for each y ∈ Y (because f is surjective), intuition says we should be able to make these choices. Sometimes the desired function s : Y → X may be constructed by techniques that depend only on the other axioms of set theory. One example of this is when f is a bijection. Here, s is the inverse function of f , and is constructed in the preceding section. Alternatively, if Y is ﬁnite, we may construct s by induction on the number of elements. Other examples exist where we may construct s by a particular formula. 2 Since X = ∅ and g : X → Z, we must have Z = ∅ as well. (See Proposition 1.7.2.) CHAPTER 1. A LITTLE SET THEORY 6 Nevertheless, intuition aside, for a generic surjection f : X → Y , we cannot construct the desired function s via the other axioms, and, if we wish to be able to construct s in all cases, we must simply accept the Axiom of Choice as part of the foundations of mathematics. Most mathematicians are happy to do this without question, but some people prefer models of set theory in which the Axiom of Choice does not hold. Thus, while we shall use it freely, we shall give some discussion of the ways in which it enters our arguments.3 We now give the dual of Proposition 1.2.5. Proposition 1.2.6. Let f : X → Y . Then the following two statements are equivalent. 1. The map f is surjective. 2. Any function g : Z → Y factors through f . Proof Suppose that f : X → Y is surjective. Then the Axiom of Choice provides a function s : Y → X with f ◦ s = 1Y . But then if g : Z → Y is any map, s ◦ g : Z → X gives the desired factorization of g through f . Conversely, suppose that any function g : Z → Y factors through f . Applying this for g = 1Y , we obtain a function s : Y → X with f ◦ s = 1Y . But then f is surjective by Lemma 1.1.4. 1.3 Relations We shall make extensive use of both equivalence relations and order relations in our work. We ﬁrst consider the general notion of a relation on a set X. Deﬁnitions 1.3.1. A relation R on a set X is a subset R ⊂ X × X of the cartesian product of X with itself. We shall use x R y as a shorthand for the statement that the pair (x, y) lies in the relation R. We shall often discuss a relation entirely in terms of the expressions x R y, and dispense with discussion of the associated subset of the product. This is an example of what’s known as “abuse of notation”: we can think and talk about a relation as a property x R y which holds for certain pairs (x, y) of elements of X. We can then, if we choose, deﬁne an associated subset R ⊂ X × X by letting R equal the set of all pairs such that x R y. For instance, there is a relation on the real numbers R, called ≤, which is deﬁned in the usual way: we say that x ≤ y if y − x is a non-negative number. We can discuss all the properties of the relation ≤ without once referring to the subset of R × R which it deﬁnes. Not all relations on a set X have a whole lot of value to us. We shall enumerate some properties which could hold in a relation that might make it interesting for certain purposes. 3 The desired functions s : Y → X often exist for reasons that do not require the assumption of the choice axiom. In the study of the particular examples of groups, rings, ﬁelds, topological spaces, etc., that mathematicians tend to be most interested in, the theorems that we know and love tend to be true without assuming the axiom. However, the proofs of these theorems are often less appealing without the choice axiom. Moreover, if we assume the axiom, we can prove very general theorems, which then may be shown to hold without the axiom, for diﬀerent reasons in diﬀerent special cases, in the cases of greatest interest. Thus, the Axiom of Choice permits us to look at mathematics from a broader and more conceptual viewpoint than we could without it. And indeed, in the opinion of the author, a world without the Axiom of Choice is a poorer one, with less functions and narrower horizons. CHAPTER 1. A LITTLE SET THEORY 7 Deﬁnitions 1.3.2. Let R be a relation on a set X. If x R x for each x ∈ X, then we say that R is reﬂexive. Let R be a relation on a set X. Suppose that y R x if and only if x R y. Then we say that R is symmetric. Alternatively, we say that R is antisymmetric if x R y and y R x implies that x = y. Let R be a relation on a set X. Then we say that R is transitive if x R y and y R z together imply that x R z. These deﬁnitions can be used to deﬁne some useful kinds of relations. Deﬁnitions 1.3.3. A relation R on X is an equivalence relation if it is reﬂexive, symmetric, and transitive. A relation R on X is a partial ordering if it is reﬂexive, antisymmetric, and transitive. A set X together with a partial ordering R on X is called a partially ordered set. A relation R on X is a total ordering if it is a partial ordering with the additional property that for all x, y ∈ X, either x R y or y R x. A set X together with a total ordering R on X is called a totally ordered set. Examples 1.3.4. 1. The usual ordering ≤ on R is easily seen to be a total ordering. 2. There is a partial ordering, ≤, on R × R obtained by setting (x, y) ≤ (z, w) if both x ≤ z and y ≤ w. Note that this example diﬀers from the last one in that it is not a total ordering. 3. There is a diﬀerent partial ordering, called the lexicographic ordering, on R × R obtained by setting (x, y) ≤ (z, w) if either x < z, or both x ≤ z and y ≤ w. (Thus, both (1, 5) and (2, 3) are ≤ (2, 4).) Note that in this case, we do get a total ordering. 4. There is an equivalence relation, ≡, on the integers, Z, obtained by setting m ≡ n if m − n is even. 5. For any set X, there is an equivalence relation, ∼, on X deﬁned by setting x ∼ y for all x, y ∈ X (i.e., the subset of X × X in question is all of X × X). 6. For any set X, there is exactly one relation which is both a partial ordering and an equivalence relation: here x R y if and only if x = y. It is customary to use symbols such as ≤ and for a partial ordering, and to use symbols such as ∼, , and ≡ for an equivalence relation. 1.4 Equivalence Relations We now look at equivalence relations, as deﬁned in the preceding section, in greater depth. Deﬁnition 1.4.1. Let ∼ be an equivalence relation on X and let x ∈ X. The equivalence class of x in X (under the relation ∼), denoted E(x), is deﬁned by E(x) = {y ∈ X | x ∼ y}. In words, E(x) is the set of all y ∈ X which are equivalent to x under the relation ∼. CHAPTER 1. A LITTLE SET THEORY 8 Lemma 1.4.2. Let ∼ be an equivalence relation on X and let x, y ∈ X. Then y is in the equivalence class E(x) of x if and only if E(x) = E(y). Proof Suppose y ∈ E(x). This just says x ∼ y. But if z ∈ E(y), then y ∼ z, and hence x ∼ z by the transitivity of ∼. Therefore, E(y) ⊂ E(x). But equivalence relations are also symmetric. Thus, x ∼ y if and only if y ∼ x, and hence y ∈ E(x) if and only if x ∈ E(y). In particular, the preceding argument then shows that x ∼ y also implies that E(x) ⊂ E(y), and hence E(x) = E(y). Conversely, if E(x) = E(y), then reﬂexivity shows that y ∈ E(y) = E(x). In consequence, we see that equivalence classes partition the set X into distinct subsets: Corollary 1.4.3. Let ∼ be an equivalence relation on X and let x, y ∈ X. Then the equivalence classes of x and y have a common element if and only if they are equal. In symbols, E(x) ∩ E(y) = ∅ if and only if E(x) = E(y). Proof If z ∈ E(x) ∩ E(y), then both E(x) and E(y) are equal to E(z) by Lemma 1.4.2. The global picture is now as follows. Corollary 1.4.4. Let ∼ be an equivalence relation on X. Then every element of X belongs to exactly one equivalence class under ∼. Proof By Corollary 1.4.3, every element belongs to at most one equivalence class. But reﬂexivity shows that x ∈ E(x) for each x ∈ X, and the result follows. Given a set with an equivalence relation, the preceding observation allows us to deﬁne a new set from the equivalence classes. Deﬁnitions 1.4.5. Let ∼ be an equivalence relation on a set X. By the set of equivalence classes under ∼, written X/∼, we mean the set whose elements are the equivalence classes under ∼ in X. In particular, each element of X/∼ is a subset of X. We deﬁne π : X → X/∼ by π(x) = E(x). Thus, π takes each element of X to its equivalence class. We call π the canonical map from X to X/∼. Example 1.4.6. Let X = {1, 2, 3, 4} and let ∼ be the equivalence relation on X in which 1 ∼ 3, 2 ∼ 4, and the other related pairs are given by reﬂexivity and symmetry. Then there are two elements in X/∼: {1, 3} and {2, 4}. We have π(1) = π(3) = {1, 3} and π(2) = π(4) = {2, 4}. We shall show that the canonical map π characterizes the equivalence relation completely. Lemma 1.4.7. Let ∼ be an equivalence relation on X and let π : X → X/∼ be the canonical map. Then π is surjective, and satisﬁes the property that π(x) = π(y) if and only if x ∼ y. Proof Surjectivity of π follows from the fact that every equivalence class has the form E(x) for some x ∈ X, and E(x) = π(x). Now π(x) = π(y) if and only if E(x) = E(y). By Lemma 1.4.2, the latter condition is equivalent to the statement that y ∈ E(x), which is deﬁned to mean that x ∼ y. CHAPTER 1. A LITTLE SET THEORY 9 Intuitively, this says that X/∼ is the set obtained from X by identifying two elements if they lie in the same equivalence class. The canonical map has an important universal property.4 Proposition 1.4.8. Let ∼ be an equivalence relation on X. Then a function g : X → Y factors through the canonical map π : X → X/∼ if and only if x ∼ x implies that g(x) = g(x ) for all x, x ∈ X. Moreover, if g factors through π, then the factorization is unique. Proof Suppose that g = g ◦ π for some g : X/∼ → Y . If x ∼ x ∈ X, then π(x) = π(x ), and hence g (π(x)) = g (π(x )). Since g = g ◦ π, this gives g(x) = g(x ), as desired. Also, since π is surjective, every element of X/∼ has the form π(x) for some x ∈ X. Since g (π(x)) = g(x), the eﬀect of g on the elements of X/∼ is determined by g. In other words, the factorization g , if it exists, is unique. Conversely, suppose given a function g : X → Y with the property that x ∼ x implies that g(x) = g(x ). Now deﬁne g : X/∼ → Y by setting g (π(x)) = g(x) for all x ∈ X. Since π(x) = π(x ) if and only if x ∼ x , this is well deﬁned by the property assumed for the function g. By the construction of g , g ◦ π = g, and hence g gives the desired factorization of g through π. We have seen that an equivalence relation determines a surjective function: the canonical map. We shall give a converse to this, and show that specifying an equivalence relation on X is essentially equivalent to specifying a surjective function out of X. Deﬁnition 1.4.9. Let f : X → Y be any function. By the equivalence relation determined by f , we mean the relation ∼ on X deﬁned by setting x ∼ x if f (x) = f (x ). The reader may easily verify that the equivalence relation determined by f is indeed an equivalence relation. We now show that if ∼ is the equivalence relation determined by f : X → Y , then there is an important relationship between the canonical map π : X → X/∼ and f . Proposition 1.4.10. Let f : X → Y be any function and let ∼ be the equivalence relation determined by f . Then f factors uniquely through the canonical map π : X → X/∼, via a commutative diagram f / Y. XD DD z= z DD zz D zzf π DD z z ! X/∼ Moreover, the map f : X/∼ → Y gives a bijection from X/∼ to the image of f . In particular, if f is surjective, then f gives a bijection from X/∼ to Y . Proof Recall that the deﬁnition of ∼ was that x ∼ x if and only if f (x) = f (x ). Thus, the universal property of the canonical map (i.e., Proposition 1.4.8) provides the unique factorization of f through π: just set f (π(x)) = f (x) for all x ∈ X. Since every element of X/∼ has the form π(x), the image of f is the same as the image of f . Thus, it suﬃces to show that f is injective. Suppose that f (π(x)) = f (π(x )). Then f (x) = f (x ), so by the deﬁnition of ∼, x ∼ x . But then π(x) = π(x ), and hence f is injective as claimed. 4 The expression “universal property” has a technical meaning which we shall explore in greater depth in Chapter 6 on category theory. CHAPTER 1. A LITTLE SET THEORY 10 In particular, if f : X → Y is surjective, and if ∼ is the equivalence relation determined by f , then we may identify Y with X/∼ and identify f with π. In particular, the dual of Proposition 1.2.4 is now almost immediate from the universal property of the canonical map π. Proposition 1.4.11. Let f : X → Y . Then the following two statements are equivalent. 1. The map f is surjective. 2. Let g : X → Z be such that f (x) = f (x ) implies that g(x) = g(x ) for all x, x ∈ X. Then g factors uniquely through f . Proof Suppose that f is surjective, and let ∼ be the equivalence relation determined by f . Then we may identify f with the canonical map π : X → X/∼. But under this identiﬁcation, the second statement is precisely the universal property of π. Conversely, suppose that the second condition holds. Then the hypothesis says that there is a unique map g : Y → Y such that the following diagram commutes. [email protected] @@ @@ f @@ f Y /Y ~? ~ ~ ~~ ~~ g Of course, g = 1Y will do, but the key is the uniqueness: if f were not surjective, we could alter g on those points of Y which are not in im f without changing the fact that the diagram commutes, contradicting the uniqueness of the factorization. Exercises 1.4.12. † 1. Let n be a positive integer. Deﬁne a relation ≡ on the integers, Z, by setting i ≡ j if j − i is a multiple of n. Show that ≡ is an equivalence relation. (We call it congruence modulo n.) 2. Let ∼ be any equivalence relation. Show that the equivalence relation determined by the canonical map π : X → X/∼ is precisely ∼. 1.5 Generating an Equivalence Relation There are many situations in mathematics where we are given a relation R which is not an equivalence relation, and asked to form the “smallest” equivalence relation, ∼, with the property that x R y implies that x ∼ y. Here, “smallest” means that if ≡ is another equivalence relation with the property that x R y implies that x ≡ y, then x ∼ y also implies that x ≡ y. In other words, the equivalence classes of ∼ are subsets of the equivalence classes of ≡. This smallest equivalence relation is called the equivalence relation generated by R, and its canonical map π : X → X/∼ will be seen to have a useful universal property with respect to R. Of course, it is not immediately obvious that such an equivalence relation exists, so we shall have to construct it. Deﬁnition 1.5.1. Let R be any relation on X. The equivalence relation, ∼, generated by R is the relation obtained by setting x ∼ x if there is a sequence x1 , . . . , xn ∈ X such that x1 = x, xn = x , and for each i with 1 ≤ i < n, either xi = xi+1 , or xi R xi+1 , or xi+1 R xi . CHAPTER 1. A LITTLE SET THEORY 11 The intuition is that we take the relation R to be a collection of “basic equivalences,” and declare that x ∼ x if they may be connected by a chain of basic equivalences, some in one direction and some in the other. Example 1.5.2. Let n be a positive integer, and deﬁne a relation R on the integers, Z, by setting i R j if j = i + n. Note then that if ∼ is the equivalence relation generated by R and if i ∼ j, then j = i + kn for some k ∈ Z. But the converse is also clear: in fact, i ∼ j if and only if j = i + kn for k ∈ Z. Thus, ∼ is the equivalence relation known as “congruence modulo n,” which was given in Problem 1 of Exercises 1.4.12. We now verify that the equivalence relation generated by R has the desired properties. Proposition 1.5.3. Let R be any relation and let ∼ be the equivalence relation generated by R. Then ∼ is, in fact, an equivalence relation, and has the property that x R x implies that x ∼ x . Moreover, if ≡ is another equivalence relation with the property that x R x implies that x ≡ x , then x ∼ x also implies that x ≡ x . Proof If x R x , then the sequence x1 = x, x2 = x shows that x ∼ x , so that x R x implies that x ∼ x , as desired. Similarly, we see that ∼ is reﬂexive. Symmetry of ∼ may be obtained by reversing the sequence x1 , . . . , xn that’s used to show that x ∼ x , while transitivity is obtained by concatenating two such sequences. Now let ≡ be another equivalence relation with the property that x R x implies that x ≡ x , and suppose that x ∼ x . Thus, we are given a sequence x1 , . . . , xn ∈ X such that x1 = x, x2 = x , and for each i with 1 ≤ i < n, either xi = xi+1 , or xi R xi+1 , or xi+1 R xi . But in any of the three situations, this says that xi ≡ xi+1 , so that x = x1 ≡ x2 ≡ · · · ≡ xn = x . Since ≡ is transitive, this gives x ≡ x , as desired. The fact that ∼ is the smallest equivalence relation containing R translates almost immediately into a statement about the canonical map π : X → X/∼. Corollary 1.5.4. Let R be a relation on X and let ∼ be the equivalence relation generated by R. Then a map f : X → Y factors through the canonical map π : X → X/∼ if and only if x R x implies that f (x) = f (x ) for all x, x ∈ X. The factorization, if it exists, is unique. Proof Suppose that f : X → Y has the property that x R x implies that f (x) = f (x ) for all x, x ∈ X. Let be the equivalence relation induced by f (i.e., x x if and only if f (x) = f (x )). Then x R x implies that x x . Thus, by Proposition 1.5.3, x ∼ x implies that x x . By the deﬁnition of , this just says that x ∼ x implies that f (x) = f (x ). But now the universal property of the canonical map π : X → X/∼ (Proposition 1.4.8) shows that f factors uniquely through π. For the converse, if f factors through π, then Proposition 1.4.8 shows that x ∼ x implies that f (x) = f (x ). But x R x implies that x ∼ x , so the result follows. 1.6 Cartesian Products The next deﬁnition should be familiar. Deﬁnition 1.6.1. Let X and Y be sets. The cartesian product X × Y is the set of all ordered pairs (x, y) with x ∈ X and y ∈ Y . CHAPTER 1. A LITTLE SET THEORY 12 When describing a function f : X × Y → Z, it is cumbersome to use multiple parentheses, as in f ((x, y)). Instead, we shall write f (x, y) for the image of the ordered pair (x, y) under f . Deﬁnition 1.6.2. There are projection maps π1 : X × Y → X and π2 : X × Y → Y deﬁned by π1 (x, y) = x, and π2 (x, y) = y, respectively. An understanding of products with the null set is useful for the theory of functions. Lemma 1.6.3. Let X be any set and let ∅ be the null set. Then X × ∅ = ∅ × X = ∅. Proof If one of the factors of a product is empty, then there are no ordered pairs: one cannot choose an element from the empty factor. We shall make extensive use of products in studying groups and modules. The next proposition is useful for studying ﬁnite groups. Here, if X is ﬁnite, we write |X| for the number of elements in X. Proposition 1.6.4. Let X and Y be ﬁnite sets. Then X × Y is ﬁnite, and |X × Y | = |X| · |Y |, i.e., the number of elements in X × Y is the product of the number of elements in X and the number of elements in Y . Proof If either X or Y is empty, the result follows from Lemma 1.6.3. Thus, we assume that both X and Y are nonempty. Consider the projection map π1 : X ×Y → X. For each x0 ∈ X, write π1−1 (x0 ) for the set of ordered pairs that are mapped onto x0 by π1 . Thus, π1−1 (x0 ) = {(x0 , y) | y ∈ Y }.5 For each x ∈ X, there is an obvious bijection between π1−1 (x) and Y , so that each π1−1 (x) has |Y | elements. Now π1−1 (x) ∩ π1−1 (x ) = ∅ for x = x , and X × Y = −1 x∈X π1 (x). Thus, X × Y is the union of |X| disjoint subsets, each of which has |Y | elements, and hence |X × Y | = |X| · |Y | as claimed. Products have an important universal property. Let f : Z → X × Y be a map, and write f (z) = (f1 (z), f2 (z)) for z ∈ Z. In the usual parlance, f1 : Z → X and f2 : Z → Y are the component functions of f . Note that if π1 : X × Y → X and π2 : X × Y → Y are the projection maps, then fi = πi ◦ f , for i = 1, 2. Since a pair (x, y) ∈ X × Y is determined by its coordinates, we see that a function f : Z → X×Y is determined by its component functions. In other words, if g : Z → X×Y is a map, and if g1 : Z → X and g2 : Z → Y are its component functions, then f = g if and only if f1 = g1 and f2 = g2 . Finally, if we are given a pair of functions f1 : Z → X and f2 : Z → Y , then we can deﬁne a function f : Z → X × Y by setting f (z) = (f1 (z), f2 (z)) for z ∈ Z. In other words, we have shown the following. Proposition 1.6.5. For each pair of functions f1 : Z → X and f2 : Z → Y , there is a unique function f : Z → X × Y such that fi = πi ◦ f for i = 1, 2. Here, π1 : X × Y → X and π2 : X × Y → Y are the projection maps. that the subsets π1−1 (x), as x ranges over the elements of X, are precisely the equivalence classes for the equivalence relation determined by π1 . 5 Note CHAPTER 1. A LITTLE SET THEORY 1.7 13 Formalities about Functions Intuitively, a function f : X → Y is a correspondence which assigns to each element x ∈ X an element f (x) ∈ Y . The formal deﬁnition of function is generally given in terms of its graph. Deﬁnitions 1.7.1. A subset f ⊂ X × Y is the graph of a function from X to Y if for each x ∈ X there is exactly one ordered pair in f whose ﬁrst coordinate is x. If f ⊂ X × Y is the graph of a function and if (x, y) ∈ f , we write y = f (x). Thus, f = {(x, f (x)) | x ∈ X}. By the function determined by f we mean the correspondence f : X → Y that assigns to each x ∈ X the element f (x) ∈ Y . Two functions from X to Y are equal if and only if their graphs are equal. Thus, every function is determined uniquely by its graph. We obtain some immediate consequences. Proposition 1.7.2. Let Y be any set. Then there is one and only one function from the null set, ∅, to Y . On the other hand, if X is a set with X = ∅, then there are no functions from X to ∅. Proof For a set Y , we have ∅ × Y = ∅. Thus, ∅ × Y has exactly one subset, and hence there is only one candidate for a function. The subset in question is the graph of a function, because, since the null set has no elements, the condition that must be satisﬁed is automatically true. On the other hand, if X = ∅, then the unique subset of X × ∅ is not a function: for any given x ∈ X, there is no ordered pair in this subset whose ﬁrst coordinate is x. The next proposition is important in the foundations of category theory. Proposition 1.7.3. Let X and Y be sets. Then the collection of all functions from X to Y forms a set. Proof The collection of all functions from X to Y is in one-to-one correspondence with the collection of their graphs. But the latter is a subcollection of the collection of all subsets of X × Y . The fact that the collection of all subsets of a set, W , is a set (called the power set of W ) is one of the standard axioms of set theory. An important special case of this is when X is ﬁnite. For simplicity of discussion, let us establish a notation. Notation 1.7.4. Let X and Y be sets. We write F (X, Y ) for the set of functions from X to Y . Proposition 1.7.5. Let X1 and X2 be subsets of X such that X = X1 ∪ X2 and X1 ∩ X2 = ∅. Write ι1 : X1 ⊂ X and ι2 : X2 ⊂ X for the inclusions of these two subsets in X. Let Y be any set, and deﬁne α : F (X, Y ) → F (X1 , Y ) × F (X2 , Y ) by α(f ) = (f ◦ ι1 , f ◦ ι2 ). Then α is a bijection.6 6 In the language of category theory, this says that the disjoint union of sets is the coproduct in the category of sets. CHAPTER 1. A LITTLE SET THEORY 14 Proof Suppose that α(f ) = α(g). Then f ◦ ι1 = g ◦ ι1 , and hence f and g have the same eﬀect on the elements of X1 . Similarly, f ◦ ι2 = g ◦ ι2 , and hence f and g have the same eﬀect on the elements of X2 . Since a function is determined by its eﬀect on the elements of its domain, f = g, and hence α is injective. Now suppose given an element (f1 , f2 ) ∈ F (X1 , Y ) × F (X2 , Y ). Thus, f1 : X1 → Y and f2 : X2 → Y are functions. Deﬁne f : X → Y by f1 (x) if x ∈ X1 f (x) = f2 (x) if x ∈ X2 . Then f ◦ ιi = fi for i = 1, 2, and hence α(f ) = (f1 , f2 ). Thus, α is surjective. As a result, we can count the number of distinct functions between two ﬁnite sets. Corollary 1.7.6. Let X and Y be ﬁnite. Then the set of functions from X to Y is ﬁnite. Speciﬁcally, if X has n elements and Y has |Y | elements, then the set F (X, Y ) of n functions from X to Y has |Y | elements. (Here, we take 00 to be equal to 1.) Proof If Y = ∅, the result follows from Proposition 1.7.2. Thus, we assume that Y = ∅, and argue by induction on n. If n = 0, then the result is given by Proposition 1.7.2. For n = 1, say X = {x}, there is a one-to-one correspondence ν : F (X, Y ) → Y given by ν(f ) = f (x). Thus, we assume n > 1. Write X = X ∪ {x}, with x ∈ X . Then we have a bijection α : F (X, Y ) → n−1 elements, while F ({x}, Y ) has F (X , Y ) × F ({x}, Y ). By induction, F (X , Y ) has |Y | |Y | elements by the case n = 1, above. Thus, the result follows from Proposition 1.6.4. Chapter 2 Groups: Basic Deﬁnitions and Examples In the ﬁrst four sections, we deﬁne groups, monoids, and subgroups, and develop the most basic properties of the integers and the cyclic groups Zn . We then discuss homomorphisms and classiﬁcation. The homomorphisms out of cyclic groups are determined, and the abstract cyclic groups classiﬁed. Classiﬁcation is emphasized strongly in this volume, and we formalize the notion as soon as possible. We then discuss isomorphism classes and show that there are only ﬁnitely many isomorphism classes of groups of any given ﬁnite order. We then give some examples of groups, in particular the dihedral and quaternionic groups, which we shall use throughout the book as illustrations of various concepts. We close with a section on cartesian products, giving a way to create new groups out of old ones. 2.1 Groups and Monoids A binary operation on a set X is a function μ : X × X → X. That is, μ assigns to each ordered pair (x, y) of elements of X an element μ(x, y) ∈ X. We think of these as multiplications, and generally write x · y, or just xy (or sometimes x + y), instead of μ(x, y). Note that the order of x and y is important. Generally x · y and y · x are diﬀerent elements of X. Binary operations without additional properties, while they do arise in various situations, are too general to study in any depth. The condition we shall ﬁnd most valuable is the associative property: our binary operation is associative if (x · y) · z = x · (y · z) for every x, y, and z in X. Another important property is the existence of an identity element. An identity element for the operation is an element e ∈ X such that e·x=x=x·e for every x in X. The next result is basic. 15 CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 16 Lemma 2.1.1. A binary operation can have at most one identity element. Proof Suppose that e and e are both identity elements for the binary operation ·. Since e is an identity element, we have e · e = e . But the fact that e is also an identity element shows that e · e = e. Thus, e = e . Deﬁnition 2.1.2. A monoid is a set with an associative binary operation that has an identity element. Examples 2.1.3. 1. Let Z+ be the set of positive integers: Z+ = {1, 2, . . . }. Then ordinary multiplication gives Z+ the structure of a monoid, with identity element 1. 2. Let N ⊂ Z be the set of non-negative integers: N = {0, 1, 2, . . . }. Then ordinary addition gives N the structure of a monoid, with identity element 0. Monoids occur frequently in mathematics. They are suﬃciently general that their structure theory is extremely diﬃcult. Deﬁnition 2.1.4. Let X be a monoid. We say that an element x ∈ X is invertible if there is an element y ∈ X with x · y = e = y · x, where e is the identity. We call y the inverse for x and write y = x−1 . The next result shows that the above deﬁnition uniquely deﬁnes y. Lemma 2.1.5. Let X be a monoid. Then any element x ∈ X which is invertible has a unique inverse. Proof Suppose that y and z are both inverses for x. Then x · y = e = y · x, and x · z = e = z · x. Thus, y = y · e = y · (x · z) = (y · x) · z = e · z = z. Note that the statement that x · y = e = y · x, used in deﬁning inverses, is symmetric in x and y, establishing the following lemma. Lemma 2.1.6. Let X be a monoid, and let x ∈ X be invertible, with inverse y. Then y is also invertible, and y −1 = x. Finally, we come to the object of our study. Deﬁnition 2.1.7. A group is a monoid in which every element is invertible. Notice that the monoid, Z+ , of positive integers under multiplication is not a group, as, in fact, no element other than 1 is invertible: if n ∈ Z+ is not equal to 1, then n > 1. Thus, for any m ∈ Z+ , n · m > 1 · m, and hence n · m > 1. The monoid, N, of the non-negative integers under addition also fails to be a group, as the sum of two positive integers is always positive. We shall generally write G for a group, rather than X. We do have numerous examples of groups which come from numbers. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 17 Examples 2.1.8. The simplest example of a group is the group with one element, e. There is only one binary operation possible: e · e = e. We call this group the trivial group and denote the group itself by e. Let Z be the integers: Z = {0, ±1, ±2, . . . }. Then addition of numbers gives Z the structure of a group: the identity element is 0, and the inverse of n ∈ Z is −n. Similarly, the rational numbers, Q, and the real numbers, R, form groups under addition. We also know examples of groups obtained from multiplication of numbers. Write Q× and R× for the nonzero numbers in Q and R, respectively. Since the product of two nonzero real numbers is nonzero, multiplication provides Q× and R× with an associative operation with identity element 1. And if x ∈ R× , the reciprocal, 1/x, provides an inverse to x in R× . Moreover, if x = m/n ∈ Q× , with m, n ∈ Z, then 1/x = n/m ∈ Q× , and hence x has an inverse in Q× . × Now write Q× + and R+ for the positive rational numbers and the positive real numbers, respectively. Since the product of two positive numbers is positive and the reciprocal × of a positive number is positive, the above arguments show that Q× + and R+ are groups under multiplication. The examples above are more than just groups. They also satisfy the commutative law : x · y = y · x for all x and y in G. In fact, the ubiquity of the commutative law in our background can exert a pernicious inﬂuence on our understanding of groups, as most groups don’t satisfy it. The ones that do merit a special name: Deﬁnitions 2.1.9. A group or monoid is abelian if it satisﬁes the commutative law. A group or monoid in which the commutative law is known to fail is said to be nonabelian. We will often use + for the operation in an abelian group. In fact, we shall restrict the use of + to these groups. Notice that the groups in the examples above all have inﬁnitely many elements. Our main focus here will be on ﬁnite groups: those with ﬁnitely many elements. Deﬁnition 2.1.10. If G is a ﬁnite group, then the order of G, written |G|, is the number of elements in G. If G is inﬁnite, we say |G| = ∞. Let X be a monoid. Write Inv(X) ⊂ X for the collection of invertible elements in X. Lemma 2.1.11. Inv(X) is a group under the multiplication inherited from that of X. Proof Let x and y be elements of Inv(X). Thus, x has an inverse, x−1 , and y has an inverse y −1 . Now the element y −1 x−1 may easily be seen to be an inverse for xy, and hence xy ∈ Inv(X), also. Since the operation on Inv(X) is inherited from that on X, it must be associative. Since e−1 = e, e is an element of Inv(X), and acts as an identity element there. And for x ∈ Inv(X), x−1 is invertible, with inverse x, and hence x−1 ∈ Inv(X), and is the inverse of x in Inv(X). In particular, every element of Inv(X) is invertible, so that Inv(X) is a group. A primary source for nonabelian groups is from matrices. We assume familiarity with the basic properties of matrix multiplication shown in a beginning course in linear algebra. We shall state the properties we are to use, and leave them as exercises to the reader. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 18 Deﬁnitions 2.1.12. We write Mn (R) for the set of n×n matrices over the real numbers, R. We shall make use of the binary operation of matrix multiplication, which is deﬁned as follows. If ⎛ ⎞ ⎛ ⎞ a11 . . . a1n b11 . . . b1n ⎜ ⎟ ⎜ ⎟ .. .. A=⎝ and B=⎝ ⎠ ⎠, . . an1 ... ann bn1 ... bnn then the product AB is the matrix whose ij-th coordinate is cij = n aik bkj = ai1 b1j + · · · + ain bnj . k=1 We write I = In for the n × n identity matrix: the matrix whose diagonal entries are all equal to 1 and whose oﬀ-diagonal entries are all equal to 0. 1 0 0 Thus, I2 = 10 01 , I3 = 0 1 0 , etc. 001 The properties we need here regarding matrix multiplication are few. We shall discuss matrices in greater detail in Section 7.10 and Chapter 10. Lemma 2.1.13. Multiplication of n × n matrices gives an associative binary operation on Mn (R). Moreover, In is an identity element for this operation. Thus, Mn (R) is a monoid under matrix multiplication. The invertible elements in this monoid structure on Mn (R) are precisely the invertible matrices in the usual sense. As such, they merit a name. Deﬁnition 2.1.14. We write Gln (R) for Inv(Mn (R)), the group of invertible elements of Mn (R). We call it the n-th general linear group of R. Later in this chapter, we shall construct two inﬁnite families of ﬁnite nonabelian groups, the dihedral groups and the quaternionic groups, as explicit subgroups of Gln (R) for n = 2 and 4, respectively. We shall show in Chapter 10 that every ﬁnite group is a subgroup of Gln (R) for some value of n. It is sometimes useful to study partial inverses in a monoid. Deﬁnitions 2.1.15. Let X be a monoid with identity element e. If x, y ∈ X with xy = e, then we say that x is a left inverse for y and that y is a right inverse for x. Exercises 2.1.16. 1. Let M be the set of all nonzero integers. Then M is a monoid under multiplication. What is Inv(M )? † 2. Let G be a group and let x, y ∈ G. Show that x and y commute if and only if x2 y 2 = (xy)2 . † 3. Let G be a group such that x2 = e for all x ∈ G. Show that G is abelian. −1 4. Let G be a group and let x1 , . . . , xk ∈ G. Show that (x1 . . . xk )−1 = x−1 k . . . x1 . 5. Let G be a group and let x, y ∈ G. Show that x and y commute if and only if (xy)−1 = x−1 y −1 . 6. Let G be a group and let x, y ∈ G. Suppose there are three consecutive integers n such that xn y n = (xy)n . Show that x and y commute. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 19 7. Let G be an abelian group and let x, y ∈ G. Show that xn y n = (xy)n for all n ∈ Z. 8. Verify that multiplication of n × n matrices is associative. 9. Show that Gl1 (R) is isomorphic to R× , the group of non-zero real numbers under multiplication. 10. Show that Gln (R) is nonabelian for n ≥ 2. 11. Let G= a b 0 c b ∈ R, a, c ∈ Q and ac = 0 . Show that G is a group under matrix multiplication. Is G abelian? 12. Let X be a monoid. Suppose that x ∈ X has a left inverse, y, and a right inverse, z. Show that y = z, and that x is invertible with inverse y. † 13. Let X be a monoid with the property that every element of X has a left inverse. Show that X is a group. 14. Let X be a ﬁnite monoid with the left cancellation property: if xy = xz, then y = z. Show that X is a group. 15. Let X be a ﬁnite set with an associative binary operation. Suppose this operation has both the left and the right cancellation properties. Show that X is a group. 2.2 Subgroups One can tell quite a bit about about a group by knowing its subgroups. Deﬁnitions 2.2.1. A subset S of a group G is said to be closed under multiplication if for x and y in S, the product xy is also in S. A subset H of G is said to be a subgroup if it is nonempty and closed under multiplication, and if for each x ∈ H, the inverse element x−1 is also in H. Examples 2.2.2. 1. The groups Z and Q are subgroups of R. 2. The inclusions below are all inclusions of subgroups. Q× + ∩ R× + ⊂ Q× ∩ ⊂ R× 3. Z+ ⊂ R× + is closed under multiplication, but is not a subgroup, because the inverses in R× + of the non-identity elements of Z+ do not lie in Z+ . 4. Any group G is a subgroup of itself. 5. For any group G, consider the subset {e} ⊂ G, consisting of the identity element alone. Because e · e = e and e−1 = e, {e} is a subgroup of G, called the trivial subgroup, or identity subgroup. By abuse of notation (i.e., for convenience), we shall generally write e in place of {e} for this subgroup. (When the identity element is called 1 or 0, we shall write 1 or 0 for the trivial subgroup as well.) CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 20 The next lemma is immediate from the deﬁnitions. Lemma 2.2.3. Let H ⊂ G be a subgroup. Then H is a group under the operation inherited from that of G. We also have what is sometimes called Gertrude Stein’s Theorem.1 We leave the proof as an exercise to the reader. Lemma 2.2.4. A subgroup of a subgroup is a subgroup. (I.e., if H is a subgroup of G and K is a subgroup of H, then K is a subgroup of G.) Identifying the subgroups of a given group is a deep and interesting activity, and the result says a lot about the group. As we shall see presently, the easiest subgroups to ﬁnd are the smallest ones. Deﬁnition 2.2.5. Let G be a group and let g ∈ G. For any integer n, we deﬁne the n-th power of g, g n , as follows. For n = 0, 1, g n is given by e and g, respectively. For n > 1, we deﬁne g n by induction on n: g n = g n−1 · g. For negative exponents, we require that g −1 be the inverse of g, and if n is positive, then g −n = (g −1 )n . If G is abelian and written in additive notation, we write n · g, or ng for the n-th power of G in the above sense. The power laws for a group are fundamental: Lemma 2.2.6. Let G be a group and let g ∈ G. Then for any integers m and n, gm · gn m n (g ) = g m+n , = g mn and . If the group operation in G is written additively, this translates to the following: mg + ng = (m + n)g, n(mg) = (nm)g. and Proof We use the multiplicative notation. Consider the ﬁrst statement ﬁrst. For n = 0, it reduces to g m · e = g m which is certainly true. For n = 1, the statement depends on the value of m: if m > 0 it just gives the deﬁnition of g m+1 , and hence is true. For m = 0, it says e · g = g. For m < 0, we have m = −k, say, and g m = (g −1 )k = (g −1 )k−1 · g −1 . Thus, g m · g = (g −1 )k−1 (g −1 · g) = (g −1 )k−1 = g −(k−1) = g m+1 , because −k = m. For n > 1, we argue by induction on n. We have g n = g n−1 · g, so that g m · g n = m (g · g n−1 ) · g. By induction, this is g m+n−1 · g, which is g m+n by the case n = 1 above. The case of n < 0 follows from that of n > 0 by exchanging the roles of g and g −1 . We also use induction for the second statement. As with the ﬁrst, we may assume n ≥ 0. For n = 0, 1 it says nothing more than e = e, or g m = g m , respectively. For n > 1, we have (g m )n = (g m )n−1 · g m . By induction, this is g mn−m · g m , which is g mn by an application of the ﬁrst statement. 1 This usage was coined by Arunas Liulevicius. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 21 Notice that there’s a potential conﬂict between the additive notation for powers and the notation for ordinary multiplication if G = Z. This is resolved by the following lemma. Lemma 2.2.7. Let m and n be integers. Then mn (as deﬁned by ordinary multiplication of integers) is the m-th power of n in the group Z. Proof First note the deﬁnition of multiplication in Z: if m and n are both positive, then mn is the sum of m copies of n and (−1)k m · (−1)l n = (−1)k+l mn. The sum of m copies of n is by deﬁnition the m-th power of n. The remaining cases follow from Lemma 2.2.6. Lemma 2.2.6 gives us a convenient source of subgroups. Deﬁnition 2.2.8. Let G be a group and let g ∈ G. Deﬁne g ⊂ G to be the set of all integer powers of g: g = {g n | n ∈ Z}. Our next result will show that g is a subgroup of G. We call it the cyclic subgroup generated by g. Note that if the group operation in G is written additively, then g is deﬁned by g = {ng | n ∈ Z}. In particular, in the case G = Z, for n ∈ Z, n = {nk | k ∈ Z} is the set of all integer multiples of n. In this case, there are a couple of alternate notations that come out of ring theory: for n ∈ Z we may write either (n) or nZ for n. Lemma 2.2.9. Let G be a group and let g ∈ G. Then g is a subgroup of G. Moreover, g has the property that it is the smallest subgroup of G which contains g. In other words, if H ⊂ G is a subgroup, and if g ∈ H, then g ⊂ H. Proof Since g m · g n = g m+n , g is closed under multiplication in G. Also, the inverse of g m is g −m ∈ g, so that g is a subgroup of G. If g ∈ H ⊂ G, with H a subgroup of G, then g n must lie in H for all n ∈ Z by the closure properties of a subgroup. Thus, g ⊂ H. Indeed, viewing g as an element of H, g is the subgroup of H generated by g. We shall see in the next section that the cyclic subgroups n ⊂ Z give all of the subgroups of Z. Cyclic subgroups have a very special kind of group structure. They are cyclic groups: Deﬁnition 2.2.10. A group G is cyclic if there is an element g ∈ G such that G = g. Cyclic groups are simplest kind of group there is. For this reason, we can answer some very complicated questions about them, questions we could not answer for more complicated groups. Nevertheless, there are many deep and interesting questions about cyclic groups (some of them connected with algebraic number theory) which have yet to be answered. We can also talk about the subgroup generated by a family of elements. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 22 Deﬁnitions 2.2.11. Let G be a group. For any subset S ⊂ G, the subgroup, S, generated by S is the set of all products (of arbitrary length) of powers of elements of S: S = {g1n1 . . . gknk | k ≥ 1, gi ∈ S and ni ∈ Z for 1 ≤ i ≤ k}. Warning: In the products g1n1 . . . gknk , we do not, and cannot, assume that the g1 , . . . , gk are distinct. Indeed, if g, h ∈ S do not commute, then, for instance, g 2 h2 and ghgh are distinct elements of S, by Problem 2 of Exercises 2.1.16. If S is ﬁnite, say S = {x1 , . . . , xn }, then by abuse of notation, we shall write x1 , . . . , xn for S. If S = G, we say that G is generated by S. If G is generated by a ﬁnite set, we say that G is ﬁnitely generated. The following lemma is straightforward. Lemma 2.2.12. S is the smallest subgroup of G which contains S. We now give a method for detecting ﬁnite subgroups. Proposition 2.2.13. Let G be a group and let H be a nonempty ﬁnite subset of G which is closed under multiplication. Then H is a subgroup of G. Proof We must show H is closed under inverses. Let x ∈ H and let S = {xk | 1 ≤ k ∈ Z}. Then S is contained in H, and hence S must be ﬁnite. Thus, the elements xk with k ≥ 1 cannot all be distinct. We must have xk = xl for some k = l. Say l > k. But then multiplying both sides by x−k , we get e = xl−k = xl−k−1 · x. Thus, xl−k−1 = x−1 . But l − k − 1 ≥ 0 since l > k. If l − k − 1 = 0, we have x = e = x−1 ∈ H. Otherwise, xl−k−1 ∈ S ⊂ H, and hence H is closed under inverses. We can give one useful operation on subgroups at this point. The proof is left to the reader. Lemma 2.2.14. Let H and K be subgroups of G. Then H ∩ K is also a subgroup of G. Warning: The union of two subgroups is almost never a subgroup. The collection of all subgroups of a group G is easily seen to form a partially ordered set as deﬁned in Deﬁnition 1.3.3. Here the order relation is given by inclusion as subsets of G; explicitly, H is less than or equal to K if H ⊂ K. Deﬁnition 2.2.15. Let G be a group. The partially ordered set of subgroups of G is known as the lattice of subgroups of G. It is often useful to diagram the lattice of subgroups of a group G. Right now we have neither enough examples of groups nor enough theory about subgroups to be able to give a good example and verify its accuracy. However, we shall give one anyway, for a group we shall encounter in this chapter. The reader should reconsider the following example as the theory develops. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 23 Example 2.2.16. We write D6 for the dihedral group of order 6. As will become clear later, the following is the lattice of subgroups of D6 . D6 ?? oo o o ?? oo o ?? o o o ? oo o o o b O a ab ab2 OOO ? ? OOO OOO ??? OOO ?? OOO? e Here, b has order 3, while a, ab, and ab2 have order 2. The upward-slanted lines represent the inclusions of subgroups. Because the elements of monoids don’t necessarily have inverses, the deﬁnition of a submonoid will have to be diﬀerent from that of a subgroup. Deﬁnition 2.2.17. A submonoid of a monoid M is a subset which is closed under multiplication and contains the identity element. Exercises 2.2.18. 1. Show that in the real numbers R, the cyclic subgroup generated by 1 is the integers. In particular, Z is cyclic. 2. In Z, show that n = Z if and only if n = ±1. 3. Consider the group, Q× + , of positive rational numbers under multiplication. What are the elements of 2 ⊂ Q× +? 4. Consider the group Q× of nonzero rational numbers under multiplication. What are the elements of −1 ⊂ Q× ? What are the elements of −2? † 5. Let M be a monoid. We can deﬁne the positive powers of the elements in M in exactly the same way that positive powers in a group are deﬁned. We have m1 = m for all m ∈ M , and the higher powers are deﬁned by induction: mk = mk−1 m. Show that for m ∈ M and for i, j ≥ 1, we have (a) mi · mj = mi+j , and (b) (mi )j = mij . 6. In Z, show that 2, 3 = Z. 7. In Z, show that 3n, 5n = n for any n ∈ Z. 8. In the group, Q× + , of positive rational numbers under multiplication, show that 2, 3 is not a cyclic subgroup. In other words, there is no rational number q such that 2, 3 = q. 9. Show that Q× + is generated by the set of all prime numbers. 10. Show that the group Q× of nonzero rational numbers is generated by the set consisting of −1 and all of the prime numbers. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 24 11. Let G be a group and let a, b ∈ G such that aba−1 ∈ b. Show that H = {ai bj | i, j ∈ Z} is a subgroup of G. Deduce that H = a, b. 12. In Z, show that 2 ∩ 3 = 6. 13. In Z, show that m ∩ n = k, where k is the least common multiple of m and n. 14. In the group Q× + of positive rational numbers under multiplication, show that 2 ∩ 3 = 1. Here, 1 is the trivial subgroup of Q× +. 15. Let M be a monoid. Show that the group of invertible elements Inv(M ) is a submonoid of M . 16. Show that not every submonoid of a group is a group. 17. Show that every submonoid of a ﬁnite group is a group. 2.3 The Subgroups of the Integers One of the simplest yet most powerful results in mathematics is the Euclidean Algorithm. We shall use it here to identify all subgroups of Z and to derive the properties of prime decomposition in Z. Theorem 2.3.1. (The Euclidean Algorithm2 ) Let m and n be integers, with n > 0. Then there are integers q and r, with 0 ≤ r < n, such that m = qn + r. Proof First, we assume that m ≥ 0, and argue by induction on m. If m < n, we may take q = 0 and r = m. If m = n, we take q = 1 and r = 0. Thus, we may assume that m > n and that the result holds for all non-negative integers less than m. In particular, the induction hypothesis gives m − 1 = q n + r , for integers q and r with 0 ≤ r < n. If r < n − 1, we may take q = q and r = r + 1 for the desired result. Otherwise, m = (q + 1)n, and the proof for m ≥ 0 is complete. If m < 0, then −m is positive, and hence −m = q n + r with 0 ≤ r < n. If r = 0, this gives m = −q n. Otherwise, we have m = −q n − r . Subtracting and adding a copy of n on the right of the equation gives m = (−q − 1)n + (n − r ), and since 0 < r < n, we have 0 < n − r < n as well. We shall use this to characterize the subgroups of Z. The subgroups we already know are the cyclic ones: n = {qn | q ∈ Z}. We shall next show that these are all the subgroups of Z. First, note that if H is a nonzero subgroup of Z, then there must be a nonzero element in it, and hence, by closure under inverses, a positive element. Since the set of positive integers less than or equal to a given one is ﬁnite, there is a unique smallest positive element in it. 2 The terminology that we’ve chosen here is not universal. There are some mathematicians who refer to Theorem 2.3.1 as the Division Algorithm, and use the term Euclidean Algorithm for the procedure for calculating greatest common divisors, outlined in Problem 3 of Exercises 2.3.18. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 25 Proposition 2.3.2. Let H ⊂ Z be a subgroup, and suppose that H = 0. Let n be the smallest positive element of H. Then H = n. Proof Since n ∈ H, n ⊂ H. Thus, it suﬃces to show that H ⊂ n. For m ∈ H, write m = qn + r with q and r integers and 0 ≤ r < n. Now m is in H, as is qn. Thus, m − qn = r must be in H. But r is non-negative and is less than n. Since n is the smallest positive integer in H, we must have r = 0. Thus, m = qn ∈ n. The language of divisibility is useful for discussing subgroups of Z: Deﬁnition 2.3.3. Let m and n be integers. We say n divides m (or that m is divisible by n) if there is an integer q such that m = nq. The next lemma is now immediate from the fact that n is the set of all multiples of n. Lemma 2.3.4. Let m and n be integers. Then m ∈ n if and only if n divides m. By Lemma 2.2.9, Lemma 2.3.4 may be restated this way. Lemma 2.3.5. Let m and n be integers. Then n divides m if and only if m ⊂ n. We shall also need a uniqueness statement about generators of cyclic subgroups. Proposition 2.3.6. Let m and n be integers such that m = n. Then m = ±n. Proof Clearly, if either of m and n is 0, so is the other. By passage to negatives, if necessary, we may assume that m and n are both positive. By Lemma 2.3.5, they must divide each other. Say m = qn and n = rm. But then q and r are positive. If q > 1, then m > n, and hence n = rm > rn ≥ n, i.e., n > n, which is impossible. Thus, q = 1, and hence m = n. Corollary 2.3.7. Let H be a subgroup of Z. Then there is a unique non-negative integer which generates H. Deﬁnition 2.3.8. Let m and n be integers. We write m + n = {rm + sn | r, s ∈ Z}. Lemma 2.3.9. m + n is a subgroup of Z. Proof The inverse of rn + sm is (−r)m + (−s)n. The result follows since (rm + sn) + (r m + s n) = (r + r )m + (s + s )n ∈ m + n. In fact, it is easy to see that m + n is the subgroup of Z generated by the set {m, n}. Deﬁnition 2.3.10. Let m, n ∈ Z. We write (m, n) for the unique non-negative integer which generates the subgroup m + n of Z: m + n = (m, n). We shall refer to (m, n) as the greatest common divisor, or g.c.d., of m and n. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 26 We shall now derive the familiar properties of the g.c.d. and of prime decomposition from the above results. Proposition 2.3.11. Let m and n be integers. Then an integer k divides both m and n if and only if it divides (m, n). In particular, if m and n are not both 0, then (m, n) is the largest integer which divides both m and n. In addition, there exist integers r and s with rm + sn = (m, n). Proof The last statement follows because (m, n) ∈ m + n = {rm + sn | r, s ∈ Z}. Now let k divide both m and n. Then it must divide any number of the form rm+sn, including (m, n). Conversely, to show that the divisors of (m, n) must divide both m and n, it suﬃces to show that (m, n) divides both m and n. But m = 1 · m + 0 · n and n = 0 · m + 1 · n both lie in (m, n), so the result follows from Lemma 2.3.4. Deﬁnition 2.3.12. We say that the integers m and n are relatively prime if (m, n) = 1. Note that if 1 ∈ m + n, then m + n = Z. The following lemma is immediate. Lemma 2.3.13. The integers m and n are relatively prime if and only if there exist integers r and s with rm + sn = 1. Deﬁnition 2.3.14. We say that an integer p > 1 is prime if its only divisors are ±1 and ±p. By Lemma 2.3.5, this says that p ⊂ k if and only if either k = p or k = Z. Lemma 2.3.15. Let p and a be integers, with p prime. Then p and a are relatively prime if and only if p does not divide a. Proof The greatest common divisor of p and a must be either p or 1. In the former case, p divides a. In the latter, it does not. Proposition 2.3.16. Let p be a prime number. Then p divides a product ab if and only if it divides at least one of a and b. Proof Clearly, if p divides a or b, it must divide their product. Conversely, suppose p divides ab, but does not divide a. We shall show that it must divide b. Since p does not divide a, we have (p, a) = 1, so that qp + ra = 1 for some integers q and r. But then qpb + rab = b. Since p divides both qpb and rab, it must divide b. We can now give the fundamental theorem regarding prime factorization. Theorem 2.3.17. Every integer m > 1 has a unique prime decomposition: there is a factorization m = pr11 . . . prkk , where pi is prime and ri > 0 for 1 ≤ i ≤ k, and pi < pj for i < j. Uniqueness means that if m = q1s1 . . . qlsl with qi prime and si > 0 for 1 ≤ i ≤ l and qi < qj for i < j, then we must have k = l and have pi = qi and ri = si for 1 ≤ i ≤ k. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 27 Proof The existence of a prime decomposition amounts to showing that any integer m > 1 is a product of primes. Clearly, this is true for m = 2. We argue by induction on m, assuming that any integer k with 1 < k < m has a prime decomposition. If m is prime, we are done. Otherwise, m = kl, where both k and l are greater than 1 and less than m. But then k and l are products of primes by induction, and hence so is m. For uniqueness, suppose given decompositions pr11 . . . prkk and q1s1 . . . qlsl of m as above. We argue by induction on t = r1 + · · · + rk . If t = 1, then m = p1 , a prime. But then p1 is divisible by q1 , and since both are prime, they must be equal. We then have p1 (1 − q1s1 −1 . . . qlsl ) = 0, so that (1 − q1s1 −1 . . . qlsl ) = 0, and hence q1s1 −1 . . . qlsl = 1. Since the qi are prime and since no positive number divides 1 other than 1 itself, we must have l = 1 and s1 = 1. Suppose t > 1. Since p1 divides m, Proposition 2.3.16, together with an induction on s1 + · · · + sl , shows that p1 must divide qi for some i. But since p1 and qi are prime, we must have p1 = qi . A similar argument shows that q1 = pj for some j, and our order assumption then shows that i = j = 1, so that p1 = q1 . Now p1 (p1r1 −1 . . . prkk − q1s1 −1 . . . qlsl ) = 0, so that pr11 −1 . . . prkk = q1s1 −1 . . . qlsl . But our inductive hypothesis applies here, and the resulting equality is exactly what we wanted. Exercises 2.3.18. 1. Let p1 , . . . , pk be distinct primes and let ri ≥ 0 for 1 ≤ i ≤ k. Show that an integer n divides pr11 . . . prkk if and only if n = ±ps11 . . . pskk with 0 ≤ si ≤ ri for 1 ≤ i ≤ k. 2. Show that there are inﬁnitely many prime numbers. † 3. Let n = qm + r with 0 ≤ r < m. Show that (m, n) = (r, m). Use this to obtain an iterative procedure to calculate (m, n). 4. Implement the iterative procedure above by a computer program to calculate greatest common divisors. † 5. A subgroup G ⊂ R is called discrete if for each g ∈ G there is an open interval (a, b) ⊂ R such that (a, b) ∩ G = {g}. Show that every discrete subgroup of R is cyclic. 2.4 Finite Cyclic Groups: Modular Arithmetic Here, we develop Zn , the group of integers modulo n. Deﬁnition 2.4.1. Let n > 0 be a positive integer. We say that integers k and l are congruent modulo n, written k ≡ l mod n, or sometimes k ≡ l (n), if k − l is divisible by n (i.e., k − l ∈ n). Lemma 2.4.2. Congruence modulo n is an equivalence relation. Proof Clearly, k − k ∈ n for any n, so congruence modulo n is reﬂexive. It is also symmetric, since l − k = −(k − l) is contained in any subgroup containing k − l. If k ≡ l mod n and l ≡ m mod n, then both k − l and l − m are in n, and hence so is k − m = (k − l) + (l − m). Thus, the relation is transitive as well. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 28 Deﬁnition 2.4.3. We write Zn for the set of equivalence classes of integers modulo n, and write m ∈ Zn for the equivalence class containing m ∈ Z. 3 We shall also use 0 to denote 0. Zn is called the group of integers modulo n. Lemma 2.4.4. There are exactly n elements in Zn , represented by the elements 0 = 0, 1, . . . , n − 1. Proof The elements listed are precisely the classes r for 0 ≤ r < n. For m ∈ Z, the Euclidean algorithm provides an equation m = qn + r, with 0 ≤ r < n. But then m = r, and hence m is in the stated list of elements. To see that these classes are all distinct, suppose that 0 ≤ k < l < n. Then 0 < l − k < n, so that l − k is not divisible by n, and hence k = l in Zn . There are two important binary operations on Zn . Deﬁnitions 2.4.5. The operations of addition and multiplication in Zn are deﬁned by setting k + l = k + l and k · l = kl, respectively, for all k, l ∈ Z. We must show that these operations are well deﬁned. Lemma 2.4.6. The operations of addition and multiplication are well deﬁned binary operations on Zn . Proof Suppose that k = q and l = r in Zn . In other words, k ≡ q mod n and l ≡ r mod n. We must show that k + l ≡ q + r mod n and kl ≡ qr mod n. Addition is easy: k − q ∈ n and l − r ∈ n. Now, (k + l) − (q + r) = (k − q) + (l − r). This lies in n, as n closed under addition. For multiplication, we have kl − qr = kl − kr + kr − qr = k(l − r) + r(k − q). This last is divisible by n because l − r and k − q are. The chief properties of addition and multiplication are inherited from those in the integers: Proposition 2.4.7. Addition gives Zn the structure of an abelian group with identity 0. Multiplication makes Zn into an abelian monoid with identity 1. The two operations satisfy the distributive law: a(b + c) = ab + a c. Proof (a + b) + c = a + b + c = a + b + c. Since a + (b + c) has the same expansion, addition is associative. The other assertions follow by similar arguments. In the language of abstract algebra, Proposition 2.4.7 says, simply, that the operations of addition and multiplication give Zn the structure of a commutative ring. When we refer to Zn as a group, we, of course, mean with respect to addition. By Lemma 2.4.4, Zn has order n. Corollary 2.4.8. Let n be a positive integer. Then there are groups of order n. 3 In the notation of Section 1.4, Zn = Z/ ≡, where ≡ denotes equivalence modulo n. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 29 Deﬁnition 2.4.9. The canonical map π : Z → Zn is deﬁned by setting π(k) = k for all k ∈ Z. Given that Zn is the set of equivalence classes of integers with respect to the equivalence relation of congruence modulo n, and that π takes each integer to its equivalence class, we see that this coincides with the canonical map deﬁned in Section 1.4 for an arbitrary equivalence relation. We shall see in the next section that the canonical map π : Z → Zn is what’s known as a homomorphism of groups. Exercises 2.4.10. 1. Show that Zn is generated as a group by 1. Thus, Zn is a cyclic group. 2. Show that 2 does not generate Z4 . Deduce that there are ﬁnite cyclic groups that possess nontrivial proper subgroups. 3. Let p be a prime. Show that the nonzero elements of Zp form a group under multiplication modulo p. 4. If n is not prime, show that the nonzero elements of Zn do not form a group under multiplication modulo n. 2.5 Homomorphisms and Isomorphisms There are many ways to construct a ﬁnite set, but the thing that describes the essence of the structure of such a set is the number of elements in it. Two sets with the same number of elements may be placed in one-to-one correspondence, at which point their properties as sets are indistinguishable. In the case of groups, the structure is too complicated to be captured by the number of elements alone. We shall see examples of ﬁnite groups with the same order whose group properties are quite diﬀerent. For instance, one may be abelian and the other not. But the idea of a one-to-one correspondence makes a good start at deciding when two groups may be identiﬁed, provided that the correspondence respects the multiplications of the groups. Deﬁnition 2.5.1. An isomorphism between two groups is a function f : G → G which is one-to-one and onto, such that f (x · y) = f (x) · f (y) for all x, y ∈ G. Here, · is used for the multiplications of both G and G . We say that the groups G and G are isomorphic if there exists an isomorphism from G to G , and write G ∼ = G . This turns out to be enough to capture the full structure of the groups in question. A glance at the deﬁnition shows that there is a weaker notion which is part of it: Deﬁnition 2.5.2. Let G and G be groups. A homomorphism from G to G is a function f : G → G such that f (x · y) = f (x) · f (y) for all x, y ∈ G. Thus, an isomorphism is a homomorphism which is one-to-one and onto. Recall that one-to-one functions are sometimes called injective functions, or injections, and that onto functions are known as surjective functions, or surjections. Also, a function which is both one-to-one and onto is called bijective. Clearly, an understanding of homomorphisms will contribute to our understanding of isomorphisms, but in fact, homomorphisms will turn out to say a lot more about the CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 30 structure of a group than may be apparent at this point. Before we dive into theory, let’s consider some examples. Examples 2.5.3. 1. For any two groups, G and G , there is a homomorphism f : G → G obtained by setting f (g) = e for all g ∈ G. We call this the trivial homomorphism from G to G . 2. Let H be a subgroup of the group G. Then the inclusion i : H ⊂ G is a homomorphism, because the multiplication in H is inherited from that of G. 3. Recall that the canonical map π : Z → Zn is deﬁned by π(k) = k for all k ∈ Z. Thus, π(k + l) = k + l = k + l = π(k) + π(l), and hence π is a homomorphism. For some pairs G, G of groups, the trivial homomorphism is the only homomorphism from G to G . The relationship between a homomorphism and the power laws in a group is an important one: Lemma 2.5.4. Let f : G → G be a homomorphism of groups and let g ∈ G. Then f (g n ) = f (g)n for all n ∈ Z. In particular, f (e) = e (recall that e = g 0 ) and f (g −1 ) = f (g)−1 . Proof For n > 0, this is a quick induction from the deﬁnition of powers. For n = 0, we have f (e) = f (e · e) = f (e) · f (e). Multiplying both sides of the resulting equation f (e) = f (e) · f (e) by f (e)−1 , we get e = f (e). (Any group element which is its own square must be the identity element of the group.) For n < 0, we have e = f (e) = f (g n g −n ) = f (g n )f (g −n ) = f (g n )f (g)−n by the two cases above, so that f (g n ) is the inverse of f (g)−n . Since the canonical map π : Z → Zn is a homomorphism, Lemma 2.2.7 gives us the next corollary. Corollary 2.5.5. Let m ∈ Zn and let k be any integer. Then the k-th power of m with respect to the group operation in Zn is km. Another consequence of Lemma 2.5.4 is a determination of all of the homomorphisms from Z to an arbitrary group G. Proposition 2.5.6. Let G be a group and let g ∈ G. Then there is a unique homomorphism fg : Z → G with fg (1) = g. Proof Deﬁne fg by fg (n) = g n for all n ∈ Z. This is a homomorphism by the power laws of Lemma 2.2.6. For uniqueness, suppose that f : Z → G is an arbitrary homomorphism with f (1) = g. We’d like to show that f (n) = g n . Since n is the n-th power of 1 with respect to the group operation in Z, this follows from Lemma 2.5.4. The most basic properties of homomorphisms are the following. We leave the proof to the reader. Lemma 2.5.7. Let f : G → G and g : G → G be homomorphisms. Then so is the composite g ◦ f : G → G . If f and g are both isomorphisms, then so is g ◦ f . If f is an isomorphism, then so is the inverse map f −1 : G → G. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 31 There are two subgroups associated with a homomorphism that can be useful in detecting its deviation from being an isomorphism. One of them is the kernel: Deﬁnition 2.5.8. Let f : G → G be a homomorphism. Then the kernel of f , written ker f , is given by ker f = {x ∈ G | f (x) = e}. Recall that the image of a function f : X → Y is the subset im f ⊂ Y given by im f = {y ∈ Y | y = f (x) for some x ∈ X}. An easy check, using Lemma 2.5.4, shows that the kernel and image of a homomorphism f : G → G are subgroups of G and G , respectively. Lemma 2.5.9. Let f : G → G be a homomorphism. Then ker f detects the deviation of f from being injective: we have f (x) = f (x ) if and only if x−1 x ∈ ker f . In particular, f is injective if and only if the kernel of f is the trivial subgroup, e, of G. The image detects whether f is onto: it is onto if and only if im f = G . Proof Suppose that f (x) = f (x ). Then f (x−1 x ) = f (x)−1 f (x ) = e, and hence x−1 x ∈ ker f . Conversely, if x−1 x ∈ ker f , then f (x ) = f (xx−1 x ) = f (x)f (x−1 x ) = f (x) · e, and hence f (x ) = f (x). Suppose that f is injective. Since f (e) = e, no element of G other than e may be carried to e by f , and hence ker f = e. Conversely, if ker f = e, then x−1 x ∈ ker f implies that x−1 x = e, so that x = x . But then x = x whenever f (x) = f (x ), and hence f is injective. The statement regarding im f is immediate. We shall refer to an injective homomorphism as an embedding. The following lemma is immediate: Lemma 2.5.10. Let f : G → G be an embedding. Then f induces an isomorphism of G onto im f . Deﬁnitions 2.5.11. Let G be a group and let n ∈ Z. We say that g ∈ G has exponent n if g n = e. If there is a positive integer n which is an exponent for g, we say that g has ﬁnite order, and deﬁne the order, o(g), of g to be the smallest positive integer which is an exponent for g. If g does not have a positive exponent, we say that o(g) = ∞. If one knows the order of an element, one can determine all possible exponents for it. Lemma 2.5.12. Let G be a group and let g ∈ G be an element of ﬁnite order. Let n be any integer. Then g has exponent n if and only if o(g) divides n. Proof Of course, g mo(g) = (g o(g) )m = em = e, so if o(g) divides n, then g has exponent n. Conversely, if g has order k and exponent n, write n = qk + r with 0 ≤ r < k. Then e = g n = g qk g r = g r , since both n and qk are exponents for g. But this says g has exponent r. If r were positive, this would contradict the fact that k is the smallest positive exponent for g. Thus, r = 0, and n = qk, and hence k divides n, as claimed. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 32 Information about exponents sometimes leads to a determination that a particular element must be the identity. Corollary 2.5.13. Let m and n be integers that are relatively prime. Suppose that m and n are both exponents for an element g ∈ G. Then g = e. Proof The order of g has to divide both m and n, and hence it must also divide their g.c.d. Since the order of an element is positive, g has order 1, and hence g 1 = e. Let fg : Z → G be the unique homomorphism with fg (1) = g. Then fg (n) = g n for all n ∈ Z. Thus, an integer n is in the kernel of fg if and only if g has exponent n. Also, im fg is precisely g. The next result now follows immediately from Proposition 2.3.2. It includes, in part, a slicker argument for Lemma 2.5.12. Proposition 2.5.14. Let G be a group and let g ∈ G. Let fg : Z → G be the unique homomorphism with fg (1) = g. Then the kernel of fg is o(g), if o(g) is ﬁnite, and is 0 otherwise.4 The image of fg is g, the cyclic subgroup of G generated by g. Lemma 2.5.10 now gives us the following corollary. Corollary 2.5.15. If o(g) = ∞, then fg : Z → g is an isomorphism. Corollary 2.5.16. Every nonzero subgroup of Z is isomorphic to Z. Proof Since mn = 0 if and only if either m or n is 0, no nonzero element of Z has ﬁnite order. But every subgroup of Z is cyclic. Next, we study the homomorphisms out of Zn . We shall make use of the notion of commutative diagrams, as deﬁned in Deﬁnition 1.2.1. Proposition 2.5.17. Let G be a group and let g ∈ G. Then there is a homomorphism hg : Zn → G with hg (1) = g if and only if g has exponent n. If g does have exponent n, then there is only one such homomorphism, and it has the property that the following diagram commutes fg / G, [email protected] > @@ } @@ }} } @ π @@ }} }} hg Zn meaning that hg ◦ π = fg . Here π : Z → Zn is the canonical map, while fg : Z → G is the unique homomorphism that takes 1 to g. The homomorphism hg must satisfy the formula hg (m) = g m for all m ∈ Zn . If the homomorphism hg exists, then its image is g. Moreover, the induced map hg : Zn → g is an isomorphism if and only if o(g) = n. In particular, if g ∈ G has ﬁnite order, then g ∼ = Zo(g) . 4 Recall that, since we are using additive notation for Z, we write 0 for the trivial subgroup consisting of the identity element alone. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 33 Proof Suppose there exists a homomorphism hg : Zn → G with hg (1) = g. Then hg ◦ π is a homomorphism from Z to G which takes 1 to hg (π(1)) = hg (1) = g. Thus, hg ◦ π = fg , and hence hg (m) = hg (π(m)) = fg (m) = g m for all m ∈ Zn . Thus, hg is unique if it exists. Moreover, the existence of hg gives e = hg (0) = hg (n) = g n , since n = 0 in Zn , and hence g has exponent n. Clearly, im hg = g. Conversely, if g has exponent n, we wish to show that hg (m) = g m gives a well deﬁned function from Zn to G. If m = l, then m−l = sn for some integer s, and hence m = l+sn. But then g m = g l (g n )s = g l es = g l by the rules of exponents. Thus, hg is well deﬁned. But hg (q + r) = g q+r = g q g r = hg (q)hg (r), and hence hg is a homomorphism. It suﬃces to show that hg is injective if and only if o(g) = n. Suppose ﬁrst that o(g) = n. Since g has exponent n, o(g) divides n. Thus, 0 < o(g) < n, and hence 0 = o(g) ∈ Zn . Since o(g) ∈ ker hg , hg is not injective. On the other hand, if o(g) = n, let k ∈ ker hg . But this says that g k = e, and hence g has exponent k. But that means that k is divisible by o(g) = n, and hence k = 0 in Zn . In particular, ker hg = 0, as desired. Since Zn has order n, we obtain the following corollary. Corollary 2.5.18. Let G be a group and let g ∈ G. Then the order of g is equal to the order of the group g. Corollary 2.5.19. Let f : G → H be a homomorphism, and suppose that g ∈ G has ﬁnite order. Then the order of f (g) divides the order of g. Proof Let hg : Zo(g) → G be the homomorphism which carries 1 to g. Then f ◦ hg : Zo(g) → H is a homomorphism that takes 1 to f (g). By Proposition 2.5.17, f (g) has exponent o(g), so the order of f (g) divides o(g). Because the elements of monoids do not all have inverses, the deﬁnition of homomorphism must be modiﬁed for it to be useful for monoids. Deﬁnitions 2.5.20. A homomorphism of monoids is a function f : M → M such that 1. f carries the identity element of M to that of M . 2. For m, m ∈ M , f (mm ) = f (m)f (m ). An isomorphism of monoids is a homomorphism f : M → M which is bijective. Exercises 2.5.21. 1. Show that any group with one element is isomorphic to the trivial group. 2. Show that any group with two elements is isomorphic to Z2 . 3. List the homomorphisms from Z9 to Z6 . 4. List the homomorphisms from Z5 to Z6 . 5. Show that the order of m in Zn is n/(m, n). Deduce that the order of each element divides the order of Zn . Deduce that every non-identity element of Zp has order p, for any prime p. † 6. Let G be a group and let x ∈ G. Show that o(x−1 ) = o(x). (Hint: We may assume G = x.) CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 34 7. Show that a group G is cyclic if and only if there is a surjective homomorphism f : Z → G. † 8. Let f : G → G be a homomorphism. (a) Let H ⊂ G be a subgroup. Deﬁne f −1 (H ) ⊂ G by f −1 (H ) = {x ∈ G | f (x) ∈ H }. Show that f −1 (H ) is a subgroup of G which contains ker f . (b) Let H ⊂ G be a subgroup. Deﬁne f (H) ⊂ G by f (H) = {f (x) | x ∈ H}. Show that f (H) is a subgroup of G . If ker f ⊂ H, show that f −1 (f (H)) = H. (c) Suppose that H ⊂ im f . Show that f (f −1 (H )) = H . Deduce that there is a one-to-one correspondence between the subgroups of im f and those subgroups of G that contain ker f . What does this tell us when f is onto? 9. Let f : G → G be a homomorphism and let x ∈ G. Show that f (x) = f (x). 10. Show that every subgroup of Zn is cyclic. 11. Show that Zn has exactly one subgroup of order d for each d dividing n, and has no other subgroups. (In particular, among many other important implications, this shows that the order of every subgroup of Zn divides the order of Zn .) Deduce the following consequences. (a) Show that a group containing a subgroup isomorphic to Zk × Zk for k > 1 cannot be cyclic. (b) Show that if a prime p divides n, then Zn has exactly p − 1 elements of order p. (c) Let H and K be subgroups of Zn . Show that H ⊂ K if and only if |H| divides |K|. (d) Show that if d divides n, then Zn has exactly d elements of exponent d. 12. Let H and K be subgroups of Zpr , where p is prime. Show that either H ⊂ K or K ⊂ H. 13. Let p and q be distinct prime numbers and let n be divisible by both p and q. Find a pair of subgroups H, K ⊂ Zn such that neither H nor K is contained in the other. 14. Show that a group G is abelian if and only if the function f : G → G given by f (g) = g 2 is a homomorphism. 15. Show that a group G is abelian if and only if the function f : G → G given by f (g) = g −1 is a homomorphism. 16. Let G be an abelian group and let n be an integer. Show that the function f : G → G given by f (g) = g n is a homomorphism. 17. Show that the exponential function exp(x) = ex gives an isomorphism from the additive group of real numbers to R× +. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 35 18. Show that a group is ﬁnite if and only if it has ﬁnitely many subgroups. 19. Let i : Gln (R) → Mn+1 (R) be deﬁned by setting i(A) to be the matrix whose ﬁrst n rows consist of the matrix A with a column of 0’s added on the right, and whose last row consists of n 0’s followed by a 1, for all A ∈ Gln (R). Pictorially, ⎞ ⎛ 0 ⎜ .. ⎟ ⎜ . ⎟ A i(A) = ⎜ ⎟. ⎝ 0 ⎠ 0 ... 0 1 Show that i(A) lies in Gln+1 (R), and that i deﬁnes an embedding from Gln (R) into Gln+1 (R). 20. Show that there is a monoid with two elements which is not a group. Show that any other monoid with two elements which is not a group is isomorphic to this one. 21. Let f : M → M be a homomorphism of monoids. Show that f restricts to a group homomorphism f : Inv(M ) → Inv(M ) between their groups of invertible elements. In particular, if m ∈ M is invertible, then so is f (m). 22. Give an example to show that condition 1 in the deﬁnition of monoid homomorphism does not follow from condition 2. 2.6 The Classiﬁcation Problem Deﬁnition 2.6.1. By a classiﬁcation of the groups with a certain property, we mean the identiﬁcation of a set of groups {Gi | i ∈ I}5 such that the following conditions hold. 1. Each group Gi has the stated property. 2. If Gi is isomorphic to Gj , then i = j. 3. Each group with the stated property is isomorphic to one of the Gi . A classiﬁcation of groups with a given property is often referred to as a classiﬁcation “up to isomorphism” of the groups in question. The property by which we collect the groups to be classiﬁed is often their order. This is a natural criterion, as isomorphic groups must have the same order. However, other criteria are often used. For instance, the Fundamental Theorem of Finite Abelian Groups, which we give in Section 4.4, gives a classiﬁcation of all ﬁnite abelian groups. Examples 2.6.2. 1. Problem 2 of Exercises 2.5.21 shows that every group of order 2 is isomorphic to Z2 . This classiﬁes the groups of order 2. 2. Corollary 2.5.15 and Proposition 2.5.17 show that every cyclic group is isomorphic to exactly one of the groups in {Z, e, Zn | n ≥ 2}. Thus, we obtain a classiﬁcation of the cyclic groups. 5 Here the set I is called the indexing set for the set {Gi | i ∈ I} of groups. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 36 Classiﬁcation can be phrased in more general terms using the language of category theory. (See Chapter 6.) In general, questions of classiﬁcation are the most fundamental and important questions that can be asked in mathematics. It may come as a surprise to some readers that the ﬁnite groups have not yet been classiﬁed. It is often astounding that questions which may be so easily phrased can be so diﬃcult to answer. Modern mathematics abounds with such questions. A beginning student of mathematics has no conception of the number of important, yet really basic, mathematical questions that have not yet been solved, despite the number of astute mathematicians that have been working on them over the years. The surprise that comes later is how many interesting open questions are amenable to solutions, given the right mixture of ideas, luck, and persistence. The main goal of the group theory portion of this book is to be able to obtain as much information as possible about the classiﬁcation of groups of relatively small order. In particular, for a given integer n, say with n ≤ 63, we’d like to classify all groups of order n. There is some language we can use to make our discussion clearer. Deﬁnition 2.6.3. We say that two groups are in the same isomorphism class if the groups are isomorphic to one another. Isomorphism classes are analogous to equivalence classes, where the equivalence relation used is that of isomorphism. The point is that the collection of all groups of a given order is not a set, as it has too many elements. (There are too many groups of order 1, for instance, to form a set, even though there is only one isomorphism class of groups of order 1.) Thus, we are looking at an analogue for classes (as opposed to sets) of an equivalence relation. The interesting thing is that classiﬁcation, as deﬁned above, can only occur in situations where the isomorphism classes themselves form a set. Indeed, the next lemma is an immediate consequence of the deﬁnition of a classiﬁcation. Lemma 2.6.4. The set {Gi | i ∈ I} classiﬁes the collection of groups with a given property if and only if each isomorphism class of the groups with the stated property includes exactly one of the groups Gi . In particular, the collection of isomorphism classes of the groups in question forms a set which is in one-to-one correspondence with {Gi | i ∈ I}. Notice that if X is a ﬁnite set, then the set of possible group multiplications we could put on X is a subset of the set of functions from X × X to X. But the set of all functions from one ﬁnite set to another is ﬁnite, by Corollary 1.7.6. Thus, there are ﬁnitely many possible group multiplications on a ﬁnite set. Diﬀerent multiplications will in many cases give rise to isomorphic groups, but the number of isomorphism classes of groups with |X| elements must be ﬁnite. Proposition 2.6.5. Let n > 0. Then there are ﬁnitely many isomorphism classes of groups of order n. A countable union of ﬁnite sets is countable. Corollary 2.6.6. The set of isomorphism classes of ﬁnite groups is countable. Exercises 2.6.7. 1. Let X = {x, y} be a set with two elements. Which of the functions from X × X to X are group multiplications? Which are monoid multiplications? 2. Classify the groups of order 3. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 2.7 37 The Group of Rotations of the Plane Here, we describe an important subgroup of Gl2 (R), the group, SO(2), of rotations of the plane. Deﬁnition 2.7.1. For θ ∈ R, we deﬁne the rotation matrix Rθ by cos θ − sin θ Rθ = . sin θ cos θ Here, the sine and cosine are taken with respect to radian measure. Elements of M2 (R) act as transformations of the plane in the usualway. We identify the plane with the space of column vectors v = ( xy ). A matrix A = ac db acts on the plane by the usual multiplication of matrices on columns: ax + by A · v = . cx + dy To display the geometric eﬀect of the transformation induced by a rotation matrix, we make use of polar coordinates and the sum formulæ for sines and cosines, as developed in calculus. Proposition 2.7.2. The transformation of the plane induced by the rotation matrix Rθ is the counterclockwise rotation of the plane through the angle θ. Proof We write the vector v = ( xy ) in polar coordinates, setting x = r cos φ and y = r sin φ, where r = x2 + y 2 is the length of v and φ is the angle that starts at the positive x-axis and ends at v . The product formula now gives r(cos θ cos φ − sin θ sin φ) r cos(θ + φ) Rθ · v = = , r(cos θ sin φ + sin θ cos φ) r sin(θ + φ) by the sum formulæ for the sine and cosine. Thus, the length of Rθ ·v is the same as that of v , but the angle that it makes with the positive x-axis has been increased by θ. As geometric intuition would suggest, the product (composite) of two rotations is again a rotation: Lemma 2.7.3. For any real numbers θ and φ, we have Rθ · Rφ = Rθ+φ . Proof The deﬁnition of matrix multiplication gives cos θ cos φ − sin θ sin φ − cos θ sin φ − sin θ cos φ Rθ · R φ = . cos θ sin φ + sin θ cos φ cos θ cos φ − sin θ sin φ Once again, the result follows from the sum formulæ for the sine and cosine. Deﬁnition 2.7.4. Let SO(2) ⊂ Gl2 (R) be the set of all rotation matrices Rθ , as θ ranges over all real numbers. We call it the second special orthogonal group. By Lemma 2.7.3, SO(2) is closed under multiplication in Gl2 (R). Moreover, since cos 0 = 1 and sin 0 = 0, we have R0 = I2 , the identity element of Gl2 (R). Since Rθ · R−θ = R0 , R−θ is the inverse of Rθ . CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 38 Lemma 2.7.5. SO(2) is a subgroup of Gl2 (R). Notice that since addition of real numbers is commutative, Lemma 2.7.3 shows that Rθ ·Rφ = Rφ ·Rθ . Thus, unlike Gl2 (R), SO(2) is an abelian group. Lemma 2.7.3 actually shows more: Proposition 2.7.6. There is a homomorphism, exp, from R onto SO(2), deﬁned by exp(θ) = Rθ . The kernel of exp is 2π = {2πk | k ∈ Z}. Thus, Rθ = Rφ if and only if θ − φ = 2πk for some k ∈ Z. Proof That exp is a homomorphism is the content of Lemma 2.7.3. Suppose that θ ∈ ker exp. Since the identity element of SO(2) is the identity matrix, the deﬁnition of Rθ forces cos θ to be 1 and sin θ to be 0. From calculus, we know this happens if and only if θ is a multiple of 2π.6 By Lemma 2.5.9, we have that exp θ = exp φ if and only if −φ + θ ∈ ker(exp), so the result follows. We conclude with a calculation of the order of the elements of SO(2). Proposition 2.7.7. The rotation Rθ has ﬁnite order if and only if θ = 2πk/n for some integers k and n. The order of R2πk/n is n/(k, n). Proof Suppose that Rθ has ﬁnite order. Then (Rθ )n = Rnθ = I2 for some integer n. Then Proposition 2.7.6 shows that nθ = 2πk for some integer k, so that θ = 2πk/n. If θ = 2πk/n, the order of Rθ will be the smallest positive integer m for which mθ is an integer multiple of 2π. Thus, m is the smallest positive integer such that mk is divisible by n. But a glance at the prime decompositions of n and k shows that this smallest positive integer is precisely n/(n, k). In particular, R2π/n has order n for all n ≥ 1. Corollary 2.7.8. There are embeddings of Zn in SO(2) for all n ≥ 1. Since SO(2) gives transformations of the plane, this says that Zn may be used to produce geometric transformations of a familiar space. A certain school of thought would summarize this by saying that Zn “exists in nature.” 2.8 The Dihedral Groups Let n > 0, we construct a group, D2n , called the dihedral group of order 2n. These groups are nonabelian for n > 2, and provide our ﬁrst examples of nonisomorphic groups of the same order. We give a deﬁnition here via matrices, which relies on calculus. A direct deﬁnition as an abstract group is also possible, but is a little awkward notationally without some further discussion. We shall give such a derivation in Section 4.6. Let b, or bn , if more than one dihedral group is being discussed, be the rotation of the plane through the angle 2π/n: b = R2π/n ∈ SO(2) in the notation of the last section. Then by Proposition 2.7.7, b has order n in Gl2 (R). Let a ∈ Gl2 (R) be the matrix 1 0 a= . 0 −1 6 Actually, one of the nicer ways to prove this fact from calculus is to use derivatives to show that the kernel of exp is a discrete subgroup of R and apply Problem 5 of Exercises 2.3.18. One may then deﬁne 2π as the smallest positive element in ker(exp) and use numerical methods to compute its value. Here, we may take the sine and cosine to be deﬁned by their Taylor series, and derive their basic properties, such as their derivatives and their sum formulæ, by manipulating the series. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 39 The matrix a is said to reﬂect the plane through the x-axis, and we say that a matrix is a reﬂection matrix if it is a product a · c with c ∈ SO(2). (It is possible to give a direct geometric deﬁnition of reﬂection using dot products, and then to prove that these geometric operations are induced by precisely the matrices just stated. We shall treat this approach in exercises below, and content ourselves here with the algebraic deﬁnition just given.) The most important piece of information about rotations and reﬂections is the way they interrelate. They don’t commute with each other, but they come close to it: Lemma 2.8.1. For θ ∈ R, we have Rθ · a = a · R−θ . Thus, for b = R2π/n as above, bk · a = a · b−k for any integer k. Proof Multiplying matrices, we get Rθ · a = cos θ sin θ sin θ − cos θ . This is equal to the matrix product a · R−θ , since cos(−θ) = cos θ and sin(−θ) = − sin θ. The statement about powers of b follows since b±k = R±k·2π/n . The following lemma may be veriﬁed by direct computation. Lemma 2.8.2. The matrix a has order 2. Deﬁnition 2.8.3. Let n ≥ 1. Then the dihedral group of order 2n, D2n ⊂ Gl2 (R), is given as follows: D2n = {bk , abk | 0 ≤ k < n}, with a and b = bn as above. Of course, this deﬁnition needs a bit of justiﬁcation: Proposition 2.8.4. The dihedral group D2n is a subgroup of Gl2 (R). Moreover, the elements {bk , abk | 0 ≤ k < n} are all distinct, so that D2n has order 2n. Proof We ﬁrst show that the 2n elements bk and abk with 0 ≤ k < n are all distinct. If abk = bl , then multiplication by b−k on the right gives a = bl−k , a rotation matrix. But rotation matrices all commute with one another, and Lemma 2.8.1 shows that a fails to commute with most rotations. Thus, abk = bl . If abk = abl , then multiplying both sides on the left by a−1 gives bk = bl , which implies that k ≡ l mod n by Proposition 2.5.17. With 0 ≤ k, l < n, this gives k = l by Lemma 2.4.4. We now show that D2n is closed under multiplication in Gl2 (R). By Lemma 2.8.1, we have abk · abl = a2 bl−k = bl−k , as a has order 2. Since b has order n, any power of b is equal to one in the list. Similarly, bk · abl = abl−k is in D2n . The remaining cases are covered by the fact that b is closed under multiplication. The fact that D2n is closed under multiplication implies that it is a subgroup of Gl2 (R), since D2n is ﬁnite (Proposition 2.2.13). CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 40 The dihedral groups have geometric meaning as groups of transformations of regular polygons in the plane. In particular, for n ≥ 3, there is a regular n-gon whose vertices are the points cos(2πk/n) vk = sin(2πk/n) for 1 ≤ k ≤ n. This means that the n-gon is the set of convex of the vertices, combinations n where a convexcombination of v1 , . . . , vn is a vector sum i=1 aivi with 0 ≤ ai ∈ R for n 1 ≤ i ≤ n and i=1 ai = 1. If c is a matrix in D2n , then it is easy to see that for each i with 1 ≤ i ≤ n, the matrix product c · vi is equal to vj for some j with 1 ≤ j ≤ n. In other words, multiplication by c carries each vertex of our n-gon to another vertex of the n-gon. But the linearity of matrix multiplication now shows that multiplication by c carries convex combinations of the vertices to convex combinations of the vertices, and hence c gives a transformation of the n-gon. In fact, it can be shown that the elements of D2n give all possible linear transformations from the n-gon onto itself. Exercises 2.8.5. 1. What is the order of abk ∈ D2n ? 2. Show that D4 and Z4 are non-isomorphic abelian groups of order 4. 3. Show that D2n is nonabelian if and only if n ≥ 3. Deduce that D2n and Z2n are non-isomorphic groups of order 2n for all n ≥ 3. 4. Show that there is a subgroup of Gl2 (R) consisting of all rotation matrices Rθ , together with the matrices a · Rθ . (This subgroup is generally called O(2), the second orthogonal group.) What is the order of a · Rθ ? 5. Show that if k divides n, then there is a subgroup of D2n which is isomorphic to D2k . † 6. Let G be a group. Show that there is a one-to-one correspondence between the homomorphisms from D2n to G and pairs of elements x, y ∈ G such that o(x) divides 2, o(y) divides n, and x−1 yx = y −1 . 7. Show that if k divides n, then there is a homomorphism of D2n onto D2k . 8. Let be a line through the origin in the plane. Standard vector geometry shows there is a vector u of length 1 such that is the set of all vectors perpendicular to u. (Note this does not deﬁne u uniquely: −u would do as well.) Then the reﬂection through is the function f deﬁned by f (w) =w − 2(w · u)u, where · is the dot product. Show that the reﬂection through is the transformation induced by the matrix aRθ for some θ. Show also that the transformation induced by aRθ is the reﬂection through some line for any θ ∈ R. 2.9 Quaternions We construct yet another family of groups via matrices. They are the quaternionic groups Q4n ⊂ Gl4 (R) of order 4n. First, we give a generalization of the rotation matrices. For CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES θ ∈ R, let ⎛ cos θ − sin θ ⎜ sin θ cos θ Sθ = ⎜ ⎝ 0 0 0 0 0 0 cos θ sin θ 41 ⎞ 0 0 ⎟ ⎟. − sin θ ⎠ cos θ The geometric eﬀect of Sθ on an element of the space R4 of column vectors is to simultaneously rotate the ﬁrst two coordinates and rotate the last two coordinates, treating these pairs of coordinates as independent planes. The following lemma may be veriﬁed directly by multiplying the matrices. Lemma 2.9.1. Let θ and φ be real numbers. Then Sθ · Sφ = Sθ+φ . Thus, there is a homomorphism f : R → Gl4 (R) deﬁned by f (θ) = Sθ . Moreover, as in the case of rotations of the plane, Sθ = Sφ if and only if θ − φ = 2πk for some k ∈ Z. Once again, we shall give a matrix a that will interact with these “rotations” in an interesting way. It is given by ⎛ 0 ⎜ 0 a=⎜ ⎝ −1 0 0 1 0 0 0 0 1 0 ⎞ 0 −1 ⎟ ⎟. 0 ⎠ 0 Lemma 2.9.2. The matrix a has order 4. The lower powers of ⎛ ⎞ ⎛ −1 0 0 0 0 ⎜ 0 −1 ⎟ ⎜ 0 0 0 2 3 ⎟ and a = ⎜ a = −I4 = ⎜ ⎝ 0 ⎝ 1 0 −1 0 ⎠ 0 0 0 −1 0 a are given by ⎞ 0 −1 0 0 0 1 ⎟ ⎟. 0 0 0 ⎠ −1 0 0 Proof The square of a is as stated by direct calculation. We then have that a3 = a2 a = −a, which is the displayed matrix. Clearly, this is not the identity matrix, but (−I4 )2 = a4 is the identity matrix. Thus, the 4-th power of a is the smallest that will produce the identity, so that a has order 4 and a3 = a−1 . Once again, there is a deviation from commutativity between the Sθ and a. The following lemma may be veriﬁed by direct computation. Lemma 2.9.3. We have Sθ · a = a · S−θ . Because Sθ and Sφ commute for any real numbers θ and φ, we obtain the following corollary. Corollary 2.9.4. The matrix a is not equal to Sθ for any θ. Write b (or b2n if more than one quaternionic group is under discussion) for S2π/2n . Then b has order 2n. The quaternionic group of order 4n is the subgroup of Gl4 (R) generated by a and b: Deﬁnition 2.9.5. The quaternionic group of order 4n, Q4n ⊂ Gl4 (R), is given as follows: Q4n = {bk , abk | 0 ≤ k < 2n}, with a and b as above. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 42 Once again we need some justiﬁcation. Proposition 2.9.6. For each n ≥ 1, Q4n is a subgroup of Gl4 (R). Moreover, the matrices {bk , abk | 0 ≤ k < 2n} are all distinct, so that Q4n does indeed have order 4n. Proof The key here is that a2 = −I4 = Sπ = bn , which is why b was chosen to have even order. Thus, abk · abl = a2 bl−k = bn bl−k = bn+l−k . Since bk · abl = abl−k and since b has order 2n, Q4n is closed under multiplication. Since it is ﬁnite, it is a subgroup by Proposition 2.2.13. It remains to show that the stated elements are all distinct. As in the dihedral case, it suﬃces to show that abk = bl , and hence that a = bk−l for any k and l. But once again this follows from the fact that a cannot equal Sθ for any θ. Exercises 2.9.7. 1. What is the order of abk ∈ Q4n ? 2. Show that Q4 is isomorphic to Z4 . 3. For n ≥ 2, show that Q4n is isomorphic to neither D4n nor Z4n . 4. Show that if k divides n, then there is a subgroup of Q4n which is isomorphic to Q4k . † 5. Let G be a group. Show that there is a one-to-one correspondence between the homomorphisms from Q4n to G and pairs of elements x, y ∈ G such that o(x) divides 4, o(y) divides 2n, x2 = y n , and x−1 yx = y −1 . 6. Show that there is a homomorphism of Q4n onto D2n . 7. Show that if n is an odd multiple of k, then there is a homomorphism of Q4n onto Q4k . 8. Show that every subgroup of Q8 other than e contains a2 . (Hint: The cyclic subgroups are always the smallest ones in a group.) 9. Let n be any power of 2. Show that every subgroup of Q4n other than e contains a2 . 2.10 Direct Products We give a way of constructing new groups from old. It satisﬁes an important universal property with respect to homomorphisms. Deﬁnition 2.10.1. Let G and H be groups. Then the product (or direct product) of G and H is the group structure on the cartesian product G × H which is given by the binary operation (g1 , h1 ) · (g2 , h2 ) = (g1 g2 , h1 h2 ). We denote this group by G × H. Once again, we must justify our deﬁnition: Lemma 2.10.2. The direct product G × H is a group. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 43 Proof Associativity follows directly from the associativity of G and H. The identity element is the ordered pair (e, e). The inverse of (g, h) is (g −1 , h−1 ). The reader should note that while this is not the only interesting group structure we can put on the product of G and H (we shall discuss semidirect products in Section 4.6), it is the only one we shall refer to as G × H. The next result is immediate from Proposition 1.6.4. Corollary 2.10.3. Let G and H be ﬁnite groups. Then G × H is ﬁnite, and |G × H| = |G| · |H|. The reader should supply the veriﬁcation for the next lemma. Lemma 2.10.4. There are subgroups G × e = {(g, e) | g ∈ G} and e × H = {(e, h) | h ∈ H} of G × H. Any element of G × e commutes with any element of e × H. There are homomorphisms ι1 : G → G × H and ι2 : H → G × H deﬁned by ι1 (g) = (g, e) and ι2 (h) = (e, h). These induce isomorphisms G ∼ = G × e and H ∼ = e × H. There are also homomorphisms π1 : G × H → G and π2 : G × H → H deﬁned by π1 (g, h) = g and π2 (g, h) = h. The homomorphisms ιi are called the canonical inclusions of G and H in G × H. The homomorphisms π1 and π2 are called the projections of G × H onto the ﬁrst and second factors, respectively. Recall that a function into a product is deﬁned uniquely by its component functions: f : X → G × H is deﬁned by f (x) = (f1 (x), f2 (x)), where f1 : X → G and f2 : X → H. In fact, the component functions are nothing other than the composites fi = πi ◦ f , where the πi are the projections in the above lemma. The group product in G × H is the unique product that makes the following statement true: Lemma 2.10.5. Let G, H, and K be groups. A function f : K → G × H is a homomorphism if and only if its component functions are homomorphisms. We can also study homomorphisms out of a product. If f : G × H → K is a homomorphism, so are f ◦ ι1 : G → K and f ◦ ι2 : H → K. Proposition 2.10.6. Let f : G × H → K be a homomorphism. Then every element of the image of f ◦ ι1 commutes with every element of the image of f ◦ ι2 . Moreover, f is uniquely determined by its restrictions to G × e and e × H. Conversely, given homomorphisms g1 : G → K and g2 : H → K such that each element in the image of g1 commutes with each element in the image of g2 , then there is a unique homomorphism f : G × H → K with f ◦ ιi = gi for i = 1, 2. Proof Since (g, h) = (g, e) · (e, h), f (g, h) = f ◦ ι1 (g) · f ◦ ι2 (h). Thus, f is determined by the f ◦ ιi , which are determined by the restriction of f to the stated subgroups. The commutativity assertion follows since if two elements commute, so must their images under any homomorphism. Given the gi as above, deﬁne f : G × H → K by f (g, h) = g1 (g) · g2 (h). Then f (g, h) · f (g , h ) = g1 (g)g2 (h)g1 (g )g2 (h ) = g1 (g)g1 (g )g2 (h)g2 (h ) = g1 (gg )g2 (hh ), with the key step being given by the fact that g1 (g ) commutes with g2 (h). But the last term is clearly f ((g, h) · (g , h )). CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 44 It is valuable to know when a group breaks up as the product of two of its subgroups. Deﬁnition 2.10.7. Let H and K be subgroups of the group G. We say that G is the internal direct product of H and K if there is an isomorphism μ : H × K → G such that μ ◦ ι1 and μ ◦ ι2 are the inclusions of H and K, respectively, as subgroups of G. Explicitly, this says μ(h, k) = hk. We now give a characterization of internal direct products. Proposition 2.10.8. Let H and K be subgroups of the group G. Then G is the internal direct product of H and K if and only if the following conditions hold. 1. H ∩ K = e. 2. If h ∈ H and k ∈ K, then h and k commute in G.7 3. G is generated by the elements of H and K. Proof Suppose that G is the internal direct product of H and K. Then the function μ : H × K → G deﬁned by μ(h, k) = hk is an isomorphism of groups. Since μ is a homomorphism and since μ ◦ ι1 (h) = h and μ ◦ ι2 (k) = k, each h ∈ H must commute with each k ∈ K by Proposition 2.10.6. Since μ is onto, each element of G must have the form hk, with h ∈ H and k ∈ K, so the elements of H and K must generate G. Finally, let x ∈ H ∩ K. Then (x, x−1 ) ∈ H × K, and μ(x, x−1 ) = xx−1 = e. Since μ is injective, this says that (x, x−1 ) is the identity element of H × K, which is (e, e). Thus, x = e, and hence H ∩ K = e. Conversely, suppose that the three stated conditions hold. Since the elements of H commute with those of K, Proposition 2.10.6 shows that the function μ : H × K → G given by μ((h, k)) = hk is a homomorphism. Suppose that (h, k) is in the kernel of μ. Then hk = e, and hence h = k −1 is in H ∩ K. But H ∩ K = e, and hence h = e, and hence k = h−1 = e also. Thus, (h, k) = (e, e), and hence μ is injective. Since H and K generate G, any element in G may be written as a product g1 . . . gn where gi is an element of either H or K for i = 1, . . . , n. Since elements of H commute with those of K, we can rewrite this as a product h1 . . . hr k1 . . . ks with hi ∈ H for i = 1, . . . , r and ki ∈ K for i = 1, . . . , s, by moving the elements in H past those in K. But this is just μ(h1 . . . hr , k1 . . . ks ), so μ is onto. We shall see in Proposition 3.7.12 that the second condition of Proposition 2.10.8 may be replaced with the statement that both H and K are what’s known as normal subgroups (to be deﬁned in Section 3.7) of G. Now let us generalize the material in this section to the study of products of more than two groups. Deﬁnition 2.10.9. Suppose given groups G1 , . . . , Gk , for k ≥ 3. The product (or direct product), G1 × · · · × Gk , of the groups G1 , . . . , Gk is the set of all k-tuples (g1 , . . . , gk ) with gi ∈ Gi for 1 ≤ i ≤ k, with the following multiplication: (g1 , . . . , gk ) · (g1 , . . . , gk ) = (g1 g1 , . . . , gk gk ). The reader should be able to supply the proof of the following proposition. 7 Of course, neither H nor K need be abelian. Thus, if h, h ∈ H, then h and h must commute with every element of K, but they need not commute with each other. CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 45 Proposition 2.10.10. The direct product G1 × · · · × Gk is a group under the stated multiplication. There are homomorphisms πi : G1 × · · · × Gk → Gi and ιi : Gi → G1 × · · · × Gk for 1 ≤ i ≤ k, deﬁned by πi (g1 , . . . , gk ) = gi , and ιi (g) = (e, . . . , e, g, e, . . . , e), where g appears in the i-th coordinate. Given homomorphisms fi : H → Gi for 1 ≤ i ≤ k, there is a unique homomorphism f : H → G1 × · · · × Gk such that πi ◦ f = fi for 1 ≤ i ≤ k. Similarly, for each collection hi : Gi → H of homomorphisms such that each element of the image of hi commutes with each element of the image of hj whenever i = j, there is a unique homomorphism h : G1 × · · · × Gk → H with h ◦ ιi = hi for 1 ≤ i ≤ k. Similarly, we can discuss internal direct products of several subgroups. Deﬁnition 2.10.11. Let H1 , . . . , Hk be subgroups of G. We say that G is the internal direct product of H1 , . . . , Hk if the function μ : H1 × · · · × Hk → G deﬁned by μ(h1 , . . . , hk ) = h1 . . . hk is an isomorphism of groups. We shall not give a characterization here of the internal direct product of multiple subgroups. (I.e., we shall not give a generalization of Proposition 2.10.8 to the case of more than two subgroups.) The reader may wish to formulate one after reading the proof of Proposition 4.4.9. Exercises 2.10.12. 1. Show that the order of (g, h) ∈ G × H is the least common multiple of the orders of g and h. † 2. More generally, show that the order of (g1 , . . . , gk ) ∈ G1 × · · · × Gk is the least common multiple of the orders of g1 , . . . , gk . 3. Show that Z6 is isomorphic to Z3 × Z2 . † 4. Let m and n be relatively prime. Show that Zmn is isomorphic to Zm × Zn . 5. Let n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes. Show that Zn ∼ = Zpr11 × · · · × Zprkk . 6. Show that if m and n are not relatively prime, then Zm × Zn is not cyclic. 7. Show that D4 is isomorphic to Z2 × Z2 . 8. Show that D12 is isomorphic to D6 × Z2 . 9. Show that if k is odd, then D4k is isomorphic to D2k × Z2 . 10. Show that G × H ∼ = H × G. 11. Show that (G × H) × K ∼ = G × (H × K) ∼ = G × H × K. 12. Show that G × e ∼ = G, where e is the trivial group. 13. Show that the group multiplication μ : G × G → G is a homomorphism if and only if G is abelian. 14. Let fi : Gi → Hi be a homomorphism for 1 ≤ i ≤ k. Deﬁne f1 × · · · × fk : G1 × · · · × Gk → H1 × · · · × Hk by f1 × · · · × fk (g1 , . . . , gk ) = (f1 (g1 ), . . . , fk (gk )). Show that f1 × · · · × fk is a homomorphism. In addition, show that f1 × · · · × fk is an isomorphism if and only if each fi is an isomorphism. 15. Show that not every homomorphism f : G1 × G2 → H1 × H2 has the form f1 × f2 . Chapter 3 G-sets and Counting Many of the examples in Chapter 2 were obtained from matrices, and hence give symmetries of n-dimensional Euclidean space, Rn . Here, we study the symmetries of sets, obtaining a more combinatorial view of symmetry. In the ﬁrst section, we deﬁne the symmetric group on n letters, Sn , and prove Cayley’s Theorem, which shows that every ﬁnite group is a subgroup of Sn for some n. Later, in Chapter 10, we will show that Sn embeds in Gln (R). In this manner, we obtain the assertion of the last chapter that every ﬁnite group is a subgroup of Gln (R). The symmetric groups are interesting in their own right as examples of groups. Via the technique of cycle decomposition, which we give in Section 3.5, one can say quite a bit about its structure. Also, there are natural subgroups An ⊂ Sn , called the alternating groups, which are the easiest examples of nonabelian simple groups. We deﬁne An in Section 3.5 in terms of the sign homomorphism ε : Sn → {±1} to be deﬁned there. We shall deﬁne simple groups in Section 4.2, and show that An is simple for n ≥ 5. We may then deduce in Chapter 5 that Sn is not what’s known as a solvable group for n ≥ 5. The non-solvability of Sn will allow us, in Chapter 11, to prove Galois’s famous theorem that there is no formula for solving polynomials of degree ≥ 5. This should be enough to arouse one’s interest in the symmetric groups. We also introduce a very important concept in group theory: that of the set G/H of left cosets of a subgroup H in G. One of the ﬁrst consequences of the study of cosets is Lagrange’s Theorem: if G is ﬁnite, then the order of any subgroup of G must divide the order of G. We give Lagrange’s Theorem in Section 3.2, and give many more applications of the study of cosets throughout the material on group theory. The sets of cosets G/H form an important example of the notion of a group acting on a set. We formulate the notion of G-sets in Section 3.3 and give a generalization of Cayley’s Theorem in which a one-to-one correspondence is established between the actions of G on a set X and the homomorphisms from G to the group of symmetries of X. This allows us to associate to a ﬁnite G-set (such as G/H, if G is ﬁnite) a homomorphism from G into Sn for appropriate n. These homomorphisms are quite useful in determining the structure of a group G, and will ﬁgure prominently in the classiﬁcation results we get in Chapter 5. Section 3.3 also contains a couple of important counting arguments, including what’s known as the G-set Counting Formula. In the ﬁnal two sections of the chapter, we study actions of one group on another. The point is that not all symmetries of a group H preserve the group structure. The ones that do are the automorphisms of H, which are deﬁned to be the group isomorphisms 46 CHAPTER 3. G-SETS AND COUNTING 47 H → H. These form a subgroup, Aut(H), of the group of symmetries of H. We study the actions of G on H such that the action of each g ∈ G induces an automorphism of H. Actions through automorphisms play an important role in constructing and classifying the extensions of one group by another (Section 4.7), and therefore, as we shall see, in the classiﬁcation of groups. Closer at hand, there is a very important action of a group G on itself through automorphisms, called conjugation. Many of the important results in group theory follow from the study of conjugation, including the Noether Isomorphism Theorems of Chapter 4 and the Sylow Theorems of Chapter 5. We shall begin that study in this chapter. 3.1 Symmetric Groups: Cayley’s Theorem Here, we show that every ﬁnite group embeds in the group of symmetries of a ﬁnite set. Recall that a function is bijective if it is one-to-one and onto. Deﬁnitions 3.1.1. Let X be a set. By a permutation of X we mean a bijective function σ : X → X. We write S(X) for the set of all permutations of X. Note that S(X) has a binary operation given by composition of functions: for σ, τ ∈ S(X), σ · τ is the composite σ ◦ τ . As we shall see shortly, S(X) is a group under this operation, called the group of symmetries of X. Recall that the identity function 1X of a set X is the function deﬁned by 1X (x) = x for all x ∈ X. Recall from Proposition 1.1.6 that every bijective function σ : X → Y has an inverse function σ −1 : Y → X, with the property that σ −1 ◦ σ = 1X and σ ◦ σ −1 = 1Y . Here, the inverse function σ −1 is deﬁned by setting σ −1 (y) to be the unique x ∈ X such that σ(x) = y. Lemma 3.1.2. Let X be a set. Then S(X) forms a group under the operation of composition of functions. The identity element of S(X) is the identity function 1X , and the inverse of an element σ ∈ S(X) is its inverse function, σ −1 . We shall see below that if X is inﬁnite, then not only is S(X) an inﬁnite group, it also contains elements of inﬁnite order. On the other hand, if X is ﬁnite, then there are only ﬁnitely many elements in S(X), as, in fact, there are only ﬁnitely many functions from X to X by Corollary 1.7.6. We shall calculate the order of S(X), for X ﬁnite, in Corollary 3.3.11. Deﬁnition 3.1.3. Let f : X → Y be a bijection of sets. Deﬁne the induced map f∗ : S(X) → S(Y ) by f∗ (σ) = f ◦ σ ◦ f −1 for σ ∈ S(X). Since two sets are considered equivalent if they may be placed in bijective correspondence, the following lemma should come as no surprise. Lemma 3.1.4. Let f : X → Y be a bijection of sets. Then the induced map f∗ : S(X) → S(Y ) is an isomorphism of groups. CHAPTER 3. G-SETS AND COUNTING 48 Proof It is easy to see that f∗ gives a group homomorphism from S(X) to S(Y ). Moreover, (f −1 )∗ : S(Y ) → S(X) gives an inverse function for f∗ , so f∗ is an isomorphism. Note that there is no preferred isomorphism between S(X) and S(Y ), but rather an isomorphism for each choice of bijection from X to Y . Of special interest here is the case when X is ﬁnite. Our preferred model for a set with n elements is {1, . . . , n}. Deﬁnition 3.1.5. We write Sn for S({1, . . . , n}), and call it the symmetric group on n letters. Fixed points are an important feature of a permutation: Deﬁnition 3.1.6. Let σ be a permutation of X. Then σ is said to ﬁx an element x ∈ X if σ(x) = x. We call such an x a ﬁxed point of σ. We should now construct some examples of elements of Sn . Recall that if Y is a subset of X, then the complement of Y in X is the set of all elements of X that are not in Y . Deﬁnition 3.1.7. Let i1 , . . . , ik be k distinct elements of {1, . . . , n}, k ≥ 2. We write (i1 i2 · · · ik ) for the permutation of {1, . . . , n} which carries ij to ij+1 for 1 ≤ j < k, carries ik to i1 , and is the identity map on the complement of {i1 , . . . , ik } in {1, . . . , n}. We call (i1 i2 · · · ik ) a k-cycle, or a cycle of length k. Note that a 2-cycle (i j) just interchanges i and j and ﬁxes all other elements of {1, . . . , n}. Another name for a 2-cycle is a transposition. Some people consider a 1-cycle (i) to be another way of writing the identity map. While this is consistent with the notation above, our cycles will all have length ≥ 2. Note that the permutations (i1 i2 · · · ik ) and (ik i1 · · · ik−1 ) are equal, as they have the same eﬀect on all elements of {1, . . . , n}. Thus, there is more than one way to denote a given k-cycle. We can calculate the order and inverse of a k-cycle. Lemma 3.1.8. A k-cycle (i1 i2 · · · ik ) has order k. Its inverse is (ik · · · i2 i1 ). Proof Let σ = (i1 i2 · · · ik ). An easy induction shows that for 1 ≤ j < k, σ j (i1 ) = ij+1 = i1 , while σ k does ﬁx i1 . Thus, σ j = e for 1 ≤ j < k. But since σ may also be written as (i2 · · · ik i1 ), σ k ﬁxes i2 by the preceding argument. Another induction shows that σ k ﬁxes each of the ij . Since it also ﬁxes the complement of {i1 , . . . , ik }, we have σ k = e. The permutation (ik · · · i1 ) is easily seen to reverse the eﬀect of σ. One of the reasons for studying the symmetric groups is that every ﬁnite group is isomorphic to a subgroup of Sn for some n. Recall that an embedding of groups is an injective homomorphism. Theorem 3.1.9. (Cayley’s Theorem) Let G be a group. For g ∈ G write g : G → G for the function g (g ) = gg . Then g is a permutation of G. Moreover, if we deﬁne Λ : G → S(G) by Λ(g) = g , then Λ is an embedding. In consequence, if G is ﬁnite, then it is isomorphic to a subgroup of S|G| . CHAPTER 3. G-SETS AND COUNTING 49 Proof To show that g is injective, suppose g (g ) = g (g ), and hence gg = gg . Multiplying both sides on the left by g −1 and cancelling, we get g = g . And g is surjective because g = g (g −1 g ) for all g ∈ G. Thus, g is a permutation of G. Now Λ(g) · Λ(g ) is the permutation g ◦ g . We wish to show this coincides with Λ(gg ), which is the permutation gg . We show they have the same value on each g ∈ G. ( g ◦ g )(g ) = g(g g ) = (gg )g = gg (g ). Thus, Λ is a homomorphism. To show it is an injection, we consider its kernel. Suppose that g = e. Thus, g (g ) = g for all g ∈ G, i.e., gg = g for all g ∈ G. But multiplying both sides of the equation on the right by (g )−1 , we get g = e. Thus, ker Λ = e, the trivial subgroup, so Λ is an embedding by Lemma 2.5.9. Recall that a set is countably inﬁnite if it may be put in one-to-one correspondence with the positive integers, Z+ . It is easy to show that Z is countably inﬁnite. Since every nonzero element of Z has inﬁnite order, the next corollary follows from Cayley’s Theorem. Corollary 3.1.10. Let X be a countably inﬁnite set. Then S(X) contains elements of inﬁnite order. Cayley’s Theorem shows that any ﬁnite group G is isomorphic to a subgroup of S|G| . But most groups are actually isomorphic to a group of permutations of a smaller number of letters. As we shall see in a moment, there is a natural choice of embedding of Sn in Sn+1 . Thus, if G embeds in Sn , then it also embeds in Sn+1 . In fact, if Y is a subset of X, there is a natural embedding of S(Y ) in S(X): Deﬁnition 3.1.11. Let Y be a subset of X, and let i : Y ⊂ X be the inclusion map. Deﬁne i∗ : S(Y ) → S(X) by σ(x) if x ∈ Y (i∗ (σ))(x) = x otherwise. We shall refer to i∗ as the canonical inclusion of S(Y ) in S(X). Lemma 3.1.12. Let i : Y ⊂ X be the inclusion of a subset, Y , of a set X. Then the canonical inclusion i∗ : S(Y ) → S(X) is an embedding onto the subgroup of S(X) consisting of those elements of S(X) which ﬁx all elements of the complement of Y in X. Proof Write X − Y for the complement of Y in X. It is easy to see that i∗ is an embedding and that elements in the image of i∗ ﬁx all elements of Y − X. Thus, it suﬃces to show that any element of S(X) that ﬁxes the elements of Y − X lies in the image of i∗ . Thus, suppose that τ ∈ S(X) ﬁxes the elements of of X − Y . Note that it suﬃces to show that τ restricts to a bijection from Y to itself, as then this restriction deﬁnes an element of S(Y ), which we shall denote by τ . Since τ is the identity on X − Y and agrees with τ on Y , we obtain that i∗ (τ ) = τ , as desired. We ﬁrst show that τ (Y ) ⊂ Y . Suppose not. Then we can ﬁnd y ∈ Y such that τ (y) = x ∈ Y − X. But since τ ﬁxes Y − X, we have τ (x) = x, so x and y have the same image under τ . This is impossible, as τ is injective. CHAPTER 3. G-SETS AND COUNTING 50 Thus, τ restricts to a function from Y to Y , which is injective, because restrictions of injective functions are always injective. But note that if x ∈ X − Y , then τ (x) cannot lie in Y . Thus, since τ is onto, we must have τ (Y ) = Y . Thus, τ restricts to a bijection from Y to itself, and the result follows. Taking i to be the inclusion map {1, . . . , n − 1} ⊂ {1, . . . , n}, we get the canonical inclusion Sn−1 ⊂ Sn . The next corollary is now immediate from Lemma 3.1.12. Corollary 3.1.13. The canonical inclusion of Sn−1 in Sn identiﬁes Sn−1 with the subgroup of Sn consisting of those permutations that ﬁx n. Note that the inclusion of {1, . . . , n} in the positive integers, Z+ , induces a canonical inclusion Sn ⊂ S(Z+ ) which is compatible with the inclusion of Sn in Sn+1 . Deﬁnition 3.1.14. Write i∗ : Sn ⊂ S(Z+ ) for the canonical inclusion, and deﬁne S∞ ⊂ S(Z+ ) by S∞ = i∗ (Sn ). n≥1 Lemma 3.1.15. S∞ is a subgroup of S(Z+ ).1 Proof Since each (i∗ (x))−1 = i∗ (x−1 ), S∞ is closed under inverses, and it suﬃces to show that S∞ is closed under multiplication. For x, y ∈ S∞ , there are m, n ≥ 1 such that x ∈ i∗ (Sm ) and y ∈ i∗ (Sn ). Suppose, then, that m ≤ n. Since the canonical inclusions of the Sk in S(Z+ ) are compatible with their canonical inclusions in one another, we have i∗ (Sm ) ⊂ i∗ (Sn ), and hence x, y ∈ i∗ (Sn ). But i∗ (Sn ) is a subgroup of S(Z+ ), and hence xy ∈ i∗ (Sn ) ⊂ S∞ . In practice, we identify each Sn with its image under i∗ , and treat Sn as a subgroup of S∞ (and hence also of S(Z+ )). Exercises 3.1.16. 1. Show that S2 is isomorphic to Z2 . 2. Show that S(X) is nonabelian whenever X has more than two elements. 3. Show that the square of a 4-cycle is not a 4-cycle. 4. What is the order of (1 2) · (3 4) in S4 ? 5. What is the order of (1 2 3) · (4 5) in S5 ? 6. Show that there is an embedding f : D6 → S3 induced by setting f (a) = (1 3) and f (b) = (1 2 3). We shall show in Section 3.3 that S3 has order 6. We may then deduce that f is an isomorphism. 7. Deﬁne x, y ∈ S4 by x = (1 4) · (2 3) and y = (1 2 3 4). Show that there is an embedding f : D8 → S4 induced by setting f (a) = x and f (b) = y. 8. Show that D12 embeds in S5 . 9. Show that D2n embeds in Sn for all n ≥ 3. 1 Those who are acquainted with category theory should note that we can identify S∞ with the direct limit lim −→ Sn . CHAPTER 3. G-SETS AND COUNTING 51 10. Show that Z2 × Z2 embeds in S4 . 11. Show that Zm × Zn embeds in Sm+n . 12. Show that the subgroup S∞ ⊂ S(Z+ ) is an inﬁnite group whose elements all have ﬁnite order. 13. Show that an element of S(Z+ ) lies in S∞ if and only if it ﬁxes all but ﬁnitely many of the elements of Z+ . 14. Let G be a group and let g ∈ G. Deﬁne rg : G → G by rg (x) = xg. Show that rg ∈ S(G). Show that the function R : G → S(G) deﬁned by R(g) = rg−1 is an embedding. ‡ 15. Let f : Y → Z be any injective function, and deﬁne f∗ : S(Y ) → S(Z) by f (σ(y)) if z = f (y) (f∗ (σ))(z) = z if z is not in the image of f . (a) Show that f∗ is an embedding onto the subgroup of S(Z) consisting of those permutations that ﬁx all elements of Z which are not in f (Y ). (b) If f is a bijection, show that this f∗ coincides with the isomorphism f∗ given in Deﬁnition 3.1.3. (c) If f is the inclusion of a subset Y ⊂ Z, show that this f∗ coincides with the embedding f∗ given in Deﬁnition 3.1.11. (d) Let g : Z → W be another injection. Show that g∗ ◦ f∗ = (g ◦ f )∗ . (e) If X is ﬁnite, and if f : X → Z+ is injective, show that the image of f∗ lies in S∞ . Thus, there is an induced embedding f∗ : S(X) → S∞ . 16. Let X be a set. Show that the set of all functions f : X → X forms a monoid under the operation of composition of functions. Show that S(X) is its group of invertible elements. 17. Show that the set of functions from {1, 2} to itself which take 1 to itself gives, via composition of functions, the unique monoid with two elements that is not a group. 3.2 Cosets and Index: Lagrange’s Theorem We here begin the study of a very important concept in algebra: the left cosets of a subgroup H ⊂ G. One of the consequences of this consideration is the famous theorem of Lagrange: If H is a subgroup of the ﬁnite group G, then the order of H divides the order of G. Lagrange’s Theorem is the very ﬁrst step in trying to classify groups up to isomorphism. Deﬁnition 3.2.1. Let H be a subgroup of G and let x ∈ G. Then the left coset xH of H in G is the following subset of G: xH = {xh | h ∈ H}. CHAPTER 3. G-SETS AND COUNTING 52 Thus, as x ranges over the elements of G, we get a collection of diﬀerent subsets xH ⊂ G, which are the left cosets of H in G. Note that eH is just H ⊂ G. The coset H is special in that it is the only left coset of H in G which is a subgroup. (The next lemma shows that e ∈ xH if and only if xH = H.) We study the cosets to understand how H relates to the other elements of G. Example 3.2.2. Let G be a cyclic group of order 4, written multiplicatively, with generator b: G = {e, b, b2 , b3 }. Let H be the cyclic subgroup generated by b2 : H = b2 = {e, b2 }. Then an easy inspection shows that eH = b2 H = {e, b2 }, while bH = b3 H = {b, b3 }. Since we have inspected xH for each element x ∈ G, we see that these are all of the left cosets of H in G. Recall that cosets are deﬁned to be subsets of G. Thus, since bH and b3 H have the same elements, they are equal as cosets. In particular, for this choice of H and G, there are exactly two left cosets of H in G. We now show that two left cosets that have a single element in common must be equal. Lemma 3.2.3. Let H be a subgroup of G and let x, y ∈ G. Then the following statements are equivalent. 1. xH = yH. 2. xH ∩ yH = ∅. 3. x−1 y ∈ H. 4. y −1 x ∈ H. Proof Clearly, the ﬁrst statement implies the second. Suppose, then, that the second holds. Let z ∈ xH ∩ yH. Then there are h, h ∈ H with z = xh = yh . But then x−1 y = h(h )−1 ∈ H. Thus, the second statement implies the third. The third and fourth statements are equivalent, as y −1 x is the inverse of x−1 y. Suppose, then, that the fourth statement holds. Thus, y −1 x = k ∈ H, and hence x = yk. But then xh = ykh ∈ yH for all h ∈ H, and hence xH ⊂ yH. But also, y = xk −1 , so that yh = xk −1 h for all h ∈ H, and hence yH ⊂ xH. Thus, xH = yH, and we see that the fourth statement implies the ﬁrst. Since x = x · e we have x ∈ xH for each x ∈ G. But any two left cosets of H that have an element in common must be equal by Lemma 3.2.3. Corollary 3.2.4. Let H be a subgroup of G. Then each element x ∈ G is contained in exactly one left coset of H, and that coset is xH. In particular, the elements x and y lie in the same left coset of H if and only if xH = yH. If the operation of G is written additively, we have an additive notation for cosets. Notation 3.2.5. Suppose that G is abelian and that we use additive notation for the group operation in G. Then we also use additive notation for the left cosets of a subgroup H ⊂ G. For x ∈ G, we write x + H for the coset which contains x: x + H = {x + h | h ∈ H}. It is very useful in group theory to consider the set of all left cosets of H in G. CHAPTER 3. G-SETS AND COUNTING 53 Deﬁnitions 3.2.6. Let H be a subgroup of G. Write G/H for the set whose elements are the cosets xH of H in G. Thus, if xH = yH as subsets of G, then xH and yH are equal as elements of G/H. If the number of elements of G/H (which is equal to the number of distinct left cosets of H in G) is ﬁnite, we say that H has ﬁnite index in G, and write [G : H], the index of H in G, for the number of elements in G/H. If the set G/H is inﬁnite, we write [G : H] = ∞. We deﬁne the canonical map, π : G → G/H, by π(x) = xH for x ∈ G. Remarks 3.2.7. Let H be a subgroup of G. We put a relation ∼ on G by setting x ∼ y if x−1 y ∈ H. Then ∼ is an equivalence relation: it is reﬂexive because (x−1 y)−1 = y −1 x, and is transitive because x−1 y · y −1 z = x−1 z. But x−1 y ∈ H if and only if y ∈ xH, so the equivalence class of x under ∼ is precisely xH. Thus, G/H is the set of equivalence classes of G under ∼, and the canonical map π : G → G/H, as deﬁned above, takes each x ∈ H to its equivalence class under ∼. Thus, π agrees with the canonical map of Deﬁnition 1.4.5. The following corollary is immediate from Corollary 3.2.4. Corollary 3.2.8. Let H be a subgroup of G and let π : G → G/H be the canonical map. Then π(x) = π(y) if and only if x and y are in the same left coset of H. Proposition 3.2.9. (Lagrange’s Theorem) Let H be a subgroup of G and suppose that H is ﬁnite. Then every left coset of H has |H| elements. Given that H is ﬁnite, if either of G or [G : H] is ﬁnite, then so is the other, and we have |G| = [G : H] · |H|. In particular, if G is a ﬁnite group and H is a subgroup of G, then the order of H divides the order of G. Proof For x ∈ G, the permutation x deﬁned in the last section (by x (g) = xg) restricts to a bijection from H to xH (with inverse induced by x−1 ). Thus, if H is ﬁnite, each coset xH has the same number of elements as H. Since each element of G lies in exactly one left coset of H, the order of G, if ﬁnite, is equal to the sum of the “orders” of the distinct left cosets of H. Since each left coset has |H| elements, this says that |G| must be equal to the number of left cosets of H times the order of H. Of course, if both H and [G : H] are ﬁnite, then G is the union of a ﬁnite number of the ﬁnite sets xH, and hence G is ﬁnite as well. Recall that the order of a given element of a group is equal to the order of the cyclic subgroup generated by that element (Corollary 2.5.18). Corollary 3.2.10. Let G be a ﬁnite group. Then the order of any element of G must divide the order of G. We should point out that the converse of Lagrange’s Theorem is false: if k divides the order of G, it is not necessarily the case that G will have a subgroup of order k. For instance, in Problem 2 of Exercises 4.2.18, we shall see that the alternating group A4 , which has order 12, has no subgroup of order 6. We shall spend some time investigating partial converses to Lagrange’s Theorem, as they are quite useful in obtaining classiﬁcation results. In Cauchy’s Theorem, we will CHAPTER 3. G-SETS AND COUNTING 54 show that if a prime p divides the order of a group, then G has a subgroup of order p. Then, in the Sylow Theorems, we will show that G has a subgroup whose order is the highest power of p that divides the order of G. We can get even better partial converses to Lagrange’s Theorem if we put extra assumptions on the group G. For instance, Hall’s Theorem (Theorem 5.6.2) gives a result that holds for what’s known as solvable groups: if G is solvable and |G| = nk with (n, k) = 1, then G has a subgroup of order n. Even here, we do not get a full converse, as the alternating group A4 is solvable. But the full converse does hold in ﬁnite abelian groups, or, more generally, in the ﬁnite nilpotent groups, which we shall study in Section 5.7. Lagrange’s Theorem allows us to classify the groups of order 4. Corollary 3.2.11. Let G be a group of order 4 which is not cyclic. Then G is isomorphic to Z2 × Z2 . Proof Let a = b be elements of G, neither of which is the identity element. Neither a nor b can have order 4, as then G would be cyclic. Thus, both have order 2. Now ab = a, as that would imply that b = e, and ab = b, because then a would be the identity element. And ab = e, as then a and b would be inverse to one another. But an element of order 2 is its own inverse, so this would mean a = b. Thus, the four elements of G are e, a, b, and ab. A similar argument shows that ba cannot equal e, a, or b. But this forces ab = ba, so that a and b commute. But now Proposition 2.10.8 shows that G is the internal direct product of a and b. There are a couple of very important consequences of Lagrange’s Theorem. Corollary 3.2.12. Let H and K be subgroups of G such that |H| and |K| are relatively prime. Then H ∩ K = e. Proof Let x ∈ H ∩ K. Then Corollary 3.2.10 shows that o(x) must divide both |H| and |K|. Since |H| and |K| are relatively prime, this forces o(x) = 1. Corollary 3.2.13. Let G and G be groups such that |G| and |G | are relatively prime. Then the only homomorphism from G to G is the trivial homomorphism, which sends each x ∈ G to e ∈ G . Proof Let x ∈ G. Then o(f (x)) divides o(x) (Corollary 2.5.19), which in turn divides |G|. But o(f (x)) also divides |G |, as f (x) ∈ G . Since |G| and |G | are relatively prime, we must have o(f (x)) = 1. So far, we have studied the left cosets of a subgroup H of G. There is an analogous notion of right cosets, which we shall study in the exercises below: Deﬁnition 3.2.14. Let H be a subgroup of the group G. The right cosets of H in G are the subsets Hx = {hx | h ∈ H} for x ∈ G. Clearly, if G is abelian, then xH = Hx, and the right and left cosets of H coincide. For some nonabelian groups, the left and right cosets of a given subgroup can be totally diﬀerent subsets of G. However, we shall see that the left and right cosets of H agree if H is what’s known as a normal subgroup of G. CHAPTER 3. G-SETS AND COUNTING 55 Exercises 3.2.15. 1. Let p be a prime number. Show that any group of order p is isomorphic to Zp . 2. For a prime number p, show that the only subgroups of Zp are e and Zp itself. 3. Let H be a subgroup of the ﬁnite group G and let K be a subgroup of H. Show that [G : K] = [G : H] · [H : K]. † 4. Generalize the preceding problem to the case where G is inﬁnite, but [G : K] is ﬁnite. 5. State and prove the analogue of Lemma 3.2.3 for right cosets. 6. Show that every subgroup of Q8 other than Q8 itself is cyclic. List all of the subgroups. 7. Find all of the left cosets of a ⊂ D6 . Now ﬁnd all of the right cosets and show that they are diﬀerent subsets of D6 . 8. Repeat the preceding exercise with D8 in place of D6 . 9. Repeat the preceding exercise for D2n with n arbitrary. 10. What are the right cosets of b ⊂ D2n ? 11. Find all of the left cosets of a ⊂ Q12 . 12. Let H be a subgroup of G. Show that there is a one-to-one correspondence between the left cosets of H and the right cosets of H. Deduce that H has ﬁnite index in G if and only if there are ﬁnitely many distinct right cosets of H in G, and that the number of right cosets, if ﬁnite, is [G : H]. 13. Show that the order of a submonoid of a ﬁnite monoid need not divide the order of the larger monoid. 3.3 G-sets and Orbits Deﬁnitions 3.3.1. Let G be a group. An action of G on a set X is a rule that assigns to each ordered pair (g, x) with g ∈ G and x ∈ X an element g · x ∈ X,2 such that 1. e · x = x for all x ∈ X 2. The following analogue of the associative law holds: g1 · (g2 · x) = (g1 g2 ) · x for all g1 , g2 ∈ G and x ∈ X. A set X together with an action of G on X is called a G-set. If X and Y are G-sets, then a G-map, or G-equivariant map, f : X → Y is a function f from X to Y such that f (g · x) = g · f (x) for all g ∈ G and x ∈ X. A G-isomorphism is a G-map which is a bijection. 2 In other words, the passage from the ordered pair (g, x) to the product g · x deﬁnes a function from G × X to X. CHAPTER 3. G-SETS AND COUNTING 56 We will often abbreviate g · x by gx. We now note that we may identify two G-sets which are G-isomorphic. Lemma 3.3.2. Let f : X → Y be a G-isomorphism. Then so is the inverse function f −1 : Y → X. Examples 3.3.3. 1. G itself is a G-set, under the action obtained by setting g · g equal to the product gg of g and g in G. We call this the standard action of G on G. 2. Let X be a set with one element: X = {x}. Then there is a unique action of G on X: g · x = x for all g ∈ G. 3. Let H be a subgroup of G. We let G act on G/H by g · xH = (gx)H for g, x ∈ G. As the reader may verify, this gives a well deﬁned action. Moreover, the canonical map π : G → G/H is easily seen to be a G-map, where G is given the action of the ﬁrst example. The G-sets G/H are sometimes called “homogeneous spaces” of G. Here, “spaces” comes from the fact that a similar construction is studied for topological groups G, where H is a subgroup of G which is also a closed subspace with respect to the topology of G. 4. Let G be any group and let X be any set. Then there is at least one action of G on X: just set g · x = x for all g ∈ G and x ∈ X. We call this the trivial action of G on X. Clearly, it does not tell us very much about either G or X. 5. Let X be a set. Then S(X) acts on X by σ · x = σ(x). Indeed, the facts that 1X is the identity of S(X) and that the group product in S(X) is composition of functions are precisely what is needed to show that this speciﬁes an action of S(X) on X. We give a generalization of this example in Proposition 3.3.16. 6. Let X be any set. Let G be a cyclic group of order 2, written multiplicatively: G = {e, a}, where a has order 2. Then there is an action of G on X × X, obtained by setting a · (x, y) = (y, x) for each ordered pair (x, y) ∈ X × X. As Example 3 suggests, the study of G-sets encodes information about the subgroups of G. We will use G-sets as one of our main tools in understanding the structure of groups. Deﬁnitions 3.3.4. Let X be a G-set and let x ∈ X. Then the isotropy subgroup of x, denoted Gx , is the set of elements of G which ﬁx x: Gx = {g ∈ G | g · x = x}. The orbit of x, denoted G · x, is the set of all elements obtained from x via multiplication by elements of G: G · x = {g · x | g ∈ G} ⊂ X. We say that G acts transitively on X if X = G · x for some x ∈ X. The next lemma is needed to justify the preceding deﬁnitions. Lemma 3.3.5. Let X be a G-set and let x ∈ X. Then the isotropy subgroup Gx is a subgroup of G. The orbit, G · x, of x is the smallest sub-G-set of X which contains x. CHAPTER 3. G-SETS AND COUNTING 57 Proof Clearly, Gx is a submonoid of G, and we must show that Gx is closed under inverses. But if g ∈ Gx , then g −1 · x = g −1 · (g · x) = (g −1 g) · x = x, and hence g −1 ∈ Gx . We leave the rest to the reader. As a ﬁrst example of these concepts, let us consider their application to the “homogeneous spaces” G/H. Lemma 3.3.6. In the G-set G/H, the isotropy subgroup of eH is H. Also, the orbit of eH is all of G/H, and hence G/H is a transitive G-set. Proof Here g · eH = gH, so that every element in G/H is in the orbit of eH. Also, g is in the isotropy subgroup of eH if and only if gH = eH. By Lemma 3.2.3, this is the case if and only if e−1 g ∈ H. Thus, the isotropy subgroup of eH is H, as claimed. G-maps are of course important in the understanding of G-sets. The reader should verify the following lemma. Lemma 3.3.7. Let f : X → Y be a G-map and let x ∈ X. Then we have an inclusion of isotropy subgroups Gx ⊂ Gf (x) . If, in addition, f is injective, then Gx = Gf (x) . In the study of G-sets, orbits play a role similar to (but simpler than) the role played by cyclic subgroups in the study of groups. Just as the cyclic groups Zn form models for the cyclic subgroups, every orbit turns out to be G-isomorphic to one of the G-sets G/H: Lemma 3.3.8. Let X be a G-set and let x ∈ X. Let H be a subgroup of G. Then there is a G-map fx : G/H → X with fx (eH) = x if and only if H ⊂ Gx . If fx exists, it is unique, and has image equal to the orbit G · x. Moreover, fx induces a G-isomorphism from G/H to G · x if and only if H = Gx . In particular, for X any G-set and x ∈ X, G · x is G-isomorphic to G/Gx . Thus, a G-set is transitive if and only if it is G-isomorphic to G/H for some subgroup H ⊂ G. Proof Suppose given a G-map fx : G/H → X with fx (eH) = x. Then since H is the isotropy subgroup of eH in G/H, it must be contained in the isotropy subgroup of fx (eH) = x. In other words, we must have H ⊂ Gx . Also, since gH = g · eH, fx (gH) = g · fx (eH) = gx, so fx is uniquely determined by its eﬀect on eH. Clearly, its image is G · x. Conversely, if H ⊂ Gx , we use the above formula to deﬁne fx : G/H → X: fx (gH) = gx. We must show this is well deﬁned: i.e. if gH = g H, we must show that gx = g x. By Lemma 3.2.3, gH = g H if and only if g = g h for some h ∈ H. But then gx = g hx = g x, since hx = x. Thus, fx is a well deﬁned function. To see that fx is a G-map, note that fx (g · gH) = fx ((g g)H) = g gx = g fx (gH). If fx is a G-isomorphism, then the isotropy subgroups of eH and fx (eH) must be equal. But the isotropy subgroup of eH in G/H is H, while the isotropy subgroup of x is Gx , so if fx is a G-isomorphism, then H = Gx . Conversely, suppose that H = Gx . To show that fx induces an isomorphism of G/Gx onto G · x, it suﬃces to show that fx is injective. Suppose that fx (gH) = fx (g H), and hence that gx = g x. Multiplying both sides by g −1 and cancelling, we see that x = g −1 g x, so that g −1 g is in the isotropy subgroup of x. Thus g −1 g ∈ Gx = H, so that gH = g H by Lemma 3.2.3. The next corollary now follows from Lagrange’s Theorem. CHAPTER 3. G-SETS AND COUNTING 58 Corollary 3.3.9. Let X be a ﬁnite G-set and let x ∈ X. Then the number of elements in the orbit G · x is equal to [G : Gx ]. In particular, since [G : Gx ] is ﬁnite, if either G or Gx is ﬁnite, so is the other, and G · x has |G|/|Gx | elements. As noted in Example 5 of Examples 3.3.3, S(X) acts on X for any set X, via σ · x = σ(x), for σ ∈ S(X) and x ∈ X. Thus, we may use the notions of orbit and isotropy to obtain information about S(X). We are primarily concerned with the case where X is ﬁnite. Recall from Corollary 3.1.13 that the canonical inclusion of Sn−1 in Sn identiﬁes Sn−1 with the subgroup of Sn consisting of those elements that ﬁx n. We restate this in terms of isotropy groups: Lemma 3.3.10. Consider the standard action of Sn on {1, . . . , n}. Under this action, the isotropy subgroup of n is the image of the canonical inclusion Sn−1 ⊂ Sn . Corollary 3.3.9 now allows us to give an inductive calculation of the order of Sn . Corollary 3.3.11. The order of Sn is n! (n-factorial). Proof Since the 2-cycle (i n) takes n to i for each 1 ≤ i < n, each element of {1, . . . , n} is in the orbit of n under the standard action of Sn . In other words, Sn acts transitively on {1, . . . , n}. Since this orbit has n elements, and since the isotropy subgroup of n is Sn−1 , we have n = |Sn |/|Sn−1 | by Corollary 3.3.9. The result follows by induction on n, starting with the obvious fact that S1 is the trivial group. It is often desirable to count the elements in a G-set X by adding up the numbers of elements in its orbits. To do that, we shall need the following lemma. Lemma 3.3.12. Let X be a G-set and let x, y ∈ X. Then y ∈ G · x if and only if G · y = G · x. In consequence, if x, z ∈ X, then G · x = G · z if and only if G · x ∩ G · z = ∅. Proof Since y ∈ G · x, y = gx for some g ∈ G. Thus, g y = (g g)x ∈ G · x for all g ∈ G, so G · y ⊂ G · x. But x = g −1 y, so G · x ⊂ G · y by the same reasoning. Notice that the analogy between the orbits in a G-set and the cyclic subgroups of a group has just been left behind, as we can ﬁnd lots of examples (e.g., in cyclic groups) of inclusions K ⊂ H of cyclic subgroups which are not equal. Deﬁnition 3.3.13. Let X be a G-set. Then a set of orbit representatives in X is a subset Y ⊂ X such that 1. If x, y ∈ Y with x = y, then G · x ∩ G · y = ∅. 2. X = x∈Y G · x. For a general G-set, existence of a set of orbit representatives turns out to be an important special case of the Axiom of Choice. But the case for ﬁnite G-sets is much easier. Lemma 3.3.14. Any ﬁnite G-set has a set of orbit representatives. Proof Let X be our G-set. We argue by induction on the number of elements in X. Let x ∈ X. If G · x = X, then {x} is the desired set. Otherwise, let X be the complement of G · x in X. By Lemma 3.3.12, if g ∈ G and x ∈ X , gx must lie in X . Thus, X is a sub-G-set of X. By induction, X has a set, Y , of orbit representatives. But then Y = Y ∪ {x} is a set of orbit representatives for X. CHAPTER 3. G-SETS AND COUNTING 59 For inﬁnite sets, one can prove the existence of a set of orbit representatives by using Zorn’s Lemma (a form of the Axiom of Choice) to perform what amounts to an inﬁnite induction argument. We shall use similar techniques in the material on ring theory in Chapters 7 through 12. The following corollary is now immediate from Corollary 3.3.9. Corollary 3.3.15. (G-set Counting Formula) Let Y be a set of orbit representatives for the ﬁnite G-set X. Then the number, |X|, of elements in X is given by [G : Gx ]. |X| = x∈Y An understanding of G-sets is vital to an understanding of permutation groups because of the following generalization of Cayley’s Theorem. Proposition 3.3.16. Let X be a G-set. For g ∈ G, deﬁne g : X → X by g (x) = gx. Then g is a permutation of X. Moreover, the function Λ : G → S(X) deﬁned by Λ(g) = g is a group homomorphism, whose kernel is the intersection of the isotropy subgroups of the elements of X: ker Λ = Gx . x∈X Conversely, if Λ : G → S(X) is any group homomorphism, then X is a G-set under the action deﬁned by g · x = Λ(g)(x). In fact, the correspondence which takes a G-set X to the homomorphism Λ : G → S(X) gives a one-to-one correspondence between the actions of G on X and the homomorphisms from G to S(X). Proof Let X be a G-set and let g ∈ G. We wish to show that g is a permutation of X. It is onto, because x = g (g −1 x) for all x ∈ X. It is one-to-one, since if g (x) = g (y), then gx = gy, and, multiplying both sides by g −1 , we see that x = y. Now Λ is a homomorphism if and only if g ◦ g = gg for all g, g ∈ G. But clearly, ( g ◦ g )(x) = gg x = gg (x) for all x ∈ X, so Λ is indeed a homomorphism. The kernel of Λ is the set of elements in G which act on X by the identity map, and hence ﬁx each element in X. These then are the group elements which lie in the isotropy group of every element of X. Conversely, given the homomorphism Λ, and deﬁning g · x to be equal to Λ(g)(x), we have that ex = x because Λ(e) is the identity permutation, while g1 (g2 x) = Λ(g1 )(Λ(g2 )(x)) = (Λ(g1 ) ◦ Λ(g2 ))(x) = Λ(g1 g2 )(x) = (g1 g2 )x. Finally, the passage from the G-set X to the homomorphism Λ is easily seen to be inverse to the passage from a homomorphism Λ : G → S(X) to an action on X. Note that if the G-set X is G (with the usual action), then the map Λ : G → S(G) given above is precisely the one given in Cayley’s Theorem. Deﬁnitions 3.3.17. Let X be a G-set. 1. We say that the action of G on X is eﬀective, or that X is an eﬀective G-set, if x∈X Gx = e. In words, G acts eﬀectively on X if the only element of G which ﬁxes every element of X is e. CHAPTER 3. G-SETS AND COUNTING 60 2. We say that the action of G on X is free, or that X is a free G-set, if the isotropy subgroup of every element of X is e. In other words, X is free if for any g ∈ G and x ∈ X, gx = x implies that g = e. Clearly, any free action is eﬀective. However, there are many eﬀective actions that are not free. Examples 3.3.18. 1. The standard action of G on G is free. Indeed, if gx = x, then right multiplication by x−1 gives g = e. 2. The standard action of S(X) on X is eﬀective, since the only function X → X which ﬁxes every element of X is the identity function. Note, however, that if X has more than two elements, then this action is not free. For instance, a 2-cycle (i j) ﬁxes every element other than i and j. Via the map Λ : G → S(X) of Proposition 3.3.16, we may interpret eﬀective actions in terms of embeddings into symmetric groups. Corollary 3.3.19. An action of a group G on a set X is eﬀective if and only if the induced map Λ : G → S(X) is an embedding. Thus, there is an embedding of G into Sn if and only if there is an eﬀective action of G on a set with exactly n elements. Proof For any G-set X, Proposition 3.3.16 tells us that the kernel of theinduced map Λ : G → S(X) is x∈X Gx . Since eﬀective actions are those for which x∈X Gx = e, they are precisely those actions for which Λ is an embedding. Thus, the key to discovering whether a given group embeds in Sn lies in understanding the isotropy groups that appear in the various orbits of G-sets. We shall take an ad hoc approach to these questions at this time, but the reader may want to come back to this material after reading Section 3.7. Deﬁnition 3.3.20. By a minimal subgroup of G we mean a subgroup H = e with the property that the only subgroups of H are e and H. This last is, of course, an abuse of language. Such a subgroup should more properly be called a minimal nontrivial subgroup. But the trivial subgroup is rather uninteresting, so we make the above deﬁnition for the purposes of economy of language. (Similarly, in our discussion of commutative rings, we shall use the expression “maximal ideal” for an ideal which is maximal in the collection of all ideals which are not equal to the ring itself.) Note that not all groups have minimal subgroups. For instance, it is easy to see that every nontrivial subgroup of Z contains a subgroup which is strictly smaller, but nontrivial. But this cannot happen in a ﬁnite group. Lemma 3.3.21. Every nontrivial ﬁnite group G has a minimal subgroup. Proof We argue by induction on the order of G. If G has no subgroups other than e and itself (e.g. if |G| = 2), then G is a minimal subgroup of itself. Otherwise, there is a nontrivial subgroup H = G. But then |H| < |G|, so that H has a minimal subgroup by induction. However, a minimal subgroup of H is also minimal in G. CHAPTER 3. G-SETS AND COUNTING 61 Most ﬁnite groups have several diﬀerent minimal subgroups, which are not necessarily isomorphic to each other. However, in the case that there is a unique minimal subgroup, we obtain a strong consequence for the embedding problem. Proposition 3.3.22. Let G be a ﬁnite group with exactly one minimal subgroup. Then any eﬀective G-set has at least |G| elements. In particular, n = |G| is the smallest integer such that G embeds in Sn . Proof Let H be the minimal subgroup of G. We claim that every nontrivial subgroup of G contains H. To see this, let K be a nontrivial subgroup of G. Then there is a minimal subgroup K ⊂ K. But any minimal subgroup of K is also a minimal subgroup of G. Since G has only one minimal subgroup, we must have K = H, and hence H ⊂ K. Now suppose given an eﬀective G-set X. We claim there must be at least one x ∈ X with Gx = e. Suppose not. Then each Gx is a nontrivial subgroup, and hence H ⊂ Gx for each x. But then H ⊂ x∈X Gx , and hence X is not eﬀective. Thus, there is an x ∈ X with Gx = e. But then G · x is G-isomorphic to G/e, and hence X contains an orbit which has |G| elements in it. In particular, X has at least |G| elements. Exercises 3.3.23. 1. Show that the standard action on G is G-isomorphic to G/e and that the one-point G-set is G-isomorphic to G/G. ‡ 2. Let G = D6 . Show that G acts eﬀectively (but not freely) on G/a, and hence that G embeds in S3 . Deduce from Corollary 3.3.11 that this embedding is an isomorphism. 3. Generalize the preceding problem to obtain an embedding from D2n into Sn . Since the order of the two groups is diﬀerent for n > 3, this will not be an isomorphism for these values of n. 4. Let p be a prime and let r ≥ 1. Show that Zpr cannot embed in Sn for n < pr . 5. Show that Q8 cannot embed in Sn for n < 8. 6. Show that Q2r cannot embed in Sn for n < 2r . 7. Find the minimal value of n such that D2r embeds in Sn . Do the same for D2pr , where p is an odd prime. 8. Let X and Y be G-sets. By X × Y , we mean the G-set given by the action of G on the cartesian product of X and Y in which g · (x, y) = (gx, gy), for all g ∈ G, x ∈ X, and y ∈ Y . What is the isotropy subgroup of (x, y) ∈ X × Y ? 9. Let G = Z6 . Let H, K ⊂ G be the subgroups of order 2 and 3, respectively. Show that G/H × G/K is a free G-set. 10. Show that the action of G on X×Y need not be transitive, even if G acts transitively on both X and Y . 11. Show that if G acts freely on X, then it also acts freely on X × Y for any Y . CHAPTER 3. G-SETS AND COUNTING 62 12. There is a construction in set theory called the disjoint union. The disjoint union of X and Y, written X Y , is a set containing both X and Y in the following way: in X Y we have X ∩ Y = ∅, and X ∪ Y = X Y . We shall give a formal construction of the disjoint union in Chapter 6. Show that if X and Y are G-sets, then G acts on X Y in such a way that X and Y are G-subsets. 13. Let G = Z6 . Let H, K ⊂ G be the subgroups of order 2 and 3, respectively. Show that G acts eﬀectively (but not freely) on G/H G/K. Deduce that Z6 embeds in S5 . 14. Let m and n be relatively prime. Show that Zmn embeds in Sm+n . 15. Show that Q12 embeds in S7 . 16. What is the isotropy subgroup of gH ∈ G/H for an arbitrary g ∈ G? 17. Let X be a G-set and let H be a subgroup of G. The set of H-ﬁxed points of X, written X H , is deﬁned to be the set of x ∈ X such that hx = x for all h ∈ H. Show that x ∈ X H if and only if H ⊂ Gx . ‡ 18. Let X be a G-set and let H be a subgroup of G. Show that there is a one-to-one correspondence between the G-maps from G/H to X and the elements of X H . ‡ 19. Let X be a G-set. Show that there is an equivalence relation on X deﬁned by setting x ∼ y if y = gx for some g ∈ G. Show that the equivalence classes of this relation are precisely the orbits G · x for x ∈ X. The set of equivalence classes thus obtained is called the orbit space of X, and denoted X/G. (Thus, the elements of X/G are the orbits G · x in X.) We write π : X → X/G for the canonical map, which takes x to its equivalence class. 20. A G-map f : X → Y is said to be isovariant if Gx = Gf (x) for each x ∈ X. Show that a G-map f : X → Y is isovariant if and only if its restriction to each orbit is injective. 21. Recall that a G-set Y has the trivial action if gy = y for all y ∈ Y and g ∈ G. Show that if X is any G-set, and if we give the orbit space X/G the trivial action, then the canonical map π : X → X/G is a G-map. Show also that if Y is any G-set with the trivial action, and if f : X → Y is a G-map, then there is a unique G-map f : X/G → Y such that f ◦ π = f . 22. Show that if G acts trivially on X and if Y is any G-set, then the G-maps from G × X to Y are in one-to-one correspondence with the functions from X to Y . (Here, G in the above product is the standard free G-set G/e.) In this context, we call G × X the free G-set on the set X. ‡ 23. Let X be any set and write X n for the product X × · · · × X of n copies of X. Show that X n is an Sn -set via σ · (x1 , . . . , xn ) = (xσ−1 (1) , . . . , xσ−1 (n) ), for σ ∈ Sn and (x1 , . . . , xn ) ∈ X n . † 24. Let G be a ﬁnite group whose order is divisible by a prime number p. In this exercise, we shall prove Cauchy’s Theorem (also given by a diﬀerent proof as Theorem 5.1.1), which says that such a group G must have an element of order p. CHAPTER 3. G-SETS AND COUNTING 63 (a) Deﬁne σ : Gp → Gp by σ(g1 , . . . , gp ) = (gp , g1 , . . . , gp−1 ). Show that σ gives a permutation of Gp of order p (i.e., σ ∈ S(Gp ) with σ p = e). Deduce that there is an action of Zp on Gp , given by k · (g1 , . . . , gp ) = σ k (g1 , . . . , gp ) for all k ∈ Zp . (In fact, if we identify Zp with the subgroup of Sn generated by the p-cycle (1 2 · · · p), this action is just the restriction to Zp of the Sn action given in Problem 23.) (b) Deﬁne X ⊂ Gp by X = {(g1 , . . . , gp ) | g1 · · · gp = e}. In other words, the product (in order) of the gi is the identity element of G. Show that X is a sub-Zp -set (but not a subgroup, unless G is abelian) of Gp . (c) Show that the isotropy group of (g1 , . . . , gp ) ∈ X is the trivial subgroup of Zp unless g1 = g2 = · · · = gp = g for some g ∈ G with g p = e. Deduce that there is a one-to-one correspondence between the ﬁxed-point set X Zp and the set {g ∈ G | g p = e}. Deduce that G has an element of order p if and only if X Zp has more than one element. (d) Note that if x ∈ X is not in X Zp , then the orbit Zp · x has p elements. Since two orbits are either disjoint or equal, X is the disjoint union of X Zp with the orbits which lie outside of X Zp . Deduce that |X| = |X Zp | + some multiple of p. Show that |X| = |G| , so that |X| is divisible by p. Deduce that |X Zp | is divisible by p, and hence G has an element of order p. p−1 25. We say that G acts on the right on a set X (or that X is a right G-set) if for all x ∈ X and g ∈ G there is an element xg ∈ X such that (xg)g = x(gg ) and xe = x for all x ∈ X and all g, g ∈ G. Show that if X is a right G-set, then there is an induced left action of G on X given by g · x = xg −1 . Deduce that there is a one-to-one correspondence between the left G-sets and the right G-sets. 26. Let H be a subgroup of G. Then H acts on the right of G by ordinary multiplication. Show that we may identify the set of left cosets G/H with the orbit space (see Problem 19) of this action. 27. Let G act on the right on X and act on the left on Y . Then G acts on the left on X × Y by g · (x, y) = (xg −1 , gy). We call the orbit space of this action the balanced product of X and Y , which we denote X ×G Y . Let H be a subgroup of G and let X be a (left) H-set. Then using the right action of H on G by ordinary multiplication, we can form the balanced product G ×H X. Show that the standard left action of G on itself induces a left action of G on G ×H X. Write [g, x] for the element of G×H X represented by the ordered pair (g, x) ∈ G×X. (In other words, [g, x] is the orbit of (g, x) under the above action of H on G × X.) Show that the isotropy subgroup G[e,x] of [e, x] under the above G-action coincides with the isotropy subgroup Hx of x under the original action of H on X. CHAPTER 3. G-SETS AND COUNTING 64 28. Let H be a subgroup of G and let X be a (left) H-set. Let Y be a (left) G-set. Then H acts on Y by restricting the action of G. Show that there is a one-to-one correspondence between the H-maps from X to Y and the G-maps from G ×H X to Y . 29. Let M be a monoid. Deﬁne an M -set by substituting M for G in the deﬁnition of G-set. Find a monoid M and an M -set X such that the function m : X → X deﬁned by m (x) = m · x need not be a permutation for all values of m. 30. Show that if you delete the requirement that e · x = x from the deﬁnition of G-set, then the functions g need not be permutations. 3.4 Supports of Permutations We isolate here some of the more elementary considerations for the study of cycle structure in Section 3.5. We give some consequences in the exercises. Deﬁnition 3.4.1. Let σ ∈ S(X). The support of σ, Supp(σ), is the following subset of X: Supp(σ) = {x ∈ X | σ(x) = x} In particular, σ is the identity outside of its support. It is easy to calculate the support of a k-cycle. Lemma 3.4.2. The support of the k-cycle (i1 · · · ik ) is {i1 , . . . , ik }. Here is another way to think of supports. Lemma 3.4.3. Let Y be a subset of X and let σ ∈ S(X). Then Supp(σ) ⊂ Y if and only if σ is the identity on the complement of Y . In particular, if i : Y → X is the inclusion of Y in X, then Supp(σ) ⊂ Y if and only if σ is in the image of the canonical inclusion i∗ : S(Y ) ⊂ S(X). Proof The support is the complement of the subset of points ﬁxed by σ, hence the ﬁrst statement. For the second, recall from Lemma 3.1.12 that the image of i∗ is the set of permutations that ﬁx the complement of Y . Applying Lemma 3.4.3 to Y = Supp(σ), we obtain the following corollary. Corollary 3.4.4. Let σ be a permutation of the set X. Then σ restricts to a bijection σ : Supp(σ) → Supp(σ). Deﬁnition 3.4.5. We say the permutations σ, τ ∈ S(X) are disjoint if Supp(σ) ∩ Supp(τ ) = ∅. Disjoint permutations interact well with one another: Proposition 3.4.6. Let σ and τ be disjoint permutations of X. Then σ and τ commute with each other. Moreover, the product στ is given as follows: ⎧ ⎨ σ(x) if x ∈ Supp(σ) τ (x) if x ∈ Supp(τ ) (στ )(x) = ⎩ x otherwise. CHAPTER 3. G-SETS AND COUNTING 65 In particular, the support of στ is the union Supp(σ) ∪ Supp(τ ). Finally, στ has ﬁnite order if and only if both σ and τ have ﬁnite order, in which case the order of στ is the least common multiple of the orders of σ and τ . Proof We wish to show that both στ and τ σ have the eﬀect speciﬁed by the displayed formula. Away from Supp(σ) ∪ Supp(τ ), both σ and τ are the identity, and hence both στ and τ σ are also. On Supp(σ), τ is the identity, as Supp(σ) ∩ Supp(τ ) = ∅. Thus, for x ∈ Supp(σ), (στ )(x) = σ(x). Since x ∈ Supp(σ), σ(x) is also in Supp(σ) by Corollary 3.4.4, and hence (τ σ)(x) = σ(x) as well. A similar argument shows that both στ and τ σ act as τ on Supp(τ ). Thus, στ = τ σ, and the displayed formula holds. An easy induction based on the displayed formula now shows: ⎧ k ⎨ σ (x) if x ∈ Supp(σ). τ k (x) if x ∈ Supp(τ ). For k > 0, (στ )k (x) = ⎩ x otherwise. Thus, (στ )k = e if and only if both σ k = e and τ k = e, and the result follows. Exercises 3.4.7. 1. Show that an element σ ∈ S(Z+ ) lies in S∞ if and only if Supp(σ) is ﬁnite. ‡ 2. Suppose given subsets Y, Z ⊂ X, with Y ∩ Z = ∅. Show that there is an embedding μ : S(Y )×S(Z) → S(X) whose restrictions to S(Y ) = S(Y )×e and S(Z) = e×S(Z) are the canonical inclusions of S(Y ) and S(Z), respectively, in S(X). Show that the image of μ is the set of permutations σ ∈ S(X) such that the following two properties hold: (a) Supp(σ) ⊂ Y ∪ Z (b) σ(Y ) = Y . ‡ 3. Combinatorics Consider the action of Sn on {0, 1}n obtained by permuting the factors (see Problem 23 of Exercises 3.3.23). We can use it to derive some of the standard properties of the binomial coeﬃcients. First, for 0 ≤ k ≤ n, let us write xk = (1, . . . , 1, 0, . . . , 0 ) ∈ {0, 1}n . k times n−k times (a) Show that the number of elements in the orbit Sn · xk is equal to the number of ways to choose k things out of n things (which we may identify with the number of k-element subsets of an n-element set), which we shall denote by the binomial coeﬃcient nk . (b) Show that the isotropy subgroup of xk is isomorphic to the direct product Sk × Sn−k . Deduce that n n! = , k k! · (n − k)! and that the right-hand side must therefore be an integer. CHAPTER 3. G-SETS AND COUNTING 66 (c) Show that the elements x0 , . . . , xn form a set of orbit representatives for the Sn action on {0, 1}n . Deduce that n 2 = n n k=0 k . (d) Deduce that there are exactly 2n subsets of an n-element set. (e) Show that the number of ordered k-element subsets of an n-element set is k! times the number of unordered k-element subsets of that set. Deduce that the number of ordered k-element subsets is n!/(n − k)!. 4. It can be shown that if X is any inﬁnite set, then the disjoint union X X may be placed in bijective correspondence with X. Deduce from Problem 2 that for any inﬁnite set X, we may embed S(X) × S(X) in S(X). 5. Reconsider the preceding problem for X = Z+ . Show that for any choice of injection Z+ Z+ → Z+ , the induced embedding S(Z+ ) × S(Z+ ) ⊂ S(Z+ ) restricts to an embedding of S∞ × S∞ in S∞ . (These embeddings turn out to be important in studying the cohomology of the group S∞ , an endeavor which falls under the heading of homological algebra, but has applications to topology.) 6. In this problem we assume a familiarity with inﬁnite disjoint unions and with inﬁnite products, as discussed in Chapter 6. (a) Suppose given a ! family of sets {X i | i ∈ I}. Show that we may embed the inﬁnite product i∈I S(Xi ) in S( i∈I Xi ). Deduce that there is a permutation of Z+ which may be thought of as the disjoint product of inﬁnitely many cycles, with one k-cycle being given for each k ≥ 2. Deduce that there are permutations in S(Z+ ) whose orbits are all ﬁnite, but which have inﬁnite order. (b) Construct a permutation of Z+ which is a product of inﬁnitely many cycles of the same order. Deduce that there are elements of S(Z+ ) which have ﬁnite order but do not lie in S∞ . 3.5 Cycle Structure Here, using the ideas of G-sets and orbits, we develop the notion of cycle decompositions for permutations. This gives a powerful tool for studying permutation groups that has some analogies to prime decompositions in Z. Using cycle decomposition, we shall construct the sign homomorphism ε : Sn → {±1} and deﬁne the alternating groups, An . We shall encounter further applications of cycle decomposition in all of our subsequent study of permutation groups, particularly in Section 4.2. Recall that the permutations σ and τ are disjoint if Supp(σ) ∩ Supp(τ ) = ∅, where Supp(σ) is the complement of the set of points ﬁxed by σ. Deﬁnition 3.5.1. By a product of disjoint permutations, we mean a product σ1 · · · σk , such that σi and σj are disjoint for i = j. We are particularly interested in products of disjoint cycles: i.e., products σ1 · · · σk of disjoint permutations in Sn , such that each σi is a cycle. CHAPTER 3. G-SETS AND COUNTING 67 We now generalize Proposition 3.4.6. Corollary 3.5.2. Let σ1 · · · σk ∈ S(X) be a product of disjoint permutations. Then the eﬀect of σ1 · · · σk is given as follows: on Supp(σi ), the eﬀect of σ1 · · · σk is the same k as that of σi for 1 ≤ i ≤ k. Away from i=1 Supp(σi ), σ1 · · · σk is the identity. In k particular, Supp(σ1 · · · σk ) = i=1 Supp(σi ). In addition, if each σi has ﬁnite order, then the order of σ1 · · · σk is the least common multiple of the orders of the σi . Proof We argue by induction on k. The case k = 2 is given k−1by Proposition 3.4.6. For k > 2, the induction hypothesis gives Supp(σ1 · · · σk−1 ) = i=1 Supp(σi ), and hence σ1 · · · σk−1 is disjoint from σk . The result now follows by applying Proposition 3.4.6 (and the induction hypothesis) to the product (σ1 · · · σk−1 ) · σk . The following proposition gives the cycle decomposition for a permutation, σ. Proposition 3.5.3. Let σ ∈ Sn for n ≥ 2, with σ = e. Then σ may be written uniquely as a product of disjoint cycles: σ = σ1 · · · σk . Here, uniqueness means that if σ may also be written as the product τ1 · · · τl of disjoint cycles, then k = l, and there is a permutation φ ∈ Sk such that σi = τφ(i) for 1 ≤ i ≤ k. Proof Let G = σ, the cyclic subgroup of Sn generated by σ. We wish to understand X = {1, . . . , n} as a G-set. Note that if the orbit of x ∈ X has only one element, then the isotropy subgroup of x has index 1 in G by Corollary 3.3.9. This means that the isotropy subgroup of x must be all of G. If this were true for all x ∈ X, then G would act as the identity on X. Since σ = e, this cannot be the case. Suppose, then, that y ∈ X has an orbit with more than one element. Let r be the smallest positive integer with the property that σ r (y) = y. (Such an integer exists, because σ o(σ) (y) = y.) Note that r > 1, as otherwise y is ﬁxed under σ and all of its powers, and hence the orbit of y under G would have only one element. We claim ﬁrst that the elements y, σ(y), . . . , σ r−1 (y) are all distinct. To see this, note that otherwise there are integers s and t with 0 ≤ s < t < r such that σ s (y) = σ t (y) = σ s (σ t−s (y)). Thus, y and σ t−s (y) have the same image under σ s . Since σ s is injective, this gives σ t−s (y) = y. Since 0 < t − s < r, this contradicts the minimality of r. Thus, there is an r-cycle σ1 = (y σ(y) · · · σ r−1 (y)). Let Y1 = Supp(σ1 ) = {y, σ(y), . . . , σ r−1 (y)}. Then, by inspection, we see that σ has the same eﬀect as σ1 on each element of Y1 . Thus, σ restricts to a bijection Y1 → Y1 . By passage to powers, we see that Y1 is a sub-G-set of X. Since G acts transitively on Y1 , we see that Y1 is the orbit of y under the action of G. Now let Y1 , . . . , Yk be the set of all distinct orbits in X with more than one element. By Lemma 3.3.12, Yi ∩ Yj = ∅ for i = j. Repeating the argument above, we can ﬁnd cycles σi for 2 ≤ i ≤ k such that σi has the same eﬀect on Yi is σ does, and Supp(σi ) = Yi . In particular, σ1 · · · σk is a product of disjoint cycles. We claim that σ = σ1 · · · σk . By Corollary 3.5.2, the eﬀect of σ1 · · · σk on Yi = Supp(σi ) is the same as that of σi . But σi was chosen to have the same eﬀect on Yi as k σ. Thus, σ agrees with σ1 · · · σk on i=1 Yi = Supp(σ1 · · · σk ). Since σ is the identity on k all one-element orbits, it is the identity oﬀ i=1 Yi , and σ = σ1 · · · σk , as claimed. We show uniqueness by reversing the above argument. In a product τ = τ1 . . . τl of disjoint cycles, the eﬀect of τ agrees with that of τi on Supp(τi ). As the cycle notation shows, τi acts transitively on Supp(τi ), so τ acts transitively on Supp(τi ) as well. CHAPTER 3. G-SETS AND COUNTING 68 Thus, there are l distinct orbits under the action of τ which have more than one element. Thus, if τ is equal to the permutation σ, above, then k = l and, equating the orbits of σ and τ , we may relabel the τi so that Supp(τi ) = Supp(σi ) for i = 1, . . . , k. But since σ = σi and τ = τi on this common support, τi and σi are equal as permutations. Remarks 3.5.4. Note the concluding statement of the last proof. A cycle is determined by its eﬀect as a permutation, not by its notation. However, it’s easy to see that there are precisely r diﬀerent ways to write an r-cycle in cycle notation. It depends solely on which element of its support is written ﬁrst. Proposition 3.5.3 shows that every permutation is a product of cycle permutations. But we can actually write a permutation as a product of even simpler terms as long as we no longer insist that these terms be disjoint. Recall that a transposition is another word for a 2-cycle. The following lemma may be veriﬁed by checking the eﬀect of both sides of the proposed equality on the elements {i1 , . . . , ik }. Here, we should emphasize the fact that if τ, τ ∈ Sn and j ∈ {1, . . . , n}, then (τ τ )(j) = τ (τ (j)). Lemma 3.5.5. Let σ be the k-cycle σ = (i1 · · · ik ). Then σ is the product of k − 1 transpositions: σ = (i1 i2 )(i2 i3 ) · · · (ik−1 ik ). Since the cycle permutations generate Sn , we see that every element of Sn is a product of transpositions. Corollary 3.5.6. Sn is generated by the transpositions. We wish to deﬁne the sign of a permutation to be ±1 in such a way that the sign will be +1 if the permutation may be written as a product of an even number of transpositions, and is −1 otherwise. Our initial deﬁnition will be more rigid than this. Deﬁnition 3.5.7. The sign of a k-cycle is (−1)k−1 . Thus, the sign is −1 if k is even and is +1 if k is odd. The sign of a product of disjoint cycles is the product of the signs of the cycles. The sign of the identity map is +1. Thus, the sign of every permutation is deﬁned, via Proposition 3.5.3. Notice the relationship between the sign of a cycle and the number of transpositions in the product given in Lemma 3.5.5. Multiplying out the signs over a product of disjoint cycles, we see that if the sign of a permutation, σ, is +1, then there is at least one way to write σ as the product of an even number of transpositions. If the sign of σ is −1, then there is at least one way to write σ as the product of an odd number of transpositions. The next proposition is the key to the utility of signs. Notice that {±1} forms a group under multiplication. Since it has two elements, it is isomorphic to Z2 . Proposition 3.5.8. For any σ, τ ∈ Sn , n ≥ 2, the sign of στ is the product of the signs of σ and τ . Thus, there is a homomorphism ε : Sn → {±1} obtained by setting ε(σ) equal to the sign of σ. In particular, ε has the property that ε(τ ) = −1 for any transposition τ . CHAPTER 3. G-SETS AND COUNTING 69 Proof Since we may write τ as a product of m transpositions in such a way that the sign of τ is (−1)m , we may argue by induction, assuming that τ is a transposition, say τ = (i j). Write σ as a product of disjoint cycles: σ = σ1 · · · σk . There are four cases to consider, where a non-trivial orbit of σ is one which has more than one element: 1. Neither i nor j lies in a non-trivial orbit of σ. 2. j lies in a non-trivial orbit of σ and i does not. 3. i and j lie in diﬀerent non-trivial orbits of σ. 4. Both i and j lie in the same non-trivial orbit of σ. We analyze the decomposition of στ as a product of disjoint cycles in each of these cases. The ﬁrst case is simplest, as then σ and τ are disjoint. Thus, τ is one of the cycles in the decomposition of στ , and the others are those from the decomposition of σ. The result follows in this case from the deﬁnition of sign. In case 2, we may assume that σ is an r-cycle, σ = (i1 · · · ir ), with ir = j. Then στ = (i i1 · · · ir ), an (r + 1)-cycle. In case 3, we may assume that σ is the product of two disjoint cycles, σ1 = (i1 · · · ir ), with ir = i, and σ2 = (j1 · · · js ), with js = j. Here, we see that στ is the (r + s)-cycle στ = (i1 · · · ir j1 · · · js ). The signs work out correctly once again. Case 4 is the most complicated. We may assume that σ is a single cycle. The analysis falls naturally into sub-cases. In the ﬁrst, σ = τ = (i j), and here στ = e, as transpositions have order 2. The next case is where σ is an r-cycle with r ≥ 3 and i and j appear next to each other in the cycle representation of σ, say σ = (i1 · · · ir ) with ir−1 = i and ir = j. Here στ is the (r − 1)-cycle στ = (i1 · · · ir−1 ). In the ﬁnal case, i and j cannot be written next to each other in the cycle representation of σ. Thus, we may write σ = (i1 · · · ir j1 · · · js ), with ir = i, js = j, and with r and s both greater than or equal to 2. Here, στ is given by στ = (i1 · · · ir )(j1 · · · js ). The sign homomorphism gives rise to an important subgroup of Sn . Deﬁnitions 3.5.9. The n-th alternating group, An , is the kernel of the sign homomorphism ε : Sn → {±1}. We say that a permutation is even if it lies in An (i.e., if its sign is 1). A permutation is said to be odd if its sign is −1. The fact that ε is a homomorphism allows us to determine the sign of a permutation from any representation of it as a product of transpositions. Corollary 3.5.10. Let σ ∈ Sn . Then σ ∈ An if and only if σ may be written as a product of an even number of transpositions. If σ ∈ An , then whenever σ is written as a product of transpositions, the number of transpositions must be even. Proof The homomorphism ε carries any product of an odd number of transpositions to −1 and carries any product of an even number of transpositions to 1. Thus, given any two presentations of a permutation σ as a product of transpositions, the number of transpositions in the two presentations must be congruent mod 2. We have made use of the fact that Sn is generated by the transpositions. Using Corollary 3.5.10, we may obtain a similar result for An . CHAPTER 3. G-SETS AND COUNTING 70 Corollary 3.5.11. For n ≥ 3, the alternating group An is generated by 3-cycles. Thus, since each 3-cycle lies in An , a permutation σ ∈ Sn lies in An if and only if σ is a product of 3-cycles. Proof Corollary 3.5.10 shows that a permutation is in An if and only if it is the product of an even number of transpositions. Thus, it suﬃces to show that the product of any two transpositions is a product of 3-cycles. If the two transpositions are not disjoint, then either they are equal, in which case their product is the identity, or the product has the form (a b) · (b c) = (a b c), a 3-cycle. The remaining case is the product of two disjoint cycles, (a b) and (c d). But then we have (a b) · (c d) = (a b c) · (b c d), the product of two 3-cycles. Exercises 3.5.12. 1. Write the product (1 2 3 4) · (3 4 5) as a product of disjoint cycles. What is the order of this product? 2. Consider the function from Z7 to itself induced by multiplication by 2. (Recall that multiplication mod n gives a well-deﬁned binary operation on Zn .) Identifying the elements of Z7 with {1, . . . , 7} (by identifying k with k for 1 ≤ k ≤ 7), show that multiplication by 2 induces a permutation of {1, . . . , 7}. Write down its cycle structure. 3. Repeat the preceding problem using multiplication by 3 in place of multiplication by 2. 4. What are the elements of A3 ? 5. What are the elements of A4 ? 6. Recall from Problem 2 of Exercises 3.3.23 that S3 is isomorphic to the dihedral group D6 . Which subgroup of D6 corresponds to A3 under this isomorphism? 7. Let p be prime and let r ≥ 1. Show that Sk has an element of order pr if and only if k ≥ pr . (Hint: Write the element in question as a product of disjoint cycles. What can you say about the length of the cycles?) 8. For a given n, what is the smallest k such that Zn embeds in Sk ? (Hint: The answer depends on the prime decomposition of n.) 9. What is the largest value of n such that Zn embeds in S8 ? ‡ 10. Expanding on Remarks 3.5.4, show that the number of k-cycles in Sn is 1/k times the number of ordered k-element subsets of {1, . . . , n}. Deduce from Problem 3 of n! Exercises 3.4.7 that there are k·(n−k)! k-cycles in Sn . CHAPTER 3. G-SETS AND COUNTING 3.6 71 Conjugation and Other Automorphisms In addition to the study of actions of the group G on sets, it is useful to study actions of other groups on G. Of course, S(G) acts on G, but we have seen that the structure of a symmetric group S(X) depends only on the size of the set X. In particular, the action of S(G) on G loses all the information about the group structure of G, and captures only its size. To capture the group structure on G, a group should act on it through permutations σ : G → G which are group homomorphisms. These are precisely the group isomorphisms from G to itself. Deﬁnitions 3.6.1. An automorphism of a group G is a group isomorphism from G to itself. We write Aut(G) ⊂ S(G) for the set of all automorphisms of G. The reader may check the details of the following examples, using Problem 6 of Exercises 2.8.5 for the examples involving dihedral groups, and Problem 5 of Exercises 2.9.7 for the examples involving quaternionic groups. Examples 3.6.2. 1. Let G be an abelian group. Then there is an automorphism f : G → G given by f (g) = g −1 for all g ∈ G. 2. Let 0 ≤ k < n. Then there is an automorphism f ∈ Aut(D2n ) determined by f (b) = b, f (a) = abk . 3. There is an automorphism f ∈ Aut(D2n ) determined by f (b) = b−1 , f (a) = a. 4. Let 0 ≤ k < 2n. Then there is an automorphism f ∈ Aut(Q4n ) determined by f (b) = b, f (a) = abk . 5. There is an automorphism f ∈ Aut(Q4n ) determined by f (b) = b−1 , f (a) = a. We now state an immediate consequence of Lemma 2.5.7. Lemma 3.6.3. Let G be a group. Then Aut(G) is a subgroup of S(G). Deﬁnition 3.6.4. We say that the group K acts on G via automorphisms if G is a K-set in such a way that the action of each k ∈ K, k : G → G (deﬁned as usual by k (g) = k · g), is an automorphism of G. The proof of the following proposition is identical to that of Proposition 3.3.16. Proposition 3.6.5. There is a one-to-one correspondence between the actions of K on G via automorphisms and the homomorphisms Λ : K → Aut(G). Here, a given action corresponds to the homomorphism deﬁned by Λ(k) = k . This formalism is useful, as we are about to deﬁne a very important action of G on itself through automorphisms, called conjugation. But we cannot denote the resulting G-set by “G,” as we reserve this symbol for the standard action induced by left multiplication. So it is useful to be able to refer to conjugation in terms of a homomorphism from G to Aut(G). Deﬁnition 3.6.6. Let G be a group and let x ∈ G. By conjugation by x, we mean the function cx : G → G speciﬁed by cx (g) = xgx−1 . CHAPTER 3. G-SETS AND COUNTING 72 Lemma 3.6.7. Conjugation by x ∈ G is an automorphism of G. There is an action of G on itself by automorphisms given by the homomorphism Γ : G → Aut(G) deﬁned by Γ(x) = cx . Proof To see that cx is a homomorphism, we have cx (g)cx (g ) = xgx−1 xg x−1 = xgg x−1 = cx (gg ). It is a bijection for the same reason that left or right multiplication by any element of G is a bijection: There is an inverse function given by an appropriate multiplication. In this case the inverse function is conjugation by x−1 . Thus, there is a well-deﬁned function Γ : G → Aut(G) with Γ(x) = cx . To see that Γ is a group homomorphism, we have (cx ◦ cy )(g) = cx (ygy −1 ) = xygy −1 x−1 = (xy)g(xy)−1 = cxy (g), because (xy)−1 = y −1 x−1 . The elements in the image of Γ : G → Aut(G) have a special name. Deﬁnitions 3.6.8. An automorphism f ∈ Aut(G) is said to be an inner automorphism of G if f = cx , the conjugation by x, for some x ∈ G. We write Inn(G) ⊂ Aut(G) for the set of inner automorphisms. Note that since Inn(G) = im Γ, Inn(G) is a subgroup of Aut(G). The isotropy subgroup of y ∈ G under the conjugation action is by deﬁnition the set {x ∈ G | xyx−1 = y}. It is an important enough subgroup that it merits a name: Deﬁnition 3.6.9. Let y be an element of the group G. Then the centralizer of y in G is the subgroup CG (y) = {x ∈ G | xyx−1 = y}. Of course, if x ∈ H ⊂ G, then we write CH (x) for the centralizer of x in H. When there’s no ambiguity, we shall simply refer to CG (x) as the centralizer of x. The next lemma is useful in understanding the eﬀect of conjugation. The proof is left to the reader. Lemma 3.6.10. Let x, y ∈ G. Then the following are equivalent. 1. x and y commute. 2. x ∈ CG (y), i.e., xyx−1 = y. 3. xyx−1 y −1 = e. We now give two useful consequences of Lemma 3.6.10. Corollary 3.6.11. Suppose that the elements x, y ∈ G commute with each other. Then every element of x commutes with every element of y. Proof Since x ∈ CG (y) and since CG (y) is a subgroup of G, x ⊂ CG (y). Thus, y commutes with xk for all k ∈ Z, and hence y ∈ CG (xk ) as well. So y ⊂ CG (xk ), and hence y l commutes with xk for all k, l ∈ Z. CHAPTER 3. G-SETS AND COUNTING 73 Corollary 3.6.12. The element x ∈ G is in the kernel of the conjugation-induced homomorphism Γ : G → Aut(G) if and only if x commutes with every element of G. Proof We have x ∈ ker Γ if and only if x is in the isotropy subgroup of every y ∈ G. But this says that x ∈ CG (y) for every y ∈ G, and hence x commutes with every element of G. We shall refer to the kernel of Γ as the center, Z(G), of G:3 Deﬁnition 3.6.13. The center, Z(G), of a group G is given as follows: Z(G) = {x ∈ G | xy = yx for all y ∈ G}. Deﬁnition 3.6.14. The conjugacy class of g ∈ G is {xgx−1 | x ∈ G}, which is the orbit of g under the action of G on itself by conjugation. Its elements are called the conjugates of g. Since the isotropy subgroup of g is CG (g), the next result follows from Corollary 3.3.9. Corollary 3.6.15. The number of conjugates of g ∈ G is equal to [G : CG (g)], the index in G of the centralizer, CG (g), of g in G. The fact that CG (g) is the set of all elements that commute with g gives a characterization of Z(G) in terms of conjugacy classes. Corollary 3.6.16. The conjugacy class of g ∈ G consists of a single element if and only if g ∈ Z(G). Proof [G : CG (g)] = 1 if and only if CG (g) = G. The G-set Counting Formula (Corollary 3.3.15) for the conjugation action on G is extremely useful. We shall give a variant on it, called the Class Formula. We shall need some setup for it ﬁrst. Deﬁnitions 3.6.17. We say that a conjugacy class of G is nontrivial if it has more than one element. By a set of class representatives for the nontrivial conjugacy classes we mean a subset X ⊂ G such that 1. The conjugacy class of each x ∈ X is nontrivial. 2. If x, y ∈ X with x = y, then the conjugacy classes of x and y intersect in the null set. 3. Every element of G that is not in Z(G) lies in the conjugacy class of some x ∈ X. The desired Class Formula is now immediate from the G-set Counting Formula. Corollary 3.6.18. (Class Formula) Let G be a ﬁnite group and let X be a set of class representatives for the nontrivial conjugacy classes in G. Then [G : CG (x)]. |G| = |Z(G)| + x∈X 3 The German word for center is Zentrum. CHAPTER 3. G-SETS AND COUNTING 74 We can compute the conjugates and the centralizers of the elements in the dihedral and quaternionic groups by hand, using the explicit information we have about these groups. We shall leave these calculations for the next set of exercises. The next result computes the conjugates of a k-cycle in Sn . Lemma 3.6.19. Let σ be any permutation and let (i1 · · · ik ) be any k-cycle in Sn . Then σ · (i1 · · · ik ) · σ −1 is a k-cycle: σ · (i1 · · · ik ) · σ −1 = (σ(i1 ) · · · σ(ik )). Proof If j is not in {σ(i1 ), . . . , σ(ik )}, then σ −1 (j) is not in {i1 , . . . , ik }, and hence (i1 · · · ik ) acts as the identity on σ −1 (j). But then σ · (i1 · · · ik ) · σ −1 (j) = j, so that σ · (i1 · · · ik ) · σ −1 acts as the identity outside {σ(i1 ), . . . , σ(ik )}. For 1 ≤ j < k, σ · (i1 · · · ik ) · σ −1 (σ(ij )) = σ · (i1 · · · ik )(ij ) = σ(ij+1 ). Similarly, σ · (i1 · · · ik ) · σ −1 (σ(ik )) = σ(i1 ), so that σ · (i1 · · · ik ) · σ −1 = (σ(i1 ) · · · σ(ik )), as claimed. Lemma 3.6.19 gives important information about the behavior of the symmetric groups. Here is a sample. Corollary 3.6.20. Suppose that σ ∈ Sn lies in the centralizer of the k-cycle (i1 · · · ik ). Then σ({i1 , . . . , ik }) = {i1 , . . . , ik }. Proof Let σ ∈ Sn Then σ · (i1 · · · ik ) · σ −1 = (σ(i1 ) · · · σ(ik )). Thus, Supp(σ · (i1 · · · ik ) · σ −1 ) = {σ(i1 ), . . . , σ(ik )} = σ({i1 , . . . , ik }). Thus, if σ · (i1 · · · ik ) · σ −1 = (i1 · · · ik ), then σ({i1 , . . . , ik }) = Supp((i1 · · · ik )) = {i1 , . . . , ik }. There is an important action related to conjugation, in which G acts on the set of homomorphisms from another group into G. We shall use a notation consistent with that used in our discussion of category theory in Chapter 6. Notation 3.6.21. Let K and G be groups. We write MorGp (K, G) for the set of group homomorphisms from K to G. Deﬁnition 3.6.22. Let K and G be groups. For g ∈ G and f ∈ MorGp (K, G), the conjugate of f by g, written g · f , is the homomorphism (g · f ) ∈ MorGp (K, G) deﬁned by (g · f )(k) = gf (k)g −1 for k ∈ K. Two homomorphisms f, f ∈ MorGp (K, G) are said to be conjugate if f = g · f for some g ∈ G. Note that g · f is equal to the composite cg ◦ f , where cg : G → G is conjugation by g. Thus, g · f is a homomorphism, as claimed. The following lemma is now immediate. CHAPTER 3. G-SETS AND COUNTING 75 Lemma 3.6.23. Let K and G be groups. Then there is an action of G on MorGp (K, G) obtained by setting the action of g ∈ G on f ∈ MorGp (K, G) to be equal to the conjugate homomorphism g · f deﬁned above. Conjugate homomorphisms are important in a number of questions, including the classiﬁcation of semidirect products and the cohomology of groups. Exercises 3.6.24. 1. What are all the conjugacy classes of elements of D6 ? Write down all the elements in each class. 2. What are all the conjugacy classes of elements of D8 ? Write down all the elements in each class. 3. What are all the conjugacy classes of elements of Q8 ? Write down all the elements in each class. 4. What are all the conjugacy classes of elements of D2n for n ≥ 3? (Hint: The answer depends in a crucial way on the value of n mod 2.) 5. What are all the conjugacy classes of elements of Q4n ? 6. Find the centralizers of each of the elements of Q8 . 7. Find the centralizers of the elements a, b ∈ D2n . 8. Find the centralizers of the elements a, b ∈ Q4n . 9. Let H be a subgroup of G and let x ∈ H. Show that CH (x) = H ∩ CG (x). 10. Suppose that Sn has an element of order k. Show that D2k embeds in Sn . 11. Consider the 2-cycle (n − 1 n) ∈ Sn . Show that the centralizer of (n − 1 n) in Sn is isomorphic to Sn−2 × Z2 . ‡ 12. Let a, b, c, d be distinct elements of {1, . . . , n}. Show that any element of Sn which centralizes both (a b c) and (b c d) must ﬁx both a and d. Deduce that if n ≥ 4, then the only element which centralizes every 3-cycle is the identity. 13. Show that the centralizer of any k-cycle in Sn is isomorphic to Sn−k × Zk . We shall see in Section 4.2 (using a straightforward application of Lemma 3.6.19 above) that n! any two k-cycles in Sn are conjugate. Assuming this, deduce that there are k·(n−k)! k-cycles in Sn . (See Problem 10 of Exercises 3.5.12 for a diﬀerent line of argument.) 14. What is the centralizer of (1 2 3) in A4 ? How many conjugacy classes of 3-cycles are there in A4 ? 15. Let σ, τ ∈ S(X). Show that Supp(τ στ −1 ) = τ (Supp(σ)). 16. What is the center of D8 ? 17. What is the center of D2n for n ≥ 3? (Hint: The answer depends in a crucial way on the value of n mod 2.) CHAPTER 3. G-SETS AND COUNTING 76 18. What is the center of Q8 ? 19. What is the center of Q4n ? 20. What is the center of Sn ? What is the center of An ? ‡ 21. Let H be any group and write H n for the product H × · · · × H of n copies of H. Show that Sn acts on H n through automorphisms via σ · (h1 , . . . , hn ) = (hσ−1 (1) , . . . , hσ−1 (n) ). 22. Verify that the speciﬁcations given in Examples 3.6.2 do in fact give automorphisms of the groups in question. 23. Show that the only automorphism of Z4 other than the identity is the one which takes each element to its inverse. 24. Let G be a ﬁnite abelian group and let f : G → G be given by f (g) = g n for all g ∈ G, where n is an integer that is relatively prime to |G|. Show that f is an automorphism of G. If G is cyclic of order 7 and n = 2, what is the order of f in Aut(G)? 25. Show that not every automorphism of D8 is an inner automorphism. 26. Show that every automorphism of D6 is an inner automorphism. (Hint: The composite of two inner automorphisms is an inner automorphism.) Deduce that Aut(D6 ) ∼ = D6 . 27. Show that any two nontrivial homomorphisms from Z2 to D6 are conjugate. 28. Show that any two nontrivial homomorphisms from Z3 to D6 are conjugate. 29. What are the conjugacy classes of homomorphisms from Z4 to D8 ? 30. Let G be a group and let f : G → Sn be a homomorphism. Write {1, . . . , n}f for the G-set obtained from the action on {1, . . . , n} induced by f under the correspondence of Proposition 3.3.16: for g ∈ G and i ∈ {1, . . . , n}f , g · i = f (g)(i). Let σ ∈ Sn and let f = σ · f , the conjugate homomorphism obtained by conjugating f by σ. Show that σ induces a G-isomorphism of G-sets, σ : {1, . . . , n}f → {1, . . . , n}f . 31. Let G be a group. Show that there is a one-to-one correspondence between the conjugacy classes of homomorphisms from G to Sn and the G-isomorphism classes of the G-sets with n elements. Here, two G-sets are in the same G-isomorphism class if and only if they are G-isomorphic. 32. Let X and Y be sets, with X ﬁnite. Show that if i : X → Y and j : X → Y are any two injective functions, then the induced homomorphisms i∗ : S(X) → S(Y ) and j∗ : S(X) → S(Y ) (as deﬁned in Problem 15 of Exercises 3.1.16) are conjugate. (Hint: In the case that Y is inﬁnite, set theory shows that Y may be put in one-to-one correspondence with the complement of any of its ﬁnite subsets.) 33. Let i and j be injective functions from {1, . . . , n} to Z+ . Show that the induced homomorphisms i∗ , j∗ : Sn → S∞ are conjugate. CHAPTER 3. G-SETS AND COUNTING 3.7 77 Conjugating Subgroups: Normality We haven’t yet made full use of the fact that conjugation is an action by automorphisms. The key point is that if f is an automorphism of G and if H ⊂ G is a subgroup, then f (H) = {f (h) | h ∈ H} is also a subgroup. We get an induced action of Aut(G) on the set Sub(G) of all subgroups of G, via f · H = f (H). The following lemma is immediate. Lemma 3.7.1. If K acts on G via automorphisms, then there is an induced action of K on Sub(G), the set of all subgroups of G. Let’s consider the action of G on Sub(G) induced by conjugation. Here, for x ∈ G and H ∈ Sub(G), the subgroup obtained by letting x act on H is xHx−1 = {xhx−1 | h ∈ H}: Deﬁnition 3.7.2. Let H be a subgroup of G, and let x ∈ G. The conjugate of H by x is xHx−1 = {xhx−1 | h ∈ H}, the subgroup obtained by letting x act on H by conjugation. Of course, the set of all conjugates of H is the orbit of H under the action of G on Sub(G) by conjugation. The isotropy subgroup of H ∈ Sub(G) under the conjugation action is called NG (H), the normalizer of H in G: Deﬁnition 3.7.3. Let H be a subgroup of G. The normalizer in G of H, NG (H), is given as NG (H) = {x ∈ G | xHx−1 = H}. It is the isotropy subgroup of H under the action of G on Sub(G) by conjugation. Our study of G-sets now allows us to calculate the size of the orbit of H under conjugation. Corollary 3.7.4. The number of conjugates of H ⊂ G is [G : NG (H)], the index of the normalizer of H in G. Of course, if x ∈ H, then conjugation by x gives an automorphism of H, and hence xHx−1 = H. Corollary 3.7.5. Let H be a subgroup of G. Then H ⊂ NG (H). In particular, the number of conjugates of H in G divides [G : H]. Proof We just saw that xHx−1 = H for x ∈ H, so H ⊂ NG (H), as claimed. The last statement follows from Problem 4 of Exercises 3.2.15: If [G : H] is ﬁnite, then [G : H] = [G : NG (H)] · [NG (H) : H]. (The reader should be able to verify this easily in the case that G is ﬁnite, which is the most important case for our purposes.) Thus, we have H ⊂ NG (H) ⊂ G. In practice, we can ﬁnd examples where NG (H) = H, and others where NG (H) strictly contains H and is strictly contained in G. Finally, there is the case where NG (H) = G, which is important enough to merit a name. Deﬁnitions 3.7.6. A subgroup H ⊂ G is called normal if NG (H) = G (i.e., xHx−1 = H for all x ∈ G). We write H G to indicate that H is a normal subgroup of G. A subgroup H ⊂ G is called characteristic if f (H) = H for all automorphisms f of G. CHAPTER 3. G-SETS AND COUNTING 78 Clearly, a characteristic subgroup of G is normal in G. However, not all normal subgroups are characteristic. For instance, conjugation is trivial in an abelian group, and hence every subgroup of an abelian group is normal. However, as we shall see in the exercises below, the only characteristic subgroups of Z2 × Z2 are e and the group itself. Thus, abelian groups can have subgroups which are not characteristic. We shall consider many more examples of both normal and characteristic subgroups in the exercises. If H G, then we get an action of G on H through automorphisms: Lemma 3.7.7. Let H be a normal subgroup of G. Then for x ∈ G, conjugation by x gives an automorphism, cx , of H: cx (h) = xhx−1 for all h ∈ H. Moreover, there is a homomorphism Γ : G → Aut(H) deﬁned by Γ(x) = cx for all x ∈ G. Proof Since cx gives an automorphism of G, it restricts to a group isomorphism from H to cx (H) = xHx−1 . Since xHx−1 = H, this restriction is precisely an automorphism of H. Now clearly, cx ◦ cy = cxy , and the result follows. Essentially the same proof gives the following lemma. Lemma 3.7.8. Let H be a characteristic subgroup of G. Then every automorphism f : G → G restricts to an automorphism f : H → H. We obtain a restriction homomorphism ρ : Aut(G) → Aut(H), by setting ρ(f ) to be the restriction f : H → H for each f ∈ Aut(G). As we shall begin to see in the next chapter, the study of normal subgroups is one of the most important subjects in group theory. It is also a very complicated subject. For instance, as we shall see in the exercises below, Gertrude Stein’s Theorem fails for normal subgroups: we can ﬁnd subgroups K ⊂ H ⊂ G such that K H and H G, but K is not normal in G. This failure of Gertrude Stein’s Theorem helps motivate the study of characteristic subgroups: Corollary 3.7.9. Suppose given subgroups K ⊂ H ⊂ G such that H G and K is a characteristic subgroup of H. Then K G. Proof Let x ∈ G. Then conjugation by x restricts to an automorphism of H, and hence preserves K. Recognition of normal subgroups is an important endeavor. Of course, if H is ﬁnite, then any injection from H to H is a bijection, so that xHx−1 ⊂ H if and only if xHx−1 = H. However, there are examples of inﬁnite groups H ⊂ G and x ∈ G where xHx−1 is contained in but not equal to H. However, if xHx−1 ⊂ H for all x ∈ G, things work out nicely: Proposition 3.7.10. Let H be a subgroup of G. Then H G if and only if xHx−1 ⊂ H for all x ∈ G. Also, H is characteristic if and only if f (H) ⊂ H for all automorphisms f of G. Proof Suppose that xHx−1 ⊂ H for each x ∈ G. To show that H G, we must show that H ⊂ xHx−1 for each x ∈ G. But x−1 Hx ⊂ H, and hence H = x(x−1 Hx)x−1 ⊂ xHx−1 , as desired. The proof for the case of characteristic subgroups is nearly identical: if f (H) ⊂ H for all f ∈ Aut(G), then f −1 (H) ⊂ H, so H = f (f −1 (H)) ⊂ f (H). CHAPTER 3. G-SETS AND COUNTING 79 Notice that the proof of Proposition 3.7.10 actually shows more than was stated. For instance, in the case of normality, it shows that if xHx−1 ⊂ H and x−1 Hx ⊂ H, then xHx−1 = H, and hence x ∈ NG (H). Thus, we have the following proposition. Proposition 3.7.11. Let H and K be subgroups of G. Suppose that xHx−1 ⊂ H for all x ∈ K. Then K ⊂ NG (H). One application of the concept of normality is the following variant of Proposition 2.10.8. Proposition 3.7.12. Let H and K be subgroups of the group G. Then G is the internal direct product of H and K if and only if the following conditions hold. 1. H ∩ K = e. 2. Both H and K are normal in G. 3. G is generated by the elements of H and K. Proof Suppose that G is the internal direct product of H and K. By Proposition 2.10.8, it suﬃces to show that H and K are normal in G. But by deﬁnition of the internal direct product, this will follow if we know that the subgroups H × e and e × K are normal in H × K. The veriﬁcation of this is left to the reader. Conversely, suppose the three conditions hold. By Proposition 2.10.8, it suﬃces to show that for each h ∈ H and k ∈ K, the elements h and k commute. By Lemma 3.6.10, it suﬃces to show that for h ∈ H and k ∈ K, hkh−1 k −1 = e. By condition 1, this says that hkh−1 k −1 ∈ H ∩ K. Now K G, so hkh−1 ∈ K, and hence hkh−1 k −1 ∈ K. Similarly, the normality of H in G gives kh−1 k −1 ∈ H, and hence hkh−1 k −1 ∈ H, so the result follows. Normality is intimately connected with the relationship between left and right cosets. The reader may verify the following lemma. Lemma 3.7.13. Let H be a subgroup of G. Then the following conditions are equivalent. 1. H is normal in G. 2. For each x ∈ G, we have xH = Hx. 3. Every left coset of H in G is equal to a right coset of H in G; i.e., for each x ∈ G there is an element y ∈ G such that xH = Hy. Exercises 3.7.14. 1. Let G be a group and let H ⊂ G × G be given by H = {(g, g)|g ∈ G}. (H is called the diagonal subgroup of G × G.) Show that H G × G if and only if G is abelian. 2. Let G be a group and let y ∈ G. Show that y G if and only if xyx−1 ∈ y for all x ∈ G. Show that y is a characteristic subgroup of G if and only if f (y) ∈ y for each f ∈ Aut(G). 3. Show that the subgroup b ⊂ D2n is normal. CHAPTER 3. G-SETS AND COUNTING 80 ‡ 4. Show that the subgroup b ⊂ D2n is characteristic for n ≥ 3. 5. Find a non-normal subgroup of D8 . 6. Show that every subgroup of Q8 is normal. 7. Find the unique characteristic subgroup of Q8 which is equal to neither e nor Q8 . ‡ 8. Show that if n ≥ 3, then b ⊂ Q4n is characteristic. ‡ 9. Let H be a subgroup of Zn and let f : Zn → Zn be any homomorphism. Show that f (H) ⊂ H. Deduce that every subgroup of Zn is characteristic. (Hint: Show that H = d = {dk | k ∈ Zn } for some d that divides n.) † 10. Using the fact that An is the subgroup of Sn generated by the 3-cycles, show that An Sn . Thus Lemma 3.7.7 provides a homomorphism Γ : Sn → Aut(An ), where Γ(σ) is induced by conjugation by σ for all σ ∈ Sn . Deduce from Problem 12 of Exercises 3.6.24 that Γ is an injection for n ≥ 4. 11. Show that the only characteristic subgroups of G = Z2 × Z2 are the identity and G itself. † 12. Let H be an index 2 subgroup of the ﬁnite group G. Show that H G. 13. Show that any index 2 subgroup of an inﬁnite group is normal. 14. For x ∈ G, show that CG (x) ⊂ NG (x). 15. If x ∈ G has order 2, show that CG (x) = NG (x). Deduce that x G if and only if x ∈ Z(G). 16. Give an example of a group G and an x ∈ G with CG (x) = NG (x). 17. List the distinct subgroups of D2n that are conjugate to a. 18. What is the normalizer of a ⊂ Q16 ? 19. List the distinct subgroups of Q16 that are conjugate to a. 20. What is the normalizer of a ⊂ Q4n ? List the distinct subgroups of Q4n that are conjugate to a. (Hint: The answer depends in a crucial way on the value of n mod 2.) 21. Show that characteristic subgroups satisfy Gertrude Stein’s Theorem: If H is a characteristic subgroup of G, and K is a characteristic subgroup of H, show that K is a characteristic subgroup of G. 22. Show that S∞ S(Z+ ). 23. Show that S({2, 3, . . . }) and S({3, 4, . . . }) are not conjugate as subgroups of S(Z+ ). 24. Let H be a subgroup of G. Show that H is a normal subgroup of NG (H). Show that NG (H) is the largest subgroup of G that contains H as a normal subgroup; i.e., if H ⊂ K ⊂ G and H K, then K ⊂ NG (H). † 25. Let H and K be subgroups of G. Show that NG (H) ∩ NG (K) ⊂ NG (H ∩ K). CHAPTER 3. G-SETS AND COUNTING 81 26. Let H be a subgroup of G and let x ∈ G. Show that NG (xHx−1 ) = xNG (H)x−1 . 27. Show that Gertrude Stein’s Theorem fails for normal subgroups: i.e., ﬁnd a group G with a normal subgroup K G and a subgroup H of K such that H K, but H is not normal in G. (Hint: Find a group G with a subgroup H such that NG (H) is normal in G but not equal to G.) 28. Let H be a subgroup of the ﬁnite group G. Show that the number of elements in −1 is less than or equal to [G : NG (H)] · (|H| − 1) + 1. Deduce that no x∈G xHx proper subgroup of G contains an element in every conjugacy class of G. 29. Let X be a G-set and let x ∈ X. Show that the isotropy subgroup of gx is gGx g −1 . 30. (See Problem 18 of Exercises 3.3.23.) What are the elements of the H-ﬁxed-point set (G/H)H ? For which elements yH ∈ (G/H)H is the G-map from G/H to itself which takes eH to yH a G-isomorphism (i.e., which elements of (G/H)H have isotropy subgroup exactly equal to H)? ‡ 31. Let H be a subgroup of G and consider the standard action of G on G/H. Let Λ : G → S(G/H) be the homomorphism induced by this action. Show that ker Λ = xHx−1 . x∈G 32. Let H be a subgroup of G and let X be the set of all subgroups conjugate to H: X = {xHx−1 | x ∈ G}. Thus, X is the orbit of H under the action of G on Sub(G) (the set of subgroups of G) by conjugation. Since X is a G-set, we get an induced homomorphism Λ : G → S(X). Show that ker Λ = xNG (H)x−1 . x∈G Chapter 4 Normality and Factor Groups The closing section of Chapter 3 gives the deﬁnition of a normal subgroup: A subgroup H ⊂ G is normal if xHx−1 = H for all x ∈ G. We write H G to denote that H is a normal subgroup of G. The normal subgroups of a given group are absolutely vital in understanding its structure. Some groups G have no normal subgroups other than the trivial subgroup e and the group G itself. Such groups are called simple. We shall see, via the technique of composition series below, that all ﬁnite groups are built up out of the ﬁnite simple groups. The ﬁnite simple groups are all known. In one of the triumphs of mathematical history, the ﬁnite simple groups have been classiﬁed, according to a plan of attack developed by Daniel Gorenstein that played out over a sequence of papers by several authors. One of the ﬁnal steps was the construction, by Robert Griess, of a group that was then known as the “monster group,” as its order is nearly 1054 . In current parlance, the monster group is the sporadic simple group F1 .1 We shall not need the classiﬁcation by Gorenstein et. al., as we are concentrating here on the groups of relatively low order. As will become apparent in the exercises from the next two chapters, the only simple groups whose order is ≤ 60 are the cyclic groups Zp , where p is a prime < 60, together with the alternating group A5 . In any case, the study of the ways in which ﬁnite groups can be put together out of the simple groups is the most important remaining problem for the classiﬁcation of ﬁnite groups. It is known as the Extension Problem. Thus, the Extension Problem is, in the opinion of this author at least, one of the important unsolved problems in mathematics. The study of normal subgroups and factor groups is at its core. We do not yet have the language with which to give a precise statement of the Extension Problem, but we shall do so within this chapter. In fact, we shall develop techniques to study certain special cases of the Extension Problem. These special cases will turn out to be enough, together with the Sylow theorems and the material on G-sets from Chapter 3, to classify the groups in a fair number of the lower orders. 1 See Finite Simple Groups, An Introduction to Their Classiﬁcation, by Daniel Gorenstein, Plenum Press, 1982. 82 CHAPTER 4. NORMALITY AND FACTOR GROUPS 83 The most important aspect of normal subgroups is that if H is normal in G, then there is a group structure on the set of left cosets G/H such that the canonical map π : G → G/H is a homomorphism. If G is ﬁnite and H is nontrivial, then G/H has a lower order than G, and hence is, in general, easier to analyze. If G has enough normal subgroups (e.g. if G is abelian or solvable), then induction arguments on the order of G may be used to deduce facts about G from facts about the lower order factor groups G/H. In Section 4.1, we give the Noether Isomorphism Theorems, which study the properties of the factor groups G/H when H is normal in G. These are truly fundamental in the understanding of groups, and will open the door to some new classiﬁcation results. As an application, we shall show that if G is ﬁnite and p is the smallest prime dividing the order of G, then any subgroup of G of index p is normal in G. In Section 4.2, we give the technique of composition series, which shows that ﬁnite groups are built up out of simple groups. We also show that the alternating groups An are simple for n ≥ 5. Then, in Section 4.3, we give the Jordan–Hölder Theorem, which shows that the simple groups out of which a given group is built are unique, though the way in which they are put together may not be. Then, by an induction on order, using the fact that every subgroup of an abelian group is normal, we give a complete classiﬁcation of the ﬁnite abelian groups, the Fundamental Theorem of Finite Abelian Groups. In Section 4.5, we calculate the automorphism group of a cyclic group. This calculation will be of essential value in constructing some new groups by the technique of semidirect products and in classifying the extensions of a given cyclic group. Section 4.6 develops the technique of semidirect products, by which new groups may be constructed from old. Unlike the direct product, many of the semidirect products of two abelian groups are nonabelian. We shall construct some interesting new groups in this manner. Section 4.7 deﬁnes and studies the extensions of one group by another. Extensions describe some of the ways that groups are built up out of smaller groups. We shall state the Extension Problem here and shall develop some classiﬁcation techniques, particularly for groups that are semidirect products. We shall make some signiﬁcant progress in obtaining actual classiﬁcation results. 4.1 The Noether Isomorphism Theorems Here, we study normal subgroups H G and the group structure that’s induced on the set of left cosets G/H. One convenient source of normal subgroups (which will also turn out to give all of them) is from the kernels of homomorphisms: Lemma 4.1.1. Let f : G → G be a homomorphism. Then ker f is a normal subgroup of G. Proof Let K = ker f . By Proposition 3.7.10, it suﬃces to show that xKx−1 ⊂ K for each x ∈ G, i.e., that xkx−1 ∈ K for all x ∈ G and all k ∈ K. But if k ∈ K, then f (xkx−1 ) = f (x) · f (k) · f (x)−1 = f (x) · e · f (x)−1 = e, and hence xkx−1 ∈ K. We shall see that im f need not be normal in G . We next explore the relationship between normality of a subgroup H ⊂ G and group structures on the set of cosets G/H: CHAPTER 4. NORMALITY AND FACTOR GROUPS 84 Theorem 4.1.2. Let H be a subgroup of G. Then H is normal in G if and only if there is a group structure on the set G/H of left cosets of H with the property that the canonical map π : G → G/H is a homomorphism. If H G, then the group structure on G/H which makes π a homomorphism is unique: we must have gH · g H = gg H for all g, g ∈ G. Moreover, the kernel of π : G → G/H is H. Thus, a subgroup of G is normal if and only if it is the kernel of a homomorphism out of G. Proof Suppose that there is a group structure on G/H such that π is a homomorphism. Then the identity element of G/H is π(e) = eH. So g ∈ ker π if and only if gH = eH. By Lemma 3.2.3, this is true if and only if g ∈ H. Thus, H is the kernel of π, and hence is normal by Lemma 4.1.1. Clearly, there is at most one group structure on G/H which makes π a homomorphism, as, since π(g) = gH, we must have gH · g H = gg H for all g, g ∈ G. Suppose then that H is normal in G. We wish to show that gH · g H = gg H deﬁnes a group structure on G/H. We ﬁrst show that it gives a well deﬁned binary operation. Thus, if gH = xH and g H = x H, we wish to show that gg H = xx H. But Lemma 3.2.3 shows that x = gh and x = g h for some h, h ∈ H. Thus, xx H = ghg h H = ghg H. By Lemma 3.2.3, this is equal to gg H provided that (gg )−1 ghg ∈ H. But (gg )−1 ghg = (g )−1 g −1 ghg = (g )−1 hg . And (g )−1 hg ∈ H because H G. Thus, the operation is well-deﬁned. That this operation gives a group structure now follows immediately from the group laws in G: associativity in G gives associativity in G/H, and π(e) = eH is an identity element for G/H. Finally, the inverse of gH is g −1 H, so G/H is indeed a group. Deﬁnition 4.1.3. When H G, we shall simply write G/H for the group structure on G/H which makes π a homomorphism (i.e., with the multiplication gH · g H = gg H), and call it the factor group, or quotient group, of G modulo H. Example 4.1.4. We’ve already seen examples of factor groups, as the groups Zn are precisely the factor groups Z/n. The point is that for k ∈ Z, the congruence class of k modulo n is precisely the coset k + n. (To see this, note that l ≡ k mod n if and only if l − k ∈ n, i.e., l ∈ k + n.) The canonical map π : Z → Zn that we’ve been using coincides with the canonical map above, as it takes each element in Z to its coset modulo n. And, indeed, the group operation in Zn is deﬁned precisely as in the more general case above. Another way of seeing that Z/n ∼ = Zn will be given using the Noether Isomorphism Theorems, below. Since |G/H| = [G : H], Lagrange’s Theorem gives the following formula. Lemma 4.1.5. Let H G with G ﬁnite. Then |G| = |H| · |G/H|. Notation 4.1.6. Let H G. For brevity, we shall generally write g in place of gH ∈ G/H. Thus, every element of G/H has the form g = π(g) for some g ∈ G, and have g · g = gg . The identity element is e = e = h for any h ∈ H. CHAPTER 4. NORMALITY AND FACTOR GROUPS 85 Note that |G/H| = [G : H] is less than |G| whenever G is ﬁnite and H = e. In this case, the factor group G/H is less complicated than G, and we may obtain information about G from an understanding of the groups G/H as H ranges over the normal subgroups of G. We can characterize the homomorphisms out of G/H in terms of the homomorphisms out of G. Proposition 4.1.7. Let H G. Then for any group G and any homomorphism f : G → G whose kernel contains H, there is a unique homomorphism f : G/H → G that makes the following diagram commute. / G GC < CC z CC zz z C zz π CC ! zz f G/H f Conversely, if f : G → G factors by a commutative diagram as above, then H must be contained in ker f . Thus, the passage from a homomorphism f : G/H → G to the composite (f ◦π) : G → G gives a one-to-one correspondence between the homomorphisms from G/H to G and those homomorphisms from G to G whose kernels contain H. Proof Suppose given f : G → G . If a homomorphism f : G/H → G exists with f ◦ π = f , then for each h ∈ H, f (h) = f (h) = f (e) = e, and hence H ⊂ ker f . Moreover, f (g) = f ◦ π(g) = f (g), so f is unique if it exists. It suﬃces to show that such an f exists whenever H ⊂ ker f . Thus, suppose H ⊂ ker f . As we’ve just seen, f , if it exists, must be given by f (g) = f (g). The key is to show this is a well-deﬁned function from G/H to G . Thus, suppose that g = g in G/H. By Lemma 3.2.3, this means g = gh for some h ∈ H. Thus, f (g ) = f (g)f (h) = f (g), since h ∈ ker f . So f is well-deﬁned. But f (g)f (g ) = f (g)f (g ) = f (gg ) = f (gg ) = f (gg ), and hence f is a homomorphism. We call f the homomorphism induced by f . In the counting argument below, we again make use of the fact that |G/H| = [G : H]. Theorem 4.1.8. (First Noether Isomorphism Theorem) Let f : G → G be a homomorphism. Then f induces an isomorphism from G/ ker f to im f . Thus, if G is ﬁnite, |im f | = [G : ker f ], so Lagrange’s Theorem gives |G| = |ker f | · |im f |. Proof Let f : G/ ker f → im f be the homomorphism induced by f . Since f (g) = f (g), f is onto. Thus, it suﬃces (Lemma 2.5.9) to show that ker f = e. Let g ∈ ker f . Then f (g) = f (g) = e, so g ∈ ker f . But then g = e in G/ ker f . Example 4.1.9. Recall the rotation matrices Rθ ∈ Gl2 (R). We saw in Section 2.7 that the collection of all rotation matrices, {Rθ | θ ∈ R}, forms a subgroup, SO(2), of Gl2 (R). Moreover, as shown in Proposition 2.7.6, there is a surjective homomorphism exp : R → SO(2) given by exp(θ) = Rθ . The kernel of exp is shown there to be the cyclic subgroup 2π ⊂ R. Thus, by the ﬁrst Noether theorem, SO(2) is isomorphic to R/2π. CHAPTER 4. NORMALITY AND FACTOR GROUPS 86 The Noether Theorem gives us another way to recognize Zn as Z/n. Corollary 4.1.10. Let n be a positive integer. Then Zn is isomorphic to the factor group Z/n. Proof The canonical map π : Z → Zn is a surjection, with kernel n. Recall that there is a homomorphism ε : Sn → {±1} deﬁned by setting ε(σ) equal to the sign of σ for σ ∈ Sn . Since ε is onto, its image has order 2. But An is the kernel of ε, so we may calculate its order: Corollary 4.1.11. An has index 2 in Sn , and hence |An | = n!/2. The next result foreshadows the techniques we shall use in Chapter 5. Recall from Problem 12 of Exercises 3.7.14 that any index 2 subgroup of a ﬁnite group is normal. Using the First Noether Isomorphism Theorem, we shall give a generalization of this result. First, note from the example of a ⊂ D6 that a subgroup of prime index in a ﬁnite group need not be normal. Of course, 2 is the smallest prime, so the next result gives the desired generalization. Proposition 4.1.12. Let G be a ﬁnite group and let p be the smallest prime dividing the order of G. Then any subgroup of G of index p is normal. Proof Let H be a subgroup of index p in G, and consider the usual action of G on the set of left cosets G/H. As in Proposition 3.3.16, we get an induced homomorphism Λ : G → S(G/H) ∼ = Sp by setting Λ(g)(xH) = gxH for all g ∈ G and xH ∈ G/H. As shown in Proposition 3.3.16, the kernel of Λ is the intersection of the isotropy subgroups of the elements of G/H. Since the isotropy subgroup of eH is H, we see that ker Λ ⊂ H. Since |Sp | = p! and since p is the smallest prime dividing |G|, the greatest common divisor of |G| and |Sp | is precisely p. The First Noether Isomorphism Theorem shows that |im Λ| must divide |G|. Since |im Λ| also divides |Sp |, the order of im Λ must be either 1 or p. Since ker Λ ⊂ H, Λ cannot be the trivial homomorphism, so we must have |im Λ| = p. By the First Noether Isomorphism Theorem, we see that |ker Λ| = |G|/p = |H|. Since ker Λ ⊂ H, this forces ker Λ = H. Since the kernel of a homomorphism is always normal, this gives H G, as desired. The remaining isomorphism theorems are concerned with a deeper analysis of the canonical maps. The reader may easily verify the following lemma. Lemma 4.1.13. Let f : G → G be a homomorphism and let H ⊂ G. Let f : H → G be the restriction of f to H. Then ker f = H ∩ ker f . Recall from Problem 8 of Exercises 2.5.21 that if f : G → G is a surjective homomorphism, then there is a one-to-one correspondence between the subgroups of G and those subgroups of G that contain ker f . Under this correspondence, the subgroup K of G corresponds to f −1 (K ) ⊂ G, while a subgroup K ⊂ G containing ker f corresponds to f (K) ⊂ G . Let us reconsider this correspondence in the case where f is the canonical map π : G → G/H. CHAPTER 4. NORMALITY AND FACTOR GROUPS 87 Lemma 4.1.14. Let H G. Then the subgroups of G/H all have the form K/H, where K is a subgroup of G that contains H. Proof The one-to-one correspondence between the subgroups of G/H and those subgroups of G that contain H is based on the following facts (which we assume were veriﬁed by the reader). 1. If K is a subgroup of G/H, then π(π −1 (K )) = K . 2. If K is a subgroup of G which contains H, then π −1 (π(K)) = K. Given K ⊂ G/H, let K = π −1 (K ). Then K contains H, and K = π(K). Now π restricts to give a homomorphism, which we shall call π , from K onto π(K). Now, the kernel of π : K → π(K) is K ∩ ker π = H (Lemma 4.1.13). The ﬁrst Noether theorem now allows us to identify π(K) with K/H. Example 4.1.15. Let n and k be positive numbers such that k divides n. Then the unique subgroup of Zn of order n/k is k. Let π : Z → Zn be the canonical map. Clearly, k = π(k). But since k divides n, we have ker π = n ⊂ k. Thus, k = π −1 k, and the above gives an isomorphism from k/n to k ∼ = Zn/k . We wish to generalize Lemma 4.1.14 to an analysis of the subgroups π(K) where K ⊂ G does not contain H. First consider this: Deﬁnition 4.1.16. Let H and K be subgroups of G. Then there is a subset KH of G deﬁned by KH = {kh | k ∈ K, h ∈ H}. (Of course, if G is abelian, we write K + H = {k + h | k ∈ K, h ∈ H} for this subset.) Examples 4.1.17. Since any element of D2n can be written as either bk or as abk for an appropriate value of k, we see that D2n = ab. Similarly, Q4n = ab. The fact that some elements may be written in more than one way as a product ar bs with 0 ≤ r ≤ 3 and 0 ≤ s ≤ 2n doesn’t change this: the set {ar bs | r, s ∈ Z} is ab, and it includes every element in Q4n . Lemma 4.1.18. Let H be a subgroup of G and let K be a subgroup of NG (H). Then KH is a subgroup of G, and H ⊂ KH ⊂ NG (H). In addition, in this case, KH is equal to the subset HK of G, and is the smallest subgroup of G that contains both H and K. Proof Since K ⊂ NG (H), for any k ∈ K and h ∈ H, we have k −1 hk = h ∈ H, and hence hk = kh . This is what we need to show that KH is closed under multiplication: for k1 , k2 ∈ K and h1 , h2 ∈ H, we have k1 h1 k2 h2 = k1 k2 h1 h2 for some h1 ∈ H. Similarly, (kh)−1 = h−1 k −1 = k −1 h for some h ∈ H, and hence KH a subgroup of G. Since h = eh and k = ke, both H and K are contained in KH. And any subgroup containing both H and K must contain each product kh, so KH is indeed the smallest subgroup containing H and K. Since K ⊂ NG (H), this implies HK ⊂ NG (H). Finally, if h ∈ H and k ∈ K, we have khk −1 = h ∈ H, so that kh = h k ∈ HK, and hence KH ⊂ HK. Similarly, HK ⊂ KH. CHAPTER 4. NORMALITY AND FACTOR GROUPS 88 Example 4.1.19. Let K be any subgroup of Sn that is not contained in An (e.g., K = (1 2)). Since An Sn , KAn is a subgroup of Sn . Moreover, since K is not contained in An , [KAn : An ] > 1. But [KAn : An ] divides [Sn : An ]. Since [Sn : An ] = 2, we see that KAn = Sn . Lemma 4.1.18 sets up the second Noether Theorem. Theorem 4.1.20. (Second Noether Isomorphism Theorem) Let H and K be subgroups of G with H G. Write π : K → G/H for the restriction to K of the canonical map π : G → G/H. Then the image of π is KH/H, while the kernel of π is K ∩ H. Thus, the First Noether Isomorphism Theorem provides an isomorphism π : K/(K ∩ H) ∼ = / KH/H. Proof By Lemma 4.1.13, the kernel of π is K ∩ ker π = K ∩ H. Thus, it suﬃces to show that the image of π is KH/H. Of course, the image of π is just π(K). Now for k ∈ K and h ∈ H, kh = k ∈ G/H, and hence π(kh) = π(k). But this says that π(KH) = π(K). Since H ⊂ KH, π(KH) = KH/H by Lemma 4.1.14, and the result follows. Corollary 4.1.21. Let H and K be subgroups of G with H G. Then [K : H ∩ K] divides [G : H]. Proof [K : H ∩ K] is the order of K/(H ∩ K), which is thus equal to the order of the subgroup KH/H of G/H. By Lagrange’s Theorem, this divides [G : H], the order of G/H. Here is a sample application. Corollary 4.1.22. Let K be a subgroup of Sn and suppose that K is not contained in An . Then [K : K ∩ An ] = 2. Proof Corollary 4.1.21 shows that [K : K ∩ An ] divides [Sn : An ], which is 2. But [K : K ∩ An ] = 1, as K is not contained in An . The Second Noether Theorem may be used for yet other calculations of indices. Corollary 4.1.23. Let f : G → G be a surjective homomorphism between ﬁnite groups and let K be a subgroup of G. Then [G : f (K)] divides [G : K]. Proof Let H be the kernel of f . By the First Noether Isomorphism Theorem, there is an isomorphism f : G/H → G such that f ◦ π = f , where π : G → G/H is the canonical map. But then f induces an isomorphism from π(K) to f (K). Thus, we see that [G : f (K)] = [G/H : π(K)]. By the Second Noether Isomorphism Theorem, π(K) ∼ = K/(H ∩ K). Thus, we obtain [G : f (K)] = = = |G|/|H| |K|/|H ∩ K| |G|/|K| |H|/|H ∩ K| [G : K] . [H : H ∩ K] Since these indices are all integers, the result follows. CHAPTER 4. NORMALITY AND FACTOR GROUPS 89 We now give a slight reﬁnement of the Second Noether Theorem. Corollary 4.1.24. Let H and K be subgroups of G, with K ⊂ NG (H). Then H ∩K K and K/(H ∩ K) ∼ = KH/H. Proof Just replace G by NG (H) and apply the second Noether Theorem. Now, we give the ﬁnal Noether Isomorphism Theorem. Theorem 4.1.25. (Third Noether Isomorphism Theorem) Suppose given two subgroups, H and K, of G, which are both normal in G, with H ⊂ K. Then K/H G/H, and the factor group (G/H)/(K/H) is isomorphic to G/K. Proof Write πH : G → G/H and πK : G → G/K for the two canonical maps. Then H ⊂ K = ker πK . By Proposition 4.1.7, there is a homomorphism f : G/H → G/K with f ◦ πH = πK . Clearly, f is onto. It suﬃces to show that ker f = K/H. Let g = gH ∈ G/H. Since f (g) = πK (g) ∈ G/K, we see that g ∈ ker f if and only if g ∈ K = ker πK . Thus, ker f is precisely K/H. Exercises 4.1.26. 1. Give an example of a homomorphism f : G → G such that im f is not normal in G . † 2. Show that every factor group of a cyclic group is cyclic. (Suggestion: Give at least two diﬀerent proofs.) 3. Recall that b D8 . For each subgroup H ⊂ b, list all the subgroups K ⊂ D8 with K ∩ b = H. (Hint: What is the order of K/H = K/(K ∩ b)?) 4. List all the normal subgroups of D8 . What is the resulting factor group in each case? 5. List all the groups (up to isomorphism) that appear as subgroups of D8 . 6. Recall that b D2n . Let H be a subgroup of b. How many subgroups of D2n are there whose intersection with b is H? (Hint: The answer depends on [b : H].) What are these subgroups? 7. List all the normal subgroups of D2n . What is the resulting factor group in each case? 8. List all the groups (up to isomorphism) that appear as subgroups of D2n . 9. Recall that b Q4n . Let H be a subgroup of b. How many subgroups of Q4n are there whose intersection with b is H? (Hint: The answer depends on [b : H].) What are these subgroups? 10. List all the normal subgroups of Q4n . What is the resulting factor group in each case? 11. List all the groups (up to isomorphism) that appear as subgroups of Q4n . † 12. Show that every element of Q/Z has ﬁnite order. Show also that Q/Z has elements of order n for every n ≥ 1. CHAPTER 4. NORMALITY AND FACTOR GROUPS 90 † 13. Show that there is an isomorphism between R/Z and R/x for any 0 = x ∈ R. Show also that the elements of ﬁnite order in R/Z are precisely those lying in the subgroup Q/Z. † 14. Let Hi Gi for 1 ≤ i ≤ k. Show that H1 × · · · × Hk G1 × · · · × Gk and that (G1 × · · · × Gk )/(H1 × · · · × Hk ) ∼ = (G1 /H1 ) × · · · × (Gk /Hk ). 15. What are all the subgroups of Z2 × Z4 ? What is the resulting factor group in each case? 16. Suppose given an index three subgroup H ⊂ G such that H is not normal in G. Show that there is a surjective homomorphism from G to S3 . † 17. Let H and K be subgroups of G with K ⊂ NG (H). Show that NG (K) ∩ NG (H) ⊂ NG (KH). 18. Generalize Corollary 4.1.23 as follows: Show that if f : G → G is a surjective homomorphism and if [G : K] is ﬁnite, then [G : f (K)] is ﬁnite and divides [G : K]. (Hint: Show that [G : f (K)] = [G : KH], where H = ker f .) 19. Let H be a subgroup of G. Show that the factor group NG (H)/H acts on the right of the set of left cosets G/H via gH · x = gxH for all g ∈ G and x ∈ NG (H). Show also that the action of each x ∈ NG (H)/H gives a G-isomorphism from G/H to itself. Finally, show that every G-isomorphism from G/H to itself arises in this manner, and deduce that the group of G-isomorphisms from G/H to itself is isomorphic to the factor group NG (H)/H. 20. Let X be any G-set and let H be a subgroup of G. Show that there is an induced action of the factor group NG (H)/H on the H-ﬁxed point set X H . 4.2 Simple Groups The ﬁnite simple groups, which we deﬁne in this section, are the building blocks out of which ﬁnite groups are made. We show this here via the technique of composition series. We then show that the alternating groups An are simple for n ≥ 5. Deﬁnition 4.2.1. A group G is simple if it has no normal subgroups other than e and G. We can characterize the abelian simple groups immediately. We shall give more complicated examples below. Lemma 4.2.2. An abelian group is simple if and only if it is a cyclic group of prime order. Proof Since every subgroup of an abelian group is normal, we see that an abelian group is simple if and only if it has no subgroups other than e and itself. Thus, an abelian group G is simple if and only if every non-identity element in it generates G. In particular, G must be cyclic, and, since Z has nontrivial proper subgroups, it must be ﬁnite. But our calculation of the orders of the elements in Zn (Problem 5 of Exercises 2.5.21) shows that Zn is simple if and only if n is prime. CHAPTER 4. NORMALITY AND FACTOR GROUPS 91 The choice of the word “simple” for the simple groups is perhaps unfortunate, as the nonabelian simple groups can be much more complicated than, say, solvable ones (see below). Nevertheless, all ﬁnite groups are built up out of simple groups, hence the name. Lemma 4.2.3. Let G = e be a ﬁnite group. Then G has a factor group which is simple and is not equal to the trivial group. Proof By the ﬁrst Noether theorem, the statement is equivalent to saying that each group G = e admits a surjective homomorphism onto a nontrivial simple group. We argue by induction on |G|. If |G| = 2, then G itself is simple. Thus, suppose |G| > 2. If G is simple, we’re done, as G = G/e is a factor group of itself. Otherwise, G has a normal subgroup H, with H not equal to either e or G. But then 1 < |G/H| < |G|, so by induction, there is a surjective homomorphism f : G/H → G , where G is a nontrivial simple group. But then the composite f ◦ π : G → G gives the desired surjection. The way in which ﬁnite groups are built up from simple groups is as follows. Deﬁnition 4.2.4. Let G be a ﬁnite group. A composition series for G is a sequence e = H0 ⊂ H1 ⊂ · · · ⊂ Hk = G, such that Hi−1 Hi , and Hi /Hi−1 is a nontrivial simple group for 1 ≤ i ≤ k. Recall that this does not imply that the groups Hi are normal in G. Proposition 4.2.5. Every nontrivial ﬁnite group has a composition series. Proof Again, we argue by induction on the order of G. Once again, if G is simple, then there is nothing to prove. Otherwise, Lemma 4.2.3 provides a subgroup H G such that G/H is a nontrivial simple group. But then |H| < |G|, and hence H has a composition series e = H0 ⊂ · · · ⊂ Hk−1 = H. But then tacking on Hk = G gives a composition series for G. The groups Hi /Hi−1 appearing in a composition series are subquotient groups of G: Deﬁnition 4.2.6. A subquotient of a group G is a group of the form H/K, where K H ⊂ G. Composition series are far from unique in many cases. We shall discuss their uniqueness properties in Section 4.3, and shall utilize them here simply to motivate the discussion of simple groups. Our remaining goal for this section is to prove the following theorem, which produces our ﬁrst examples of nonabelian simple groups. Theorem 4.2.7. The alternating groups An are simple for all n ≥ 5. In the process of proving this, we shall derive some useful information about symmetric groups. We shall start by identifying the conjugacy classes in Sn . First consider this: Deﬁnition 4.2.8. We say that the permutations σ, τ ∈ Sn have the same cycle structure if when you write each of them as a product of disjoint cycles, both have the same number of k-cycles for each k ≥ 2. Recall that Lemma 3.6.19 computes the conjugates of a k-cycle. Proposition 4.2.9. Two permutations in Sn are conjugate if and only if they have the same cycle structure. CHAPTER 4. NORMALITY AND FACTOR GROUPS 92 Proof Write σ ∈ Sn as a product of disjoint cycles: σ = σ1 . . . σk . Then for τ ∈ Sn , τ στ −1 = τ σ1 τ −1 . . . τ σk τ −1 . By Lemma 3.6.19, each τ σi τ −1 is a cycle of the same length as σi . Thus, σ and its conjugate τ στ −1 have the same cycle structure, provided we show that for two disjoint cycles σ1 and σ2 , and any permutation τ , the cycles τ σ1 τ −1 and τ σ2 τ −1 are disjoint. But that follows from the explicit formula of Lemma 3.6.19. Thus, it suﬃces to show that any two permutations with the same cycle structure are conjugate. Let σ and τ be permutations with the same cycle structure. Thus, we can write σ and τ as products of disjoint cycles, σ = σ1 . . . σk , and τ = τ1 . . . τk , in such a way that σj and τj have the same length, length rj , for 1 ≤ j ≤ k. Say σj = (ij1 . . . ijrj ), and τj = (ij1 . . . ijrj ), for 1 ≤ j ≤ k. Suppose we can ﬁnd a permutation μ with μ(ijs ) = ijs for 1 ≤ j ≤ k and 1 ≤ s ≤ rj . Then μσj μ−1 = τj by Lemma 3.6.19 for all j. Thus, μσμ−1 = τ , so it will suﬃce to ﬁnd such a μ. Since the permutations σj are disjoint for 1 ≤ j ≤ k, an element of {1, . . . , n} may occur in at most one of the sets {iji , . . . , ijrj }. Thus, setting μ(ijs all of the ) = ijs for indices js gives a well-deﬁned function μ : 1≤j≤k {ij1 , . . . , ijrj } → 1≤j≤k {ij1 , . . . , ijrj }. Moreover, the disjointness ofthe τj shows that this function is bijective. Now the complement of 1≤j≤k {ij1 , . . . , ijrj } in {1, . . . , n} may be placed in one-toone correspondence with the complement of 1≤j≤k {ij1 , . . . , ijrj } in {1, . . . , n}. But any such one-to-one correspondence extends μ to a permutation of {1, . . . , n}, which then gives the desired conjugation. It would make our lives easier in the arguments below if we knew that any two elements of An with the same cycle structure were conjugates in An (i.e., if the conjugating permutation μ could be chosen to lie in An ). Unfortunately, this is not always true, so some care will be required. Lemma 4.2.10. Let σ, σ ∈ An be conjugate in Sn . Suppose that σ commutes with an element of Sn which is not in An . Then σ and σ are conjugate in An . Proof Since σ and σ are conjugate in Sn , there is a permutation τ with τ στ −1 = σ . If τ ∈ An , there’s nothing to prove. Otherwise, the hypothesis provides a τ ∈ Sn , with τ not in An (hence ε(τ ) = −1), which commutes with σ. Now ε(τ τ ) = ε(τ )ε(τ ) = +1, −1 so τ τ ∈ An . And (τ τ )σ(τ τ )−1 = τ (τ στ )τ −1 = τ στ −1 = σ . And here are some situations where the hypothesis is satisﬁed. Lemma 4.2.11. Let σ ∈ An . Suppose that either of the following conditions hold. 1. There are at least two elements of {1, . . . , n} which are left ﬁxed by σ. 2. At least one of the cycles in the cycle structure of σ has even length. Then σ commutes with an element of Sn which is not in An . Proof Suppose the ﬁrst conditions holds, and, say, σ ﬁxes i and j. Then σ is disjoint from the transposition (i j), and hence commutes with it. Suppose now that the second condition holds. Then σ commutes with the even length cycle in question. (Said cycle is disjoint from all of the other cycles in the cycle decomposition of σ, and hence commutes with them, but it also commutes with itself.) But even length cycles have sign −1. CHAPTER 4. NORMALITY AND FACTOR GROUPS 93 Our strategy for proving Theorem 4.2.7 may now be laid out. Recall from Corollary 3.5.11 that An is generated by the 3-cycles. Since n ≥ 5, any 3-cycle commutes with a transposition in Sn . Thus, by Lemma 4.2.10, any two 3-cycles are conjugate in An . In particular, if H An and if H contains a 3-cycle, it must contain all of its conjugates in An , and hence must contain every 3-cycle. Since the 3-cycles generate An , we must have H = An . Thus, we have veriﬁed the following lemma. Lemma 4.2.12. Let H An for n ≥ 5. If H contains a 3-cycle, then H = An .2 Thus, it suﬃces to show that if n ≥ 5, then any normal subgroup of An which is not the identity must contain a 3-cycle. Given a subgroup e = H G, the only information we have for free is that there is a non-identity element in H. We shall consider various possibilities for the cycle structure of such an element, and verify, in the end, that the presence of any non-identity element in H implies the existence of a 3-cycle in H. We may now generalize Lemma 4.2.12. Lemma 4.2.13. Let H An for n ≥ 5. Suppose that H contains an element of the form στ such that 1. σ is a cycle of length ≥ 3. 2. σ and τ are disjoint. (Note that τ may be the identity here.) 3. στ commutes with an element of Sn which is not in An . Then H = An . Proof Let σ = (i1 . . . ik ). Since the inverse of an r-cycle is an r-cycle, τ −1 has the same cycle structure as τ . Thus, στ is conjugate to σ τ −1 , where σ = (ik−1 ik ik−2 . . . i1 ). But then σ τ −1 στ is in H. Since σ and τ are disjoint, they commute, so this is just σ σ, which is easily seen to be equal to the 3-cycle (ik−2 ik ik−1 ). So H = An by Lemma 4.2.12. We would like to be able to eliminate the condition in the preceding lemma which requires that στ commutes with an element of Sn which is not in An . We shall ﬁrst consider another special case, which is actually already covered by Lemma 4.2.13 except in the cases of n = 5 or 6. Lemma 4.2.14. Let H An , and suppose that H contains a 5-cycle. Then H = An . Proof Let σ = (a b c d e) be a 5-cycle in H. Then μ = (a b)(c d) is in An . Since H is normal in An , μσμ−1 ∈ H, and hence σ · μσμ−1 is also in H. By Lemma 3.6.19, σ · μσμ−1 = (a b c d e) · (b a d c e) = (a e c), a 3-cycle. So H = An by Lemma 4.2.12. We can now improve on Lemma 4.2.13, provided the cycle σ in the statement there has length > 3. Lemma 4.2.15. Let H An . Suppose that H contains an element of the form στ , where σ is a cycle of length > 3 and σ and τ are disjoint. Then H = An . 2 This result is still true for n < 5, but with diﬀerent argumentation. We shall treat these cases in the exercises below. CHAPTER 4. NORMALITY AND FACTOR GROUPS 94 Proof First note that the hypothesis forces n ≥ 5, as if σ has length 4, then τ cannot be the identity element, as 4-cycles are not in An . Note also that if σ is a 4-cycle, then στ commutes with an element (σ) which is in Sn but not An , and hence the hypothesis of Lemma 4.2.13 is satisﬁed. Thus, we may assume that σ has length ≥ 5. Write σ = (i1 . . . ik ), and let μ = (i1 i2 )(i3 i4 ). Then μ ∈ An . Moreover, since the support of μ is contained in the support of σ and since σ and τ are disjoint, μ commutes with τ . Since μ ∈ An and H An , στ · μ(στ )−1 μ−1 is in H. We shall show that στ · μ(στ )−1 μ−1 is a 5-cycle. The result will then follow from Lemma 4.2.14. Expanding (στ )−1 and associating, we get στ · μ(στ )−1 μ−1 = σ(τ μτ −1 )σ −1 μ−1 . But τ commutes with μ, so that τ μτ −1 = μ. Simplifying, we see that στ · μ(στ )−1 μ−1 = σ · μσ −1 μ−1 = (i1 . . . ik )(ik . . . i5 i3 i4 i1 i2 ) = (i5 i4 i2 i1 i3 ). Here, the second equality follows from Lemma 3.6.19, and the last by direct calculation. Thus, there is a 5-cycle in H as claimed, and the result follows. We can now go one step farther and give a version of Lemma 4.2.13 that eliminates the third condition. Proposition 4.2.16. Let H An for n ≥ 5. Suppose there is an element σ ∈ H such that the decomposition of σ as a product of disjoint cycles contains a cycle of length ≥ 3. Then H = An . Proof Suppose that the decomposition of σ as a product of disjoint cycles contains a cycle of length > 3. Then the hypothesis of Lemma 4.2.15 is satisﬁed, and H = An as desired. Thus, we may assume that σ is the product of disjoint cycles whose lengths are all ≤ 3. But if any of them is a 2-cycle, then σ commutes with an element of Sn which is not in An , and hence the hypothesis of Lemma 4.2.13 is satisﬁed. Thus, we may assume that σ is a product of disjoint 3-cycles. If σ consists of a single 3-cycle, then H = An by Lemma 4.2.12. Thus, we may assume there are at least two 3-cycles in the decomposition of σ as a product of disjoint cycles. Write σ = (a b c)(d e f )τ , where τ is disjoint from the stated 3-cycles. But an immediate application of Lemma 3.6.19 shows that σ commutes with the permutation (a d)(b e)(c f ), which lies in Sn but not in An . Thus, σ satisﬁes the hypothesis of Lemma 4.2.13, and hence H = An . Thus, the only remaining possibility for an H An with H = An is for H to consist only of products of disjoint transpositions (2-cycles). Indeed, A4 has such a normal subgroup, as we shall see in the exercises below. So n ≥ 5 will be essential in what is to follow. We begin with an important special case. Lemma 4.2.17. Let H An for n ≥ 5. Suppose that H contains a product of two disjoint transpositions. Then H = An . CHAPTER 4. NORMALITY AND FACTOR GROUPS 95 Proof Let σ = (a b)(c d) ∈ H. Since n ≥ 5, there is a ﬁfth letter, e, to play with. Since σ commutes with a 2-cycle, it is conjugate in An to any element of the same cycle structure, by Lemma 4.2.10. In particular, it is conjugate to (b c)(d e). Thus, (a b)(c d)(b c)(d e) = (a b d e c) is in H. The result now follows from Lemma 4.2.14. We may now complete the proof of the theorem. Proof of Theorem 4.2.7 Let e = H An for n ≥ 5. Let σ be any non-identity element of H. If the decomposition of σ as a product of disjoint cycles contains a cycle of length ≥ 3, then H = An by Proposition 4.2.16. Thus, we may assume that σ is a product of disjoint transpositions. Since transpositions have sign −1, there must be an even number of disjoint transpositions in the decomposition of σ. But if σ is the product of two disjoint transpositions, then H = An by Lemma 4.2.17. Thus, we may assume that σ is the product of four or more disjoint transpositions. Thus, we can write σ = (a b)(c d)(e f )τ , where τ is disjoint from the displayed transpositions. Once again, since the decomposition of σ contains a cycle of even length, it is conjugate in An to any permutation with the same cycle structure. In particular, σ is conjugate to (b c)(d e)(f a)τ −1 . Thus, (a b)(c d)(e f )(b c)(d e)(f a)τ τ −1 = (a e c)(b d f ) lies in H. The result now follows from Proposition 4.2.16. Exercises 4.2.18. † 1. Show that A4 has exactly four elements whose order is a power of 2. Show that these elements form a normal subgroup of A4 . † 2. Show that A4 has no subgroup of order 6. Deduce that A4 is the only subgroup of S4 which has order 12. (Hint: One way to proceed is as follows. Use the Noether Theorems to show that if H ⊂ A4 has order 6, then H must have a normal subgroup, K H of order 2. Show that every element of H outside of K must have order 3 or 6, and that K ⊂ Z(H). Deduce that H must have an element of order 6 and derive a contradiction.) ‡ 3. Show that An is a characteristic subgroup of Sn for n ≥ 3. (Hint: Use the simplicity of An for n ≥ 5. For n = 3 or 4, give a separate argument.) 4. Show that any normal subgroup of A4 containing a 3-cycle must be all of A4 . 5. Show that any normal subgroup of A3 containing a 3-cycle must be all of A3 . 6. Let σ ∈ An . Show that the conjugates of σ in Sn are all conjugate to σ in An if and only if the centralizer of σ in Sn is not contained in An . (Hint: Recall that the centralizer, CAn (σ), of σ in An satisﬁes CAn (σ) = An ∩ CSn (σ). What are the possible values for [CSn (σ) : CAn (σ)]? What does this say about [Sn : CSn (σ)]/[An : CAn (σ)]?) CHAPTER 4. NORMALITY AND FACTOR GROUPS 4.3 96 The Jordan–Hölder Theorem Recall that a composition series for a group G is a sequence e = H0 ⊂ · · · ⊂ Hk = G such that Hi−1 Hi for i = 1, . . . , k and the factor groups Hi /Hi−1 are all nontrivial simple groups. Proposition 4.2.5 shows that every nontrivial ﬁnite group has a composition series. But composition series are not unique. However, the Jordan–Hölder Theorem shows that the simple subquotient groups Hi /Hi−1 that appear in a composition series for G are independent of the choice of composition series, though they may occur in a diﬀerent order in diﬀerent composition series: Theorem 4.3.1. (Jordan–Hölder Theorem) Suppose given two diﬀerent composition series, e = H0 ⊂ · · · ⊂ H k = G and e = G0 ⊂ · · · ⊂ Gl = G of the ﬁnite group G. Then k = l and there is a permutation σ ∈ Sk such that Hi /Hi−1 is isomorphic to Gσ(i) /Gσ(i)−1 for 1 ≤ i ≤ k. Example 4.3.2. We have two diﬀerent composition series for Z6 : e ⊂ 2 ⊂ Z6 and e ⊂ 3 ⊂ Z6 . The successive subquotient groups are Z3 followed by Z2 in the ﬁrst case, and Z2 followed by Z3 in the second. Rather than present the proof from the bottom up, which would lose the intuition in this case, we shall argue conceptually, leaving the proof (and statement) of the key lemma for the end. We ﬁrst generalize the notion of composition series: Deﬁnition 4.3.3. A subnormal series for a group G is a sequence e = H0 ⊂ · · · ⊂ Hk = G such that Hi−1 Hi for 1 ≤ i ≤ k. Some treatments would call this a normal series, but we prefer subnormal since it is not required that Hi G. Note that a composition series is a subnormal series for which each of the successive factor groups Hi /Hi−1 is a nontrivial simple group. Note also that it is possible in a subnormal series (but not a composition series) to have Hi−1 = Hi . Deﬁnition 4.3.4. A reﬁnement of a subnormal series is a subnormal series obtained by inserting more subgroups into the sequence. Thus, every subnormal series is a reﬁnement of the trivial series e ⊂ G. CHAPTER 4. NORMALITY AND FACTOR GROUPS 97 Deﬁnition 4.3.5. Two subnormal series e = H0 ⊂ · · · ⊂ Hk = G and e = G0 ⊂ · · · ⊂ G l = G are equivalent if k = l and there is a permutation σ ∈ Sk such that Hi /Hi−1 is isomorphic to Gσ(i) /Gσ(i)−1 for 1 ≤ i ≤ k. Thus, the Jordan–Hölder Theorem simply states that any two composition series are equivalent. It will follow rather quickly from the next result. Theorem 4.3.6. (Schreier’s Theorem) Suppose given two subnormal series for the group G. Then we may reﬁne the two series in such a way that the reﬁned series are equivalent. This immediately implies the Jordan–Hölder Theorem because reﬁning a composition series does nothing other than adding trivial groups to the list of factor groups. So two composition series equivalent after reﬁnement must have been equivalent in the ﬁrst place. Let us now consider the proof of Schreier’s Theorem. We are given subnormal series e = H0 ⊂ · · · ⊂ H k = G and e = G0 ⊂ · · · ⊂ G l = G of G. We shall insert between each pair Hi−1 ⊂ Hi a collection of inclusions derived from the G0 ⊂ · · · ⊂ Gl in such a way that the resulting sequence of k · l inclusions is a subnormal series. Similarly, we shall insert groups between Gi−1 and Gi . Since Hi−1 Hi , the group Hi ∩ Gj normalizes Hi−1 for each j. Thus, the product Hi−1 (Hi ∩ Gj ) is a subgroup of G, and we have inclusions Hi−1 = Hi−1 (Hi ∩ G0 ) ⊂ · · · ⊂ Hi−1 (Hi ∩ Gl ) = Hi . Similarly, we have inclusions of subgroups Gj−1 = Gj−1 (H0 ∩ Gj ) ⊂ · · · ⊂ Gj−1 (Hk ∩ Gj ) = Gj . Schreier’s Theorem now follows by setting Gj−1 = B, Hi−1 = A, Gj = K, and Hi = H in the next lemma. Lemma 4.3.7. (Zassenhaus’ Butterﬂy Lemma) Let H and K be subgroups of G, and suppose given normal subgroups A H and B K of H and K. Then A(H ∩ B) A(H ∩ K), B(A ∩ K) B(H ∩ K) and there is an isomorphism A(H ∩ K)/A(H ∩ B) ∼ = B(H ∩ K)/B(A ∩ K). Proof We shall show that both of the quotient groups are isomorphic to a third quotient: H ∩K/(A∩K)(H ∩B). (We’ll justify presently that this is a group.) The diagram below, CHAPTER 4. NORMALITY AND FACTOR GROUPS 98 which gives this result its name, illustrates the relations between the various subgroups we shall consider. All arrows are the natural inclusions. A(H ∩ K) O iRRRR RRRR R H ∩O K l5 llll llll B(H ∩ K) O A(H ∩ B) B(A ∩ K) iRRRR iRRR l5 l ll5 l RRR l R l l RRR RRR lll lll l l RR l ll (A ∩ K)(H ∩ B) A iRRR ll5 B RRR iRRRR lll ll5 RRR l l l R l RRRR RRR ll ll lll lll A∩K H ∩B We shall give the proof that H ∩ K/(A ∩ K)(H ∩ B) ∼ = A(H ∩ K)/A(H ∩ B). The argument that H ∩ K/(A ∩ K)(H ∩ B) ∼ = B(H ∩ K)/B(A ∩ K) is analogous. First note that since H ∩ B ⊂ H ∩ K ⊂ H ⊂ NG (A), the subsets A(H ∩ K) and A(H ∩ B) are subgroups of G. We argue by a straightforward application of the Second Noether Isomorphism Theorem, which we restate as follows (using the form given in Corollary 4.1.24) to avoid conﬂicting notation: If X and Y are subgroups of G with X ⊂ NG (Y ), then X ∩ Y X, and X/(X ∩ Y ) ∼ = XY /Y . We apply this with X = H ∩ K and Y = A(H ∩ B). We must ﬁrst verify that for this choice of X and Y that X ⊂ NG (Y ), and then must show that the quotients stated in the Noether theorem are the ones that concern us. First recall from Problem 25 of Exercises 3.7.14 that for subgroups H1 , H2 ⊂ G, we have NG (H1 ) ∩ NG (H2 ) ⊂ NG (H1 ∩ H2 ). Also, Problem 17 of Exercises 4.1.26 shows that if H1 ⊂ NG (H2 ), then NG (H1 ) ∩ NG (H2 ) ⊂ NG (H1 H2 ). Thus, since Y = A(H ∩ B), the inclusion of X in NG (Y ) will follow if we show that X ⊂ NG (A) ∩ NG (H) ∩ NG (B). But X ⊂ H, which is contained in both NG (A) and NG (H), and X ⊂ K, which is contained in NG (B), so X normalizes Y as desired. It suﬃces to show that X ∩ Y = (A ∩ K)(H ∩ B) and that XY = A(H ∩ K). For the former equality, we have X ∩ Y = [H ∩ K] ∩ [A(H ∩ B)]. Clearly, this contains (A ∩ K)(H ∩ B), so it suﬃces to show the opposite inclusion, i.e., that [H ∩ K] ∩ [A(H ∩ B)] ⊂ (A ∩ K)(H ∩ B). So let a ∈ A and let b ∈ H ∩ B, and suppose that ab ∈ H ∩ K. It suﬃces to show that a ∈ K. But since ab ∈ H ∩ K ⊂ K, and since b ∈ B ⊂ K, we must have a ∈ K. Finally, we have XY = [H ∩ K][A(H ∩ B)]. This group clearly contains A(H ∩ K), so it suﬃces to show the opposite inclusion. Let a ∈ A, b ∈ H ∩ B, and c ∈ H ∩ K. Since H ∩ K normalizes A, cab = a cb with a ∈ A. But cb ∈ H ∩ K and the result follows. 4.4 Abelian Groups: the Fundamental Theorem Here, we classify all ﬁnite abelian groups. Recall that an element x of a group G has exponent n if xn = e. Recall also that if x has exponent n, then the order of x divides n. CHAPTER 4. NORMALITY AND FACTOR GROUPS 99 Deﬁnitions 4.4.1. We say that a group G has exponent n if every element of G has exponent n. Given an arbitrary element x ∈ G, we say that x is a torsion element if x has ﬁnite order. If every element of G has ﬁnite order, we call G a torsion group. At the other extreme, if no element of G other than the identity has ﬁnite order, we call G torsion-free. Finally, let p be a prime number. We say that x ∈ G is a p-torsion element if the order of x is a power of p. If every element of G is a p-torsion element, we call G a p-group. Recall that an element has ﬁnite order if and only if it has exponent n for some n > 0. Since the order then divides all exponents for x, we see that x is a p-torsion element if and only if it has exponent pr for some r ≥ 0. Examples 4.4.2. 1. Let G be a ﬁnite group. By the corollary to Lagrange’s Theorem (Corollary 3.2.10), the order of any element of G divides |G|. Thus, the group G has exponent |G|. 2. S∞ (the ascending union of the symmetric groups Sn for all n ≥ 1) has been shown to be an inﬁnite torsion group. But it does not have an exponent, as it contains elements of every ﬁnite order. 3. As shown in Problem 12 of Exercises 4.1.26, Q/Z is an example of an inﬁnite abelian torsion group which does not have an exponent. 4. ! The reader acquainted with inﬁnite products should have no trouble verifying that ∞ i=1 Z2 , the product of inﬁnitely many copies of Z2 , has exponent 2. Thus, in addition to having an exponent, it is an inﬁnite abelian 2-group. Let us now specialize to the study of abelian groups. To emphasize this specialization, we shall use additive notation for the group operation in the remainder of this section. Lemma 4.4.3. Let G be abelian and let x, y ∈ G be torsion elements. Then the least common multiple of the orders of x and y is an exponent for x + y. Proof Let n be the least common multiple of the orders of x and y. Then n is an exponent for both x and y: nx = ny = 0. But since G is abelian, n(x + y) = nx + ny = 0. It is immediate from Lemma 4.4.3 and the fact that the inverse of an element of ﬁnite order also has ﬁnite order (Problem 6 of Exercises 2.5.21) that the torsion elements of an abelian group form a subgroup of it: Deﬁnition 4.4.4. Let G be an abelian group. Write Tors(G) ⊂ G for the set of all torsion elements of G. We call it the torsion subgroup of G. Example 4.4.5. As shown in Problem 13 of Exercises 4.1.26, the torsion subgroup of R/Z is Q/Z. Surprisingly, the set of torsion elements in a nonabelian group is not necessarily a subgroup. For instance, Problem 2 of Exercises 10.8.13 shows that Sl2 (Z), the group of 2 × 2 matrices with integer coeﬃcients that have determinant 1, is generated by torsion elements, but the matrix 1 1 0 1 is easily seen to have inﬁnite order. CHAPTER 4. NORMALITY AND FACTOR GROUPS 100 Deﬁnition 4.4.6. Let G be an abelian group and let p be prime. The p-torsion subgroup, Gp , of G is the set of p-torsion elements of G. The justiﬁcation is, in this case, immediate from Lemma 4.4.3: Corollary 4.4.7. Let G be an abelian group and let p be prime. Then the p-torsion subgroup Gp is, in fact, a subgroup of G. The analogue for nonabelian groups of the preceding corollary is false even for ﬁnite nonabelian groups: Example 4.4.8. Recall from Corollary 3.5.11 that for n ≥ 3, An is generated by 3cycles. But while 3-cycles are 3-torsion elements, if n ≥ 4 there are elements of An (e.g., (1 2)(3 4)) that are not 3-torsion elements. Thus, the 3-torsion elements of An do not form a subgroup if n ≥ 4. Let us now specialize to the study of ﬁnite abelian groups. Proposition 4.4.9. Let G be a ﬁnite abelian group. Let p1 , . . . , pk be the prime numbers that divide |G|. Then G is the internal direct product of Gp1 , . . . , Gpk , i.e., the function μ : Gp1 × · · · × Gpk → G deﬁned by μ(g1 , . . . , gk ) = g1 + · · · + gk is an isomorphism of groups. Proof Of course, μ restricts on each Gpi to its inclusion into G. Since G is abelian, the images of these inclusions commute with each other, so that μ is a homomorphism by Proposition 2.10.10. Suppose that g = (g1 , . . . , gk ) lies in the kernel of μ. If each gi = 0, then g is the identity element of Gp1 × · · · × Gpk . Otherwise, let i be the smallest index with gi = 0. Then gi + (gi+1 + · · · + gk ) = 0, and we must have i < k, as otherwise gi = μ(g) = 0. By inductive application of Lemma 4.4.3, the element gi+1 + · · · + gk has an exponent which is a product of powers of pi+1 , . . . , pk . But gi+1 + · · · + gk is the inverse of gi , and hence has order equal to a power of pi . This forces the order of −gi to be 1 = p0i , and hence gi = 0. So g had to be 0 in the ﬁrst place, and hence μ is injective. It suﬃces to show μ is surjective. Let g ∈ G, and let n = o(g). We argue by induction on the number of primes that divide n. If there is only one prime dividing n, then n is a power of pi for some i, and hence g ∈ Gpi ⊂ im μ. Otherwise, n = piri m for some i, where ri ≥ 1, and (pi , m) = 1. Thus, we can ﬁnd integers s and t with spri i + tm = 1. But then g = (spri i + tm)g = s(pri i g) + t(mg). By Problem 5 of Exercises 2.5.21, mg has order pri i , and hence lies in Gpi ⊂ im μ, and pri i g has order m. Since m has one less prime divisor than n, pri i g ∈ im μ by induction. Thus, g is the sum of two elements of im μ, and hence lies in im μ as claimed. Note that if p doesn’t divide the order of G, then Gp = 0 by Lagrange’s Theorem. Since the cartesian product of any group H with the trivial group is canonically isomorphic to H, we obtain the following immediate extension of Proposition 4.4.9, which is useful when considering homomorphisms between abelian groups whose orders have diﬀering prime divisors. CHAPTER 4. NORMALITY AND FACTOR GROUPS 101 Corollary 4.4.10. Let G be a ﬁnite abelian group. Let p1 , . . . , pk be a collection of primes including all the prime divisors of |G|. Let μG : Gp1 × · · · × Gpk → G be deﬁned by μG (g1 , . . . , gk ) = g1 + · · · + gk . Then μG is an isomorphism. The proof of Proposition 4.4.9 is an elaboration of the proof of the next result, which could have been used inductively to obtain it. We include both proofs as it may be instructive to the beginner. Proposition 4.4.11. Let G be a ﬁnite group of order n. Suppose that n = mk, where m and k are relatively prime. Let H ⊂ G be the set of all elements of exponent m and let K ⊂ G be the set of all elements of exponent k. Then H and K are subgroups of G, and G is the internal direct product of H and K. Proof That H and K are subgroups is a consequence of Lemma 4.4.3. For instance, if x, y ∈ H, then the prime divisors of o(x + y) must all divide m. Since o(x + y) must divide |G|, this forces o(x + y) to divide m. By Proposition 2.10.8, since G is abelian, it suﬃces to show that H ∩ K = 0, and that H and K generate G. Since m and k are relatively prime, any element which has both m and k as an exponent must have order 1 (Corollary 2.5.13), so that H ∩ K is indeed the trivial subgroup. But if g ∈ G with o(g) = n , then we may write n = m k , where m divides m and k divides k. But then m and k are relatively prime, so that sm + tk = 1 for some integers s and t. But now g = s(m g) + t(k g). Since m g and k g have orders k and m , respectively, g is in H + K as desired. We shall now give a special case of Cauchy’s Theorem, showing that the groups Gp are nontrivial for all primes p dividing the order of G. Cauchy’s Theorem has already been given as Problem 24 of Exercises 3.3.23, and will be given again by a diﬀerent argument, using the case we’re about to give, as Theorem 5.1.1. Both proofs are instructive. Proposition 4.4.12. (Cauchy’s Theorem for Abelian Groups) Let G be a ﬁnite abelian group and let p be a prime number dividing the order of G. Then G has an element of order p. Proof We argue by induction on |G|, with the case |G| = p a triviality. Let 0 = g ∈ G. If o(g) is divisible by p, say o(g) = pq, then qg has order p. But if o(g) is not divisible by p, then |G/g| is divisible by p. Since |G/g| < |G|, the induction hypothesis gives an element x ∈ G/g of order p. But since x is the image of x ∈ G under the canonical homomorphism, Corollary 2.5.19 shows that p must divide the order of x. This now serves to clarify the situation of Proposition 4.4.11: Corollary 4.4.13. Let G be a ﬁnite abelian group of order n = mk, with (m, k) = 1. Let H ⊂ G be the set of elements in G of exponent m and let K ⊂ G be the set of elements in G of exponent k. Then H has order m and K has order k. Proof By Proposition 4.4.12, the order of H cannot be divisible by any prime that doesn’t divide m, as then H would have an element whose order does not divide m. But then |H| must divide m by Lagrange’s Theorem. Similarly, |K| divides k. But since G ∼ = H × K, |G| = |H| · |K|. Since n = mk, the result must follow. CHAPTER 4. NORMALITY AND FACTOR GROUPS 102 As a special case, Corollary 4.4.13 gives a calculation of the order of the p-torsion subgroup of a ﬁnite abelian group for any prime p dividing the order of the group. Corollary 4.4.14. Let G be a ﬁnite abelian group of order n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes. Then the pi -torsion subgroup Gpi has order pri i for 1 ≤ i ≤ k. The analogue of the preceding statement for nonabelian groups is the second Sylow Theorem. Its proof is a little deeper than that of the abelian case. We now give another useful consequence of Corollary 4.4.13. Corollary 4.4.15. Let G be a ﬁnite abelian group and let H be a subgroup. Suppose that the orders of H and G/H are relatively prime. Then H is the subgroup of G consisting of all elements of exponent |H|, and G is isomorphic to H × G/H. Proof Write n, m and k for the orders of G, H, and G/H, respectively. Then n = mk. Let H ⊂ G be elements of exponent m and let K ⊂ G be the elements of exponent k. We shall show that H = H and that K is isomorphic to G/H. To see the former, let i : H ⊂ G be the inclusion. Since the elements of H have exponent m, we must have i(H) ⊂ H . But then i : H → H is an injective homomorphism. Since H and H both have order m, it must be an isomorphism. Let π : G → G/H be the canonical map. Then ker π = H. Since |K| and |H| are relatively prime, we have H ∩ K = 0, and hence π restricts to an injection from K to G/H. But since K and G/H have the same order, π must induce an isomorphism from K to G/H. Next, we show that for any prime p that any ﬁnite abelian p-group is a direct product of cyclic groups. Together with Proposition 4.4.9, this will give the existence part of the Fundamental Theorem: Theorem 4.4.16. (Fundamental Theorem of Finite Abelian Groups) Every ﬁnite abelian group is a direct product of cyclic groups of prime power order. We shall restate it with a uniqueness statement below. Lemma 4.4.17. Let G be an abelian group of exponent m. Suppose that g ∈ G has order m. Let x ∈ G/g. Then there exists y ∈ G with y = x, such that the order of y in G is equal to the order of x in G/g. Proof Let k = o(x) and let n = o(x). Because x is the image of x under the canonical map π : G → G/g k divides n (Corollary 2.5.19). So n = kl for l ∈ Z. Note then that kx has order l. But since x has order k, kx = kx = 0, and hence kx ∈ ker π = g. Say kx = sg. Since g has order m, the order of sg is m/(m, s). Since kx = sg, this says l = m/(m, s), and hence l · (m, s) = m. Since n = kl is the order of x, it must divide the exponent of G, which is m. Thus, k divides (m, s). But then s = kt, and hence y = x − tg has order k. For the next result, we only use the special case of Lemma 4.4.17 in which the exponent m is a prime power. The reader is invited to re-examine the proof in this simpler case. Proposition 4.4.18. Let p be a prime. Then any ﬁnite abelian p-group is a direct product of cyclic groups. CHAPTER 4. NORMALITY AND FACTOR GROUPS 103 Proof By Proposition 4.4.12, |G| is a power of p. We argue by induction on |G|. The induction begins with |G| = p, in which case G itself is cyclic. By Lagrange’s Theorem, |G| is an exponent for G. Let pj be the smallest power of p which is an exponent for G. Then there must be an element g ∈ G of order pj . Let H = G/g, with π : G → H the canonical map. Then |H| < |G|, so our induction hypothesis gives an isomorphism ν : Zpr1 × · · · × Zprk → H. Let xi = ν ◦ ιi (1), where ιi : Zpri → Zpr1 × · · · × Zprk is the inclusion of the i-th factor of Zpr1 × · · · × Zprk . Then xi has order pri . By Lemma 4.4.17, there are elements yi ∈ G with π(yi ) = xi , such that yi has order pri as well. Deﬁne ν : Zpr1 × · · · × Zprk × Zpj → G by ν(m1 , . . . , mk+1 ) = m1 y1 + · · · + mk yk + mk+1 g. Since the orders of the yi and of g are as stated, ν ◦ ιi is a homomorphism for 1 ≤ i ≤ k + 1 by Proposition 2.5.17. Thus, ν is a homomorphism by Proposition 2.10.10. We claim that ν is an isomorphism. Suppose that α = (m1 , . . . , mk+1 ) is in the kernel of ν. Then 0 = π ◦ ν(m1 , . . . , mk+1 ) = ν(m1 , . . . , mk ). Since ν is an isomorphism, m1 = · · · = mk = 0. But then 0 = ν(α) = mk+1 g. Since o(g) = pj , mk+1 = 0, and hence α = 0. Thus, ν is injective. Let x ∈ G, and suppose that ν −1 ◦ π(x) = (m1 , . . . , mk ). Then π(x − (m1 y1 + · · · + mk yk )) = 0. Since ker π = g, x − (m1 y1 + · · · + mk yk ) = mk+1 g for some integer mk+1 . But then x = ν(m1 , . . . , mk+1 ), and hence ν is onto. This completes the proof of the existence part of the Fundamental Theorem. We shall now discuss uniqueness. First consider this. Lemma 4.4.19. Let f : G → H be a homomorphism of abelian groups. Then for each prime p, f restricts to a homomorphism fp : Gp → Hp . Proof The order of f (g) must divide the order of g. If g ∈ Gp , then the order of g is a power of p, so the same must be true of f (g). Thus, f (g) ∈ Hp . For those who understand the language of category theory, this implies that passage to the p-torsion subgroup provides a functor from abelian groups to abelian p-groups. Even better, the next result shows that the isomorphisms μG of Corollary 4.4.10 are natural transformations. The proof is obtained by chasing elements around the diagram. Lemma 4.4.20. Let G and H be ﬁnite abelian groups and let p1 , . . . , pk be the collection of all distinct primes that divide either |G| or |H|. Let f : G → H be a homomorphism. Then the following diagram commutes. Gp1 × · · · × Gpk μG ∼ = fp1 ×···×fpk Hp1 × · · · × Hpk / G f μH ∼ = / H Here, μG and μH are the isomorphisms of Corollary 4.4.10. Note that if p divides |G| but not |H|, then Hp = 0, so that fp is the constant homomorphism to 0. Similarly, if p divides |H| but not |G|, then Gp = 0, and fp is CHAPTER 4. NORMALITY AND FACTOR GROUPS 104 the unique homomorphism from 0 to Hp . Note that since μG and μH are isomorphisms, the above diagram shows that f is an isomorphism if and only if fp1 × · · · × fpk is an isomorphism. We obtain an immediate corollary. Corollary 4.4.21. Let f : G → H be a homomorphism of ﬁnite abelian groups. Then f is an isomorphism if and only if fp : Gp → Hp is an isomorphism for each prime p dividing either |G| or |H|. Thus, a uniqueness statement for the Fundamental Theorem will follow if we can prove the uniqueness of the decomposition (from Proposition 4.4.18) of an abelian p-group as a product of cyclic groups. Fix a prime p. Then, by ordering the exponents, we can write any ﬁnite abelian p-group as a product Zpr1 × · · · × Zprk , where 1 ≤ r1 ≤ · · · ≤ rk . Proposition 4.4.22. Suppose that ν : Zpr1 × · · · × Zprk ∼ = / Zps1 × · · · × Zpsl , where 1 ≤ r1 ≤ · · · ≤ rk , and 1 ≤ s1 ≤ · · · ≤ sl . Then k = l and ri = si for 1 ≤ i ≤ k. Proof We argue by induction on the order of the group. There is only one way in which we can get a group of order p, so the induction starts there. Write Zp ⊂ Zps for the cyclic subgroup ps−1 . Note that the elements of Zp are the only ones in Zps of exponent p. By Problem 2 of Exercises 2.10.12, the elements of Zpr1 × · · · × Zprk which have exponent p are precisely those in the product Zp × · · · × Zp of the subgroups of order p in each factor. Thus, Zpr1 × · · · × Zprk has pk elements of exponent p, all but one of which has order p. Similarly, Zps1 ×· · ·×Zpsl has pl elements of exponent p. Since isomorphisms preserve order, k = l. Moreover, ν restricts to an isomorphism ν : Zp × · · · × Zp → Zp × · · · × Zp . Passing to factor groups, we obtain an isomorphism ν : (Zpr1 × · · · × Zprk )/(Zp × · · · × Zp ) → (Zps1 × · · · × Zpsk )/(Zp × · · · × Zp ). By Problem 2 of Exercises 4.1.26, Zps /Zp ∼ = Zps−1 . Thus, Problem 14 of the same exercises shows that we may identify ν with an isomorphism ν : Zpr1 −1 × · · · × Zprk −1 ∼ = / Zps1 −1 × · · · × Zpsk −1 . The result now follows from induction on order. The full statement of the Fundamental Theorem is now immediate: Theorem 4.4.23. (Fundamental Theorem, with Uniqueness) Any ﬁnite abelian group may be written uniquely as a direct product of cyclic groups of prime power order. Uniqueness means that given any two such decompositions of a given group, the number of factors of a given order in the two decompositions must be the same. Exercises 4.4.24. 1. Let G be an abelian group. Show that G/Tors(G) is torsion-free. ‡ 2. Show that a ﬁnite abelian group G is cyclic if and only if the p-torsion subgroup Gp is cyclic for each prime p dividing the order of G. (Hint: This does not require the Fundamental Theorem.) CHAPTER 4. NORMALITY AND FACTOR GROUPS 105 3. Show that a ﬁnite abelian group is cyclic if and only if it has no subgroup of the form Zp × Zp for any prime p. (Hint: Apply the Fundamental Theorem.) † 4. Use the Fundamental Theorem to show that a ﬁnite abelian group is cyclic if and only if it has at most d elements of exponent d for each d dividing the order of G. 5. Give a direct proof (not using the Fundamental Theorem) of the preceding problem. (Hint: One direction has already been done in Problem 11 of Exercises 2.5.21. For the other direction, ﬁrst show via Problem 2 above that we may assume that the group has order pr for some prime p and some r ≥ 1. Then show that there must be a generator.) 6. Show that the full converse to Lagrange’s Theorem holds for ﬁnite abelian groups G: if k divides the order of G, then G has a subgroup of order k. 7. Show that any ﬁnite abelian group may be written as a product Zn1 × · · · × Znk , where ni divides ni−1 for 1 < i ≤ k. 8. Show that the decomposition of the preceding problem is unique. 4.5 The Automorphisms of a Cyclic Group We calculate the automorphism group of a cyclic group. Recall that for any group G and for any g ∈ G whose order divides n, there is a unique homomorphism hg : Zn → G with hg (1) = g. Lemma 4.5.1. For n > 1 and m ∈ Z, the homomorphism hm : Zn → Zn is an automorphism if and only if m and n are relatively prime. Proof The image of hm is the cyclic subgroup generated by m. Since the order of m is n/(m, n), hm is onto if and only if (m, n) = 1. But any surjection from a ﬁnite set to itself is an injection. Thus, we’ve identiﬁed which homomorphisms from Zn to Zn are automorphisms, but we have yet to identify the group structure on Aut(Zn ). Recall that there are two operations deﬁned on Zn , addition and multiplication, and that multiplication gives Zn the structure of a monoid, with unit element 1. Recall that the collection of invertible elements in a monoid M forms a group (which we have denoted Inv(M )) under the monoid operation of M . Deﬁnition 4.5.2. For n > 1 we write Z× n for the group of invertible elements in the multiplicative monoid structure on Zn . Thus, m ∈ Z× n if and only if there is a k ∈ Zn with km = 1 ∈ Zn . Z× n is an example of a general construction in rings. The set of elements in a ring A which have multiplicative inverses forms a group under the operation of multiplication. It is called the group of units of A, and may be denoted A× . We wish to identify the elements of Z× n . The answer turns out to be quite relevant to our discussion of automorphisms. Lemma 4.5.3. Let n > 1 and let m ∈ Z. Then m ∈ Z× n if and only if m and n are relatively prime. CHAPTER 4. NORMALITY AND FACTOR GROUPS 106 Proof Suppose that m is invertible in Zn , with inverse k. Since km is congruent to 1 mod n, we obtain that km − 1 = sn for some integer s. But then km − sn = 1, and hence (m, n) = 1. Conversely, if rm + sn = 1, then r is the multiplicative inverse for m in Zn . Thus, there is a one-to-one correspondence between the elements of Aut(Zn ) and Z× n. We next show that this correspondence is an isomorphism of groups. Proposition 4.5.4. Deﬁne ν : Z× n → Aut(Zn ) by ν(m) = hm , where hm : Zn → Zn is the homomorphism carrying 1 to m. Then ν is an isomorphism of groups. Proof It suﬃces to show that hm ◦ hm = hmm . But it’s easy to see (e.g., by Corollary 2.5.5 and Proposition 2.5.17) that hm (k) = mk, and the result follows. Since multiplication in Zn is commutative, we obtain the following corollary. Corollary 4.5.5. For any positive integer n, Aut(Zn ) ∼ = Z× is an abelian group. n In contrast, most other automorphism groups, even of abelian groups, are nonabelian. It will require a bit more work to display the group structures on these groups of units Z× n . It will be easiest to do so in the case that n only has one prime divisor. We will then be able to put this information together, via methods we are about to discuss. First recall from Section 4.4 that if G is abelian and if Gp is its p-torsion subgroup, then any homomorphism f : G → G restricts to a homomorphism fp : Gp → Gp (i.e., fp (g) = f (g) for all g ∈ Gp ). Moreover, Corollary 4.4.21 shows that f : G → G is an automorphism if and only if fp is an automorphism for each p dividing |G|. The next result follows easily. Lemma 4.5.6. Let G be an abelian group and let p1 , . . . pk be the primes dividing the order of G. Let Gpi be the pi -torsion subgroup of G for 1 ≤ i ≤ k. Then there is an isomorphism ∼ = ϕ : Aut(G) / Aut(Gp1 ) × · · · × Aut(Gpk ) deﬁned by ϕ(f ) = (fp1 , . . . , fpk ). Let n = pr11 . . . prkk with p1 , . . . , pk distinct primes. Then the pi -torsion subgroup of Zn is isomorphic to Zpri . We obtain the following corollary, which may also be proven i using the Chinese Remainder Theorem, below. Corollary 4.5.7. With n as above, there is an isomorphism ∼ = ρ : Z× n / Z×r1 × · · · × Z×rk p p 1 k given by ρ(m) = (m, . . . , m). Proof As the reader may easily check, there is a commutative diagram Z× n ν ∼ = Aut(Zn ) ρ ϕ ∼ = / Z×r1 × · · · × Z×rk p1 pk ∼ = ν×···×ν / Aut(Zpr1 ) × · · · × Aut(Zprk ) 1 k CHAPTER 4. NORMALITY AND FACTOR GROUPS 107 Thus, the calculation of the automorphism groups of cyclic groups is reduced to the × calculation of Z× pr for p a prime and r ≥ 1. We begin by calculating the order of Zpr . Lemma 4.5.8. For any prime p and any integer r > 0, the group of units Z× pr has order pr−1 (p − 1). Proof The elements of Zpr that do not lie in Z× pr are the elements m such that (p, m) = 1. But since p is prime, this means that p divides m, and hence m ∈ p ⊂ Zpr . But p r r−1 = pr−1 (p − 1). has order pr−1 , and hence |Z× pr | = p − p Corollary 4.5.7 now allows us to calculate |Z× n |. Corollary 4.5.9. Let n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes and the exponents ri are all positive. Then r1 −1 |Aut(Zn )| = |Z× . . . (pk − 1)prkk −1 . n | = (p1 − 1)p1 In particular, Z× p has order p − 1. In Corollary 7.3.15, we shall show, using some elementary ﬁeld theory, that Z× p is always a cyclic group. However, the proof is theoretical, and doesn’t provide any hint as to what elements generate this group. And indeed there is no known algorithm to obtain a generator other than trial and error. (Some of the deeper aspects of number theory are lurking in the background.) So for the practical purpose of identifying the orders of individual elements of Z× p , direct calculation is necessary. The reader with some computer expertise may ﬁnd it interesting to write a program for this purpose. On the other hand, when r > 1, the orders of the elements of Z× p will be the only open question in our understanding of Z× . We need to understand what happens to Z× r p pr s r as r varies. Thus, let 1 ≤ s < r. Since p divides p , there is a homomorphism, which we shall denote by π : Zpr → Zps , with π(1) = 1. In fact, π(m) = m for all m ∈ Z. We call π the canonical homomorphism. We have m ∈ Z× pr if and only if (m, p) = 1, which × is true if and only if m ∈ Zps . The next lemma is now immediate. Lemma 4.5.10. For 1 ≤ s < r and for p a prime, the canonical homomorphism π : Zpr → Zps induces a surjective homomorphism of multiplicative groups × π : Z× pr → Zps . At this point, the cases p > 2 and p = 2 diverge. One diﬀerence that is immediately × apparent is that Z× 2 = e, while Zp is nontrivial for p > 2. Thus, for p > 2 we shall deﬁne × K(p, r) to be the kernel of π : Z× pr → Zp , while K(2, r) is deﬁned to be the kernel of × × π : Z2r → Z4 (and hence K(2, r) is only deﬁned for r ≥ 2). Note that Z× 4 = {±1} has order 2. Lemma 4.5.10 and the First Noether Theorem now give us the order of K(p, r). Corollary 4.5.11. For p > 2 and r ≥ 1, K(p, r) has order pr−1 . Also, for r ≥ 2, K(2, r) has order 2r−2 . For p > 2, the orders of K(p, r) and Z× p are relatively prime, so Corollary 4.4.15 gives a splitting of Z× . r p CHAPTER 4. NORMALITY AND FACTOR GROUPS 108 Corollary 4.5.12. For p > 2 and r ≥ 1, K(p, r) is the p-torsion subgroup of Z× pr , and we have an isomorphism × ∼ Z× pr = K(p, r) × Zp . The prime 2 is a little diﬀerent, in that K(2, r) and Z× 4 are both 2-groups. But we get a product formula regardless. Lemma 4.5.13. For r ≥ 2, we have an isomorphism × ∼ Z× 2r = K(2, r) × Z4 . × Proof Let H = {±1} ⊂ Z× 2r . Then H is a subgroup of Z2r , and is carried isomorphically × × onto Z4 by π. We shall show that Z2r is the internal direct product of K(2, r) and H, i.e., the map μ : H × K(2, r) → Z× 2r given by μ(h, k) = hk is an isomorphism. The key is that since −1 is not congruent to 1 mod 4, −1 is not in K(2, r), and hence H ∩ K(2, r) = 1. (The unit element here is denoted 1.) This easily implies that μ is injective. But since m · k = mk in both Z2r and Z4 , either m or −m is in K(2, r) for each m ∈ Z× 2r . Thus, μ is onto. We shall show that K(p, r) is cyclic for all primes p and all r ≥ 2. (Unlike the case of Z× p , we shall provide a criterion to determine the order of each element.) The key is to understand the order of the elements 1 + ps . As the group operation is multiplication modulo pr , we can get a handle on this if we can calculate the powers (x + y)n of the sum of two integers x and y. The formula for (x + y)n is given by what’s called the Binomial Theorem, which in turn depends on an understanding of the binomial coeﬃcients. Deﬁnition 4.5.14. Let 0 ≤ k ≤ n ∈ Z. We deﬁne the binomial coeﬃcient nk by n n! = . k!(n − k)! k Here, the factorial numbers are deﬁned as usual: 0! = 1, 1! = 1, and n! is deﬁned inductively by n! = (n − 1)! · n for n > 1. The next lemma is useful for manipulating the binomial coeﬃcients. Lemma 4.5.15. The binomial coeﬃcients are all integers. Moreover, if 1 ≤ k < n we have n−1 n−1 n + = . k−1 k k Proof An easy inspection of the deﬁnition shows that n0 = nn = 1 for all n. But the only other binomial coeﬃcients are the nk with 1 ≤ k < n. Thus, if the displayed equation is true, then the binomial coeﬃcients are all integers by induction on n, starting with the case n = 1, which we’ve just veriﬁed by inspection. The left-hand side of the displayed equation expands to (n − 1)! (n − 1)! (n − 1)![k + (n − k)] + = , (k − 1)!(n − k)! k!(n − k − 1)! k!(n − k)! and the result follows. CHAPTER 4. NORMALITY AND FACTOR GROUPS 109 Our next result is the Binomial Theorem, which is true, with the same proof, not only in the integers, but in any commutative ring. Theorem 4.5.16. (Binomial Theorem) Let x and y be integers, and n ≥ 1 be another. Then n n k n−k (x + y)n = x y k k=0 n 2 n−2 = y n + nxy n−1 + x y + · · · + nxn−1 y + xn 2 Proof The second linejust gives n an expansion of the preceding expression, using the = n, in addition to the already known calculations (easy) calculations that n1 = n−1 n n of 0 and n . We prove the theorem by induction on n. For n = 1, it just says that x + y = x + y. Thus, assume that n > 1 and the theorem is true for n − 1. Then # "n−1 n − 1 xk y n−k−1 (x + y). (x + y)n = (x + y)n−1 (x + y) = k k=0 Applying the distributive law to the last expression and collecting terms, we get $ % n−1 n−1 n−1 + . xn + y n + xk y n−k k−1 k k=1 But this is exactly the desired sum, by Lemma 4.5.15. We wish to calculate the order of 1 + ps in K(p, r). By the deﬁnitions, the order will be the smallest integer m such that (1 + ps )m is congruent to 1 mod pr . Since the order of K(p, r) is a power of p, m must be also. Thus, we shall study the binomial expansion k for (1 + ps )p . For this purpose, we shall need to know the largest power of p that divides k the binomial coeﬃcients pj . Lemma 4.5.17. Let p be prime and let k ≥ 1. Let 1 ≤ j ≤ pk and write j = pa b, with (p, b) = 1. Then k p = pk−a c, j with (p, c) = 1. Proof After making appropriate cancellations, we have k p pk (pk − 1) . . . (pk − j + 1) = . j 1 · 2 . . . (pa b) Consider the right-hand side as a rational number and rearrange the terms slightly. We get the product of pk /pa b with the product k k k p −2 p − (j − 1) p −1 · ... 1 2 j−1 But for 1 ≤ i < pk , pk − i and i have the same p-divisibility, so that the only contribution to the p-divisibility of our binomial coeﬃcient comes from pk /pa b. But this gives the desired result. CHAPTER 4. NORMALITY AND FACTOR GROUPS 110 We shall now calculate the order of the generic element of K(p, r). Note that if p > 2, then x ∈ K(p, r) if and only if x = 1 + y for some y divisible by p. Similarly, x ∈ K(2, r) if and only if x = 1 + y for some y divisible by 4. Proposition 4.5.18. Let p be a prime and let x = 1 + ps a, where (p, a) = 1. If p = 2, assume that s > 1. Then x has order pr−s in K(p, r). Proof As discussed above, the order of x in K(p, r) is the smallest power, pk , of p for k which xp is congruent to 1 mod pr . So we want to ﬁnd the smallest value of k such k k that (ps a + 1)p − 1 is divisible by pr . Consider the binomial expansion for (ps a + 1)p . k k We have (ps a + 1)p = 1 + pk ps a plus terms of the form pj psj aj with j > 1. But then k k (ps a + 1)p − 1 is equal to ps+k a plus the terms pj psj aj with j > 1. k It suﬃces to show that if j > 1, then pj psj aj is divisible by ps+k+1 . k By Lemma 4.5.17, if j = pb c, with (p, c) = 1, then pj is divisible by pk−b . Thus, pk sj j b sj+k−b = psp c+k−b , so it suﬃces to show that spb c+k−b > s+k. j p a is divisible by p But this is equivalent to showing that s[pb c − 1] > b. If b = 0, then c = j > 1, and the result holds. Otherwise, b > 0, and we may and shall assume that c = 1. For p > 2, we must show that s[pb − 1] > b for all s ≥ 1. But the general case will follow from that of s = 1, where we show that pb − 1 > b for all b > 0. For p = 2, our hypothesis is that s > 1, so it suﬃces to show that 2b − 1 ≥ b for all b > 0. For any p, pb − 1 = [(p − 1) + 1]b − 1, so we may use the Binomial Theorem again. We see that pb − 1 is equal to b(p − 1) plus some other non-negative terms. But this is greater than b if p > 2, and is ≥ b if p = 2, precisely as desired. Corollary 4.5.19. For p > 2 and r ≥ 1, K(p, r) is the cyclic group (of order pr−1 ) generated by 1 + p. For p = 2 and r ≥ 2, K(2, r) is the cyclic group (of order 2r−2 ) generated by 5. Proof The orders of the groups in question were calculated in Corollary 4.5.11. By Proposition 4.5.18, the 1 + p has the same order as K(p, r) for p > 2, while 5 has the same order as K(2, r). Lemma 4.5.13 now gives an explicit calculation of Z× 2r . Corollary 4.5.20. For r ≥ 2, Z× 2r is isomorphic to Z2r−2 × Z2 . In particular, Z× 2r is not cyclic for r ≥ 3. On the other hand, when we have established Corollary 7.3.15, we will know that Z× p is a cyclic group of order p − 1 for all primes p. We shall state the consequence here, though the reader is advised to read ahead for the proof. This time we make use of Corollary 4.5.12: Corollary 4.5.21. Let p be an odd prime and let r ≥ 1. Then Z× pr is a cyclic group of order (p − 1)pr−1 . Exercises 4.5.22. 1. Find generators for Z× p for all primes p ≤ 17. 2. Find a generator for Z× 3r for any r. 3. Find a generator for Z× 5r for any r. CHAPTER 4. NORMALITY AND FACTOR GROUPS 111 4. Find a generator for Z× 49 . 5. Find all elements of order 2 in Z× 2r . × × 6. Find all elements of order 2 in Z× 15 , Z30 , and Z60 . 7. Write a computer program to calculate the order of an element in Z× pr , where the input is p, r, and the element in question. Be sure it checks that the element is a unit. 8. Write a computer program to calculate the order of an element in Z× n . Be sure it checks that the element is relatively prime to n. 4.6 Semidirect Products The simplest way to construct new groups from old is to take direct products. But there are ways to construct more interesting groups if we can calculate the automorphisms of one of the groups in question. Suppose given an action of K on H through automorphisms. We may represent this by a homomorphism α : K → Aut(H), so that the action of k ∈ K takes h ∈ H to α(k)(h). We shall deﬁne a group multiplication on the set H × K, to obtain a new group, which we denote H α K, called the semidirect product of H and K given by α: Deﬁnition 4.6.1. Let α : K → Aut(H) be a homomorphism. By the semidirect product of H and K with respect to α, written H α K, we mean the set H × K with the binary operation given by setting (h1 , k1 ) · (h2 , k2 ) = (h1 · α(k1 )(h2 ), k1 k2 ). Proposition 4.6.2. H α K, with the above product, is a group. There are homomorphisms π : H α K → K and ι : H → H α K, given by π(h, k) = k, and ι(h) = (h, e). The homomorphism ι is an injection onto the kernel of π. Finally, there is a homomorphism s : K → H α K given by s(k) = (e, k). This homomorphism has the property that π ◦ s is the identity map of K. Proof We ﬁrst show associativity. ((h1 , k1 ) · (h2 , k2 )) · (h3 , k3 ) = (h1 · α(k1 )(h2 ), k1 k2 ) · (h3 , k3 ) = (h1 · α(k1 )(h2 ) · α(k1 k2 )(h3 ), k1 k2 k3 ) = (h1 · α(k1 )(h2 ) · α(k1 )(α(k2 )(h3 )), k1 k2 k3 ). Here, the last equality holds because α is a homomorphism, so that α(k1 k2 ) = α(k1 ) ◦ α(k2 ). Since α(k1 ) is an automorphism of H, we have α(k1 )(h2 ) · α(k1 )(α(k2 )(h3 )) = α(k1 )(h2 · α(k2 )(h3 )), and hence ((h1 , k1 ) · (h2 , k2 )) · (h3 , k3 ) = (h1 · α(k1 )(h2 · α(k2 )(h3 )), k1 k2 k3 ). But this is precisely what one gets if one expands out (h1 , k1 ) · ((h2 , k2 ) · (h3 , k3 )). CHAPTER 4. NORMALITY AND FACTOR GROUPS 112 The pair (e, e) is clearly an identity element for this product, since α(e) is the identity homomorphism of H. We claim that the inverse of (h, k) is (α(k −1 )(h−1 ), k −1 ). The reader is welcome to verify this fact, as well as the statements about π, ι, and s. Note that if α(k) is the identity automorphism for all k ∈ K (i.e., α : K → Aut(H) is the trivial homomorphism), then H α K is just the direct product H × K. Notation 4.6.3. In the semidirect product H α K, we shall customarily identify H with ι(H) and identify K with s(K), so that the element (h, k) is written as hk. Thus, H α K = {hk | h ∈ H, k ∈ K}. Lemma 4.6.4. With respect to the above notations, the product in H α K is given by h1 k1 · h2 k2 = h1 α(k1 )(h2 )k1 k2 for h1 , h2 ∈ H and k1 , k2 ∈ K. In particular, this gives khk −1 = α(k)(h). Thus, H α K is abelian if and only if H and K are both abelian and α : K → Aut(H) is the trivial homomorphism. Proof The multiplication formula is clear. But this gives kh = α(k)(h)k, which in turn implies the stated conjugation formula. Thus, when α is nontrivial, the new group H α K may have some interesting features. Indeed, this method of constructing groups will give us a wide range of new examples. But it also gives new ways of constructing several of the groups we’ve already seen. For instance, as we shall see below, the dihedral group D2n is the semidirect product obtained from an action of Z2 on Zn , and A4 is the semidirect product obtained from an action of Z3 on Z2 × Z2 . We wish to analyze the semidirect products obtained from actions of cyclic groups on other cyclic groups. Since these are, in general, nonabelian, we shall use multiplicative notation for our cyclic groups. Thus, we shall begin with a discussion of notation. × Recall ﬁrst the isomorphism ν : Z× n → Aut(Zn ) of Proposition 4.5.4. Here, Zn is the group of invertible elements with respect to the operation of multiplication in Zn , and, for m ∈ Z× n , ν(m) is the automorphism deﬁned by ν(m)(l) = ml. In multiplicative notation, we write Zn = b = {e, b, . . . , bn−1 }. Thus, the element of Zn written as l in additive notation becomes bl . In other words, in this notation, we have an isomorphism ν : Z× n → Aut(b), where ν(m) is the automorphism deﬁned by ν(m)(bl ) = (bml ). Now write Zk = a in multiplicative notation, and recall from Proposition 2.5.17 that there is a homomorphism hg : Zk → G with hg (a) = g ∈ G if and only if g has exponent k in G. Moreover, if g has exponent k, there is a unique such homomorphism, and it is given by the formula hg (aj ) = g j . Thus, suppose given a homomorphism α : Zk = a → Z× n , with α(a) = m. Then m × k has exponent k in Zn , meaning that m is congruent to 1 mod n. Moreover, α(aj ) = mj . By abuse of notation, we shall write α : Zk → Aut(Zn ) for the composite of this α with the isomorphism ν above. Thus, as an automorphism of Zn = b, α(aj ) is given j by α(aj )(bi ) = bm i . × Corollary 4.6.5. Let m have exponent k in Z× n , and let α : Zk → Zn be the homomorphism that takes the generator to m. Then writing Zn = b, Zk = a, and identifying Z× n with Aut(Zn ), we obtain Zn α Zk = {bi aj | 0 ≤ i < n, 0 ≤ j < k}, CHAPTER 4. NORMALITY AND FACTOR GROUPS 113 where b has order n, a has order k, and the multiplication is given by j bi aj bi aj = bi+m i aj+j . Moreover, the nk elements {bi aj | 0 ≤ i < n, 0 ≤ j < k} are all distinct. Proof The multiplication formula comes via Lemma 4.6.4 from the formula α(aj )(bi ) = j bm i . Examples 4.6.6. 1. Let α : Z2 → Z× n be the homomorphism that carries the generator to −1. Then in Zn α Z2 , with the notations above, we see that b has order n, a has order 2, and aba−1 = b−1 . Thus, Problem 6 of Exercises 2.8.5 provides a homomorphism f : D2n → Zn α Z2 with f (a) = a and f (b) = b. Since a and b generate Zn α Z2 , f is onto, and hence an isomorphism, as the two groups have the same order. Thus, semidirect products provide a construction of the dihedral groups that does not require trigonometry. 2. Let α : Z2 → Z× 8 be the homomorphism that carries the generator to 3. We obtain a group of order 16, Z8 α Z2 = {bi aj | 0 ≤ i < 8, 0 ≤ j < 2}, where a has order 2, b has order 8, and the multiplication is given by bi aj bi aj = j bi+3 i aj+j . 3. Let β : Z2 → Z× 8 be the homomorphism that carries the generator to 5. We obtain a group of order 16, Z8 β Z2 = {bi aj | 0 ≤ i < 8, 0 ≤ j < 2}, where a has order 2, b has order 8, and the multiplication is given by bi aj bi aj = j bi+5 i aj+j . 4. Let α : Z4 → Z× 5 be the homomorphism that carries the generator to 2. We obtain a group of order 20, Z5 α Z4 = {bi aj | 0 ≤ i < 5, 0 ≤ j < 4}, where a has order 4, b has order 5, and the multiplication is given by bi aj bi aj = j bi+2 i aj+j . 5. Let α : Z3 → Z× 7 be the homomorphism that carries the generator to 2. We obtain a group of order 21, Z7 α Z3 = {bi aj | 0 ≤ i < 7, 0 ≤ j < 3}, where a has order 3, b has order 7, and the multiplication is given by bi aj bi aj = j bi+2 i aj+j . 6. Let p be an odd prime. Recall from Proposition 4.5.18 that p + 1 has order p in × Z× p2 . Thus, there is a homomorphism α : Zp → Zp2 that carries the generator of Zp to p + 1. We obtain a group of order p3 , Zp2 α Zp = {bi aj | 0 ≤ i < p2 , 0 ≤ j < p}, where a has order p, b has order p2 , and the multiplication is given by bi aj bi aj = j bi+(p+1) i aj+j . CHAPTER 4. NORMALITY AND FACTOR GROUPS 114 Other than the ﬁrst example, these groups are all new to us. Indeed, the groups of order 16 in Examples 2 and 3 turn out to be the lowest-order groups that we had not constructed previously. The group of Example 5 turns out to be the smallest odd order group that is nonabelian, while for each odd prime p, the group of Example 6 represents one of the two isomorphism classes of nonabelian groups of order p3 . We may also construct examples of semidirect products where at least one of the two groups is not cyclic. For instance, since the groups Z× n are not generally cyclic, we can obtain a number of groups of the form Zn α K, where K is not cyclic. We are also interested in the case where group “H” in the construction of semidirect products is not cyclic. The simplest case is where H = Zn × Zn . As above, write Zn × Zn = {bi cj | 0 ≤ i, j < n}, where b and c have order n and commute with each other. Since every element of Zn × Zn has exponent n, Propositions 2.10.6 and 2.5.17 show that there is a homomorphism f : Zn × Zn → Zn × Zn with the property that f (b) = bc and f (c) = c: explicitly, f (bi cj ) = bi ci+j . Lemma 4.6.7. The homomorphism f just described is an automorphism of Zn × Zn . It has order n in the group of automorphisms of Zn × Zn . Proof For k ≥ 1 the k-fold composite of f with itself is given by f k (c) = c and f k (b) = f k−1 (bc) = f k−1 (b)f k−1 (c) = (f k−1 (b))c. An easy induction now shows that f k (b) = bck . Thus, f k is the identity if and only if k is congruent to 0 mod n. Since f n is the identity, f is indeed an automorphism, with inverse f n−1 . Thus, if we write Zn = a, there is a homomorphism α : Zn → Aut(Zn × Zn ) with α(a) = f . We obtain a construction which is of special interest when n is prime: Corollary 4.6.8. There is a group G of order n3 given by G = {bi cj ak | 0 ≤ i, j, k < n}, where a, b, and c all have order n, and b commutes with c, a commutes with c, and aba−1 = bc. Thus, bi cj ak · bi cj ak = bi+i cj+j +ki k+k a . There are, of course, at least as many examples of semidirect products as there are groups with nontrivial automorphism groups. Thus, we shall content ourselves for now with these examples. In our studies of the dihedral and quaternionic groups, we have made serious use of our knowledge of the homomorphisms out of these groups. There is a similar result for the generic semidirect product. First, we need some discussion. Let ι : H → H α K and s : K → H α K be the structure maps: in the ordered pairs notation, ι(h) = (h, e) and s(k) = (e, k). Then every element of H α K may be written uniquely as ι(h) · s(k) with h ∈ H and k ∈ K. This is, indeed, the basis for the HK notation. In particular, H α K is generated by the elements of ι(H) and s(K), so that any homomorphism f : H α K → G is uniquely determined by its restriction to the subgroups ι(H) and s(K). Since ι and s are homomorphisms, this says that f is uniquely determined by the homomorphisms f ◦ ι and f ◦ s. CHAPTER 4. NORMALITY AND FACTOR GROUPS 115 Thus, we will understand the homomorphisms out of H α K completely if we can give necessary and suﬃcient conditions for a pair of homomorphisms f1 : H → G and f2 : K → G to be obtained as f1 = f ◦ ι and f2 = f ◦ s for a homomorphism f : H α K → G. Proposition 4.6.9. Let α : K → Aut(H) be a homomorphism. Suppose given homomorphisms f1 : H → G and f2 : K → G for a group G. Then there is a homomorphism f : H α K → G with f1 = f ◦ ι and f2 = f ◦ s if and only if f2 (k)f1 (h)(f2 (k))−1 = f1 (α(k)(h)) for all h ∈ H and k ∈ K. Such an f , if it exists, is unique, and is given in the HK notation by f (hk) = f1 (h)f2 (k). Proof Suppose ﬁrst that f exists with f ◦ ι = f1 and f ◦ s = f2 . In the HK notation we have that hk = ι(h)s(k), so that f (hk) = (f ◦ ι)(h) · (f ◦ s)(k) = f1 (h)f2 (k) for all h ∈ H and k ∈ K. As shown in Lemma 4.6.4, khk −1 = α(k)(h) in H α K, for all h ∈ H and k ∈ K. Applying f to both sides, we obtain f2 (k)f1 (h)(f2 (k))−1 = f1 (α(k)(h)), as claimed. Conversely, suppose that f2 (k)f1 (h)(f2 (k))−1 = f1 (α(k)(h)) for all h ∈ H and k ∈ K. We deﬁne f : H α K → G by f (hk) = f1 (h)f2 (k). It suﬃces to show that f is a homomorphism. But f (h1 k1 )f (h2 k2 ) = f1 (h1 )f2 (k1 )f1 (h2 )f2 (k2 ) = f1 (h1 )f1 (α(k1 )(h2 ))f2 (k1 )f2 (k2 ) = f1 (h1 α(k1 )(h2 ))f2 (k1 k2 ) = f (h1 k1 h2 k2 ), for all h1 , h2 ∈ H and k1 , k2 ∈ K, where the second equality follows from the formula f2 (k1 )f1 (h2 )(f2 (k1 ))−1 = f1 (α(k1 )(h2 )). This permits a slight simpliﬁcation when H and K are cyclic. Corollary 4.6.10. Let α : Zk → Z× n be a homomorphism, with α(a) = m, where Zk = a in multiplicative notation. Write Zn α Zk in multiplicative notation, with Zn = b. Then for any group G, there is a homomorphism f : Zn α Zk → G with f (a) = x and f (b) = y if and only if the elements x, y ∈ G satisfy the following conditions: 1. x has exponent k 2. y has exponent n 3. xyx−1 = y m . Proof The conditions are clearly necessary, as a has order k, b has order n, and aba−1 = bm in Zn α Zk . Conversely, if the three conditions hold, there are homomorphisms f1 : b → G and f2 : a → G given by f1 (bi ) = y i and f2 (aj ) = xj . Thus, by Proposition 4.6.9, it suﬃces j to show that xj y i x−j = y m i . j Since conjugation by x is a homomorphism, xj y i x−j = (xj yx−j )i , so it suﬃces by the j law of exponents to show that xj yx−j = y m . But since xj yx−j = x(xj−1 yx−(j−1) )x−1 , the desired result follows from the third condition by an easy induction argument. The above is an important ingredient in calculating the automorphism groups of the semidirect products of cyclic groups. But it does not give good criteria for showing CHAPTER 4. NORMALITY AND FACTOR GROUPS 116 when the homomorphism f is an isomorphism. For this it will be convenient to use the language and concepts of extensions, as developed in Section 4.7. We shall now begin to consider the question of the uniqueness of the semidirect product: When are two diﬀerent semidirect products isomorphic? We will not answer this question completely, but will study some aspects of it. First, what happens if we switch to a diﬀerent homomorphism K → Aut(H)? Recall from Section 3.6 that G acts on the set of homomorphisms from K to G as follows: if f : K → G is a homomorphism and g ∈ G, then g · f is the homomorphism deﬁned by (g · f )(k) = gf (k)g −1 for all k ∈ K. By abuse of notation, we shall also write gf g −1 for g · f . We say that the homomorphisms f and gf g −1 are conjugate homomorphisms. So it makes sense to ask whether H α K and H α K are isomorphic when α and α are conjugate homomorphisms of K into Aut(H). Another fairly minor way in which we could alter a homomorphism α : K → Aut(H) is by replacing it with α ◦ g, where g ∈ Aut(K). We wish to study the eﬀect of this sort of change as well. Since we’re considering two diﬀerent semidirect products of the same groups, we shall use diﬀerent notations for their structure maps. Thus, we write ι : H → H α K, π : H α K → K, and s : K → H α K for the structure maps of H α K, and write ι : H → H α K, π : H α K → K, and s : K → H α K for the structure maps of H α K. Proposition 4.6.11. Let α and α be homomorphisms of K into Aut(H). Then the following conditions are equivalent: 1. α is conjugate to α ◦ g, with g ∈ Aut(K). 2. There is an isomorphism ϕ : H α K → H α K such that im(ϕ ◦ ι) = im ι and im(ϕ ◦ s) = im s . 3. There is a commutative diagram H f H ι ι / H α K o ϕ / H α K o s s K g K in which the vertical maps are all isomorphisms. Proof Suppose that the second condition holds. Then ϕ restricts to an isomorphism from ι(H) to ι (H), and hence there exists an isomorphism f : H → H making the left-hand square of the diagram commute. Similarly, ϕ restricts to an isomorphism from s(K) to s (K) and there exists an automorphism g of K such that the right-hand square of the diagram commutes. Thus, the second condition implies the third. Suppose then that the third condition holds. Then identifying H and K with their images under ι and s , we see that ϕ ◦ ι = f and ϕ ◦ s = g. By Proposition 4.6.9, this says that g(k)f (h)(g(k))−1 = f (α(k)(h)) for all h ∈ H and k ∈ K. But since f (h) ∈ H and g(k) ∈ K, we have g(k)f (h)(g(k))−1 = α (g(k))(f (h)) by the conjugation formula that holds in H α K. CHAPTER 4. NORMALITY AND FACTOR GROUPS 117 Putting these two statements together, we see that α (g(k))(f (h)) = (f ◦ α(k))(h), and hence α (g(k)) ◦ f = f ◦ α(k) as automorphisms of H. But then as maps of K into Aut(H), this just says that α ◦ g is the conjugate homomorphism to α obtained by conjugating by f ∈ Aut(H). Thus, the third condition implies the ﬁrst. Now suppose that the ﬁrst condition holds, and we’re given g ∈ Aut(K) such that α ◦g is conjugate to α. Thus, there is an f ∈ Aut(H) such that (α ◦ g)(k) = f ◦ α(k) ◦ f −1 for all k ∈ K. Reversing the steps above, this implies that g(k)f (h)(g(k))−1 = f (α(k)(h)) for all h ∈ H and k ∈ K. Thus, by Proposition 4.6.9, There is a homomorphism ϕ : H α K → H α K given by ϕ(hk) = f (h)g(k). Clearly, ϕ restricts to an isomorphism, f , from ι(H) to ι (H) and restricts to an isomorphism, g, from s(K) to s (K), so it suﬃces to show that ϕ is an isomorphism. Since the elements of H α K may be written uniquely as an element of H times an element of K, f (h)g(k) = e if and only if f (h) = e and g(k) = e. Since f and g are both injective, so is ϕ. Finally, since f and g are both surjective, every element of H α K may be written as a product f (h)g(k), and hence ϕ is onto. Thus, the ﬁrst condition implies the second. This is far from giving all the ways in which semidirect products can be isomorphic, but we shall ﬁnd it useful nevertheless, particularly when H is abelian. As a sample application, consider this. Corollary 4.6.12. Let α, α : Z5 → Z× 11 be homomorphisms such that neither α nor α is the trivial homomorphism. Then the semidirect products Z11 α Z5 and Z11 α Z5 are isomorphic. Proof By Corollary 7.3.15, Z× 11 is a cyclic group of order 10. Alternatively, we can verify it by hand: 2k is not congruent to 1 mod 11 for k ≤ 5. Since the order of 2 in Z× 11 divides 10, we see that 2 has order 10, and hence generates Z× 11 . Thus, Z× 11 has a unique subgroup H of order 5. Since the images of α and α must have order 5, they induce isomorphisms of Z5 onto H. Thus, the composite Z5 α / H α−1 / Z5 gives an isomorphism, which we’ll denote by g : Z5 → Z5 . But then α = α ◦ g, and hence Z11 α Z5 ∼ = Z11 α Z5 by Proposition 4.6.11. Using the Sylow Theorems in Chapter 5, we’ll be able to show that every group of order 55 is a semidirect product of the form Z11 α Z5 . Together with the above corollary, this shows that there are exactly two isomorphism classes of groups of order 55: the product Z11 ×Z5 (which is cyclic), and the semidirect product Z11 α Z5 obtained from the homomorphism α : Z5 → Z× 11 that sends the generator of Z5 to 4. Exercises 4.6.13. 1. Show that the subgroup s(K) ⊂ H α K is normal if and only if α : K → Aut(H) is the trivial homomorphism. 2. Consider the groups of Examples 4.6.6. For the groups of Examples 2, 3, 4, and 5, ﬁnd the orders of all the elements not contained in b. Deduce that the groups of Examples 2 and 3 cannot be isomorphic. 3. Show that the group of Example 4 of Examples 4.6.6 contains a subgroup isomorphic to D10 . CHAPTER 4. NORMALITY AND FACTOR GROUPS 118 4. Let α : Z4 → Z× 3 be the homomorphism that carries the generator of Z4 to the unique nontrivial element of Z× 3 . Construct an isomorphism f : Q12 → Z3 α Z4 . 5. Consider the group G = Z8 β Z2 of Example 3 of Examples 4.6.6. Show that the subgroup b is not characteristic in G, by constructing an automorphism of G which carries b outside of itself. (Of course, b is normal in G, as it is the group being acted on in the semidirect product.) 6. Show that if m divides the order of an element x ∈ G, then o(x) = m · o(xm ). Deduce that if f : G → G is a homomorphism and if x ∈ G, then xo(f (x)) ∈ ker f and o(x) = o(f (x)) · o(xo(f (x)) ). Deduce that if hk ∈ H α K, then (hk)o(k) ∈ H, and o(hk) = o(k) · o((hk)o(k) ). Thus, the determination of the order of hk depends only on calculating the element (hk)o(k) ∈ H and on understanding the orders of the elements in H. † 7. Let x, y ∈ G, such that xyx−1 = y m . Show that (yx)k = y 1+m+···+m k ≥ 2. k−1 xk for all † 8. Let m > 1 and k > 1 be integers. Show that (1 + m + · · · + mk−1 )(m − 1) = mk − 1, and hence 1 + m + · · · + mk−1 = (mk − 1)/(m − 1). 9. Find the orders of the elements in the group of Example 6 of Examples 4.6.6. Show also that the subgroup b is not characteristic. (Hint: Take a look at the proof of Proposition 4.5.18.) ‡ 10. Show that the group constructed in Corollary 4.6.8 for n = 2 is isomorphic to D8 . 11. Consider the groups constructed in Corollary 4.6.8 for n a prime number p. When p > 2, show that every element in this group has exponent p. (Why is this diﬀerent from the case n = 2?) Recall that in Problem 3 of Exercises 2.1.16, it was shown that any group of exponent 2 must be abelian. As we see here, this property is not shared by primes p > 2. 12. Now consider the groups of Corollary 4.6.8 for general values of n. For which values of n does the group in question have exponent n? If the exponent is not equal to n, what is it? 13. In this problem we assume a familiarity with the group of invertible 3 × 3 matrices over the ring Zn . Let ⎫ ⎧⎛ ⎞ ⎬ ⎨ 1 x y G = ⎝ 0 1 z ⎠ x, y, z ∈ Zn . ⎭ ⎩ 0 0 1 Show that G forms a group under matrix multiplication. Show, using Proposition 4.6.9, that G is isomorphic to the group constructed in Corollary 4.6.8. † 14. There are three non-identity elements in Z2 × Z2 , which must be permuted by any automorphism. Show that the automorphism group of Z2 × Z2 is isomorphic to the group of permutations on these elements, so that Aut(Z2 × Z2 ) ∼ = S3 . (Hint: Take a look at the proof of Corollary 3.2.11.) 15. Show that any two nontrivial homomorphisms from Z2 to S3 are conjugate. Deduce from Problem 14 that if α, α : Z2 → Aut(Z2 × Z2 ) are nontrivial, then (Z2 ×Z2 )α Z2 ∼ = (Z2 ×Z2 )α Z2 . Now deduce from Problem 10 that if α : Z2 → Aut(Z2 × Z2 ) is nontrivial, then (Z2 × Z2 ) α Z2 ∼ = D8 . CHAPTER 4. NORMALITY AND FACTOR GROUPS 119 16. Show that Aut(Z) ∼ = {±1} and write α : Z2 → Aut(Z) for the nontrivial homomorphism. The semidirect product Z α Z2 is called the inﬁnite dihedral group. Show that the dihedral groups D2n are all factor groups of the inﬁnite dihedral group. 17. Let α : Z → Aut(Z) be the unique nontrivial homomorphism. The semidirect product Z α Z is called the Klein bottle group, as it turns out to be isomorphic to the fundamental group of the Klein bottle. Show that the inﬁnite dihedral group is a factor group of the Klein bottle group. 18. Let Z[ 12 ] ⊂ Q be the set of rational numbers that may be written as 2mk with m ∈ Z and k ≥ 0. In other words, if placed in lowest terms, these elements have denominators that are powers of 2. Show that Z[ 12 ] is a subgroup of the additive group of rational numbers and that multiplication by 2 induces an automorphism of Z[ 12 ]. Deduce the existence of a homomorphism μ2 : Z → Aut(Z[ 12 ]) such that μ2 (k) is multiplication by 2k for all k ∈ Z. ‡ 19. Let G = Z[ 12 ] μ2 Z. Let H ⊂ Z[ 12 ] be the standard copy of the integers in Q. Find an element g ∈ G such that gHg −1 is contained in H but not equal to H.3 4.7 Extensions Here, loosely, we study all groups G that contain a given group H as a normal subgroup, such that the factor group G/H is isomorphic to a particular group K. This is the ﬁrst step in trying to classify composition series. Deﬁnitions 4.7.1. Let H and K be groups. An extension of H by K consists of a group G containing H as a normal subgroup, together with a surjective homomorphism f : G → K with kernel H.4 Examples 4.7.2. 1. Let f : D2n → Z2 be induced by f (a) = 1 and f (b) = 0. Then f is an extension of Zn = b by Z2 . 2. The projection map π : Zn × Z2 → Z2 is an extension of Zn by Z2 . 3. The unique nontrivial homomorphism Z2n → Z2 is yet another extension of Zn = 2 by Z2 . 4. Let f : Q4n → Z2 be induced by f (a) = 1 and f (b) = 0. Then f is an extension of Z2n = b by Z2 . 5. Let α : K → Aut(H) be a homomorphism and let π : H α K → K be the projection. Then π is an extension of H by K. 3 We could have used the rational numbers Q here in place of Z[ 1 ], obtaining a construction that 2 would be slightly simpler for the beginner. However, by using Z[ 12 ], we obtain a group that is ﬁnitely generated (say, by the generator of H and the generator of the Z that acts), whereas we would not have if we’d used Q. 4 It is sometimes useful to think of an extension of H by K as a group G, together with an embedding ι : H → G and a surjective homomorphism f : G → K with kernel ι(H). In this language, the entire f ι sequence H −→ G −→ K constitutes the extension in question. Note that given such data, we can form an extension in the sense of Deﬁnition 4.7.1 by replacing G by the group G obtained by appropriately renaming the elements in im ι. CHAPTER 4. NORMALITY AND FACTOR GROUPS 120 Clearly, for appropriate choices of H and K, there may be a number of nonisomorphic extensions of H by K. Moreover, an understanding of the extensions of one group by another is a key to understanding groups in general. Recall from Proposition 4.2.5 that every nontrivial ﬁnite group G has a composition series, which is a sequence e = H0 ⊂ H1 ⊂ · · · ⊂ Hk = G, such that Hi−1 Hi , and Hi /Hi−1 is a nontrivial simple group for 1 ≤ i ≤ k. But this says that Hi is an extension of Hi−1 by a nontrivial simple group. In other words, every ﬁnite group may be obtained by a sequence of extensions of lower order groups by simple groups. But the ﬁnite simple groups are all known. Thus, the next step in a systematic study of ﬁnite group theory is to classify all possible extensions of one ﬁnite group by another. The search for such a classiﬁcation is known as the Extension Problem, and constitutes one of the major unsolved problems in mathematics. Notice that there are various ways in which we could desire to classify the extensions of H by K. For our purposes here, we’d be most interested in simply determining the isomorphism classes of groups G that occur as extensions of H by K. Such a classiﬁcation takes no account of the precise embedding of H in G, nor of the map f : G → K that determines the extension. Thus, such a classiﬁcation forgets some of the data that’s part of the deﬁnition of an extension. A classiﬁcation that would forget less data would be to classify the extensions of H by K up to the equivalence relation obtained by identifying extensions f : G → K and f : G → K if there is a commutative diagram H ⊂ / K f g2 g1 H / G ⊂ / G g3 f / K in which the vertical maps are all isomorphisms. An even more rigid equivalence relation is the one given by such diagrams where the maps g1 and g3 are required to be the identity maps of H and K, respectively. All three equivalence relations are relevant for particular questions. We shall discuss these issues further. In any case, we shall not be able to solve all the extension questions that might interest us, but we can make considerable progress in special cases. These cases will be suﬃcient to classify all groups of order ≤ 63. Here is a useful result for obtaining equivalences between extensions. Lemma 4.7.3. (Five Lemma for Extensions) Let f : G → K be an extension of H by K and let f : G → K be an extension of H by K . Suppose given a commutative diagram of group homomorphisms H ⊂ g1 H / G f g2 ⊂ / G / K g3 f / K where g1 and g3 are isomorphisms. Then g2 is also an isomorphism. CHAPTER 4. NORMALITY AND FACTOR GROUPS 121 Proof Let x ∈ ker g2 . Then (f ◦ g2 )(x) = e, and hence g3 (f (x)) = e by commutativity of the diagram. Since g3 is injective, f (x) = e, and hence x ∈ ker f = H. But on H ⊂ G, g2 acts as g1 : H → H , and hence x ∈ ker g1 . But g1 is injective, so we must have x = e. Thus, ker g2 = e, and hence g2 is injective. Thus, it suﬃces to show that each y ∈ G lies in the image of g2 . Since g3 is surjective, f (y) = g3 (z ) for some z ∈ K. But f is also surjective, so z = f (z) for some z ∈ G. Thus, f (g2 (z)) = g3 (f (z)) = g3 (z ) = f (y), so that g2 (z) and y have the same image under f . But then y(g2 (z))−1 lies in the kernel of f , which is H . Thus, since g1 is surjective, there is an x ∈ H such that g1 (x) = y(g2 (z))−1 . Thus, y = g1 (x)g2 (z) = g2 (xz), and hence y ∈ im g2 as desired. The Five Lemma for Extensions is a special case of a more general result which is given for abelian groups as the Five Lemma in Lemma 7.7.48. The statement of the general case involves ﬁve vertical maps, rather than the three displayed here, hence the name. In both cases, the technique of proof is called a diagram chase. As noted above, if α : K → Aut(H) is a homomorphism, then the projection map π : H α K → K is an extension of H by K. As a general rule, the semidirect products are the easiest extensions to analyze, so it is useful to have a criterion for an extension to be one. Deﬁnition 4.7.4. A section, or splitting, for f : G → K is a homomorphism s : K → G, such that f ◦ s is the identity map of K. A homomorphism f : G → K that admits a section is said to be a split surjection. An extension of H by K is called a split extension if f : G → K admits a section. The section s is said to split f . The homomorphism s : K → H α K is clearly a section for π. We also have a converse. Proposition 4.7.5. Split extensions are semidirect products. Speciﬁcally, let f : G → K be a split extension of H by K, with splitting map s : K → G. Then s induces a homomorphism α : K → Aut(H) via α(k)(h) = s(k)h(s(k))−1 . Moreover, there is an isomorphism ϕ : H α K → G such that the following diagram commutes. H H ⊂ ι / H α K o ϕ ⊂ / G o s1 s K K Here, s1 is the standard section for the semidirect product. Explicitly, ϕ is deﬁned by ϕ(hk) = hs(k). Proof Recall from Lemma 3.7.7 that, since H G, we have a homomorphism Γ : G → Aut(H), given by setting Γ(g)(h) = ghg −1 for all g ∈ G and h ∈ H. The above speciﬁcation for α is just the composite Γ ◦ s : K → Aut(H), and hence is a homomorphism as claimed. Recall from Proposition 4.6.9 that if f1 : H → G and f2 : K → G are homomorphisms such that f2 (k)f1 (h)(f2 (k))−1 = f1 (α(k)(h)) for all h ∈ H and k ∈ K, then there CHAPTER 4. NORMALITY AND FACTOR GROUPS 122 is a unique homomorphism f : H α K → G such that f ◦ ι = f1 and f ◦ s1 = f2 : speciﬁcally, f (hk) = f1 (h)f2 (k). We apply this with f1 equal to the inclusion of H in G and with f2 equal to the section s : K → G. But then the condition we need to verify to construct our homomorphism is that s(k)h(s(k))−1 = α(k)(h), which holds by the deﬁnition of α. Thus, there is a homomorphism ϕ : H α K → G with ϕ◦ι equal to the inclusion of H in G, and ϕ◦s1 = s: speciﬁcally, ϕ(hk) = hs(k). But this just says that the stated diagram commutes, so it suﬃces to show that ϕ is an isomorphism. Since ϕ(hk) = hs(k), we obtain that (f ◦ ϕ)(hk) = (f ◦ s)(k) = k. But this says that f ◦ ϕ = π, where π : H α K → K is the projection map. Thus, we have an additional commutative diagram. H H ⊂ ι / H α K ϕ ⊂ / G π f / K / K But this shows ϕ to be an isomorphism by the Five Lemma for Extensions. Of course, if α : K → Aut(H) is the trivial homomorphism, then the semidirect product H α K is simply H × K. Thus, the corollary below is immediate from the way that α depends on s in the statement of Proposition 4.7.5. Corollary 4.7.6. Let f : G → K be a split extension of H by K, with splitting map s : K → G. Suppose that s(k) commutes with each element of H for all k ∈ K. Then G∼ = H × K. In particular, if f : G → K is a split surjection and if G is abelian, then G ∼ = ker f × K. Semidirect products are the easiest extensions to construct, and hence may be viewed as the most basic of the extensions of one group by another. For this reason, it is useful to be able to recognize whether an extension splits. The easiest case is when the two groups have orders that are relatively prime. Proposition 4.7.7. Let H and K be groups whose orders are relatively prime and let f : G → K be an extension of H by K. Then the extension splits if and only if G has a subgroup of order |K|. Proof If s : K → G is a section of f , then s is injective, so s(K) has the same order as K. Conversely, suppose that K ⊂ G has order |K| and write f1 : K → K for the restriction of f to K . Then ker f1 = K ∩ H. Since the orders of K and H are relatively prime, K ∩ H = e, by Lagrange’s Theorem. Thus, f1 is injective. Since K and K have f −1 1 the same order, f1 is an isomorphism, and the composite K −− → K ⊂ G gives a section for f . In fact, under the hypotheses above, G always has a subgroup of order |K|. Theorem (Schur–Zassenhaus Lemma) Let f : G → K be an extension of H by K, where H and K have relatively prime order. Then G has a subgroup of order |K|. CHAPTER 4. NORMALITY AND FACTOR GROUPS 123 The proof requires the Sylow Theorems and could be given in Chapter 5, but we shall not do so. In practice, when considering a particular extension, one can generally construct the section by hand. Let us now consider the issue of classifying the split extensions of one group by another. Note that the deﬁnition of the homomorphism α : K → Aut(H) constructed in Proposition 4.7.5 makes use of the speciﬁc section s. Thus, we might wonder whether a diﬀerent homomorphism α would have resulted if we’d started with a diﬀerent section. Indeed, this can happen when the group H is nonabelian. Example 4.7.8. Let ε : Sn → {±1} take each permutation to its sign. Then ε is an extension of An by Z2 . Moreover, setting s(−1) = τ ∈ Sn determines a section of ε if and only if τ has order 2 and is an odd permutation. Since the order of a product of disjoint cycles is the least common multiple of the orders of the individual cycles, we see that τ determines a section of ε if and only if it is the product of an odd number of disjoint 2-cycles. For such a τ , we write sτ for the section of ε with sτ (−1) = τ . Given the variety of such τ , we see that there are many possible choices of section for ε. Write ατ : Z2 → Aut(An ) for the homomorphism induced by sτ . By deﬁnition, ατ = Γ ◦ sτ , where Γ : Sn → An is the homomorphism induced by conjugation. Thus, ατ (−1) = Γ(τ ). But Problem 10 in Exercises 3.7.14 shows that Γ is injective for n ≥ 4. Thus, if n ≥ 4, then each diﬀerent choice for a section of ε induces a diﬀerent homomorphism from Z2 to Aut(An ). If τ and τ have the same cycle structure, then they are conjugate in Sn , and hence sτ and sτ are conjugate as homomorphisms into Sn . But then ατ and ατ are conjugate as homomorphisms into Aut(An ). However, as we shall see in the exercises below, if τ is a 2-cycle and τ is a product of three disjoint 2-cycles, then ατ and ατ are not conjugate as homomorphisms into Aut(An ) if n > 6. Thus, in this case, Proposition 4.6.11 would not have predicted that An ατ Z2 and An ατ Z2 would be isomorphic; nevertheless they are isomorphic, as each one of them admits an isomorphism to Sn . However, the criterion in Proposition 4.6.11 is the most general suﬃcient condition we shall give for the isomorphism of two semidirect products. Thus, the classiﬁcation of even the split extensions of a nonabelian group oﬀers substantial diﬃculty. But we shall be mainly concerned here with the extensions of an abelian group, a problem for which we can develop some reasonable tools. Lemma 4.7.9. Let f : G → K be an extension of H by K, where H is abelian. For x ∈ G, write cx : H → H for the automorphism obtained from conjugation by x. Suppose that f (x) = f (y). Then cx = cy as automorphisms of H. Moreover, there is a homomorphism αf : K → Aut(H) deﬁned by setting αf (k) equal to cx for any choice of x with f (x) = k. In particular, if f happens to be a split extension and if s : K → G is a section of f , then the homomorphism from K to Aut(H) induced by s coincides with the homomorphism αf above. Thus, if H is abelian, the homomorphism from K to Aut(H) induced by a section of f is independent of the choice of section. Proof If f (x) = f (y), then y −1 x ∈ ker f = H, so x = yh for some h ∈ H. But then cx = cy ◦ ch . Since H is abelian, conjugation of H by h is the identity map, so cx = cy as claimed. CHAPTER 4. NORMALITY AND FACTOR GROUPS 124 Thus, there is a well deﬁned function αf : K → Aut(H) obtained by setting αf (k) equal to cx for any x with f (x) = k. But if f (x) = k and f (y) = k , then f (xy) = kk , so that αf (kk ) = cxy = cx ◦ cy = αf (k)αf (k ), so αf is a homomorphism. If s : K → G is a section of f , then the homomorphism α : K → Aut(H) induced by this section is obtained by setting α(k) = cs(k) : H → H. Since f (s(k)) = k, this just says that α = αf . Thus, we may make the following deﬁnition. Deﬁnition 4.7.10. Let f : G → K be an extension of H by K, where H is abelian. Then the action of K on H induced by f is the homomorphism αf : K → Aut(H) obtained by setting αf (k) equal to the automorphism obtained by conjugating H by any given element of f −1 (k). The existence of the homomorphisms αf gives a valuable method of studying the extensions of an abelian group: ﬁx a homomorphism α : K → Aut(H) and classify the extensions whose induced action is α. Do this for all possible homomorphism α, and look to see if a given group can be written as an extension in more than one way. We shall illustrate this process by classifying the groups of order 8. Lemma 4.7.11. Let f : G → Z2 be an extension of Z4 by Z2 . Suppose that the induced action αf : Z2 → Z× 4 carries the generator of Z2 to −1. Then G is isomorphic to either D8 or Q8 . Proof Let Z4 = b and Z2 = a. For any x ∈ G with f (x) = a, we have xbk x−1 = αf (a)(bk ) = b−k for all k. Also, f (x2 ) = a2 = e, so x2 ∈ ker f = b. Thus, x2 = bk for some k. Note that x fails to commute with b and b−1 , so neither of these can equal x2 . Thus, either x2 = e or x2 = b2 . If x2 = e, then x has order 2, and hence there is a section of f , speciﬁed by s(a) = x. In this case, G ∼ = Z4 αf Z2 , which, for this value of αf , is isomorphic to the dihedral group D8 by the ﬁrst example of Examples 4.6.6. The remaining case gives x2 = b2 . But then there is a homomorphism h : Q8 → G with h(a) = x and h(b) = b by Problem 5 of Exercises 2.9.7. As the image of h contains more than half the elements of G, h must be onto, and hence an isomorphism. (Alternatively, we could use the Five Lemma for Extensions to show that h is an isomorphism.) Q8 , of course, is not a split extension of b by Z2 , for two reasons. First, we know that every element of Q8 which is not in b has order 4, and hence it is impossible to deﬁne a section from Q8 /b to Q8 . Second, if it were a split extension, then it would be isomorphic to D8 , which we know is not so. We have classiﬁed the extensions of Z4 by Z2 corresponding to the homomorphism × from Z2 to Z× 4 that takes the generator of Z2 to −1. Since Z4 is isomorphic to Z2 , there × is only one other homomorphism from Z2 to Z4 : the trivial homomorphism. Let us begin with some general observations about extensions whose induced action is trivial. Lemma 4.7.12. Let f : G → K be an extension of H by K, where H is abelian. Then the induced homomorphism αf : K → Aut(H) is the trivial homomorphism if and only if H ⊂ Z(G). CHAPTER 4. NORMALITY AND FACTOR GROUPS 125 Proof Suppose that αf is the trivial homomorphism. Then for any x ∈ G, let y = f (x). By the deﬁnition of αf , we have xhx−1 = αf (y)(h) for all h ∈ H. But αf is the trivial homomorphism, so that αf (y) is the identity map of H. Thus xhx−1 = h for all h ∈ H, and hence each h ∈ H commutes with every element of G. Thus, H ⊂ Z(G). But the converse is immediate: if H ⊂ Z(G), then conjugating H by any element of G gives the trivial homomorphism, and hence αf (y) is the identity element of Aut(H) for every y ∈ K. The extensions of H containing H in their centers are important enough to merit a name. Deﬁnition 4.7.13. An extension f : G → K of H by K is a central extension if H ⊂ Z(G). Example 4.7.14. As one may easily see, the center of D8 is b2 . Moreover, D8 /b2 is isomorphic to Z2 × Z2 , with generators a and b. Thus, D8 is a central extension of Z2 by Z2 × Z2 . In particular, a central extension of a cyclic group by an abelian group is not necessarily abelian. But a central extension of an abelian group by a cyclic group is abelian: Lemma 4.7.15. Let f : G → Zk be a central extension of the abelian group H by the cyclic group Zk . Then G is abelian. Proof Write Zk = a, and let x ∈ G with f (x) = a. Let g ∈ G. Then f (g) = ai = f (xi ) for some i, and hence gx−i ∈ ker f = H. But then g = hxi for some h ∈ H. If g is any other element of G, we may write g = h xj for h ∈ H and j ∈ Z. But then gg = hxi h xj = hh xi+j , as h ∈ Z(G). But multiplying out g g gives the same result, and hence g and g commute. We now return to the groups of order 8. Lemma 4.7.16. Let f : G → Z2 be a central extension of Z4 by Z2 . Then G is isomorphic to either Z8 or Z4 × Z2 . Proof Since Z2 is cyclic, Lemma 4.7.15 shows that G is abelian. So we could just appeal to the Fundamental Theorem of Finite Abelian Groups, which tells us that Z8 , Z4 × Z2 , and Z2 × Z2 × Z2 are the only abelian groups of order 8. Clearly, the ﬁrst two are the only ones that are extensions of Z4 . Alternatively, we can argue directly. Write Z4 = b and Z2 = a and let f (x) = a. Then f (x2 ) = e, and hence x2 ∈ b. Now x has even order, so that o(x) = 2o(x2 ). Thus, if x2 = b or b−1 , then x has order 8, and hence G is cyclic. If x has order 2, then s(a) = x gives a splitting of f , and hence G is the semidirect product obtained from the action of Z2 on Z4 . But in this case, the action is trivial, so that the semidirect product thus obtained is just the direct product. If x has order 4, then x2 = b2 . But then b−1 x has order 2, and setting s(a) = b−1 x gives a section of f , and hence G ∼ = Z4 × Z2 as above. Once again, we have a nonsplit extension: the unique nontrivial homomorphism f : Z8 → Z2 is a nonsplit extension of Z4 by Z2 . We now classify the groups of order 8. Proposition 4.7.17. Every group of order 8 is isomorphic to exactly one of the following groups: Z8 , D8 , Q8 , Z4 × Z2 , and Z2 × Z2 × Z2 . CHAPTER 4. NORMALITY AND FACTOR GROUPS 126 Proof If G has an element of order 4, say o(b) = 4, then b has index two in G, and hence is normal in G by Proposition 4.1.12 (or by the simpler argument of Problem 12 of Exercises 3.7.14). But then G is an extension of Z4 by Z2 . As Z× 4 = {±1}, there are , the homomorphism that carries the only two diﬀerent homomorphisms from Z2 to Z× 4 generator of Z2 to −1 and the trivial homomorphism. Thus, if G is an extension of Z4 by Z2 , then G is isomorphic to one of Z8 , D8 , Q8 , and Z4 × Z2 by Lemmas 4.7.11 and 4.7.16. If G does not have an element of order 4, then every element of G has exponent 2. But then G is abelian by Problem 3 of Exercises 2.1.16. But the fundamental theorem of ﬁnite abelian groups shows Z2 × Z2 × Z2 to be the only abelian group of order 8 and exponent 2. The reader may easily check that no two of the listed groups are isomorphic. We shall classify all groups of order p3 for odd primes p as well, but this will require theoretical results to be given in the next chapter. We now return to the general theory of extensions. Lemma 4.7.18. Let f : G → K and f : G → K be extensions of H by K, where H is abelian. Suppose given a commutative diagram H ⊂ f ⊂ / G / K g3 g2 g1 H / G f / K, where g1 , g2 , and g3 are isomorphisms. Then the actions αf , αf : K → Aut(H) induced by f and f are related by g1 ◦ αf (k) ◦ g1−1 = αf (g3 (k)), so that αf is conjugate to αf ◦ g3 as a homomorphism from K to Aut(H). Proof For k ∈ K, let x ∈ G with f (x) = k. Then f (g2 (x)) = g3 (k), so αf (g3 (k))(g1 (h)) = g2 (x)g1 (h)g2 (x)−1 = g2 (xhx−1 ) = g1 (αf (k)(h)) for all h ∈ H. Thus, αf (g3 (k)) ◦ g1 = g1 ◦ αf (k), and the result follows. Remarks 4.7.19. There is a general method that may be used to classify the extensions of an abelian group H by a group K up to the following equivalence relation: The extensions f : G → K and f : G → K are equivalent if there is a commutative diagram H ⊂ / G f / K f / K. g H ⊂ / G Here, g is forced to be an isomorphism, by the Five Lemma for Extensions. Notice that Lemma 4.7.18 implies that if f and f are equivalent, then the homomorphisms αf and αf are equal. Thus, the method under discussion will give a separate classiﬁcation for each diﬀerent homomorphism α : K → Aut(H). Thus, ﬁxing such an α, we shall write Hα to denote H with this particular action of K. The method of classiﬁcation is given using the cohomology of groups, a subtopic of homological algebra. The classiﬁcation theorem gives a one-to-one correspondence CHAPTER 4. NORMALITY AND FACTOR GROUPS 127 between the equivalence classes of extensions corresponding to the homomorphism α and the elements of the cohomology group H 2 (K; Hα ).5 Nevertheless, a couple of deﬁciencies in this approach should be pointed out. First, given an element in H 2 (K; Hα ), it is quite tedious to construct the associated extension. And it comes in a form that is diﬃcult to analyze. Second, the classiﬁcation is too rigid for many of the questions of interest: there are extensions whose associated actions are both α which are not equivalent in the above sense, but are equivalent under weaker hypotheses. Thus, there may be fewer isomorphism classes of groups G that may be written as extensions of H by K with action α than there are elements of the cohomology group H 2 (K; Hα ). Nevertheless, the homological approach can be useful. It is sometimes useful to be able to go back and forth between the existence of a commutative diagram such as that in Lemma 4.7.18 and the existence of a suitable isomorphism from G to G . Lemma 4.7.20. Let f : G → K and f : G → K be extensions of H by K, and write ι : H → G and ι : H → G for the inclusions. Then the following conditions are equivalent. 1. There is a commutative diagram H ⊂ ι f ⊂ ι / G / K g3 g2 g1 H / G f / K, where g1 , g2 , and g3 are isomorphisms. 2. There is an isomorphism g : G → G with g(im ι) = im ι . Proof If the ﬁrst condition holds, then the second one does also, as we may take g = g2 . Thus, suppose that the second condition holds, and we are given an isomorphism g : G → G with g(im ι) = im ι . Then g induces an isomorphism from im ι to im ι . Since ι and ι are embeddings, there is an isomorphism g1 : H → H such that ι ◦ g1 = g ◦ ι. Thus, if we set g2 = g, then the left square of the displayed diagram commutes. We must now construct g3 , which comes from a couple of applications of the First Noether Theorem. First, we have surjective maps f : G → K and f : G → K, with kernels given by ker f = H and ker f = H. Thus, the First Noether Theorem gives ∼ ∼ = = isomorphisms f : G/H −→ K and f : G /H −→ K such that the following diagrams commute. /K GD DD z< z DD zz D zzf π DD z ! z G/H f f /K G E y< EE y EE yy E yyf π EE y y " G /H 5 See, for instance, A Course in Homological Algebra, by P.J. Hilton and U. Stammbach, SpringerVerlag, 1971, for an exposition of this classiﬁcation. CHAPTER 4. NORMALITY AND FACTOR GROUPS g 128 π → G − → G /H is precisely H, so Since g(im ι) = im ι , the kernel of the composite G − ∼ = the First Noether Theorem provides an isomorphism k : G/H −→ G /H that makes the following diagram commute. H g1 ∼ = H Now set g3 = f ◦ k ◦ f −1 ⊂ ι ⊂ ι / G ∼ g = / G π π / G/H k ∼ = / G /H f ∼ = f ∼ = / K / K . Note that the group H of Lemma 4.7.20 need not be abelian. We now wish to explore the question of when two diﬀerent homomorphisms, α, α : K → Aut(H) produce isomorphic semidirect products. In this regard, the above lemmas permit an improvement of the result in Proposition 4.6.11 in the case that H is abelian. Proposition 4.7.21. Let H be an abelian group and let α and α be homomorphisms of K into Aut(H). Then α is conjugate to a composite α ◦ g with g ∈ Aut(K) if and only if there is an isomorphism ϕ : H α K → H α K such that ϕ(im ι) = im ι , where ι and ι are the canonical inclusions of H in H α K and H α K, respectively. Proof If α is conjugate to a composite α ◦ g with g ∈ Aut(K), then Proposition 4.6.11 provides an isomorphism ϕ : H α K → H α K with ϕ(im ι) = im ι . Thus, it suﬃces to show the converse. So suppose given an isomorphism ϕ : H α K → H α K with ϕ(im ι) = im ι . Then Lemma 4.7.20 provides a commutative diagram of the sort appearing in the hypothesis of Lemma 4.7.18. By Lemma 4.7.9, the action map απ : K → Aut(H) induced by π : H α K → K is precisely α. A similar result holds for α . Thus, the output of Lemma 4.7.18 is precisely that α is conjugate to α ◦ g for some g ∈ Aut(K). We would like to eliminate the hypothesis that ϕ(im ι) = im ι from the above, and obtain a necessary and suﬃcient condition for the semidirect products H α K and H α K to be isomorphic. We can do this if H and K are ﬁnite groups whose orders are relatively prime. Corollary 4.7.22. Let H and K be ﬁnite groups whose orders are relatively prime, and suppose that H is abelian. Let α and α be two homomorphisms from K to Aut(H). Then the semidirect products H α K and H α K are isomorphic if and only if α is conjugate to α ◦ g for some automorphism g of K. Proof It suﬃces to show that any homomorphism from H α K to H α K carries ι(H) into ι (H), and that the analogous statement holds for homomorphisms in the opposite direction. But this will follow if we show that in either one of the semidirect products, the only elements of exponent |H| are those in H itself. If |H| is an exponent for x, then it is also an exponent for f (x) for any homomorphism f . Since any element whose order divides both |H| and |K| must be the identity, we see that any element of either semidirect product which has exponent |H| must be carried to the identity element by the projection map (π or π ) into K. Since H is the kernel of the projection map, the result follows. CHAPTER 4. NORMALITY AND FACTOR GROUPS 129 We now give a generalization of Corollary 4.7.22. Corollary 4.7.23. Let H be an abelian group and let α, α : K → Aut(H). Suppose that α has the property that any embedding f : H → H α K must have image equal to ι(H). Then H α K is isomorphic to H α K if and only if α is conjugate to α ◦ g for some g ∈ Aut(K). Proof Again, it suﬃces to show that ϕ(im ι) = im ι , or, equivalently, that im ι = im(ϕ−1 ◦ ι ), for any isomorphism ϕ : H α K → H α K. But since ϕ−1 ◦ ι is an embedding, this follows from the hypothesis. Corollary 4.7.23 may be applied, for instance, if ι(H) is the only subgroup of H α K which is isomorphic to H. An example where this occurs is for H = b ⊂ D2n for n ≥ 3. Here, we make use of the fact that every element of D2n outside of b has order 2. Since b has order n > 2, every embedding of b in D2n must have image in b. Since b is ﬁnite, this forces the image to be equal to b. Finally, let us consider the extensions of a nonabelian group H. As shown in Example 4.7.8, if f : G → K is even a split extension of H by K, diﬀerent choices of splittings may deﬁne diﬀerent homomorphisms from K into Aut(H). In particular, we do not have a homomorphism from K to Aut(H) which is associated to the extension itself. But we can deﬁne something a little weaker. As usual, we write ch : H → H for the automorphism induced by conjugating by h ∈ H: ch (h ) = hh h−1 for all h ∈ H. Recall that an automorphism f ∈ Aut(H) is an inner automorphism if f = ch for some h ∈ H. The inner automorphisms form a subgroup, Inn(H) ⊂ Aut(H). Lemma 4.7.24. The group of inner automorphisms of a group G is a normal subgroup of Aut(G). Proof Let x ∈ G and write cx for conjugation by x. Let f ∈ Aut(G). We wish to show that f ◦ cx ◦ f −1 is an inner automorphism. But f ◦ cx ◦ f −1 (g) = f (xf −1 (g)x−1 ) = f (x)f (f −1 (g))f (x)−1 , and hence f ◦ cx ◦ f −1 = cf (x) . Thus, we may make the following deﬁnition. Deﬁnition 4.7.25. We write Out(G) for the quotient group Aut(G)/Inn(G) and call it the outer automorphism group of G. Proposition 4.7.26. Let f : G → K be an extension of H by K. Then f determines a homomorphism α : K → Out(H). Here, for k ∈ K, α(k) is represented by the automorphism of H given by conjugation by g, for any g ∈ G with f (g) = k. In particular, if f is a split extension and s is a splitting of f , then α = π ◦ αs , where αs : K → Aut(H) is the homomorphism induced by s and π is the canonical map from Aut(H) to Out(H) = Aut(H)/Inn(H). Thus, if s and s are two diﬀerent sections of f , then the homomorphisms αs and αs agree in Out(H). CHAPTER 4. NORMALITY AND FACTOR GROUPS 130 Proof If f (g) = f (g ) = k, then g = gh for some h ∈ H. But then conjugating by g is the same as conjugating ﬁrst by h and then by g. But conjugation by h is an inner automorphism of H, and hence represents the identity in Out(H). Thus, conjugation by g represents the same element of Out(H) as conjugation by g, and hence the prescription above gives a well-deﬁned function α : K → Out(H). It suﬃces to show that α is a homomorphism. But for k, k ∈ K, if f (g) = k and f (g ) = k , then f (gg ) = kk . But conjugation by gg is the composite of the conjugations by g and by g . We should also note that the above is not always very useful. For instance, for n = 6, Out(Sn ) is the trivial group. Exercises 4.7.27. † 1. Problem 1 of Exercises 4.2.18 provides a normal subgroup H ⊂ A4 of order 4. Show that H ∼ = Z2 × Z2 and that A4 is a split extension of H by Z3 . 2. Show that any nonabelian group G of order 12 which contains a normal subgroup of order 4 must be isomorphic to A4 . (Hint: Find a G-set with 4 elements and analyze the induced homomorphism from G to S4 . Alternatively, show that G is a split extension of a group of order 4 by Z3 , and, using the calculation of Aut(Z2 × Z2 ) given in Problem 14 of Exercises 4.6.13, show that there is only one such split extension which is nonabelian.) † 3. Consider the subgroup H ⊂ An of Problem 1. Show that H is normal in S4 and that S4 is isomorphic to the semidirect product (Z2 × Z2 ) α S3 , where α : S3 → Aut(Z2 × Z2 ) is an isomorphism. (Recall the calculation of Aut(Z2 × Z2 ) given in Problem 14 of Exercises 4.6.13. In the situation at hand, you may enjoy giving a new proof of this calculation, based on showing that the map α above is injective.) 4. Show that if k is odd, then Q4k is isomorphic to Zk α Z4 for some α : Z4 → Aut(Zk ). Calculate α explicitly. 5. Find all the central extensions of Z2 by Z2 × Z2 . 6. Let f : G → Z2 be an extension of Z8 by Z2 and suppose that the induced homomorphism αf : Z2 → Z× 8 carries the generator of Z2 to 5. Show that f is a split extension, and hence that G is isomorphic to the group Z8 β Z2 of Example 3 in Examples 4.6.6. (Hint: Write Z8 = b and Z2 = a and let x ∈ G with f (x) = a. Note that x2 = bk for some k, and use this to calculate the order of each of the elements bi x.) 7. Let f : G → Z2 be an extension of Z8 by Z2 and suppose that the induced homomorphism αf : Z2 → Z× 8 carries the generator of Z2 to 3. Show that f is a split extension, and hence that G is isomorphic to the group Z8 α Z2 of Example 2 in Examples 4.6.6. (Hint: Argue as in the preceding problem. Here, if x2 = bk , note that the fact that x must commute with x2 places restrictions on which values of k can occur.) 8. Let f : G → Z2 be an extension of Z8 by Z2 and suppose that the induced homomorphism αf : Z2 → Z× 8 carries the generator of Z2 to −1. Show that f is isomorphic to either D16 or Q16 . 9. Classify those groups of order 16 that contain an element of order 8. CHAPTER 4. NORMALITY AND FACTOR GROUPS 131 10. Let H be any group and let f : G → Z be an extension of H by Z. Show that f splits. 11. We show that the inner automorphisms of Sn induced by diﬀerent sections of the sign homomorphism are not always conjugate in Aut(Sn ). (a) Let f ∈ Aut(G) and let Gf = {g ∈ G | f (g) = g}, the ﬁxed points of G under f . Show that Gf is a subgroup of G. Show also that if f and f are conjugate in Aut(G), then Gf and Gf are isomorphic as groups. (b) Let τ ∈ Sn be a 2-cycle and write An τ for the ﬁxed points of An under the automorphism obtained from conjugating by τ . Show that An τ is isomorphic to Sn−2 . (c) Let H ⊂ S6 be the subgroup consisting of those elements of S6 that commute with (1 2)(3 4)(5 6). Show that H is a split extension of Z2 × Z2 × Z2 by S3 . (d) Let τ ∈ Sn be a product of three disjoint 2-cycles. Show that the subgroup An τ ﬁxed by τ , is isomorphic to an index 2 subgroup of H × Sn−6 , where H ⊂ S6 is the group of the preceding part. Deduce that if n > 6, then the automorphism of An obtained from conjugation by a 2-cycle cannot be conjugate in Aut(An ) to the automorphism obtained from conjugation by the product of three disjoint 2-cycles. ‡ 12. Let α : K → Aut(H) and β : K → Aut(H ) be homomorphisms. Let f : H → H be a group homomorphism such that f (α(k)(h)) = β(k)(f (h)) for all k ∈ K and h ∈ H (i.e., f is a K-map between the K-sets H and H ). Deﬁne f∗ : H α K → H β K by f∗ (hk) = f (h)k. Show that f∗ is a homomorphism. 13. Let α : K → Aut(H) be a homomorphism. Suppose given f ∈ Aut(H) that commutes with every element of the image of α. Show that the homomorphism f∗ : H α K → H α K of Problem 12 is deﬁned and gives an automorphism of H α K. ‡ 14. In this problem, we calculate the automorphisms of a dihedral group. The case of D4 is exceptional, and indeed has been treated in Problem 14 in Exercises 4.6.13. Thus, we shall consider D2n for n ≥ 3. A key fact is given by either Problem 6 of Exercises 2.8.5 or by Proposition 4.6.10: The homomorphisms from D2n to a group G are in one-to-one correspondence with the choices of x = f (a) and y = f (b) such that x has exponent 2, y has exponent n, and xyx−1 = y −1 . The reader should ﬁrst recall from Problem 4 of Exercises 3.7.14 that since n ≥ 3, the subgroup b ⊂ D2n is characteristic. Thus, Lemma 3.7.8 provides a restriction homomorphism ρ : Aut(D2n ) → Aut(b), given by setting ρ(f ) equal to its restriction f : b → b. As usual, by abuse of notation, we shall identify Aut(b) with × Z× n , and write ρ : Aut(D2n ) → Zn . (a) Show that any homomorphism f : D2n → D2n whose restriction to b gives an automorphism of b must have been an automorphism of D2n in the ﬁrst place. (b) Show that the restriction homomorphism ρ : Aut(D2n ) → Z× n is a split sur× jection (i.e., ρ : Aut(D2n ) → Z× n is a split extension of ker ρ by Zn ). CHAPTER 4. NORMALITY AND FACTOR GROUPS 132 (c) Show that an automorphism f is in the kernel of ρ if and only if f (b) = b and f (a) = bi a for some i. Deduce that the kernel of ρ is isomorphic to Zn . (d) Show that the action of Z× n on Zn = ker ρ induced by the split extension is the same as the usual action of Z× n = Aut(Zn ). Deduce that Aut(D2n ) is isomorphic to the semidirect product Zn id Aut(Zn ), where id is the identity map of Aut(Zn ). 15. Show that every automorphism of S3 is inner. 16. Recall from Problem 8 of Exercises 3.7.14 that if n ≥ 3, the subgroup b ⊂ Q4n is characteristic. Using this as a starting point, give a calculation of Aut(Q4n ), n ≥ 3, by methods analogous to those of Problem 14 above. Note that the lack of a semidirect product structure on Q4n necessitates greater care in the construction of a splitting for ρ than was required in the dihedral case. (We shall treat the exceptional case of Aut(Q8 ) in Problem 9 of Exercises 5.4.9.) ‡ 17. Recall from Problem 3 of Exercises 4.2.18 that An is a characteristic subgroup of Sn for n ≥ 3. Thus, Lemma 3.7.8 provides a restriction homomorphism ρ : Aut(Sn ) → Aut(An ). Let Γ : Sn → Aut(An ) be the homomorphism induced by conjugation. By Problem 10 in Exercises 3.7.14, Γ is injective for n ≥ 4. Use this fact, together with Proposition 4.6.9, to show that ρ is injective for n ≥ 4. 18. We compute the automorphism groups of A4 and S4 . (a) Let H ⊂ A4 be the subgroup of order 4. Show that H is characteristic in A4 . Thus, Lemma 3.7.8 provides a restriction homomorphism ρ : Aut(A4 ) → Aut(H). (b) Show that the kernel of ρ is isomorphic to H, and hence to Z2 × Z2 . (c) Show that ρ is a split surjection. Deduce that the homomorphism Γ : S4 → Aut(A4 ) induced by conjugation is an isomorphism. (d) Deduce from Problem 17 that the homomorphism Γ : S4 → Aut(S4 ) induced by conjugation is an isomorphism. ‡ 19. We consider the automorphism groups of the semidirect products G = Zn α Zk for selected homomorphisms α : Zk → Aut(Zn ). As usual, we write ι : Zn → G and s : Zk → G for the standard inclusions and write π : G → Zk for the standard projection. The one assumption we shall insist on is that ι(Zn ) be a characteristic subgroup of G. As the reader may easily verify, this is always the case if n and k are relatively prime. But we have seen other examples, e.g., D2n = Zn α Z2 or Example 2 of Examples 4.6.6, where Zn is characteristic in G, and our arguments will apply to these cases as well. In particular, the material here generalizes that in Problem 14. As usual, we use multiplicative notation, with Zn = b and Zk = a. As discussed j in Section 4.6, we have α(aj )(bi ) = bm i for an integer m with the property that mk ≡ 1 mod n. The elements of G may be written uniquely as bi aj with 0 ≤ i < n j and 0 ≤ j < k, and bi aj bi aj = bi+m i aj+j . Since b is assumed characteristic in G, we have a restriction homomorphism ρ : Aut(G) → Aut(b) ∼ = Z× n. CHAPTER 4. NORMALITY AND FACTOR GROUPS 133 (a) Let f ∈ Aut(G). Show that there is a unique element f ∈ Aut(a) such that the following diagram commutes: G f π a / G π f / a Show that the passage from f to f gives a homomorphism η : Aut(G) → Aut(a) ∼ = Z× k. (b) Show, via Corollary 4.6.10, that the image of η consists of those f ∈ Aut(Zk ) such that the following diagram commutes: Zk α / Z× n α / Z× . n f Zk Show that such automorphisms f ∈ Aut(Zk ) correspond to elements j ∈ Z× k such that mj ≡ m mod n. Deduce that if α is injective (i.e., if m has order k in Z× n ), then η is the trivial homomorphism. (c) Consider the map (ρ, η) : Aut(G) → Aut(b) × Aut(a) given by (ρ, η)(f ) = (ρ(f ), η(f )). Show that (ρ, η) induces a split surjection of Aut(G) onto Aut(b)× (im η). (d) Show that an automorphism f is in the kernel of (ρ, η) if and only if f (b) = b and f (a) = bi a, where i is chosen so that bi a has order k in G. Show, via Problems 7 and 8 of Exercises 4.6.13, that bi a has order k if and only if i · (mk − 1)/(m − 1) is divisible by n. Show also that passage from f to bi induces an isomorphism from the kernel of (ρ, η) onto a subgroup H ⊂ b. (e) Let β : Aut(Zn ) × (im η) → Aut(H) be the action induced by the extension (ρ, η). Show that β restricts on Aut(Zn ) to the restriction map r : Aut(Zn ) → Aut(H) induced by the fact that H is characteristic in Zn . (Recall from Problem 9 of Exercises 3.7.14 that every subgroup of a cyclic group is characteristic.) Deduce that if α is injective, then Aut(G) ∼ = H r Aut(Zn ). (f) Let β : Aut(Zn ) × (im η) → Aut(H) be as above. Compute the restriction of β to im η. (g) Show that if n is prime, then H = Zn . (h) Give an example of α, n, and k where H is not all of Zn . Chapter 5 Sylow Theory, Solvability, and Classiﬁcation The main theorems of this chapter, e.g., Cauchy’s Theorem, the Sylow Theorems, and the characterization of solvable and nilpotent groups, could have been given much earlier in the book. In fact, a proof of Cauchy’s Theorem has appeared as Problem 24 in Exercises 3.3.23. Indeed, these theorems depend on very little other than the theory of G-sets (which it might be advisable to review at this time) and the Noether Isomorphism Theorems. Our main reason for delaying the presentation of these theorems until now was so that we could develop enough theory to support their applications. Given our analysis of abelian groups, extensions, and some automorphism groups, we will be able to use the results in this chapter to classify the groups of most orders less than 64, and, hopefully, to bring the study of ﬁnite groups into a useful perspective. Lagrange’s Theorem tells us that if the group G is ﬁnite, then the order of every subgroup must divide the order of G. But there may be divisors of |G| that do not occur in this manner. For instance, Problem 2 of Exercises 4.2.18 shows that A4 has no subgroup of order 6. In this chapter, we shall give some existence theorems for subgroups of particular orders in a given ﬁnite group. The ﬁrst and most basic such result is Cauchy’s Theorem, which says that if G is a ﬁnite group whose order is divisible by a prime p, then G has an element (hence a subgroup) of order p. We give this in Section 5.1, along with applications to classiﬁcation. In Section 5.2, we study ﬁnite p-groups, which, because of Cauchy’s Theorem, are precisely the ﬁnite groups whose order is a power of the prime p. By a proof similar to the one given for Cauchy’s Theorem, we show that any p-group has a nontrivial center. We give some applications, including most of the pieces of a classiﬁcation of the groups of order p3 for p odd. The last step will be given via the Sylow Theorems in Section 5.3. Section 5.3 is devoted to the Sylow Theorems, which study the p-subgroups of a ﬁnite group. Extending the result of Cauchy’s Theorem, the Sylow Theorems show that if p is prime and if |G| = pj m with (p, m) = 1, then G must have a subgroup of order pj . Such a subgroup is the highest order p-group that could occur as a subgroup of G, and is known as a p-Sylow subgroup of G (though that is not the initial deﬁnition we shall give for a p-Sylow subgroup). 134 CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 135 The Sylow Theorems are actually a good bit stronger than a simple existence proof for subgroups whose order is the highest power of p dividing the order of G. They also show that any two p-Sylow subgroups of G are conjugate and that the number of p-Sylow subgroups of G is congruent to 1 mod p. This leads to some useful theoretical results regarding p-Sylow subgroups which are normal, as well as to some valuable counting arguments for the classiﬁcation of groups of particular orders. In particular, Sylow theory can be useful for ﬁnding normal subgroups in a particular group. The Sylow Theorems give a useful point of view for considering ﬁnite groups. Finite p-groups are more tractable than general ﬁnite groups, and a useful ﬁrst step in understanding a given group is to understand its Sylow subgroups and their normalizers. For many low order groups, an understanding of the entire structure of the group will follow from this analysis. Given the Sylow Theorems and the material on semidirect products, extensions, and the automorphisms of cyclic groups, we are able to classify the groups of many of the orders less than 64. We give numerous classiﬁcations in the text and exercises for Section 5.3. Section 5.4 studies the commutator subgroup and determines the largest abelian factor group that a group can have. Then, in Section 5.5, and using the commutator subgroups, we study solvable groups. Solvable groups are a generalization of abelian groups. In essence, in the ﬁnite case, they are the groups with enough normal subgroups to permit good inductions on order. One characterization we shall give of the ﬁnite solvable groups is that they are the ﬁnite groups whose composition series have no nonabelian simple groups in the list of successive simple subquotients. As expected, the induction arguments available for solvable groups provide for stronger results than we get for arbitrary groups. In Section 5.6, we prove Hall’s Theorem, which shows that if G is a ﬁnite solvable group and if |G| = mn with (m, n) = 1, then G has a subgroup of order m. This fact turns out to characterize the solvable groups, though we won’t give the converse here. The nilpotent groups form a class of groups that sits between the abelian groups and the solvable groups. They are deﬁned by postulating properties that always hold in p-groups, as we showed in Section 5.2. By the end of Section 5.7, we show that the ﬁnite nilpotent groups are nothing other than direct products of ﬁnite p-groups over varying primes p. In the process we will learn some more about the p-groups themselves. We shall also see that the full converse of Lagrange’s Theorem holds for nilpotent groups: A nilpotent group G has subgroups of every order that divides the order of G. We close the chapter with a discussion of matrix groups over commutative rings. Here, we assume an understanding of some of the material in Chapters 7 and 10. The reader who has not yet studied this material should defer reading Section 5.8 until having done so. One of the reasons for studying matrix groups is that the automorphism group of a direct product, Zkn , of k copies of Zn is isomorphic to Glk (Zn ), so by studying the latter, we can obtain classiﬁcation results for groups containing normal subgroups isomorphic to Zkn . This situation arises for some low order groups. We also calculate the order of Gln (Fq ), where Fq is the ﬁnite ﬁeld with q = pr elements with p prime. We describe some important subquotient groups, PSln (Fq ) of Gln (Fq ). Most of these projective special linear groups turn out to be simple, but we shall not show it here. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 5.1 136 Cauchy’s Theorem We now prove Cauchy’s Theorem. An alternative proof was given as Problem 24 of Exercises 3.3.23. Theorem 5.1.1. (Cauchy’s Theorem) Let G be a ﬁnite group and let p be a prime dividing the order of G. Then G has an element of order p. Proof Note that we’ve already veriﬁed the theorem in the case that G is abelian: Cauchy’s Theorem for abelian groups was given as Proposition 4.4.12. We argue by induction on the order of G. The induction starts with |G| = p, in which case G ∼ = Zp , and the result is known to be true. Recall that the nontrivial conjugacy classes in G are those with more than one element and that the conjugacy class of x is nontrivial if and only if x ∈ Z(G). Recall also that a set of class representatives for the nontrivial conjugacy classes in G is a set X ⊂ G such that 1. X ∩ Z(G) = ∅. 2. If y ∈ G is not in Z(G), then y is conjugate to some x ∈ X. 3. If x, x ∈ X with x = x , then x and x are not conjugate in G. Then the class formula (Corollary 3.6.18), shows that if X ⊂ G is a set of class representatives for the nontrivial conjugacy classes in G, then [G : CG (x)], |G| = |Z(G)| + x∈X where CG (x) is the centralizer of x in G. As stated, we already know Cauchy’s Theorem to be true for abelian groups. Thus, if p divides the order of Z(G), then Z(G) has an element of order p, and hence G does also. Thus, we may assume that p does not divide the order of Z(G). Since p does divide the order of G, the class formula tells us that there must be some x ∈ Z(G) such that p does not divide [G : CG (x)]. Since |G| = [G : CG (x)] · |CG (x)|, this says that p must divide the order of CG (x). Since x ∈ Z(G), [G : CG (x)], which is equal to the number of conjugates of x in G, cannot equal 1. Thus, |CG (x)| < |G|, so the induction hypothesis provides an element of order p in CG (x), and hence also in G. As a sample application, we classify the groups of order 6. Corollary 5.1.2. Let G be a group of order 6. Then G is isomorphic to either Z6 or D6 . Proof By Cauchy’s Theorem, G has an element, b, of order 3. But then b has index 2 in G, and hence b G by Proposition 4.1.12 (or by the simpler argument of Problem 12 of Exercises 3.7.14). Let π : G → G/b be the canonical map. A second application of Cauchy’s Theorem given an element a ∈ G of order 2. But then a ∈ b, and hence π(a) = a is not the identity element of G/b. Thus, a generates G/b. Since a and a both have order 2, setting s(a) = a gives a section of π. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 137 ∼ Z× = {±1}, there are Thus, G is a split extension of Z3 by Z2 . Since Aut(Z3 ) = 3 exactly two homomorphisms from Z2 to Aut(Z3 ): the trivial homomorphism and the homomorphism that sends the generator of Z2 to −1. The semidirect product given by the trivial homomorphism is Z3 × Z2 , which is Z6 , because 2 and 3 are relatively prime (e.g., by Problem 4 of Exercises 2.10.12). The semidirect product induced by the nontrivial homomorphism from Z2 to Aut(Z3 ) was shown to be isomorphic to D6 in the ﬁrst example of Examples 4.6.6. Exercises 5.1.3. 1. Show that any group of order 10 is isomorphic to either Z10 or D10 . 2. Assuming that Z× p is cyclic for any odd prime p (as shown in Corollary 7.3.15), show that any group of order 2p is isomorphic to either Z2p or D2p . 3. Recall that Cauchy’s Theorem for abelian groups was available prior to this chapter. Give a proof of the preceding problem that doesn’t use Cauchy’s Theorem for nonabelian groups. (Hint: If G is a group of order 2p, then all of its elements must have order 1, 2, p, or 2p. If all nonidentity elements have order 2, then G must be abelian, and hence Cauchy’s Theorem for abelian groups gives a contradiction. So G must have an element, b, of order p. Now study the canonical map G → G/b, and show that G must have an element of order 2.) 5.2 p-Groups Recall that a p-group is a group in which the order of every element is a power of p. Lagrange’s Theorem shows that any group whose order is a power of p must be a pgroup. But by Cauchy’s Theorem, if |G| has a prime divisor q = p, then it must have an element of order q, and hence it is not a p-group. We obtain the following corollary. Corollary 5.2.1. A ﬁnite group is a p-group if and only if its order is a power of p. The proof used above for Cauchy’s Theorem may be used, with suitable modiﬁcations, to prove some other useful facts. For instance: Proposition 5.2.2. A ﬁnite p-group has a nontrivial center (i.e., Z(G) = e). Proof Let X be a set of class representatives for the nontrivial conjugacy classes in G and consider the class formula: |G| = |Z(G)| + [G : CG (x)]. x∈X If x ∈ Z(G), then [G : CG (x)], which is equal to the number of conjugates of x in G, cannot equal 1. Thus, since [G : CG (x)] divides |G|, we must have [G : CG (g)] = pr for some r > 0, and hence p divides [G : CG (g)]. Thus, p must divide every term in the class formula other than |Z(G)|, so it must divide |Z(G)| as well. In particular, |Z(G)| = 1, and hence Z(G) = e. We ﬁrst give an immediate consequence of Proposition 5.2.2 to classiﬁcation. Corollary 5.2.3. Every group of order p2 is abelian. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 138 Proof We argue by contradiction. If G is not abelian, then Z(G) = G. But then if |G| = p2 , we must have |Z(G)| = p. But conjugation acts trivially on Z(G), so that Z(G) G. Thus, G/Z(G) has order p, and hence is cyclic. But now G must be abelian by Lemma 4.7.15, which gives the desired contradiction. We now consider some more general corollaries to Proposition 5.2.2. Corollary 5.2.4. The only ﬁnite p-group which is simple is Zp . Proof Let G be a ﬁnite simple p-group. The center of any group is a normal subgroup. Since the center of G is nontrivial and G is simple, G must equal its center. Thus, G is abelian. But the only abelian simple groups are the cyclic groups of prime order, by Lemma 4.2.2. Recall that a composition series for a group G is a sequence e = H0 ⊂ H1 ⊂ · · · ⊂ H j = G such that Hi−1 Hi and Hi /Hi−1 is a nontrivial simple group for 1 ≤ i ≤ j. By Proposition 4.2.5, every nontrivial ﬁnite group has a composition series. Since a quotient of a subgroup of a p-group is again a p-group, the simple groups Hi /Hi−1 associated to the composition series for a nontrivial ﬁnite p-group must all be Zp . But a direct proof of this fact will give us a stronger statement: We can construct a composition series for a p-group G such that all the subgroups in the series are normal in G itself, and not just normal in the next group in the series. Corollary 5.2.5. Let G be a ﬁnite group of order pj , with p prime and j > 0. Then there is a sequence e = H0 ⊂ H1 ⊂ · · · ⊂ Hj = G, with Hi G and Hi /Hi−1 ∼ = Zp for 1 ≤ i ≤ j. Proof We argue by induction on |G|, with the result being trivial for |G| = p. By Proposition 5.2.2, Z(G) = e, so Cauchy’s Theorem provides a subgroup H1 ⊂ Z(G) with |H1 | = p. But any subgroup of Z(G) is normal in G, so we can form the factor group G/H1 . Since |G/H1 | < |G|, the induction hypothesis provides a sequence e = K1 ⊂ · · · ⊂ Kj = G/H1 , such that Ki G/H1 and Ki /Ki−1 ∼ = Zp for 2 ≤ i ≤ j. Let π : G → G/H1 be the canonical map and set Hi = π −1 (Ki ) for 2 ≤ i ≤ j. Since the pre-image under a homomorphism of a normal subgroup is always normal, we have Hi G for all i. Also, since Ki = Hi /H1 for i ≥ 1, the Third Noether Theorem shows that Hi /Hi−1 ∼ = Ki /Ki−1 ∼ = Zp for 2 ≤ i ≤ j, and the result follows. Since the subgroup Hi above has order pi , we see that the full converse of Lagrange’s Theorem holds in p-groups. Corollary 5.2.6. Let G be a ﬁnite p group. Then every divisor of |G| is the order of a subgroup of G. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 139 We shall now begin to try to classify the groups of order p3 for primes p > 2. We shall not yet succeed entirely, because not all the groups we must consider are p-groups. We must also understand the automorphism groups of the groups of order p2 , which in the case of Zp × Zp will require the Sylow theorems. As in the case of p = 2, the abelian groups of order p3 are Zp3 , Zp2 × Zp , and Zp × Zp × Zp . For the nonabelian groups of order p3 , the whole story will depend on whether there exists an element of order p2 . Proposition 5.2.7. Let G be a nonabelian group of order p3 where p is an odd prime. Suppose that G contains an element of order p2 . Then G is isomorphic to the semidirect 1 product Zp2 α Zp , where α : Zp → Z× p2 is induced by α(1) = p + 1. Proof Let b ∈ G have order p2 . Then b has index p in G, so b G by Proposition 4.1.12. Thus, G is an extension of Zp2 by Zp , via f : G → Zp . Since G is not abelian, 2 Lemma 4.7.15 shows that the induced homomorphism α : Zp → Z× p2 = Aut(Zp ) is nontrivial. We shall use strongly the fact that Z× p2 is abelian. Because of this, the p-torsion elements form a subgroup, which, according to Corollary 4.5.12, is K(p, 2), the subgroup consisting of those units congruent to 1 mod p. Corollary 7.3.15 then shows that K(p, 2) is a cyclic group of order p, generated by 1 + p. × Since K(p, 2) is the p-torsion subgroup of Z× p2 , the image of α : Zp → Zp2 must lie in K(p, 2). Since α is nontrivial and Zp is simple, α must be an isomorphism from Zp to K(p, 2). But then p + 1 = α(x) for some generator x of Zp . But we may change our identiﬁcation of Zp to make x the canonical generator of Zp . Thus, we may assume that α carries the canonical generator of Zp to 1 + p. Thus, it suﬃces to show that f is a split extension, as then G ∼ = Zp2 α Zp , with α as claimed. Let us continue to write x for our preferred generator of Zp , so that α(x) = p + 1. Since α is the action induced by the extension f : G → Zp , Lemma 4.7.9 shows that ghg −1 = h1+p for any h ∈ ker f and any g with f (g) = x. To show that f splits, it suﬃces to ﬁnd g ∈ G of order p such that f (g) = x: The splitting is then deﬁned by setting s(x) = g. Let a ∈ G with f (a) = x. Then a cannot have order p3 , as G would then be cyclic. Alternatively, if a has order p, then a itself gives the desired splitting of f . Thus, we shall assume that a has order p2 . For simplicity of notation, let m = p + 1, so that aha−1 = hm for any h ∈ ker f . j Moreover, since α(aj ) = mj , aj ha−j = hm for h ∈ ker f . Thus, as shown in Problem 7 of Exercises 4.6.13, an easy induction on k shows that k−1 (ha)k = h1+m+···+m ak for k ≥ 1. Thus, taking k = p, Problem 8 of Exercises 4.6.13 gives (ha)p = h mp −1 m−1 ap . Recall that m = p + 1. Then m − 1 = p, while the proof of Proposition 4.5.18 shows that mp is congruent to 1 + p2 mod p3 . But then (mp − 1)/(m − 1) is congruent to p mod p2 , so that (ha)p = hp ap . Now ap has order p and lies in b = ker f . Thus, ap = bjp for some j relatively prime to p. But then if we take h = b−j , then the element ha has order p, and gives the desired splitting for f . 1 Thus, G is the group of Example 6 of Examples 4.6.6. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 140 Thus, there is no analogue at odd primes of Q8 : Every nonabelian extension of Zp2 by Zp is split when p > 2. Lemma 5.2.8. Let p be an odd prime and let G be a nonabelian group of order p3 that does not contain an element of order p2 . Then G is a split extension of Zp × Zp by Zp . Proof By Corollary 5.2.5, G has a normal subgroup H of order p2 . Since groups of order p2 are abelian, and since G contains no element of order p2 , H must be isomorphic to Zp × Zp . Since H has index p, G is an extension of H by Zp . But all elements of G have exponent p, so any lift of the generator of G/H to G provides a splitting for this extension. We have constructed an example of a nonabelian group of order p3 in which every element has exponent p, in Corollary 4.6.8. We would like to show that every nonabelian split extension of Zp × Zp by Zp is isomorphic to this one. To show this we ﬁrst need to understand the automorphism group of Zp × Zp . We will then need to understand the conjugacy classes of homomorphisms from Zp into this group of automorphisms. But the automorphism group in question is nonabelian, so that a complete picture of the conjugacy classes of its subgroups of order p will depend on the Sylow Theorems of Section 5.3. We will content ourselves here with a calculation of the order of Aut(Zp × Zp ). Together with the Sylow theorems, this will be suﬃcient to complete our classiﬁcation. Given a basic knowledge of linear algebra, the following result becomes a little more transparent, and may be extended to a calculation of the order of the automorphism group of a product of n copies of Zp . Lemma 5.2.9. The automorphism group of Zp × Zp has order (p2 − 1)(p2 − p). In particular, this group has order jp, where (j, p) = 1. Proof Write Zp × Zp = {ai bj | 0 ≤ i, j ≤ p}. Of course, a and b commute and each has order p. Since every element of Zp × Zp has exponent p, the homomorphisms, f , from Zp × Zp to itself are in one-to-one correspondence with all possible choices of the elements f (a) and f (b). Since there are p2 choices for f (a) and p2 choices for f (b), this gives p2 · p2 = p4 diﬀerent homomorphisms from Zp × Zp to itself. We’d like to know which of these are isomorphisms. Let f : Zp × Zp → Zp × Zp be a homomorphism. Then the image of f has order 1, p or p2 . It is an isomorphism if and only if the image has order p2 . So one way to argue at this point is to count up the homomorphisms whose images have order 1 or p, and subtract this number from p4 . Since Zp × Zp has exponent p, every e = x ∈ Zp × Zp is contained in exactly one subgroup of order p. That subgroup is x. Thus, the number of subgroups of order p is the number of nonidentity elements in Zp × Zp divided by the number of nonidentity elements of Zp : we have (p2 − 1)/(p − 1) = p + 1, subgroups of order p in Zp × Zp . But if H ⊂ Zp × Zp has order p, then the homomorphisms from Zp × Zp to H are in one-to-one correspondence with the choices of pairs of elements of H, so there are p2 diﬀerent homomorphisms from Zp × Zp to H. Only one of these is the trivial homomorphism, and hence there are p2 − 1 homomorphisms from Zp × Zp onto H. Multiplying this by the number of subgroups of order p, we see that there are p3 + 2 p − p − 1 homomorphisms from Zp × Zp to itself with an image of order p. Since the trivial homomorphism is the only homomorphism whose image has order 1, this gives exactly p3 + p2 − p homomorphisms from Zp × Zp to itself that are not automorphisms. But then there are exactly p4 − p3 − p2 + p automorphisms of Zp × Zp . But this is easily seen to factor as (p2 − 1)(p2 − p), as desired. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 141 A perhaps simpler argument is as follows. Choose f (a). Any nonidentity choice will do, so there are p2 − 1 choices for f (a). Now consider the choice of f (b). As long as f (b) is not in f (a), f will be an automorphism. Thus, there are p2 − p possible choices for f (b) once f (a) has been chosen. In particular, there are (p2 − 1)(p2 − p) ways to choose an automorphism f . Exercises 5.2.10. 1. Show that every nonabelian group of order p3 is a central extension of Zp by Zp ×Zp . 2. Construct a group of order p4 whose center is isomorphic to Zp2 . 3. Find a group of order 16 whose center has order 2. 4. Construct a group of order 81 whose center has order 3. 5. Let p be an odd prime. Construct a group of order p4 whose center has order p. 6. Show that every group of order pn may be generated by n elements. 5.3 Sylow Subgroups Cauchy’s Theorem shows that if p divides the order of a ﬁnite group G, then G has a subgroup of order p. But what if |G| is divisible by p2 ? Can we then ﬁnd a subgroup of order p2 ? We cannot argue here by Cauchy’s Theorem and induction, as there may not be a normal subgroup of order p. (E.g., the group G may be simple.) So we need a new approach to the question. Note that there are examples of groups G and numbers n such that |G| is divisible by n, but G has no subgroup of order n. For instance, we’ve seen in Problem 2 of Exercises 4.2.18 that A4 has no subgroup of order 6. Deﬁnitions 5.3.1. Let G be a ﬁnite group and let p be a prime. A p-subgroup of G is a subgroup which is a p-group. A p-Sylow subgroup P of G is a maximal p-subgroup of G, i.e., P is a p-subgroup of G, and any p-subgroup of G containing P must be equal to P. Examples 5.3.2. In D6 , b has order 3. Since no higher power of 3 divides the order of D6 , b cannot be properly contained in a 3-subgroup of D6 . Thus, b is a 3-Sylow subgroup of D6 . Similarly, a, ab, and ab2 are all 2-subgroups of D6 . Since 2 itself is the highest 2-power that divides |D6 |, each of these is a 2-Sylow subgroup of D6 . Lemma 5.3.3. Let H be a p-subgroup of a ﬁnite group G. Then H is contained in a p-Sylow subgroup of G. Proof Let |G| = pi m with p relatively prime to m, and let |H| = pj . If i = j, then H is a p-Sylow subgroup of G by Lagrange’s Theorem. Otherwise, we argue by induction on i − j. From what we know at this point, H itself might be a p-Sylow subgroup of G even if j < i. If this is the case, we are done. Otherwise, there is a p-subgroup K of G that strictly contains H. But then the index of K in G is smaller than that of H, so by induction, K is contained in a p-Sylow subgroup. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 142 Lemma 5.3.4. Let G be a ﬁnite group and let p be a prime dividing the order of G. Let P be a p-Sylow subgroup of G. Then the index of P in its normalizer is relatively prime to p. Proof Recall that P NG (P ). Thus, the order of NG (P )/P is the index of P in NG (P ). If this order is divisible by p, then Cauchy’s Theorem gives a subgroup H ⊂ NG (P )/P with |H| = p. But then if π : NG (P ) → NG (P )/P is the canonical map, then π −1 H is a p-subgroup of G that strictly contains P . Corollary 5.3.5. Let G be a ﬁnite group and let p be a prime dividing the order of G. Let P be a p-Sylow subgroup of G. Then any p-subgroup of NG (P ) is contained in P . Proof Let K ⊂ NG (P ) be a p-group. Then K/(K ∩ P ) is isomorphic to a subgroup of NG (P )/P . Since K is a p-group, so is K/(K ∩ P ). Since |NG (P )/P | is prime to p, K/(K ∩ P ) = e, and hence K ⊂ P . Lemma 5.3.6. Let P be a p-Sylow subgroup of the ﬁnite group G. Then any conjugate of P is also a p-Sylow subgroup of G. Proof Since conjugation gives an automorphism of G, any conjugate of a p-subgroup of G is again a p-subgroup of G. Now for x ∈ G, suppose that xP x−1 not a p-Sylow subgroup of G. Then xP x−1 must be properly contained in a p-subgroup P of G. But then P is properly contained in x−1 P x, contradicting the statement that P is a p-Sylow subgroup of G. We now give yet another variant of the proof of Cauchy’s Theorem. Theorem 5.3.7. (First Sylow Theorem) Let G be a ﬁnite group and let p be a prime dividing the order of G. Then any two p-Sylow subgroups of G are conjugate, and the number of p-Sylow subgroups of G is congruent to 1 mod p. Proof Let P be a p-Sylow subgroup of G and let X ⊂ Sub(G) be the orbit of P under the G-action given by conjugation. (Recall that Sub(G) is the set of subgroups of G.) By Lemma 5.3.6, each element of X is a p-Sylow subgroup of G. The G-set X may be considered as a P -set by restricting the action. For P ∈ X, the isotropy subgroup of P under the action of P on X is given by PP = {x ∈ P | xP x−1 = P } = P ∩ NG (P ). In particular, P is ﬁxed by P if and only if P ⊂ NG (P ). By Corollary 5.3.5, this can happen if and only if P = P , as both P and P are p-Sylow subgroups of G. Thus, the action of P on X has exactly one trivial orbit: {P }. (Here, an orbit is trivial if it has only one element.) The number of elements in the orbit of P ∈ X under the P -action on X is [P : P ∩ NG (P )], which divides |P |. Thus, since P is a p-group, the number of elements in any nontrivial orbit is divisible by p (hence congruent to 0 mod p). In particular, if Y ⊂ X is a set of orbit representatives for the nontrivial orbits of the action of P on X (i.e., the intersection of each nontrivial P -orbit in X with Y consists of exactly one element), then the G-set Counting Formula (Corollary 3.3.15) gives |X| = |{P }| + [P : P ∩ NG (P )]. P ∈Y CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 143 Since each [P : P ∩ NG (P )] is congruent to 0 mod p and since |{P }| = 1, we see that the number of elements in X is congruent to 1 mod p. It remains to show that X contains every p-Sylow subgroup of G. Let P be an arbitrary p-Sylow subgroup of G, and consider the action of P on X by conjugation. If P is not in X, then every orbit of this P -action must be nontrivial. But that would make the number of elements in X congruent to 0 mod p, rather than 1 mod p. So we must have P ∈ X. Theorem 5.3.8. (Second Sylow Theorem) Let G be a ﬁnite group and let p be a prime dividing the order of G. Let P be a p-Sylow subgroup of G. Then the index of P in G is prime to p. Thus, if |G| = pi m, with p and m relatively prime, then |P | = pi . Moreover, the number of p-Sylow subgroups in G divides m in addition to being congruent to 1 mod p. Proof The index, m, of P in G is the product of [NG (P ) : P ] and [G : NG (P )]. The former is prime to p by Corollary 5.3.5. The latter is equal to the number of conjugates of P , and hence is 1 mod p by the First Sylow Theorem. Recall from Corollary 5.2.6 that if p is prime and if P is a ﬁnite group of order pi , then P has a subgroup of every order dividing the order of P . Applying this to a p-Sylow subgroup of a group G, the Second Sylow Theorem produces the following corollary. Corollary 5.3.9. Let G be a ﬁnite group and let q be any prime power that divides the order of G. Then G has a subgroup of order q. We now give the ﬁnal Sylow Theorem. Theorem 5.3.10. (Third Sylow Theorem) Let G be a ﬁnite group and let p be a prime dividing the order of G. Let P be a p-Sylow subgroup of G. Then the normalizer of P is its own normalizer: NG (NG (P )) = NG (P ). Proof We claim that P is a characteristic subgroup of NG (P ). The point here is that if x ∈ P and if f is any automorphism of NG (P ), then f (x) must be a p-torsion element. But by Corollary 5.3.5, every p-torsion element of NG (P ) must be contained in P . Now suppose that NG (P ) H. Then conjugation by an element of H gives an automorphism of NG (P ). But this must preserve P , so P is normal in H. But then H ⊂ NG (P ). We now summarize some important facts implicit in the above discussion. Corollary 5.3.11. Let G be a ﬁnite group and let p be a prime dividing the order of G. Suppose there is a p-Sylow subgroup P of G which is normal in G. Then P is the only p-Sylow subgroup of G, and P is the set of all p-torsion elements in G. Indeed, P is a characteristic subgroup of G. Conversely, if there is only one p-Sylow subgroup of G, then it is normal in G. Proof The p-Sylow subgroups of G are all conjugate. But if P G, then it has no conjugates other than itself. Moreover, since NG (P ) = G, P contains all the p-torsion elements of G by Corollary 5.3.5. And since P is a p-group, all its elements are p-torsion, so that P is, in fact, the set of p-torsion elements of G. But then P is characteristic, since any automorphism preserves the p-torsion elements. If there is only one p-Sylow subgroup, P , of G, then P has no conjugates other than itself. Thus, [G : NG (P )] = 1, and hence P G. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 144 In many cases a normal p-Sylow subgroup is suﬃcient to unfold the entire structure of a group. The point is that if P is a normal p-Sylow subgroup of G, then |P | and |G/P | are relatively prime. Thus, the argument of Proposition 4.7.7 gives the following corollary. Corollary 5.3.12. Suppose that G has a normal p-Sylow subgroup, P , and that G has a subgroup K of order [G : P ]. Then G is the semidirect product P α K, where α : K → Aut(P ) is induced by conjugation in G. According to the Schur–Zassenhaus Lemma, which we will not prove, such a subgroup K always exists. However, if [G : P ] is a prime power, the existence of such a K is automatic from the second Sylow Theorem. Corollary 5.3.13. Let G be a group of order pr q s where p and q are distinct primes. Suppose that the p-Sylow subgroup, P , of G is normal. Let Q be a q-Sylow subgroup of G. Then G is isomorphic to the semidirect product P α Q, where α : Q → Aut(P ) is induced by conjugation in G. The Sylow Theorems are extremely powerful for classifying low-order ﬁnite groups. We shall give a few of their applications now. First, let us complete the classiﬁcation of groups of order p3 . Proposition 5.3.14. Let p be an odd prime and let G be a nonabelian group of order p3 that contains no element of order p2 . Then G is isomorphic to the group constructed in Corollary 4.6.8 for n = p. Proof By Lemma 5.2.8, G is a semidirect product (Zp × Zp ) α Zp for some α : Zp → Aut(Zp × Zp ). Moreover, α cannot be the trivial homomorphism, as then G would be abelian by Lemma 4.7.15. But by Lemma 5.2.9, the p-Sylow subgroups of Aut(Zp × Zp ) have order p, so α must be an isomorphism onto one of these p-Sylow subgroups. But the p-Sylow subgroups are all conjugate, and hence an isomorphism onto one of them is conjugate as a homomorphism into Aut(Zp × Zp ) to an isomorphism onto any of the others. Moreover, if α and α are isomorphisms onto the same p-Sylow subgroup of Aut(Zp × Zp ), then because they are isomorphisms, we have α = α ◦ g for an automorphism g of Zp . Thus, any two nonabelian semidirect products (Zp ×Zp )α Zp and (Zp × Zp ) α Zp are isomorphic by Proposition 4.6.11. Let us return to the theme of discovering normal Sylow subgroups and using them to split the group as a semidirect product. Corollary 5.3.15. Let G be a group of order pr m, where m < p, with p prime. Then the p-Sylow subgroup of G is normal. Proof The number of p-Sylow subgroups of G is congruent to 1 mod p and divides m. Since m < p, there must be only one p-Sylow subgroup, which is thus normal. When used in conjunction with Corollary 5.3.13, we can get some concrete results: Corollary 5.3.16. Let G be a group of order pr q s , where p and q are distinct primes. Suppose that q s < p. Then G is a semidirect product P α Q for some α : Q → Aut(P ), where P and Q are the p-Sylow and q-Sylow subgroups of G, respectively. This becomes more powerful when used in conjunction with what we know about automorphisms: CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 145 Corollary 5.3.17. Every group of order 15 is isomorphic to Z15 . Proof Let |G| = 15. By Corollary 5.3.16, G is a semidirect product Z5 α Z3 for some × homomorphism α : Z3 → Z× 5 . But Z5 has order 4, which is relatively prime to 3, and hence α must be the trivial homomorphism. But then G ∼ = Z5 × Z3 , which is isomorphic to Z15 . Now let us consider a deeper sort of application of the Sylow Theorems. Sometimes, under the assumption that a given Sylow subgroup is nonnormal, we can use the count of the number of its conjugates to ﬁnd the number of elements of p-power order. This will sometimes permit us to ﬁnd other subgroups that are normal. Corollary 5.3.18. Let G be a group of order 12 whose 3-Sylow subgroups are not normal. Then G is isomorphic to A4 . Proof We begin by showing that the 2-Sylow subgroup of G is normal. Let P be a 3-Sylow subgroup of G. Then the number of conjugates of P is 1 mod 3 and divides 4. Thus, P has either 1 or 4 conjugates. But we’ve assumed that P is nonnormal, so it must have 4 conjugates. Note that if P and P are distinct 3-Sylow subgroups, then P ∩ P = e, since P and P both have order 3. But since each 3-Sylow subgroup has exactly two elements of order 3, there must be 8 elements of order 3 in G. The Second Sylow Theorem, on the other hand, shows that G has a subgroup, H, of order 4. Since there are 8 elements of order 3 in G, every element outside of H must have order 3. Since the elements of H are all 2-torsion, none of their conjugates may lie outside of H, and hence H must be normal. By Corollary 5.3.16, G ∼ = H α Z3 for some α : Z3 → Aut(H). Since there are no nontrivial homomorphisms of Z3 into Z× 4 = {±1} and since G is nonabelian, H must be isomorphic to Z2 × Z2 . As shown in either Problem 14 of Exercises 4.6.13 or Problem 3 of Exercises 4.7.27, the automorphism group of Z2 × Z2 is isomorphic to S3 . It’s easy to see that there are only two nontrivial homomorphisms from Z3 into S3 ∼ = D6 , and that one is obtained from the other by precomposing with an automorphism of Z3 . Thus, Proposition 4.6.11 shows that there is exactly one isomorphism class of semidirect products (Z2 × Z2 ) α Z3 which is nonabelian. As shown in Problem 1 of Exercises 4.7.27, A4 is a semidirect product (Z2 ×Z2 )α Z3 , and the result follows. Thus, the classiﬁcation of groups of order 12 depends only on classifying the split extensions of Z3 by the groups of order 4. We shall leave this to the exercises. Here is a slightly more complicated application of the Sylow Theorems. Lemma 5.3.19. Let G be a group of order 30. Then the 5-Sylow subgroup of G is normal. Proof We argue by contradiction. Let P5 be a 5-Sylow subgroup of G. Then the number of conjugates of P5 is congruent to 1 mod 5 and divides 6. Thus, there must be six conjugates of P5 . Since the number of conjugates is the index of the normalizer, we see that NG (P5 ) = P5 . Since the 5-Sylow subgroups of G have order 5, any two of them intersect in the identity element only. Thus, there are 6 · 4 = 24 elements in G of order 5. This leaves CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 146 6 elements whose order does not equal 5. We claim now that the 3-Sylow subgroup, P3 , must be normal in G. The number of conjugates of P3 is congruent to 1 mod 3 and divides 10. Thus, if P3 is not normal, it must have 10 conjugates. But this would give 20 elements of order 3, when there cannot be more than 6 elements of order unequal to 5, so that P3 must indeed be normal. But then P5 normalizes P3 , and hence P5 P3 is a subgroup of G. Moreover, the Second Noether Theorem gives (P5 P3 )/P3 ∼ = P5 /(P5 ∩ P3 ). But since |P5 | and |P3 | are relatively prime, P5 ∩ P3 = e, and hence P5 P3 must have order 15. Thus, P5 P3 ∼ = Z15 , by Corollary 5.3.17. But then P5 P3 normalizes P5 , which contradicts the earlier statement that NG (P5 ) has order 5. Corollary 5.3.20. Every group of order 30 is a semidirect product Z15 α Z2 for some α : Z2 → Z× 15 . Proof Since the 5-Sylow subgroup P5 is normal, it is normalized by any 3-Sylow subgroup P3 , and hence the product P3 P5 is a subgroup of order 15, and hence index 2. Since index 2 subgroups are normal, G is an extension of a group of order 15 by Z2 . By Cauchy’s Theorem, this extension must be split. Since there is only one group of order 15, the result follows. We shall classify these semidirect products in Exercises 5.3.22. Often in mathematics the statement of a theorem is insuﬃcient, by itself, to derive all the applications of its method of proof. This is, in general, a good thing. It means that the methods have enough substance to be applied in more varied settings. (The alternative would be to continually learn or write proofs that one never needed to see again, which, in the end, would become boring.) Here is an example of a proof based on the method of proof of the Sylow Theorems. Proposition 5.3.21. Let G be a group of order 24 that is not isomorphic to S4 . Then one of its Sylow subgroups is normal. Proof Suppose that the 3-Sylow subgroups are not normal. The number of 3-Sylow subgroups is 1 mod 3 and divides 8. Thus, if there is more than one 3-Sylow subgroup, there must be four of them. Let X be the set of 3-Sylow subgroups of G. Then G acts on X by conjugation, so we get a homomorphism f : G → S(X) ∼ = S4 . As we’ve seen in the discussion on G-sets, the kernel of f is the intersection of the isotropy subgroups of the elements of X. Moreover, since the action is that given by conjugation, the isotropy subgroup of H ∈ X is NG (H). Thus, ker f = NG (H). H∈X For H ∈ X, the index of NG (H) is 4, the number of conjugates of H. Thus, the order of NG (H) is 6. Suppose that K is a diﬀerent element of X. We claim that the order of NG (H) ∩ NG (K) divides 2. To see this, note that the order of NG (H) ∩ NG (K) cannot be divisible by 3. This is because any p-group contained in the normalizer of a p-Sylow subgroup must be contained CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 147 in the p-Sylow subgroup itself (Corollary 5.3.5). Since the 3-Sylow subgroups have prime order here, they cannot intersect unless they are equal. But if the order of NG (H) ∩ NG (K) divides 6 and is not divisible by 3, it must divide 2. In consequence, we see that the order of the kernel of f divides 2. If the kernel has order 1, then f is an isomorphism, since G and S4 have the same number of elements. Thus, we shall assume that ker f has order 2. In this case, the image of f has order 12. But by Problem 2 of Exercises 4.2.18, A4 is the only subgroup of S4 of order 12, so we must have im f = A4 . By Problem 1 of Exercises 4.2.18, the 2-Sylow subgroup, P2 , of A4 is normal. But since ker f has order 2, f −1 P2 has order 8, and must be a 2-Sylow subgroup of G. As the pre-image of a normal subgroup, it must be normal, and we’re done. The results we have at this point are suﬃcient to classify the groups of most orders less than 64. Almost all of them are semidirect products. We shall treat many examples in the exercises below. But for some of the orders less than 64, we must understand more about the subnormal series of p-groups. We shall develop the material to do this in the next three sections. Then, we must make a deeper analysis of some automorphism groups. For this, we will need some information from our analysis of matrix groups over ﬁnite ﬁelds below. Exercises 5.3.22. 1. Show that the 2-Sylow subgroup of S4 is isomorphic to D8 . Is it normal in S4 ? 2. What is the 2-Sylow subgroup of S6 ? 3. What is the 2-Sylow subgroup of S8 ? 4. Let H be any group and write H n for the product H × · · · × H of n copies of H. Recall from Problem 21 of Exercises 3.6.24 that Sn acts on H n through automorphisms via σ · (h1 , . . . , hn ) = (hσ−1 (1) , . . . , hσ−1 (n) ). For any subgroup K ⊂ Sn we get an induced semidirect product H n α K, called the wreath product of H and K, and denoted H K. Write Pr for the 2-Sylow subgroup of S2r . (a) Show that Pr ∼ = Pr−1 S2 for r ≥ 2. (b) Suppose that 2r−1 < n < 2r . Show that the 2-Sylow subgroup of Sn is isomorphic to the product of Pr−1 with the 2-Sylow subgroup of Sn−2r−1 . (c) Express the 2-Sylow subgroup of Sn in terms of the binary expansion of n. 5. Prove the analogue of the preceding problem for primes p > 2. 6. Let p and q be primes, with q < p. Assume, via Corollary 7.3.15, that Z× p is cyclic. Show that there are at most two groups of order pq. When are there two groups, and when only one? 7. Let p and q be primes. Show that every group of order p2 q has a normal Sylow subgroup. 8. Give a simpler, direct argument for Corollary 5.3.18 by studying the homomorphism Λ : G → S(G/P3 ) obtained via Proposition 3.3.16 from the standard action of G on the left cosets of a 3-Sylow subgroup of G. 9. Classify the groups of order 12. How does Q12 ﬁt into this classiﬁcation? CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 148 10. Classify the groups of order 20. 11. Classify the groups of order 24 whose 2-Sylow subgroup is either cyclic or dihedral. 12. Classify the groups of order 30. 13. We show here that any group G of order 36 without a normal 3-Sylow subgroup must have a normal 2-Sylow subgroup. Thus, assume that G has a non-normal 3-Sylow subgroup P3 . Then the action of G on the set of left cosets G/P3 induces a homomorphism Λ : G → S(G/P3 ) ∼ = S4 . (a) Show, using Problem 31 of Exercises 3.7.14, that the kernel of Λ must have order 3. (b) Deduce from Problem 2 of Exercises 4.2.18 that the image of Λ is A4 . Write V ⊂ A4 for the 2-Sylow subgroup of A4 and let H = Λ−1 V ⊂ G. (c) Show that H, as an extension of Z3 by Z2 × Z2 , must be isomorphic to either D12 or to Z3 × Z2 × Z2 . (d) Show that every 2-Sylow subgroup of G must be contained in H. Deduce that if H ∼ = Z3 × Z2 × Z2 , then the 2-Sylow subgroup of G is normal. (e) Show that if H ∼ = D12 , then G has exactly three 2-Sylow subgroups. Deduce that if P2 is a 2-Sylow subgroup of G, then |NG (P2 )| = 12. Deduce that Λ induces an isomorphism from NG (P2 ) to A4 . (f) Show that the only split extension of Z3 by A4 is the trivial extension Z3 ×A4 . Deduce that H must have been isomorphic to Z3 × Z2 × Z2 . 14. Classify those groups of order 40 whose 2-Sylow subgroup is isomorphic to one of the following: (a) Z8 (b) D8 (c) Z4 × Z2 . 15. Show that any group of order 42 is a split extension of Z7 by a group of order 6. Classify all such groups. 16. Alternatively, show that any group of order 42 is a split extension of a group of order 21 by Z2 . Classify these, and relate this classiﬁcation to the preceding one. (To do this, you should calculate the automorphism group of the nonabelian group of order 21. See Problem 19 of Exercises 4.7.27.) 17. Classify the groups of order 45. 18. Show that any group G of order 60 whose 5-Sylow subgroups are nonnormal must be simple. (Hint: First calculate the orders of the normalizers of the 5-Sylow and the 3-Sylow subgroups of G and then go through all possible orders for the subgroups of G and show that no subgroup of that order can be normal. In some cases, it is possible to show that no subgroup of the stated order can exist at all. The classiﬁcations above of the groups whose order properly divides 60 will be useful.) CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 149 19. Show that any simple group G of order 60 is isomorphic to A5 . (Hint: You deﬁnitely want to use the information derived in the solution of the previous exercise. The key is ﬁnding a subgroup of index 5. It may help to calculate the order of the centralizer of an element of order 2 in G.) 20. Classify the groups of order 60. 21. Show that every group G of order 90 has a normal 5-Sylow subgroup. Deduce that G has a subgroup of index two. 22. Classify the groups of order 90 whose 3-Sylow subgroup is cyclic. 5.4 Commutator Subgroups and Abelianization The commutator of two group elements, a and b, may be thought of as a measurement of how far a and b are from commuting with each other. The commutators of all pairs of elements in a group generate a useful normal subgroup called the commutator subgroup of G, and denoted [G, G]. We shall see that G/[G, G] is the largest abelian quotient group of G. Deﬁnition 5.4.1. Let a and b be elements of the group G. The commutator of a and b, written [a, b], is deﬁned by [a, b] = aba−1 b−1 . We’ve already used the next lemma repeatedly. Lemma 5.4.2. The elements a and b commute if and only if [a, b] = e. The next lemma may be seen by inspection. Lemma 5.4.3. For a, b ∈ G, [a, b]−1 = [b, a]. Deﬁnition 5.4.4. The commutator subgroup of G, written [G, G], is the subgroup generated by all the commutators of the elements of G. Thus, an element of G is in [G, G] if and only if it is a product of commutators. Note, however, that not every element of [G, G] is actually a commutator itself. Lemma 5.4.5. Let G be any group. Then the commutator subgroup is a characteristic subgroup of G, and hence is normal. Proof Let f be an automorphism of G. Then clearly, f ([a, b]) = [f (a), f (b)]. Thus, since f is a homomorphism, it must take a product of commutators to a product of commutators. Proposition 5.4.6. The quotient group G/[G, G] is abelian. In addition, any homomorphism f : G → H with H abelian factors through the canonical map π : G → G/[G, G]. Proof In G/[G, G], we have ab(a)−1 (b)−1 = aba−1 b−1 = e, so that each pair of elements a and b of G/[G, G] commutes. Let f : G → H with H abelian. To see that f factors through the canonical map to G/[G, G], it suﬃces to show that the commutator subgroup [G, G] is contained in the kernel of f . But f ([a, b]) = [f (a), f (b)], and the latter is the identity element of H, since f (a) and f (b) commute. But then the generators of [G, G] are contained in the kernel of f , and hence [G, G] must be also. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 150 Deﬁnition 5.4.7. We write Gab = G/[G, G], and call it the abelianization of G. Corollary 5.4.8. Every subgroup of G that contains [G, G] is normal. Moreover, if H G, then H contains [G, G] if and only if G/H is abelian. Proof If [G, G] ⊂ H, then H = π −1 π(H), where π is the canonical map from G to Gab . But every subgroup of the abelian group Gab is normal, and hence H is normal as the pre-image of a normal subgroup. If G/H is abelian, then the canonical map from G to G/H factors through Gab . But this happens if and only if [G, G] ⊂ H. Conversely, if [G, G] ⊂ H, then G/H is a quotient group of Gab by the third Noether Theorem, and hence is abelian. Exercises 5.4.9. † 1. Show that a group G is abelian, if and only if [G, G] = e. † 2. Show that if G is nonabelian and simple, then [G, G] = G, and Gab = e. 3. Show that for nonabelian groups of order p3 , p prime, the commutator subgroup and center coincide. 4. What is the commutator subgroup of D2n ? What is D2n ab ? 5. What is the commutator subgroup of Q4n ? What is Q4n ab ? 6. What is the commutator subgroup of A4 ? What is A4 ab ? 7. What is the commutator subgroup of S4 ? What is S4 ab ? 8. What is the commutator subgroup of Sn for n ≥ 5? What is Sn ab ? ‡ 9. We calculate the automorphism group of Q8 . (a) Show that any automorphism f of Q8 induces an automorphism f of Q8 ab . Show that passage from f to f induces a surjective homomorphism η : Aut(Q8 ) → Aut(Q8 ab ). (b) Show η is a split extension whose kernel is isomorphic to Z2 × Z2 . (c) Deduce that Aut(Q8 ) is isomorphic to S4 . 10. Classify the groups of order 24 whose 2-Sylow subgroups are quaternionic. Show that there is a unique such group whose 3-Sylow subgroups are not normal. We shall meet this group again later as the matrix group Sl2 (Z3 ). 5.5 Solvable Groups The most tractable groups to analyze are the abelian groups. Every subgroup is normal, and hence lots of induction arguments are available. This is amply illustrated by the classiﬁcation of ﬁnite abelian groups given in Section 4.4. At the other extreme, we have the nonabelian simple groups. With no nontrivial proper normal subgroups available, it is diﬃcult to get our hands on the structure of the group. The solvable groups give a generalization of abelian groups that share some of the nice properties found in the abelian setting. In particular, knowing that a group is solvable will open up a collection of tools for revealing its structure. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 151 Solvability has long been known to be an important idea. For instance, solvability plays a key role in Galois’ famous proof that there does not exist a formula based on the operations of arithmetic and n-th roots for ﬁnding solutions of polynomials of degree 5 or higher. We shall give an argument for this in Section 11.11. Thus, it is of interest to determine whether a given group is solvable. Of particular importance in this regard is the famous Feit–Thompson Theorem: Theorem (Feit–Thompson Theorem) Every group of odd order is solvable. The Feit–Thompson Theorem is considered to be the ﬁrst major breakthrough along the way to the classiﬁcation of the ﬁnite simple groups. Its proof is quite lengthy and is well beyond the scope of this book.2 But the fact that odd order groups are solvable is an important one to keep in mind for intuition’s sake. We shall need a more compact notation for commutator subgroups, as we shall be discussing commutator subgroups of commutator subgroups, etc. Deﬁnition 5.5.1. For i ≥ 0, deﬁne G(i) inductively by G(0) = G and G(i) = [G(i−1) , G(i−1) ] for i ≥ 1. We call G(i) the i-th derived subgroup of G. Thus, G(1) = [G, G]. Given this, let us make the following deﬁnition. Deﬁnition 5.5.2. A group G is solvable if G(k) = e for some k ≥ 0. The virtue of this deﬁnition is that it is easy to work with. We shall give an equivalent deﬁnition below that may be more intuitive. In the meantime, we can give a quick workup of some basic properties. The next lemma is immediate from Problems 1 and 2 of Exercises 5.4.9. Lemma 5.5.3. Every abelian group is solvable. A simple group, on the other hand, is solvable if and only if it is abelian. Lemma 5.5.4. Let f : G → H be a surjective homomorphism. Then f (G(i) ) = H (i) for all i ≥ 0. Proof We shall show that f ([G, G]) = [H, H]. This then shows that f : [G, G] → [H, H] is surjective, and the result follows by induction on k. But since f ([a, b]) = [f (a), f (b)], f ([G, G]) ⊂ [H, H]. And since f is surjective, the generators of [H, H] must lie in f ([G, G]). Corollary 5.5.5. A quotient group of a solvable group is solvable. Proof If G(k) = e, so is H (k) for any quotient group H of G. Lemma 5.5.6. Let H be a subgroup of G. Then H (i) ⊂ G(i) for all i ≥ 0. Proof Once again, by induction, it suﬃces to treat the case of i = 1. But since H ⊂ G, if h, k ∈ H, then [h, k] ∈ [G, G]. Of course, H (i) need not equal G(i) ∩ H. Corollary 5.5.7. A subgroup of a solvable group is solvable. 2 The proof takes up the entirety of a 255-page journal issue. See W. Feit and J. Thompson, Solvability of groups of odd order, Paciﬁc J. Math. 13 (1963), 775–1029. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 152 Recall that a subnormal series for G consists of a sequence of subgroups e = Gk Gk−1 · · · G0 = G. Here, each Gi+1 is normal in Gi , but need not be normal in G if i > 1. (Note that we’ve reversed the order of the indices from that used previously.) Deﬁnition 5.5.8. A subnormal series is abelian if the quotient groups Gi /Gi+1 are abelian for 0 ≤ i < k. Proposition 5.5.9. A group G is solvable if and only if it has an abelian subnormal series. Proof Suppose that G is solvable, with G(k) = e. Consider the sequence e = G(k) G(k−1) · · · G(0) . ab Since G(i+1) = [G(i) , G(i) ], the quotient group Gi /Gi+1 is just (G(i) ) , which is certainly abelian, so that the sequence above is an abelian subnormal series. Conversely, given an abelian subnormal series e = Gk Gk−1 · · · G0 = G, we claim that G ⊂ Gi for all i, which then forces G(k) = e. Clearly, G(0) = G0 , so assume inductively that G(i−1) ⊂ Gi−1 . But then G(i) ⊂ [Gi−1 , Gi−1 ] by Lemma 5.5.6, so it suﬃces to see that [Gi−1 , Gi−1 ] ⊂ Gi . But this follows immediately from Corollary 5.4.8, since Gi−1 /Gi is abelian. (i) The above characterization of solvable groups can be quite useful. Corollary 5.5.10. Let H be a normal subgroup of G. Then G is solvable if and only if both H and G/H are solvable. Proof Suppose that G is solvable. Then H is solvable by Corollary 5.5.7, and G/H is solvable by Corollary 5.5.5. Conversely, suppose that H and G/H are solvable. Let e = Hk · · · H0 = H be an abelian subnormal series for H and let e = Kl · · · K0 = G/H be an abelian subnormal series for G/H. Then if π : G → G/H is the canonical map, we have an abelian subnormal series e = Hk · · · H0 = π −1 Kl π −1 Kl−1 · · · π −1 K0 = G for G. Also, the following corollary is immediate from Corollary 5.2.5. Corollary 5.5.11. Let p be a prime. Then any ﬁnite p-group is solvable. Note that other than this last result, we have made no use of ﬁniteness in the discussion of solvability. Recall that a composition series for a group G is a subnormal series e = Gk Gk−1 · · · G0 = G where the subquotient groups Gi /Gi+1 of G are all nontrivial simple groups. By Proposition 4.2.5, every nontrivial ﬁnite group has a composition series. The Jordan–Hölder Theorem shows that any two composition series for a group G have the same length, and give the same collection of subquotient groups Gk−1 /Gk , . . . , G/G1 , though possibly in a diﬀerent order. However, for economy of means, we shall not use it in the following proof. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 153 Proposition 5.5.12. A ﬁnite group is solvable if and only if it has a composition series in which the simple subquotient groups are all cyclic of prime order. If G is solvable, then every composition series for G has that form. Proof A composition series whose simple subquotient groups are cyclic is certainly an abelian subnormal series, so any group that has such a composition series is solvable by Proposition 5.5.9. Conversely, suppose that G has a composition series in which one of the quotients Gi /Gi+1 is not cyclic of prime order. Since the only abelian simple groups have prime order (Lemma 4.2.2), Gi /Gi+1 must be a nonabelian simple group. But if G were solvable, the subgroup Gi would be also. And this would force the quotient Gi /Gi+1 to be solvable as well, contradicting Lemma 5.5.3. Exercises 5.5.13. 1. Show that Sn is not solvable if n ≥ 5. 2. Show that A5 is the only group of order ≤ 60 that is not solvable. 3. Show that the Feit–Thompson Theorem is equivalent to the statement that every nonabelian simple group has even order. 5.6 Hall’s Theorem There is a generalization of Sylow theory that works in solvable groups. It is due to Philip Hall. Deﬁnition 5.6.1. Let G be a ﬁnite group. A subgroup H ⊂ G is a Hall subgroup if |H| and [G : H] are relatively prime. Our purpose in this section is to prove the following theorem, which shows that a ﬁnite solvable group G has Hall subgroups of all possible orders n dividing |G| such that (n, |G|/n) = 1. Theorem 5.6.2. (Hall’s Theorem) Let G be a ﬁnite solvable group, and let |G| = nk with (n, k) = 1. Then G has a subgroup of order n. Moreover, any two subgroups of order n are conjugate. In fact, Hall also showed a converse to this. We shall state it here, but its proof is beyond the scope of this book. Theorem (The Hall Converse) Let G be a ﬁnite group with |G| = pr11 . . . prkk with p1 , . . . , pk distinct primes. Suppose that G has a Hall subgroup of order |G|/pri i for each i = 1, . . . , k. Then G is solvable. Thus, the existence of Hall subgroups of every order n dividing G with (n, |G|/n) = 1 is equivalent to solvability. There is an interesting special case of the Hall Converse. Suppose that |G| = pr q s with p and q prime. Then a p-Sylow subgroup of G is a Hall subgroup of order |G|/q s and a q-Sylow subgroup of G is a Hall subgroup of order |G|/pr . The Hall Converse produces an immediate corollary. Corollary (Burnside’s Theorem) Let G be a ﬁnite group whose order has only two prime divisors. Then G is solvable. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 154 In fact, Burnside’s Theorem is one of the ingredients in the proof of the Hall Converse, and, like the converse itself, is beyond the scope of this book. We shall now concentrate on proving Hall’s Theorem. Lemma 5.6.3. Let G be a ﬁnite group and let H be a Hall subgroup of G of order n. Let K G with |K| = m. Then H ∩ K is a Hall subgroup of K of order (m, n) and HK/K is a Hall subgroup of G/K of order n/(m, n). Proof Write |G| = nk. Since H is a Hall subgroup, this gives (n, k) = 1. Write m = n k with n dividing n and k dividing k. Thus, n = (m, n). Of course, |H ∩ K| divides (m, n) = n , so the order of H/(H ∩ K) is divisible by n/(m, n) = n/n . The Noether Isomorphism Theorem gives H/(H ∩ K) ∼ = HK/K, so |H/(H ∩ K)| divides |G/K| = (n/n ) · (k/k ). But also, |H/(H ∩ K)| divides |H| = n. Since n and k are relatively prime, |H/(H ∩ K)| must divide n/n . Thus, |H/(H ∩ K)| = n/(m, n), which forces |H ∩ K| = (m, n). Recall that a “minimal subgroup” of G was deﬁned to be a nontrivial subgroup H ⊂ G such that H has no nontrivial proper subgroups. Thus, H is minimal among the nontrivial subgroups of G. We deﬁne minimal normal subgroups by analogy. Deﬁnition 5.6.4. We say that H is a minimal normal subgroup of G if e = H G and if no nontrivial proper subgroup of H is normal in G. If G is ﬁnite, then the existence of minimal normal subgroups in G is easy. Just choose the nontrivial normal subgroup of the lowest order. Lemma 5.6.5. Let G be a ﬁnite solvable group and let H be a minimal normal subgroup of G. Then H is an abelian group of exponent p for some prime p. Proof The commutator subgroup [H, H] is a characteristic subgroup of H. Since H G, this gives [H, H] G by Corollary 3.7.9. Thus, since H is a minimal normal subgroup of G, either [H, H] = H or [H, H] = e. Suppose that [H, H] = H. Then the derived subgroups H (i) = H for all i > 0. But solvable groups are by deﬁnition those for which H (i) = e for some i. Since H is nontrivial, it thus cannot be solvable, contradicting the fact that a subgroup of a solvable group is solvable (Corollary 5.5.7). Thus, [H, H] = e, and hence H is abelian. Let p be a prime dividing |H| and let Hp be the p-torsion subgroup of G. Then Hp is a characteristic subgroup of H (Lemma 4.4.19), and hence Hp G. Since Hp is nontrivial by Cauchy’s Theorem, we see that H must be an abelian p-group for some prime p. But then the elements of exponent p form a nontrivial characteristic subgroup of H, so H must have exponent p as well. We are now ready to prove Hall’s Theorem. Proof of Theorem 5.6.2 We wish to show that for any divisor n of |G| such that (n, |G|/n) = 1 there are Hall subgroups of G of order n and that any two such subgroups are conjugate. We argue by induction on |G|. The proof breaks up into two parts. First, we assume that G contains a nontrivial normal subgroup, K, whose order is not divisible by k. Write |K| = n k with n dividing n and k dividing k. The induction hypothesis gives a Hall subgroup H/K of G/K of order n/n . Thus, H ⊂ G has order nk . Since CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 155 k < k, the induction hypothesis shows that there’s a Hall subgroup H ⊂ H of order n. This gives the desired subgroup of G. Suppose now that H and H are two subgroups of G of order n. By Lemma 5.6.3, HK/K and H K/K are Hall subgroups of order n/n in G/K, and hence are conjugate by the induction hypothesis. But if x conjugates HK/K to H K/K, then x conjugates HK to H K. But then x conjugates H to a Hall subgroup of H K of order n. And induction shows that xHx−1 is conjugate to H in H K. Thus, we are left with the case where the order of every nontrivial normal subgroup of G is divisible by k. Let P G be a minimal normal subgroup. Then Lemma 5.6.5, shows that P is an abelian group of exponent p for a prime number, p. Since |P | = pr is divisible by k and since (n, k) = 1, we must have k = pr . Thus, P is a p-Sylow subgroup of G. Since P G, there is only one subgroup of G of order k = pr , and hence P is the unique minimal normal subgroup of G. The quotient group G/P is solvable, so it has a minimal normal subgroup Q/P , whose order is q s for some prime number q. Let Q be a q-Sylow subgroup of Q. Then the canonical map π : G → G/P restricts to an isomorphism Q ∼ = Q/P (Corollary 5.3.12), so Q = QP . We claim that NG (Q) is the desired subgroup of G of order n. To show this, it suﬃces to show that [G : NG (Q)] = pr . To see this, ﬁrst note that, since Q is normal in G, every conjugate of Q in G must lie in Q. Thus, every conjugate of Q in G is a q-Sylow subgroup of Q, and is conjugate to Q in Q by the First Sylow Theorem. So Q has the same number of conjugates in Q as in G, and hence [G : NG (Q)] = [Q : NQ (Q)]. We claim now that NQ (Q) = Q. Since [Q : Q] = pr , this will complete the proof that |NG (Q)| = n. Since P is the p-Sylow subgroup of Q and is normal there, it contains every p-torsion element of Q. Since Q ⊂ NQ (Q), the only way the two could diﬀer is if |NQ (Q)| is divisible by p. Thus, it suﬃces to show that NQ (Q) ∩ P = e. Note ﬁrst that if x ∈ P normalizes Q and if y ∈ Q, then xyx−1 y −1 ∈ P ∩ Q = e. Thus, x commutes with every element of Q. Since P is abelian and Q = QP , we see that NQ (Q) ∩ P ⊂ Z(Q), so it suﬃces to show that Z(Q) = e. Recall that Q/P G/P , and hence Q G. Since the center of a group is a characteristic subgroup, we see that Z(Q) G. Suppose that Z(Q) is nontrivial. Since P is the unique minimal normal subgroup of G, this implies that P ⊂ Z(Q). But then P normalizes Q. Since Q = QP , we must have NQ (Q) = Q, so that Q Q. But Q is the q-Sylow subgroup of Q, so this implies that Q is characteristic in Q and hence normal in G, contradicting the assumption that G has no normal subgroups whose order is not divisible by k. Thus, Z(Q) = e, and hence NG (Q) is a Hall subgroup of G of order n, as desired. It now suﬃces to show that if H is another subgroup of G of order n, then H is conjugate to NG (Q). Since |H| = |G/P | is prime to |P |, the canonical map induces an isomorphism H → G/P . Thus, H ∩ Q = Q maps isomorphically to Q/P under the canonical map, and hence is a q-Sylow subgroup of Q. Thus Q and Q are conjugate in Q, hence in G. But then NG (Q) and NG (Q ) are conjugate as well. Now Q = (H ∩ Q) H, so H ⊂ NG (Q ). Since the two groups have the same order, the result follows. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 5.7 156 Nilpotent Groups Nilpotent groups form a class of groups that lie between the solvable and the abelian groups. They form a generalization of the properties found among the ﬁnite p-groups. The fact that a ﬁnite p-group, P , has a nontrivial center is a very useful one, and it demonstrates that ﬁnite p-groups are, in fact, better behaved than the general run of solvable groups. And since P/Z(P ) is again a p-group, the center of P/Z(P ) must be nontrivial as well. Thus, induction arguments are possible. This forms the basic idea that we will generalize to deﬁne nilpotency. In the end, we shall see that the ﬁnite nilpotent groups are nothing more than cartesian products of p-groups. But the abstractions we will develop along the way turn out to be useful for our understanding of the p-groups, if nothing else. Deﬁnition 5.7.1. Let G be a group. Deﬁne the subgroups Z i (G) inductively as follows. Z 0 (G) = e and Z 1 (G) = Z(G). Suppose given Z i−1 (G), which is assumed to be normal in G, and let π : G → G/Z i−1 (G) be the canonical map. Then Z i (G) = π −1 (Z(G/Z i−1 (G))). Of course, as the pre-image of a normal subgroup of G/Z i−1 (G), Z i (G) G, and the induction continues. Deﬁnition 5.7.2. A group G is nilpotent if Z k (G) = G for some k ≥ 0. Just so the abstraction doesn’t get away from us here, we prove the following lemma. Lemma 5.7.3. Finite p-groups are nilpotent. Proof Suppose that Z i−1 (G) = G. Then G/Z i−1 (G) is a nontrivial p-group, and must have a nontrivial center. Thus, Z i (G) is strictly larger than Z i−1 (G). But then if |G| = pj , we must have Z k (G) = G for some k ≤ j. Since the quotient group Z i (G)/Z i−1 (G) is the center of G/Z i−1 (G), and hence abelian, the sequence e = Z 0 (G) · · · Z k (G) = G is an abelian subnormal series when G is nilpotent. Proposition 5.7.4. Nilpotent groups are solvable. The next result gives a useful property of nilpotent groups, which we have not yet veriﬁed for p-groups. (We dealt with a special case in Proposition 4.1.12 by a diﬀerent, but related, argument.) Proposition 5.7.5. Let H be a subgroup of the nilpotent group G, with H = G. Then H cannot be its own normalizer in G. Proof Let H ⊂ G be its own normalizer. We shall show that H must equal G. Since G is nilpotent, it suﬃces to show that Z i (G) ⊂ H for all i. We show this by induction on i, with the case i = 0 immediate. Suppose that Z i−1 (G) ⊂ H, and let π : G → G/Z i−1 (G) be the canonical map. Since Z i−1 (G) ⊂ H, we have H = π −1 (π(H)). This is easily seen to imply that NG (H) = π −1 (NG/Z i−1 (G) (π(H))), so π(H) must be its own normalizer in G/Z i−1 (G). Since Z i (G) = π −1 (Z(G/Z i−1 (G))), it suﬃces to show that Z(G/Z i−1 (G)) ⊂ π(H). In other words, it suﬃces to assume that i = 1, and show that Z(G) ⊂ H. But the elements of Z(G) normalize every subgroup of G, so that Z(G) ⊂ NG (H) = H, and the result follows. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 157 Corollary 5.7.6. Every subgroup of a ﬁnite nilpotent group is part of a subnormal series. Proof Let H be the subgroup and G the group. If H is not already normal in G, then its normalizer at least strictly contains it, so that the index of the normalizer is strictly smaller than the index of H. Thus, the result follows by induction on the index. We can use this in our eﬀorts to classify the low order groups. Corollary 5.7.7. Let G be a nonabelian group of order 16 that does not contain an element of order 8. Then G has an index 2 subgroup H such that H has an element of order 4. Proof If every element of G has order 2, then G is abelian by Problem 3 of Exercises 2.1.16. Thus, G must have an element x of order 4. If x is not normal in G, then we may take H to be its normalizer. Otherwise, take H to be the pre-image of a subgroup of order 2 in G/x. Thus, we shall be able to classify the groups of order 16 without ﬁrst understanding the structure of Aut(Z2 × Z2 × Z2 ). This is useful, because Aut(Z2 × Z2 × Z2 ) turns out to be a simple group of order 168. Lemma 5.7.8. A product H × K is nilpotent if and only if both H and K are nilpotent. Proof This will follow once we show that Z i (H × K) = Z i (H) × Z i (K) for all i ≥ 0. We show this by induction on i, with the case i = 0 immediate. Suppose Z i−1 (H × K) = Z i−1 (H) × Z i−1 (K). Then the product of canonical maps π × π : H × K → H/Z i−1 (H) × K/Z i−1 (K) has kernel Z i−1 (H × K). Thus, it suﬃces to show that Z( H/Z i−1 (H) × K/Z i−1 (K) ) = Z(H/Z i−1 (H)) × Z(K/Z i−1 (K)). But it is easy to see that for any groups H and K , Z(H × K ) = Z(H ) × Z(K ). But now we can characterize ﬁnite nilpotent groups. Proposition 5.7.9. A ﬁnite group is nilpotent if and only if it is the internal direct product of its Sylow subgroups, meaning that if p1 , . . . , pk are the primes dividing |G|, then the pi -Sylow subgroup, Gpi , of G is normal for all i, and G is the internal direct product of Gp1 , . . . , Gpk . Proof If G is isomorphic to Gp1 × · · · × Gpk , then G is nilpotent by Lemmas 5.7.3 and 5.7.8. Thus, we prove the converse. Note that for abelian groups, the converse is given as Proposition 4.4.9. Let G be nilpotent and let Gpi be a pi -Sylow subgroup of G. Then the third Sylow Theorem shows that the normalizer of Gpi is its own normalizer in G. But since G is nilpotent, Proposition 5.7.5 shows that the normalizer of Gpi must be G itself, so that Gpi G. We shall show that this is suﬃcient to obtain the stated product formula. If 1 ≤ i, j ≤ k and if i = j, let x ∈ Gpi and let y ∈ Gpj . Consider the commutator [x, y] = xyx−1 y −1 . Since Gpj is normal in G, xyx−1 ∈ Gpj , and hence the commutator [x, y] is also. But the normality of Gpi shows that yx−1 y −1 and hence also [x, y] are in Gpi as well. But because the identity is the only element that has prime power order for more than one prime, Gpi ∩ Gpj = e, and hence [x, y] = e. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 158 Thus, the elements of Gpi commute with the elements of Gpj whenever i = j, and we have a homomorphism μ : Gp1 × · · · × Gpk → G given by μ(g1 , . . . , gk ) = g1 · · · gk . We claim that μ is an isomorphism. By the second Sylow Theorem, Gp1 × · · · × Gpk has the same order as G, so it suﬃces to show that μ is injective. Suppose that μ(g1 , . . . , gk ) = e. We wish to show that gi = e for all i. Assume by contradiction that i is the smallest index for which gi = e. We cannot have i = k, as then gi = e as well. Thus, gi−1 = gi+1 · · · gk . But the order of (e, . . . , e, gi+1 , · · · , gk ) in Gp1 × · · · × Gpk is the least common multiple of the orders of gi+1 , . . . , gk (by Problem 2 of Exercises 2.10.12), and hence the order of its image, under μ is divisible only by the primes pi+1 , . . . , pk . But the image in question happens to be gi−1 , which is a pi -torsion element, and we have our desired contradiction. We obtain a converse to Lagrange’s Theorem for nilpotent groups. Corollary 5.7.10. Let G be a ﬁnite nilpotent group. Then every divisor of the order of G is the order of a subgroup of G. Proof Write |G| = pr11 . . . prkk where p1 , . . . , pk are distinct primes. Then any divisor of |G| has the form m = ps11 . . . pskk with 0 ≤ si ≤ ri for i = 1, . . . , k. Let Gpi be the pi -Sylow subgroup of G. Then there is a subgroup Hi of Gpi of order psi i by Corollary 5.2.6. By Proposition 5.7.9, G has a subgroup isomorphic to H1 × · · · × H k . Exercises 5.7.11. 1. Give an example of a solvable group that’s not nilpotent. 2. Classify the groups of order 16 containing a copy of D8 . 3. Classify the groups of order 16 containing a copy of Q8 . 4. Classify the groups of order 16 containing a copy of Z4 × Z2 . 5. Which groups appear in more than one of the three preceding problems? Which of them contain an element of order 8? 6. Classify the groups of order 24 whose 2-Sylow subgroups are isomorphic to Z4 ×Z2 . 5.8 Matrix Groups Perhaps the most important missing piece at this point in being able to classify low order groups is an understanding of the automorphism groups of the Sylow subgroups that arise in them. So far, we know how to calculate the automorphisms only of cyclic groups, and groups such as dihedral and quaternionic groups, whose automorphisms are calculated in Exercises 4.7.27 and 5.4.9. Here, we shall investigate the automorphisms of groups of the form Zn × · · · × Zn , with special attention paid to the case where n is prime. There are a number of groups of order < 64 which have Sylow subgroups of this form. Unlike the preceding material, we shall need to make use of some theory we have not yet developed: basic ring theory and linear algebra. The reader who has not yet been exposed to this material should come back to this section after reading the appropriate chapters. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 159 Of course, for any ring A, the cartesian product A × · · · × A of n copies of A is the underlying abelian group of the free A-module of rank n. Consonant with this, we shall write A × · · · × A = An . n times Linear algebra gives techniques of studying the A-module homomorphisms from An to itself. Recall that if N is an A-module, then EndA (N ) denotes the set of A-module homomorphisms from N to itself, while AutA (N ) denotes the group (under the operation of composition of homomorphisms) of A-module isomorphisms from N to itself. Recall also that an A-module homomorphism is an isomorphism if and only if it is bijective. Since every A-module isomorphism is an isomorphism of abelian groups, forgetting the A-module structure gives an injective group homomorphism ϕ : AutA (N ) → Aut(N ). For A = Zn , Corollary 7.7.16 shows that every group homomorphism between Zn modules is a Zn -module homomorphism. Thus, the next lemma is immediate. Lemma 5.8.1. For all k ≥ 1, forgetting the Zn -module structure gives an isomorphism ∼ = ϕ : AutZn (Zkn ) −→ Aut(Zkn ). Recall that the group of invertible n × n matrices over a ring A is denoted by Gln (A). Here, the group structure comes from matrix multiplication. Corollary 7.10.9 shows that allowing matrices to act on the left on column vectors gives a group isomorphism from Gln (A) to AutA (An ).3 Thus, for any A, allowing matrices to act on the left on column vectors gives an injective ring homomorphism ιn : Gln (A) → Aut(An ). For A = Zn , Lemma 5.8.1 allows us to do even better. Corollary 5.8.2. The standard left action of matrices on column vectors induces a group isomorphism ιk : Glk (Zn ) → Aut(Zkn ). In particular, the study of matrix groups will give us calculations of automorphism groups of groups that we care about. Recall that the transpose, M t , of an n × n matrix M = (aij ) is the matrix whose ij-th entry is aji . Lemma 5.8.3. For any commutative ring A, there is a group automorphism α : Gln (A) → Gln (A) deﬁned by α(M ) = (M t )−1 . Proof It is easy to check that for any pair of matrices M1 , M2 ∈ Mn (A), we have (M1 M2 )t = M2t M1t . This is enough to show that the transpose of an invertible matrix is invertible, with (M t )−1 = (M −1 )t . Since inversion of matrices also reverses order, the result follows. 3 Here, if A is not commutative, then this applies only to the standard right module structure on An . CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 160 For ﬁnite rings A, the set of n × n matrices is ﬁnite, and hence so is Gln (A). (If A is inﬁnite, it is not hard to show that Gln (A) is inﬁnite for n ≥ 2.) The next result will assist in calculating the order of Gln (A) when A is ﬁnite. Proposition 5.8.4. Let A be a commutative ring. Consider the standard action of Gln (A) on columns, and let en be the n-th canonical basis vector. Then the isotropy subgroup of en under this action is isomorphic to An−1 ιn−1 Gln−1 (A). Proof The isotropy subgroup, Gln (A)en , of en consists of those invertible matrices whose lastcolumn is en . Thus, we are concerned with the invertible matrices of the form M 0 , where M is (n−1)×(n−1), X is a 1×(n−1) row matrix, 0 is the (n−1)×1 X 1 column matrix whose coordinates are all 0, and the 1 in the lower corner is the 1 × 1 matrix 1. Note that M must lie in Gln−1 (A), as otherwise the ﬁrst n − 1 rows of the full matrix would be linearly dependent. if M is any element of Gln−1 (A) and X Conversely, is any 1 × (n − 1) row matrix, M 0 M −1 0 is invertible, with inverse , and hence lies in the isotropy X 1 −XM −1 1 group Gln (A)en . Thus, if α is the automorphism of Gln (A) replacing a matrix withthe inverseof its M X transpose, then H = α(Gln (A)en ) is the set of all matrices of the form with 0 1 M ∈ Gln−1 (A) and X ∈ An−1 , where we identify An−1 with the set of (n − 1) × 1 column matrices. Of course, Gln (A)en is isomorphic to H, so we may study the latter. Note that the elements of H multiply according to the formula M X M Y MM X + MY = . 0 1 0 1 0 1 M X Thus, there is an isomorphism f : An−1 ιn−1 Gln−1 (A) → H, via f (X, M ) = . 0 1 We can apply this to calculate the order of the general linear group of a ﬁnite ﬁeld. It is shown in Section 11.3 that the order of any ﬁnite ﬁeld must have the form q = pr for p prime, and that for any such q, there is a unique ﬁeld, which we shall denote by Fq , of order q. Of course, Fp = Zp . Corollary 5.8.5. Let Fq be the ﬁeld of order q = pr . Then for n ≥ 1, |Gln (Fq )| = (q n − 1)(q n − q) . . . (q n − q n−1 ). Proof We argue by induction on n. For n = 1, Gl1 (Fq ) ∼ = F× q has order q − 1, and the assertion holds. Suppose the result is true for n − 1. Then the formula for the size of an orbit gives |Gln (Fq )| = |Gln (Fq ) · en | · |Fn−1 ιn−1 Gln−1 (Fq )| q = |Gln (Fq ) · en | · q n−1 · (q n−1 − 1) . . . (q n−1 − q n−2 ) = |Gln (Fq ) · en | · (q n − q) . . . (q n − q n−1 ). Thus, it suﬃces to show that the orbit of en under the action of Gln (Fq ) has q n − 1 elements, and hence that for any nonzero element v ∈ Fnq , there is an M ∈ Gln (Fq ) with M en = v. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 161 But this follows immediately from elementary linear algebra: v is linearly independent over Fq , so there is a basis of Fnq that contains v. Matrix groups over ﬁnite ﬁelds have another important use in the theory of ﬁnite groups. They are a source of ﬁnite simple groups. Recall that if A is a commutative ring, then the determinant induces a surjective group homomorphism det : Gln (A) → A× for n ≥ 1. The special linear group Sln (A) is deﬁned to be the kernel of det. Since a matrix is invertible if and only if its determinant is a unit, we see that Sln (A) consists of all n × n matrices over A of determinant 1. Recall that aIn is the matrix whose diagonal entries are all equal to a and whose oﬀ-diagonal entries are all 0. It’s an easy exercise to show that det(aIn ) = an . Deﬁnition 5.8.6. Let A be a commutative ring and deﬁne H ⊂ Sln (A) by H = {aIn | a ∈ A with an = 1}. Then the projective special linear group is deﬁned as PSln (A) = Sln (A)/H. It turns out that the subgroup H above is the center of Sln (A). Theorem Let Fq be the ﬁeld with q = pr elements. The projective special linear group PSln (Fq ) is simple for all q if n ≥ 3 and is simple for q > 3 if n = 2. The proof is accessible to the techniques of this volume, but we shall not give it. It is somewhat analogous to the proof that An is simple for n ≥ 5. There, the cycle structure gave us a handle on the conjugacy classes. Here, the conjugacy classes are determined by the rational canonical form. Exercises 5.8.7. 1. Give a new proof that Aut(Z22 ) = Gl2 (Z2 ) is isomorphic to S3 . 2. Show that Gl3 (Z2 ) has a subgroup of index 7 that is isomorphic to S4 . 3. What are the normalizers of all the Sylow subgroups of Gl3 (Z2 )? 4. Show that Gl3 (Z2 ) is simple. 5. Complete the classiﬁcation of the groups of order 24. 6. Let q = pr with p prime. Show that one of the p-Sylow subgroups of Gln (Fq ) is the set of matrices with 1’s on the diagonal and 0’s below it (i.e., the upper triangular matrices with 1’s on the diagonal). 7. Show that the p-Sylow subgroups of Gl3 (Zp ) are isomorphic to the group constructed in Corollary 4.6.8 for n = p. 8. Write Tn ⊂ Gln (Fq ) for the subgroup of upper triangular matrices with 1’s on the diagonal. Then Tn acts on Fnq through automorphisms by an action ιn obtained by restricting the action of the full general linear group. Show that Tn is isomorphic to Fn−1 ιn−1 Tn−1 . q 9. Show that D8 embeds in Gl2 (A) whenever A has characteristic = 2. CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 162 10. Show that any element of order 4 in Z24 is part of a Z4 -basis. Calculate the order of Gl2 (Z4 ). 11. Classify the groups of order 48 whose 2-Sylow subgroup is Z4 × Z4 . 12. What are the orders of Sln (Fq ) and PSln (Fq )? 13. Show that the 2-Sylow subgroup of Sl2 (Z3 ) is isomorphic to Q8 and is normal in Sl2 (Z3 ). (Hint: Look at the elements of Sl2 (Z3 ) of trace 0.) 14. Deduce that the 2-Sylow subgroup of Gl2 (Z3 ) is a group of order 16 containing both D8 and Q8 . 15. Classify the groups of order 18. 16. What is the 2-Sylow subgroup of Sl2 (Z5 )? 17. What is the 2-Sylow subgroup of Sl2 (Z4 )? 18. Show that the subgroup H of Deﬁnition 5.8.6 is the center of Sln (A). Chapter 6 Categories in Group Theory Category theory, a.k.a. abstract nonsense, is a branch of mathematics that attempts to identify phenomena that crop up in many diﬀerent types of mathematical structures. One may think of it as the study of those things that happen for reasons which are purely formal. For this reason, the facts that hold for categorical reasons are often not the deepest ones in a given ﬁeld. But categorical patterns repeat in so many contexts that they form a collection of truths that may be instantly recognized and used. What category theory gives us is a language by which to describe patterns common to diﬀerent areas of mathematics. Verifying that these patterns hold in a given context can take some work. For instance, we will give some concepts regarding commonly occurring universal properties. In each category of interest, one must make a construction, often unique to that category, to show that objects satisfying the universal property actually exist. Once that construction has been made, one can revert to formal arguments in using it. Categorical language is well suited to describing the overview of the relationships between objects in a given context. For instance, classiﬁcation results may be expressed in categorical terms. For reasons like this, some of the more important mathematical questions can be phrased in terms of categorical language. Formal study of category theory will not generally be useful in solving them, but categorical thinking can sometimes suggest a viable approach. In the ﬁrst two sections, we give the most general categorical concepts: categories and functors. The next three sections introduce some important kinds of universal properties: initial and terminal objects, products, coproducts, pushouts, and pullbacks. We give constructions realizing these universal properties in various categories of interest, particularly in the categories of sets, groups, and abelian groups. The constructions realizing these universal properties in these three categories can be quite diﬀerent. Thus, the reader can see that a general categorical notion can behave quite diﬀerently in diﬀerent categories. The insight coming from the categorical concept tells us nothing about whether or how that concept might be realized in a particular category. In Section 6.6, we deﬁne the notion of a free object on a set in a given category and construct the free groups and the free abelian groups. Then in Section 6.7, we give the technique of describing a group by generators and relations. This technique can be valuable in studying inﬁnite groups, and arises frequently in topology, as the fundamental group of a CW complex with one vertex is expressed naturally in this form. We then deﬁne direct and inverse limits, giving constructions for them in various categories. Direct limits will be used in the construction of algebraic closures of ﬁelds. 163 CHAPTER 6. CATEGORIES IN GROUP THEORY 164 p. Inverse limits are used to construct the ring of p-adic integers, Z Then, generalizing free functors, we deﬁne adjoint functors, giving examples and showing their utility for deducing general results. We then close with a description of colimits and limits indexed on a partially ordered set or a small category. This material has applications to sheaves and to various of the constructions in algebraic geometry and homotopy theory, and generalizes many of the universal properties described earlier in the chapter. Because of the number of concepts and constructions here that may be new to the reader, Sections 6.2, 6.3, 6.4, 6.5, and 6.8 have been divided into subsections, each with its own set of exercises. 6.1 Categories The notion of a category is very abstract and formal, and is best understood by going back and forth between the deﬁnition and some examples. The point is that the word “object” has to be taken as a black box at ﬁrst, as there are categories with many diﬀerent sorts of objects. The objects in a category are the things you want to study. So the category of sets has sets as objects, the category of groups has groups as objects, etc. The morphisms in a category determine the ways in which the objects are related. For instance, when studying sets, one generally takes the functions between two sets as the morphisms. With this in mind, here is the abstraction. Deﬁnition 6.1.1. A category, C, consists of a collection of objects, ob(C), together with morphisms between them. Speciﬁcally, if X and Y are objects of C (written X, Y ∈ ob(C)) there is a set of morphisms from X to Y , denoted MorC (X, Y ). Morphisms may be composed, just as functions or homomorphisms may be composed: If Z is another object of C, then for any f ∈ MorC (Y, Z) and any g ∈ MorC (X, Y ) there is a composite morphism f ◦ g ∈ MorC (X, Z). This composition rule is required to satisfy two properties. First, there’s an associativity law: if h ∈ MorC (W, X), then for f and g as above, we have f ◦ (g ◦ h) = (f ◦ g) ◦ h as elements of MorC (W, Z). Second, each object has an identity morphism. We write 1X ∈ MorC (X, X) for the identity of X. The deﬁning property of the identity morphisms is that g ◦ 1X = 1Y ◦ g = g for all g ∈ MorC (X, Y ). Example 6.1.2. One of the most ubiquitous categories is the category of sets and functions, which we denote by Sets. The objects of Sets consist of all possible sets. The morphisms between X and Y in Sets are all the functions from X to Y . The identity function is the usual one, and the properties of composition of functions are easily seen to satisfy the requirements for this to be a category. A very large number of the categories we will look at can be thought of as subcategories of the category of sets and functions: Deﬁnition 6.1.3. A category D is a subcategory of C if ob(D) is a subcollection of ob(C), MorD (X, Y ) is a subset of MorC (X, Y ) for all objects X and Y of D, and the laws of composition for the two categories coincide, as do the identity morphisms. It may come as a surprise that not all categories are equivalent (in a formal sense we shall give below, but which includes the intuitive notion of two categories being CHAPTER 6. CATEGORIES IN GROUP THEORY 165 isomorphic) to subcategories of Sets.1 But the structure of Sets does provide a very good intuition for what a category is. Of course, one can deﬁne categories from now until doomsday, and very few of the ones chosen randomly will be mathematically interesting. In particular, if one wants to study something like groups, one should not take the morphisms between two groups to be all functions between their underlying sets. The morphisms in a given category should convey information about the basic structure of the objects being considered. Without that, a study of the category will not shed light on the mathematics one wants to do. Examples 6.1.4. 1. The category in which we’ve been working for the most part here is the category Gp, of groups and homomorphisms. Thus, the objects are all groups and the morphisms between two groups are the group homomorphisms between them. 2. An interesting subcategory of Gp is the category of abelian groups and homomorphisms, which we denote by Ab. Thus, the objects of Ab are all abelian groups, while the morphisms between two abelian groups are the group homomorphisms between them. The inclusion of Ab in Gp is an example of what’s called a full subcategory: Deﬁnition 6.1.5. A subcategory D of C is a full subcategory if for any objects X, Y ∈ ob(D), the inclusion MorD (X, Y ) ⊂ MorC (X, Y ) is bijective. It is sometimes useful to restrict the morphisms in a category, rather than the objects. Thus, we can form a subcategory of Gp whose objects are all groups, but whose morphisms are the injective homomorphisms. Note that in many cases here, the set of morphisms between two objects is the null set. But that doesn’t cause any problems as far as the deﬁnition of a category is concerned. It is sometimes useful to make new categories out of old ones. Deﬁnition 6.1.6. Let C and D be categories. The product category C × D is deﬁned as follows. The objects of C × D are the ordered pairs (X, Y ) with X ∈ ob(C) and Y ∈ ob(D). The morphisms of C × D are given by MorC×D ((X, Y ), (X , Y )) = MorC (X, X ) × MorD (Y, Y ). Here, the latter product is the cartesian product of sets. We write f ×g : (X, Y ) → (X , Y ) for the morphism corresponding to f ∈ MorC (X, X ) and g ∈ MorD (Y, Y ), and deﬁne composition by (f × g) ◦ (f × g ) = (f ◦ f ) × (g ◦ g ) for f × g : (X , Y ) → (X , Y ) in C × D. One of the most important ideas about the relationships between objects in a category is that of isomorphism: Deﬁnition 6.1.7. An isomorphism in a category C is a morphism f : X → Y in C which admits an inverse. Here, an inverse for f is a morphism g : Y → X such that f ◦ g = 1Y and g ◦ f = 1X . 1 One example of a category not equivalent to a subcategory of Sets is the homotopy category encountered in topology. CHAPTER 6. CATEGORIES IN GROUP THEORY 166 Clearly, a morphism in Sets is an isomorphism in Sets if and only if it is a bijection. Also, a morphism in Gp (or Ab) is an isomorphism in Gp (resp. Ab) if and only if it is a group isomorphism in the sense we have been studying. As in the case of ﬁnite group theory, one of the most important problems in studying a given category is being able to classify the objects. Deﬁnitions 6.1.8. The isomorphism class containing a given object consists of the collection of all objects isomorphic to it. Clearly, any two isomorphism classes having an object in common are equal. A classiﬁcation of the objects that have a given property consists of a determination of all the individual isomorphism classes with that property. One example of a classiﬁcation is the determination of a set of representatives of the isomorphism classes with a given property. This means a set S = {Xi | i ∈ I} of objects such that every isomorphism class in C having the given property contains exactly one of the objects in S. Not every collection of isomorphism classes has a set of representatives. For instance, the isomorphism classes of torsion-free abelian groups do not form a set. Even if there is a classiﬁcation of the objects with a given property, it can be diﬃcult to determine which isomorphism class contains a given object. In particular, given two objects in a given category, there may not be any algorithmic procedure to determine whether the objects are isomorphic. For instance, think about the classiﬁcations we’ve seen of the groups in diﬀerent orders. At ﬁrst glance, it can be quite diﬃcult to decide whether two groups are isomorphic. And the classiﬁcations for diﬀerent orders can involve very diﬀerent criteria. The development of criteria to determine whether two objects are isomorphic is one of the most important problems in studying a given category. It should be noted that general category theoretic techniques are rarely useful for solving this problem. However, the notion of functors, to be studied in Section 6.2, can be useful for showing that two objects are not isomorphic. There are partial forms of invertibility which can be useful. Deﬁnitions 6.1.9. A left inverse for a morphism f : X → Y in C is a morphism g : Y → X in C such that g ◦ f = 1X . A left inverse is often called a retraction, and X is often called a retract of Y in C. Similarly, a right inverse for f : X → Y in C is a morphism g : Y → X in C such that f ◦ g = 1Y . A right inverse for f is often called a section of f . We shall also consider categories of monoids. We write Mon for the category whose objects are all monoids and whose morphisms are the homomorphisms of monoids. We write Amon for the full subcategory whose objects are the abelian monoids. Exercises 6.1.10. 1. Show that if f : X → Y is an isomorphism in C, then its inverse is unique. 2. Show that every injection in Sets admits a left inverse, and that every surjection in Sets admits a right inverse. Are these inverses unique? 3. Does every injective group homomorphism have a left inverse in Gp? Does every surjective group homomorphism have a right inverse in Gp? 4. Let g be a left inverse for f : X → Y in C. Suppose that g admits a left inverse also. Show then that f is an isomorphism with inverse g. CHAPTER 6. CATEGORIES IN GROUP THEORY 167 5. Suppose that f : G → K is a morphism in Gp that admits a right inverse. Show then that G is a split extension of ker f by K. 6. Suppose that i : H → G is a morphism in Gp that admits a left inverse. Show that i must be injective. Show that if i(H) G, then G ∼ = H × G/i(H). What can you say if i(H) is not normal in G? 7. Let G be a group and let G-sets be the category whose objects are all G-sets and whose morphisms are all G-maps. Show that the isomorphisms in G-sets are precisely the G-isomorphisms. 8. Let C be any category and let X ∈ ob(C). Show that MorC (X, X) is a monoid under composition of morphisms. We call it the monoid of endomorphisms of X, and write MorC (X, X) = EndC (X). Show that this monoid is a group if and only if every morphism from X to itself is an isomorphism in C. 9. Let M be a monoid. Show that there is a category which has one object, O, such that the monoid of endomorphisms of O is M . Thus, monoids are essentially equivalent to categories with one object. 6.2 Functors Just as we can learn about a group by studying the homomorphisms out of it, we can learn about a category by studying its relationship to other categories. But to get information back, we shall need to deﬁne ways to pass from one category to another that preserve structure. We study covariant functors here, and study contravariant functors in a subsection below. Deﬁnition 6.2.1. Let C and D be categories. A functor, F : C → D assigns to each object X ∈ ob(C) an object F (X) ∈ ob(D) and assigns to each morphism f ∈ MorC (X, Y ) a morphism F (f ) ∈ MorD (F (X), F (Y )), such that F (1X ) = 1F (X) for all X ∈ ob(C), and F (f ) ◦ F (g) = F (f ◦ g) for each pair of morphisms f and g in C for which the composite f ◦ g is deﬁned. Functors are often used in mathematics to detect when objects of a given category are not isomorphic. Lemma 6.2.2. Let F : C → D be a functor and suppose that f : X → Y is an isomorphism in C. Then F (f ) : F (X) → F (Y ) is an isomorphism in D. Thus, if Z and W are objects of C, and if F (Z) and F (W ) are not isomorphic in D, then Z and W must not be isomorphic in C. Proof Let g be the inverse of f . Then F (g) ◦ F (f ) = F (g ◦ f ) = F (1X ) = 1F (X) . Similarly, F (f ) ◦ F (g) = 1F (Y ) . Thus, F (g) is an inverse for F (f ) in D. CHAPTER 6. CATEGORIES IN GROUP THEORY 168 Indeed, in a lot of ways, the preceding lemma is the main reason for looking at functors, to the point where the deﬁnition of functor is tailored to make it true. The utility of this approach lies in the fact that we can often ﬁnd functors to categories that are much more manageable than the one we started in. Examples 6.2.3. 1. Let p be a prime. Then for an abelian group G, the set Gp of p-torsion elements of G is a subgroup of G. As shown in Lemma 4.4.19, if f : G → H is a homomorphism with H abelian, then f restricts to a homomorphism fp : Gp → Hp . Clearly, gp ◦ fp = (g ◦ f )p for any homomorphism g : H → K with K abelian, and (1G )p = 1(Gp ) , so there is a functor Fp : Ab → Ab given by setting Fp (G) = Gp for all abelian groups G, and setting Fp (f ) = fp for f : G → H a homomorphism of abelian groups. Notice that while Fp (f ) may be an isomorphism even if f is not, Corollary 4.4.21 shows that if f : G → H is a homomorphism between ﬁnite abelian groups, then f is an isomorphism if and only if Fp (f ) is an isomorphism for each prime p dividing either |G| or |H|. Thus, on the full subcategory of Ab whose objects are the ﬁnite abelian groups, the collection of functors Fp , for p prime, together detect whether a morphism is an isomorphism. 2. Let C be any category and let X be an object of C. We deﬁne a functor F : C → Sets as follows. For an object Y of C, we set F (Y ) = MorC (X, Y ). If f : Y → Z is a morphism from Y to Z in C, we deﬁne F (f ) : MorC (X, Y ) → MorC (X, Z) by setting F (f )(g) = f ◦ g for all g ∈ MorC (X, Y ). (It is often customary to write f∗ instead of F (f ).) The reader may easily check that F is a functor. 3. Deﬁne φ : Gp → Sets by setting φ(G) = G, the underlying set of G. For a homomorphism f : G → H, we set φ(f ) = f . We call φ the forgetful functor that forgets the group structure on G. More generally, any functor that simply forgets part of the structure of a given object is called a forgetful functor. Thus, there is also a forgetful functor from Gp to Mon. Deﬁnition 6.2.4. A functor C × C → D is often called a bifunctor on C, and may be thought of as a functor of two variables. Example 6.2.5. There is a bifunctor F : Sets × Sets → Sets obtained by setting F (X, Y ) = X × Y and setting F (f × g) = f × g, for f ∈ MorSets (X, X ) and g ∈ MorSets (Y, Y ). Here, the right-hand function f ×g is the usual mapping f ×g : X ×Y → X × Y given by (f × g)(x, y) = (f (x), g(y)). Exercises 6.2.6. 1. Show that functors preserve left inverses and right inverses. 2. Show that there is a functor F : Gp → Gp such that F (G) is the commutator subgroup of G for every group G. Here, if f : G → G is a group homomorphism, then F (f ) is obtained by restricting f to the commutator subgroup. 3. Give an example of groups H and G such that H is a retract of G, but Z(H) is not a retract of Z(G). Deduce that there is no functor on Gp that takes each group to its center. CHAPTER 6. CATEGORIES IN GROUP THEORY 169 4. Show that there is a functor F : Gp → Ab such that F (G) = Gab for all groups G, and if f : G → G is a group homomorphism, then F (f ) = f ab , the unique homomorphism such that the following diagram commutes. G πG Gab f f ab / G πG / ab G Here, for any group H, we write πH for the canonical map from H to H ab = H/[H, H]. 5. Show that the inclusion of a subcategory is a functor. In particular the inclusion of a category in itself is called the identity functor. 6. Let φ : Gp → Sets be the forgetful functor. Suppose that φ(f ) is an isomorphism. Is f then an isomorphism? Alternatively, suppose that φ(G) and φ(G ) are isomorphic in Sets. Are G and G then isomorphic in Gp? 7. Show that passage from a monoid M to its group of invertible elements Inv(M ) gives a functor from Mon to Gp (or from Amon to Ab). 8. Let G be a group and let G be the category with one object, O, such that the monoid of endomorphisms of O is G. Show that there is a one-to-one correspondence between G-sets and the functors from G into Sets. 9. Let M and M be monoids and let M and M be categories with one object whose monoids of endomorphisms are M and M , respectively. Show that there is a oneto-one correspondence between the monoid homomorphisms from M to M and the functors from M to M . 10. Let C and D be categories. (a) Show that there are functors πC : C × D → C and πD : C × D → D obtained by setting πC (X, Y ) = X, πD (X, Y ) = Y , πC (f × g) = f and πD (f × g) = g. (b) For an object Y of D, show that there is a functor ιY : C → C × D obtained by setting ιY (X) = X × Y and ιY (f ) = f × 1Y . Contravariant Functors Now we come to the fact that there’s more than one kind of functor. The functors we’ve considered so far are called covariant functors. The other kind of functors are called contravariant. They behave very much like covariant ones, except that they reverse the direction of morphisms. Deﬁnition 6.2.7. A contravariant functor F : C → D associates to each X ∈ ob(C) an object F (X) ∈ ob(D) and to each morphism f ∈ MorC (X, Y ) a morphism F (f ) ∈ MorD (F (Y ), F (X)), such that F (1X ) = 1F (X) CHAPTER 6. CATEGORIES IN GROUP THEORY 170 for all X ∈ ob(C), and F (f ) ◦ F (g) = F (g ◦ f ) for each pair of morphisms f and g in C for which the composite g ◦ f is deﬁned. Contravariant functors may be viewed as covariant functors out of a diﬀerent category. Deﬁnition 6.2.8. Let C be a category. The opposite category, C op , is deﬁned by setting ob(C op ) = ob(C) and setting MorC op (X, Y ) = MorC (Y, X) for all X, Y ∈ ob(C). Here, if f : X → Y is a morphism in C, we write f op : Y → X for the associated morphism in C op . We deﬁne composition in C op by setting f op ◦ g op = op op (g ◦ f ) , and set the identity morphism of X ∈ ob(C op ) equal to (1X ) . The point here is that any contravariant functor on C may be viewed as a covariant functor on C op . The reader may supply the details: Proposition 6.2.9. For any category C, there is a contravariant functor I : C → C op deﬁned by setting I(X) = X for X ∈ ob(C) and setting I(f ) = f op for f ∈ MorC (X, Y ). If F : C → D is a contravariant functor, then there is a covariant functor F op : op C → D deﬁned by setting F op (X) = F (X) for X ∈ ob(C) = ob(C op ) and setting F op (f op ) = F (f ) for f ∈ MorC (X, Y ). Thus, F itself factors as the composite I F op C− → C op −−→ D. Exercises 6.2.10. 1. Let C be any category and let Y be an object of C. For X ∈ ob(C), let F (X) = MorC (X, Y ). For f ∈ MorC (X, X ), deﬁne F (f ) : MorC (X , Y ) → MorC (X, Y ) by F (f )(g) = g ◦ f . (It is often customary to write f ∗ instead of F (f ).) Show that F deﬁnes a contravariant functor from C to Sets. 2. Let C be a category. Show that there is a functor Hom : C op × C → Sets, given by setting Hom(X, Y ) = MorC (X, Y ) for each pair of objects X, Y of C. Here, if f : X → X and g : Y → Y are morphisms in C, we deﬁne Hom(f op , g) : MorC (X, Y ) → MorC (X , Y ) by Hom(f op , g)(h) = g ◦ h ◦ f for all h ∈ MorC (X, Y ). 3. Show that a contravariant functor takes isomorphisms to isomorphisms. 4. Show that a contravariant functor takes morphisms that admit left inverses to morphisms that admit right inverses and vice versa. 5. Let G be a group and let G be the category with one object, O, such that the monoid of endomorphisms of O is G. Show that there is a one-to-one correspondence between right G-sets and the contravariant functors from G into Sets. CHAPTER 6. CATEGORIES IN GROUP THEORY 6.3 171 Universal Mapping Properties: Products and Coproducts An object, or the relationship between a family of objects connected by morphisms, is sometimes uniquely determined by what morphisms exist either in or out of the object or family. This phenomenon is known as a universal mapping property, and is perhaps best understood through examples. We treat initial and terminal objects here, and then give subsections for products, coproducts, and free products. The simplest examples of universal mapping properties are initial and terminal objects for a category: Deﬁnitions 6.3.1. Suppose that X ∈ ob(C) has the property that MorC (X, Y ) has exactly one element for each Y ∈ ob(C). Then X is called an initial object for C. Alternatively, if MorC (Y, X) has exactly one element for each Y ∈ ob(C), then X is called a terminal object for C. If X is both an initial object and a terminal object for C, we call X a zero object for C. One example that may not be immediately obvious is that the null set is an initial object for Sets. The point here is that a function is determined by where it sends the elements of its domain. Since the null set has no elements, there is exactly one way to specify where its elements go. Exercises 6.3.2. 1. Show that any two initial objects for C must be isomorphic. 2. Show that any two terminal objects for C must be isomorphic. 3. Show that a set with one element is a terminal object for Sets. 4. Show that the trivial group e is a zero object in both Gp and in Ab. 5. Let G be a group. Show that the G-set with one element is a terminal object in G-sets. 6. Let G be any nontrivial group. Show that G-sets does not have an initial object. Products Other universal mapping properties express relationships between more than one object. These properties often recur in many diﬀerent categories, and are given names appropriate to that fact. The one we’ve seen the most of is products: Deﬁnition 6.3.3. Let X, Y ∈ ob(C). A product for X and Y in C consists of an object, Z ∈ ob(C), together with two morphisms, πX : Z → X and πY : Z → Y , called projections. The universal property that makes these a product is as follows: For any object W ∈ ob(C) and any two morphisms f : W → X and g : W → Y , there is a unique CHAPTER 6. CATEGORIES IN GROUP THEORY 172 morphism (f, g) : W → Z such that the following diagram commutes: W/ ?OO // ??OOOO // ??? h OOgOO OOO // ?? OO' // Z πY / Y f // // // πX / X, Here, h = (f, g). Thus, the projection maps are part of the deﬁnition of a product, and not just the object Z itself. Examples 6.3.4. 1. The cartesian product, together with the usual projections, is the product in Sets: a function into the product is determined by its component functions, which are precisely the composites of the original function with the projection maps. Indeed, a complete argument showing that the cartesian product is the product in the category of sets is given in Proposition 1.6.5. 2. In Gp, the direct product has been shown in Lemma 2.10.5 to be the categorical product, and this remains true in Ab. In Gp and Ab, the product has some extra features (like the inclusion maps) that do not occur in Sets or more general categories. The universal mapping property that deﬁnes the product has at most one solution (up to isomorphism) for a given pair of objects X and Y . This is because of the uniqueness of the map from W to Z that makes the diagram commute in the deﬁnition of the product: Proposition 6.3.5. Suppose given objects and morphisms π π X Y X ←− − Z −−→ Y π π X Y − Z −−→ Y and X ←− in C such that both (Z, πX , πY ) and (Z , πX , πY ) satisfy the universal mapping property for the product of X and Y in C. Let f = (πX , πY ) be the unique morphism from Z to Z which makes the following diagram commute. (Here, we use the universal mapping π π X Y property of X ←− Y .) − Z −−→ Z / ?OO // ??OOOO // ??? fOOOπOY OOO // ?? OOO // '/ / Y Z πX / πY // // // πX X Then f is an isomorphism. CHAPTER 6. CATEGORIES IN GROUP THEORY π 173 π X Y Proof By the universal property of X ←− − Z −−→ Y , there is a unique morphism g : Z → Z such that πX ◦ g = πX and πY ◦ g = πY . Now g◦f : Z → Z is the unique morphism such that πX ◦g◦f = πX and πY ◦g◦f = πY . Thus, g ◦ f = 1Z . Similarly, f ◦ g = 1Z , and the result follows. Exercises 6.3.6. 1. What is the product in G-sets? Is the product of two orbits an orbit? What are its isotropy groups? 2. Suppose that for each pair (X, Y ) of objects of C we are given a product for X and Y . Show that passage from a pair of objects to their product gives a functor from C × C to C. Coproducts It is frequent in categorical mathematics to ask what happens to a deﬁnition if you turn all the arrows around. This is called dualization. The dual of a product is called a coproduct. Deﬁnition 6.3.7. Let X, Y ∈ ob(C). A coproduct for X and Y in C is an object Z together with morphisms ιX : X → Z and ιY : Y → Z, called the natural or canonical inclusions, with the following universal property: For any pair of morphisms f : X → W and g : Y → W , there is a unique morphism f, g : Z → W such that the following diagram commutes: Y/ // // ιY // /g ιX / X OO Z ? /// OOO OOO ??h? // OOO ?? // OOO?? / f O' W, Here, h = f, g. Examples 6.3.8. 1. Proposition 2.10.6, when applied strictly to abelian groups and their homomorphisms, shows that the direct product, together with the canonical inclusions, ι ι 1 2 G × H ←− H, G −→ is the coproduct in the category Ab of abelian groups. 2. The coproduct in Sets is called the disjoint union. Here, the disjoint union of X and Y is denoted X Y . Its elements are the union of those in X and those in Y , where the elements from X are not allowed to coincide with any of the elements of Y . Formally, this may be accomplished as follows. Set theory tells us that there’s some set, S, containing both X and Y . The disjoint union of X and Y is deﬁned to be the subset of S × {1, 2} consisting of the elements (x, 1) with x ∈ X and the elements (y, 2) with y ∈ Y . We then have ιX (x) = (x, 1) and ιY (y) = (y, 2). CHAPTER 6. CATEGORIES IN GROUP THEORY 174 Now if f : X → W and g : Y → W are functions, we deﬁne f, g : X Y → W by setting f, g(x, 1) = f (x) and setting f, g(y, 2) = g(y). Since ιX and ιY are disjoint and have X Y as their union, this is well deﬁned and satisﬁes the universal property for the coproduct. Thus, unlike the case of the product, the forgetful functor from Ab to Sets does not preserve coproducts. As we shall see presently, the forgetful functors Ab → Gp and Gp → Sets also fail to preserve coproducts. Indeed, the coproducts in these three categories are given by completely distinct constructions. Exercises 6.3.9. 1. What is the coproduct in G-sets? 2. Show that the direct product is not the coproduct in Gp, even if the two groups are abelian. 3. Prove the analogue of Proposition 6.3.5 for coproducts, showing that any two coproducts for a pair of objects are isomorphic via a morphism that makes the appropriate diagram commute. Free Products The coproduct in Gp is called the free product. We shall give the construction now. Let H and K be groups, and write X = H K for the disjoint union of the sets H and K. By a word in X we mean a formal product x1 x2 · · · xk , with xi ∈ X for all i. (We write for the formal product operation to distinguish it from the multiplications in H and K.) The empty word is the word with no elements. The elements in the free product H ∗ K will not be the words themselves, but rather equivalence classes of words. The equivalence relation we shall use is derived from a process called reduction of words. Words may be reduced as follows. First, if xi and xi+1 are both in the image of ιH : H → X, then x1 · · · xk reduces to x1 · · · (xi xi+1 ) · · · xk , where xi xi+1 is the product of xi and xi+1 in H. There is a similar reduction if xi and xi+1 are both in the image of ιK : K → X. Finally, if xi is the identity element of either H or K, then x1 · · ·xk reduces to x1 · · ·xi−1 xi+1 · · ·xk . In particular, for either e ∈ H or e ∈ K, then the singleton word e reduces to the empty word. e We shall refer to the reductions above as elementary reductions, and write w w or e w w if the word w is obtained from the word w by one of these elementary reductions. e e e w1 ... wn of elementary reductions, we say More generally, given a sequence w0 that w0 reduces to wn , and write w0 wn (or wn w0 ). We shall also declare that w w, so that is reﬂexive. The equivalence relation we shall use is the equivalence relation generated by the process of reduction, in the sense of Section 1.5. This means that the words w and w are equivalent, written w ∼ w , if there is a sequence w = w0 , . . . , wn = w of words such that for i = 0, . . . , n − 1, either wi wi+1 or wi wi+1 . Note that some words (e.g., h h eK k, where h, h ∈ H, k ∈ K, and eK is the identity of K) can be reduced in more than one way. (Also, as in the case of words containing an adjacent pair such as eK k above, two diﬀerent elementary reduction operations can produce exactly the same word as a result. However, both cases are e represented by the same symbols, w w .) An examination of words which admit more than one reduction shows the following important fact. CHAPTER 6. CATEGORIES IN GROUP THEORY 175 Lemma 6.3.10. Suppose given two diﬀerent elementary reduction operations on a word w, giving w e w1 and w e w2 . Then either w1 = w2 or there is a word w3 with e e w1 w3 w2 . It is convenient to be able to represent an element of the free product by what’s known as a reduced word: Deﬁnition 6.3.11. We say that a word x1 · · · xk is reduced if xi = e for 1 ≤ i ≤ k and if no adjacent pair xi , xi+1 comes from the same group (H or K). We also declare the empty word to be reduced. The next lemma is fundamental for the understanding of free products. Lemma 6.3.12. Let w be a word in X = H K. Then there is a unique reduced word, w such that w w . Proof We argue by induction on the number of letters in w. If w is reduced, then it cannot be reduced further, and we take w = w. Otherwise, there is an elementary reduction from w to a word v. By induction, there is a unique reduced word w with v w . But then w w . For the uniqueness e e e w1 ... w . statement, suppose that w is reduced and that w w , say, via w If w1 = v, then w = w by the induction hypothesis. Otherwise, by Lemma 6.3.10, there e e is a word w ) with v w ) w1 . But if w ) w with w reduced, then the uniqueness statement of the induction hypothesis gives w = w = w . Recall that ∼ is the equivalence relation generated by reduction. Proposition 6.3.13. Every equivalence class under ∼ contains exactly one reduced word. If w is a reduced word, then every word in the equivalence class of w under ∼ reduces to w. Proof Suppose that w ∼ w . Then there is a sequence w = w0 , . . . , wn = w such that for each i = 0, . . . , n − 1, either wi wi+1 or wi wi+1 . In either case, Lemma 6.3.12 tells us that the unique reduced word w to which wi reduces is equal to the unique reduced word to which wi+1 reduces. In particular, each wi reduces to w , and there can be at most one reduced word in such a sequence. We deﬁne the elements of the free product H ∗ K to be the equivalence classes of words under ∼. Thus, the elements of H ∗ K are in one-to-one correspondence with the reduced words. We deﬁne a binary operation on the words themselves by concatenation: (x1 · · · xk ) · (y1 · · · yl ) = x1 · · · xk y1 · · · yl . In word notation, we write w · w for the product in this sense of the words w and w . But clearly, if w reduces to w1 and w reduces to w1 , then w · w reduces to w1 · w1 . Thus, concatenation induces a well deﬁned binary operation on the equivalence classes of words. Lemma 6.3.14. Concatenation of words induces a group structure on the free product H ∗ K. Proof Concatenation is clearly an associative operation on the set of words themselves, with the empty word as an identity element. Since it preserves equivalence classes, the free product inherits a monoid structure from the set of words themselves. But the −1 inverse of x1 · · · xk in H ∗ K is given by x−1 k · · · x1 . CHAPTER 6. CATEGORIES IN GROUP THEORY 176 The inclusions of H and K in X induce group homomorphisms from H and K to H ∗ K, which we’ll denote by ιH and ιK . Proposition 6.3.15. The free product H ∗ K, together with the homomorphisms ιH and ιK is the coproduct of H and K in the category of groups. Proof Given group homomorphisms f : H → G and g : K → G, we deﬁne f, g on X to be f on H ⊂ X and to be g on K ⊂ X. Given a word x1 · · · xk in X, we may then deﬁne f, g(x1 · · · xk ) to be (f, g(x1 )) . . . (f, g(xk )). Stipulating that f, g carries the empty word to the identity element of G, we obtain a monoid homomorphism from the monoid of words in X into G. Since two words diﬀering by an elementary reduction are carried to the same element of G (since f and g are homomorphisms), passage to equivalence classes of words gives a group homomorphism from H ∗ K into G. Clearly, f, g ◦ ιH = f and f, g ◦ ιK = g. Since the images of ιH and ιK generate H ∗ K, f, g is the unique homomorphism with this property. Free products can be somewhat intractable, especially in comparison to the groups we have considered previously. But they are important in understanding groups in general. For instance, ﬁnitely generated free groups, which we shall study in Section 6.6, are iterated free products of copies of the integers. We shall see that any ﬁnitely generated group is a factor group of such a free group, and this opens up a new way of looking at groups, called generators and relations, which we shall study in Section 6.7. A concrete example of free products “in nature” is PSl2 (Z), which turns out to be isomorphic to Z2 ∗ Z3 . It should be noted that not all categories have products, coproducts, or other of the constructions we shall give. Exercises 6.3.16. 1. Show that if H and K are both nontrivial, then H ∗ K is inﬁnite. 2. Show that if H and K are both nontrivial, then H ∗ K is nonabelian even if H and K are both abelian. 3. Show that the homomorphisms ιH : H → H ∗ K and ιK : K → H ∗ K are injective. 4. What is H ∗ e? 5. Show that the same constructions that give the product and coproduct in Gp give a product and coproduct in Mon. 6. Show that the same constructions that give the product and coproduct in Ab give a product and coproduct in Amon. 6.4 Pushouts and Pullbacks We now give a generalization of products and coproducts. These are universal mapping properties not just of pairs of objects, but of particular collections of objects and morphisms. We treat pullbacks now, and treat pushouts in a separate subsection. In further subsections, we will construct pushouts in Gp, Ab, and Sets. CHAPTER 6. CATEGORIES IN GROUP THEORY 177 Deﬁnition 6.4.1. Suppose given a pair of morphisms into Z, f : X → Z and g : Y → Z. This gives us the following diagram: X . f g Y / Z A pullback for this diagram consists of the following data: We have an object W and morphisms g : W → X and f : W → Y such that 1. W f g / X f / Z Y g W g / X commutes. 2. For every commutative diagram f Y f g / Z there is a unique morphism h : W → W such that the following diagram commutes. W / ?OO // ?? OOOO // ??? OOOgO OOO // ??h? OOO // '/ / W X f / g // // f // f g / Z Y In other words, the pullback is a universal completion of the original 3-object diagram to a commutative square. When the context is totally unambiguous, one often refers to the object W as the pullback, but the proper usage is that “pullback” refers to the whole square. Exercises 6.4.2. 1. Show that pullbacks in Sets, Gp, and Ab may all be constructed as follows. Given a diagram X f Y g / Z , CHAPTER 6. CATEGORIES IN GROUP THEORY 178 let Y ×Z X ⊂ Y × X be given by Y ×Z X = {(y, x) | g(y) = f (x)}. Then the pullback of the original diagram is given by Y ×Z X f g Y g / X f , / Z where f (y, x) = y and g(y, x) = x. The construction Y ×Z X is sometimes called the ﬁbre product of Y and X over Z. 2. Show that pullbacks, when they exist, are unique. 3. Show that if C has a terminal object, ∗, then the pullback of the unique diagram X / ∗ Y is the product of X and Y (i.e., show that the pullback of this diagram satisﬁes the same universal property as the product). Thus, pullbacks do generalize products. 4. Suppose given a pullback diagram W f g / X f Y g / Z in either Gp or Ab. Show that ker f ∼ = ker f and that ker g ∼ = ker g. 5. In either Gp or Ab, show that the pullback of X f e / Y is ker f . Pushouts Pushout is the notion dual to pullback. In Sets and Gp, pushouts are more delicate and harder to construct than pullbacks, but they can be very important. CHAPTER 6. CATEGORIES IN GROUP THEORY 179 Deﬁnition 6.4.3. The pushout of a diagram X f / Y f / Y g Z is a commutative diagram X g g Z f / W which is universal in the sense that if f X / Y g g Z f / W is another commutative diagram, then there is a unique morphism h : W → W such that the following diagram commutes. / Y // // g g // // f /g / Z OO W ? // OOO OOO ??h? /// OOO ?? / OOO ??// f OO' W X f As in the case of pullbacks, it is customary to refer to the object W as the pushout as well. Exercises 6.4.4. 1. Show that pushouts, when they exist, are unique. 2. Show that if C has an initial object, ∗, then the pushout of the unique diagram ∗ Z is the coproduct of Y and Z. / Y CHAPTER 6. CATEGORIES IN GROUP THEORY 180 3. Suppose given a pushout diagram f X / Y g g Z f / W in C, where f is an isomorphism. Deduce that f is also an isomorphism. 4. Show, using the universal mapping property, that in Sets, Gp, or Ab, the pushout of a product diagram πY / X ×Y Y πX X is the terminal object. (In Sets, we must assume that neither X nor Y is the null set.) Pushouts in Gp and Ab Pushouts are closely related to coproducts, and are often constructed using them. We ﬁrst give the construction of pushouts in Ab. Proposition 6.4.5. The pushout in Ab of a diagram f / A B g C is the diagram f A g C / B ιB ιC / (C × B)/D, where D = im(g, −f ) = {(g(a), −f (a)) | a ∈ A} ⊂ C × B, and where the maps ι are the composites of the standard inclusions of B and C in the product with the canonical map from C × B to its factor group. Proof First, note that factoring out by D makes the square commute. Now given a commutative square f / A B g g C f / G CHAPTER 6. CATEGORIES IN GROUP THEORY 181 in Ab, there is a unique homomorphism h from the coproduct C × B into G such that h ◦ ιB = g and h ◦ ιC = f . But then h((g(a), −f (a)) = f g(a) − g f (a) = 0 by commutativity of the diagram. Thus, h factors through the factor group of C × B by D. But factorizations of homomorphisms through factor groups are unique. In order to understand pushouts in the category of groups, we will need the following concept. Deﬁnition 6.4.6. Let G be a group and let S be any subset of G. We write S ⊂ G for the subgroup generated by T = {gxg −1 | g ∈ G, x ∈ S}, the set of all conjugates of the elements of S. We call S the normal subgroup generated by S. As usual, if S = {g1 , . . . , gn }, we write S = g1 , . . . , gn . From a diﬀerent viewpoint, if S is contained in the subgroup H G and if the subgroup S = T above equals H, we say that H is normally generated by S in G. Lemma 6.4.7. Let S be a subset of the group G. Then S is the smallest normal subgroup of G that contains S. Proof Clearly, any normal subgroup containing S must contain T , and hence must contain the subgroup generated by T . Thus, it suﬃces to show that S is normal in G. A generic element of S is a product of powers of elements of T . Since any conjugate of an element of T is also in T , the result follows. The pushout in Gp is called the free product with amalgamation. The construction proceeds precisely as in the abelian case. We give it as an exercise. We shall see below that every group may be presented in terms of “generators and relations,” which is an example of a free product with amalgamation. Exercises 6.4.8. 1. Show that the pushout in Gp of the diagram f G / H g K is the diagram G g K f / H ιH ιK / (K ∗ H)/S, where S = {g(x) · f (x)−1 | x ∈ G}, and where the maps ι are the composites of the standard inclusions of H and K in the free product with the canonical map from K ∗ H to its factor group. CHAPTER 6. CATEGORIES IN GROUP THEORY 182 2. What is the pushout in Gp of a diagram of the following form? f / H G e What would the pushout be in Ab if the diagram were also in Ab? 3. Suppose given a pushout diagram A f g C / B g f / D in Ab. (a) If f is injective, show that f is also. (b) Show that f is surjective if and only if f is. 4. Suppose the diagram of the previous problem is a pushout square in Gp rather than Ab. (a) If f is surjective, show that f is also. (b) Show that f may be surjective when f is not. Pushouts in Sets In Sets, the pushout is again a quotient of the coproduct. This time we proceed by placing an equivalence relation on the coproduct and passing to the set of equivalence classes. Thus, suppose given a diagram f / X Y g Z in Sets. Our equivalence relation on the coproduct Y Z will be the equivalence relation, ∼, generated by the following notion of basic equivalence: There is a basic equivalence between ιY (y) with ιZ (z) in Y Z if there is an element x ∈ X with f (x) = y and g(x) = z. In this case, we write either ιY (y) ιZ (z) or ιZ (z) ιY (y), as the relation in question is symmetric. Since ∼ is the equivalence relation generated by , we see that two elements v, v ∈ Y Z are equivalent under ∼ if and only if there is a sequence v = v0 v1 . . . vn = v . We deﬁne the pushout object Pto be the set of equivalence classes in Y Z under ∼. We then have a function π : Y Z → P that takes each element to its equivalence class. CHAPTER 6. CATEGORIES IN GROUP THEORY 183 Exercises 6.4.9. 1. Show that the pushout in sets of the above diagram is f / X Y g Z ιY ιZ / P, where the functions ι are the composites of the standard inclusions into the coproduct with the quotient function π. 2. If f is injective, show that ιZ is also. (In this case, P is often denoted Z ∪g Y .) 3. If f is surjective, show that ιZ is also. 6.5 Inﬁnite Products and Coproducts We’ve discussed products and coproducts of pairs of objects in a category. This is suﬃcient, by induction, to provide products or coproducts for any ﬁnite collection of objects. But in some cases, as we shall see in our discussion of rings and modules, it is necessary to make use of products or coproducts of inﬁnite families of elements. We deﬁne inﬁnite products and coproducts here and construct the products. We give constructions of inﬁnite coproducts in a separate subsection. We shall consider sets of objects {Xi | i ∈ I}. In other words, I is a set, called the indexing set, and Xi is an object of our category for each i ∈ I. Deﬁnitions 6.5.1. The product of a set {Xi | i ∈ I} of objects of C is an object X, together with morphisms πi : X → Xi , called projections, with the following universal property. If fi : Y → Xi is a morphism in C for each i ∈ I, then there is a unique morphism f : Y → X such that πi ◦ f = fi for all i ∈ I. The coproduct of a set {Xi | i ∈ I} of objects of C is an object X, together with morphisms ιi : Xi → X for each i ∈ I, called the natural inclusions, with the following universal property. If fi : Xi → Y is a morphism in C for each i ∈ I, then there is a unique morphism f : X → Y such that f ◦ ιi = fi for all i ∈ I. Once again, the products in Sets, Gp, and Ab come from the same construction. We shall see in our discussion of adjoint functors in Section 6.9 that this is no coincidence. ! The construction of products in Sets is as follows. The elements of the product i∈I Xi are collections (xi | i ∈ I) such that xi ∈ Xi for all i ∈ I. In other words, the element (xi | i ∈ I) consists of a choice, for each i ∈ I, of an element of Xi . We call (xi | i ∈ I) a tuple or an I-tuple. Another way of thinking about the elements of a product is as follows. Let S be a set that contains each Xi . (The ! existence of such a set is one of the standard axioms of set theory.) Then an I-tuple in i∈I Xi is a function from I to S whose value on i ∈ I must lie in Xi . From this viewpoint, we ! can see that the collection of all such I-tuples, ! i∈I Xi , is a set. The projection πi : i∈I Xi → Xi takes (xi | i ∈ I) to xi . The act of choosing elements xi ∈ Xi for each i ∈ I suggests that the Axiom of Choice is involved here. This is indeed the case. Axiom of Choice (second form) Let {Xi | i ∈ I} be a family of nonempty sets. Then ! the product i∈I Xi is nonempty. CHAPTER 6. CATEGORIES IN GROUP THEORY 184 We shall discuss the relationship between the two given forms of the Axiom of Choice in Exercises 6.5.11. The following argument does not make ! use of the Axiom of Choice. The output, if one does not accept that axiom, is that if i∈I Xi = ∅, then ∅ is the product of {Xi | i ∈ I} in Sets. In particular, any inﬁnite family of sets has a product in Sets. ! Proposition 6.5.2. The product i∈I Xi , together with its projection maps, is the product of {Xi | i ∈ I} in Sets. Proof Suppose given a collection of functions fi : Y → Xi for each i ∈ I. Deﬁne ! f : Y → i∈I Xi by f (y) = (fi (y) | i ∈ I). Then πi ◦ f = fi for all i ∈ I. But two elements of the product are equal if their coordinates are all equal, so f is the unique function with this property. Exercises 6.5.3. 1. Show that the category of ﬁnite groups and homomorphisms does not have inﬁnite products. ! 2. Show that if i∈I Xi is the product of {Xi | i ∈ I} in an arbitrary category!C, then for each ! Y ∈ ob(C) there is an isomorphism of sets between MorC (Y, i∈I Xi ) and i∈I MorC (Y, Xi ), where the latter product is that in Sets. (In fact, this isomorphism is natural, in the sense of natural transformations, Section 6.9.) 3. Suppose given objects {Gi | i ∈ I} in either Gp or Ab. Deﬁne a!binary operation ! on i∈I Gi by (xi | i ∈ I) · (yi | i ∈ I) = (xi yi | i ∈ I). Show that i∈I Gi , with this multiplication, is the product of {Gi | i ∈ I} in the category in question. Constructions of Inﬁnite Coproducts The inﬁnite coproduct in Sets is the inﬁnite disjoint union. Deﬁnition 6.5.4. Let {Xi | i ∈ I} be a family of sets. We deﬁne i∈I Xi as follows. Let S be a set that contains each Xi . Then i∈I Xi is the subset of S × I consisting of the pairs (x, i) such that x ∈ Xi . The inclusion ιi : Xi → i∈I Xi takes x ∈ Xi to the pair (x, i). Proposition 6.5.5. The disjoint union, together with the structure maps ιi , is the coproduct in the category of sets. Thus, given a family {Xi | i ∈ I} ofsets, together with functions fi : Xi → Y for all i ∈ I, there is a unique function f : i∈I Xi → Y such that f ◦ ιi = fi for all i. Proof Given the functions fi : Xi → Y , deﬁne f : i∈I Xi → Y by f (x, i) = fi (x) for each pair (x, i) ∈ i∈I Xi . Since (x, i) = ιi (x), f satisﬁes the requirement that f ◦ ιi = fi for all i, and is the unique function that does so. The inﬁnite coproduct in Gp is called the inﬁnite free product. It is constructed by precise analogy to the free product of two groups. Thus, given a collection {Gi | i ∈ I} of groups, we set X = i∈I Gi , the disjoint union of the sets Gi . The inﬁnite free product i∈I Gi is obtained by a passage to equivalence classes from an equivalence relation, ∼, on the set of all ﬁnite words in X. Here, as in the case of the free product of two groups, the equivalence relation ∼ is the equivalence relation generated by a notion of elementary reductions. The elementary reductions are deﬁned as follows. ∗ CHAPTER 6. CATEGORIES IN GROUP THEORY Deﬁnition 6.5.6. Let X = i∈I 185 Gi and let x1 · · · xn be a word in X. 1. If xi is the identity in the group Gj that contains it, then e x1 · · · xn x1 · · · xi−1 xi+1 · · · xn . 2. If xi and xi+1 both lie in the same group Gj ⊂ X, then e x1 · · · xn x1 · · · (xi xi+1 ) · · · xn . ∗ As in the case of the free product of two groups, i∈I Gi is a group under the operation induced by concatenation of words. The canonical map ιi : Gi → i∈I Gi takes g ∈ Gi to the equivalence class of the singleton word g. The proof of the following proposition proceeds precisely as did that of Proposition 6.3.15. ∗ Proposition 6.5.7. Let {Gi | i ∈ I} be any family of groups. Then the inﬁnite free product i∈I Gi , together with the canonical maps ιi : Gi → i∈I Gi , is the coproduct of {Gi | i ∈ I} in the category of groups. ∗ ∗ Intuition might now suggest that the inﬁnite coproduct in Ab would coincide with the inﬁnite product. However, this is not the case (mainly because we cannot add an inﬁnite number of group elements). Rather, the coproduct is a subgroup of the inﬁnite product, called the direct sum. Deﬁnition 6.5.8. Let {Ai | i ∈ I}!be a family of abelian groups. We deﬁne the direct * sum i∈I Ai to be the subset of i∈I Ai consisting of * the tuples (ai | i ∈ I) for which ai = 0 for all but ﬁnitely many i. We deﬁne ιi : Ai → i∈I Ai as follows: For a ∈ Ai , the coordinates of ιi (a) are all 0except for the i-th coordinate, which *is a. We shall customarily write a for the I-tuple (a | i ∈ I) ∈ i i∈I i i∈I Ai . Note that the sum i∈I ai is actually ﬁnite, as all but ﬁnitely many of the ai are 0. * ! i∈I Ai is a subgroup of the product i∈I Ai , and the maps ιi : Ai → * Clearly, i∈I Ai are homomorphisms. The next lemma is immediate from the deﬁnitions. Lemma 6.5.9. Let {Ai | i ∈ I} be a family of abelian groups. Then in the direct sum * A , we have i i∈I ai = ιi (ai ), i∈I i∈I where the sum on the right is the sum, with respect to the group operation in of the ﬁnitely many terms ιi (ai ) for which ai = 0. * i∈I Ai , Proposition 6.5.10. Let {Ai | i ∈ I} be a family of abelian * * groups. Then the direct sum, A , together with the structure maps ι : A → i i i∈I i i∈I Ai , is the coproduct in the category of abelian groups. In other words, if B is an abelian group and if fi * : Ai → B is a homomorphism for each i ∈ I, then there is a unique homomorphism f : i∈I Ai → B such that f ◦ ιi = fi for all i ∈ I. Explicitly, the homomorphism f is given by " # f ai = fi (ai ), i∈I i∈I where the right-hand side is the sum in B of the ﬁnite number of terms fi (ai ) for which ai = 0. CHAPTER 6. CATEGORIES IN GROUP THEORY 186 Proof That f must satisfy the stated formula is immediate from Lemma 6.5.9 and the requirement that f ◦ ιi = fi for all i ∈ I. Moreover, the deﬁnition of ιi shows that if * f : i∈I → B is deﬁned by the stated formula, = fi for all i, as required. then f ◦ ιi = a Thus, it suﬃces to show that setting f i∈I i i∈I fi (ai ) deﬁnes a homo* A to B. But that is immediate from the fact that each fi is a morphism from i∈I i homomorphism. Coproducts are so important in the study of abelian groups that it is often customary to refer to the ﬁnite products as direct sums. Thus, A1 ⊕· · ·⊕Ak is an alternative notation for A1 × · · · × Ak . Exercises 6.5.11. 1. Suppose given a family {Xi | i ∈ I} of sets. Deﬁne π : i∈I Xi → I by setting π ◦ ιi equal to the constant map from Xi to i, for each i ∈ I. Show that the ! elements of X are in one-to-one correspondence with the functions s : I → i i∈I i∈I Xi such that π ◦ s = 1I . Deduce that the two forms of the Axiom of Choice are equivalent. * ! 2. Show that the direct sum i∈I Ai is the subgroup of the product i∈I Ai generated by i∈I ιi (Ai ). 3. What are the inﬁnite coproducts in Mon and Amon? 4. For any category C, show that if X is the coproduct of {Xi | i ∈ I}, ! then for any object Y , there is an isomorphism of sets between MorC (X, Y ) and i∈I MorC (Xi , Y ). 6.6 Free Functors Free functors are actually a special case of the adjoint functors discussed in Section 6.9. For simplicity, and because they constitute a very important special case, we give a separate treatment of them. We give an illustration ﬁrst. Let X be a set. Since abelian groups are sets with additional structure, we can consider the functions f : X → A, where A is an abelian group. The free abelian group on X is an abelian group, F A(X), together with a function ηX : X → F A(X) which is universal in the sense that if f : X → A is any function from X to an abelian group, then there is a unique group homomorphism f : F A(X) → A such that the following diagram commutes. f /A XG ; GG xx x GG x GG x xx ηX GG xx f # F A(X) As usual, we shall refer to F A(X) as the free abelian group on X, even though that name more properly applies to the function ηX . We shall give a construction of free abelian groups below. The following proposition shows how useful they can be. Proposition 6.6.1. Let f : A → F A(X) be an extension of the abelian group B = ker f by the free abelian group F A(X). Suppose that A is also abelian. Then the extension splits, and A is isomorphic to B × F A(X). CHAPTER 6. CATEGORIES IN GROUP THEORY 187 Proof We shall see below that the function ηX : X → F A(X) is always injective. For simplicity, we shall make use of this fact, and treat X as a subset of F A(X). Group extensions are always onto, so that X is contained in the image of f . Thus, for each x ∈ X there is an element g(x) ∈ A such that f (g(x)) = x. But this deﬁnes a function g : X → A. Since A is abelian, the universal property of free abelian groups provides a homomorphism g : F A(X) → A such that g(x) = g(x) for all x ∈ X. We claim that g is a section of f . Thus, we wish to show that f ◦ g is the identity map of F A(X). But the universal property shows that homomorphisms out of F A(X) are determined by their restriction to X. And f ◦ g(x) = f (g(x)) = x. Thus, f ◦ g has the same eﬀect on X as the identity map, so that f ◦ g is the identity as desired. Thus, the extension splits. Since A is abelian, F A(X) acts trivially on B = ker f , and hence A is the product of B and F A(X). We deﬁne free functors in general as follows. (More general deﬁnitions are possible.) Deﬁnition 6.6.2. Suppose given a category C with a forgetful functor to Sets. (In other words, the objects of C are sets with additional structure, and passage from the objects to the underlying sets and from morphisms to the underlying functions provides a functor φ : C → Sets.) Then a free C-object on a set X consists of an object F (X) ∈ ob(C) together with a function ηX : X → F (X) from X to the underlying set of F (X) with the universal property that if f : X → A is any function from X to a C-object, then there is a unique C-morphism f : F (X) → A such that the following diagram commutes. /A XE z< EE z EE zz E zzf ηX EE z " z F (X) f The free objects we’re most interested in will be very easy to construct. The key is the following lemma. Lemma 6.6.3. Let C be a category with a forgetful functor to Sets. Suppose given a family of sets {Xi | i ∈ I} such that 1. Each set Xi has a free C-object ηi : Xi → F (Xi ). 2. The collection {F (Xi ) | i ∈ I} has a coproduct in C. * Write i∈I F (Xi ) for the coproduct in C of the F (Xi ) and write η : i∈I Xi → * F (X ) for the unique function that makes the following diagram commute for each i i∈I i. ηi / F (Xi ) Xi ιi ιi , + η / Xi F (Xi ) i∈I i∈I Here, the maps ιi are the canonical maps for the coproducts, in Sets on the left and in C on the right. Then η is the free C-object on i∈I Xi . CHAPTER 6. CATEGORIES IN GROUP THEORY Proof Suppose given a function f : write fi : Xi → A for the composite i∈I ι i Xi −→ 188 Xi → A, with A and object of C. For i ∈ I, + f Xi − → A. i∈I Then since ηi is the free C-object on Xi , there is a unique C-morphism f i : F (Xi ) → A such that f i ◦ ηi = fi . Now apply * the universal property of the coproduct in C, obtaining the unique morphism f from i∈I F (Xi ) to A such that f ◦ ιi = f i . An easy diagram chase shows that f ◦ η = f , and uniqueness follows by composition of the uniqueness property of the coproduct in C and the uniqueness properties of the free C-objects ηi . For any set X, we have X = x∈X {x}. Thus, if C has arbitrary coproducts, it will have arbitrary free objects, provided that a one-point set has a free object. But we have seen free objects for the one-point set already in Gp and Ab: if G is a group and if g ∈ G, there is a unique homomorphism f : Z → G with f (1) = g. Thus, if η : {x} → Z takes x to 1, then η is the free object on {x} in both Gp and Ab. (One-point sets are the only sets for which free groups are abelian, and therefore are the only sets for which the free groups and the free abelian groups coincide.) Proposition 6.6.4. A group is a free abelian group if and only if it is a direct sum of copies of Z. A group is a free group if and only if it is a free product of copies of Z. * Proof By the preceding lemma, x∈X Z is the free abelian group on X, while x∈X Z is the free group on X. But given a direct sum or free product of copies of Z, just let X be the indexing set for the sum or free product, and this says that our sum or free product is free in the appropriate category. ∗ We can turn free objects into free functors as follows. The proof follows from the universal property and is left to the reader. Proposition 6.6.5. Let C be a category with a forgetful functor to Sets. Suppose that for every set X we are given a free C-object ηX : X → F (X). Then for any function f : X → Y there is a unique C-morphism F (f ) : F (X) → F (Y ) such that F (f ) ◦ ιX = ιY ◦ f . The correspondence that takes X to F (X) and takes f to F (f ) is a functor F : Sets → C. In practice, we identify X with its image under ιX in any of the categories under discussion. Thus, the free group on X is given by x∈X x, where each x ∈ X has inﬁnite order, and the generic element may be written xn1 1 . . . xnk k where x1 , . . . , xk ∈ X are not necessarily distinct, and n1 , . . . , nk ∈ Z. There is a similar additive notation for the free abelian group on X. Its elements may be written in the form n1 x1 + · · · + nk xk , where x1 , . . . , xk ∈ X and n1 , . . . , nk ∈ Z. In this case, because the group is abelian, we may collect terms, and assume that x1 , . . . , xk are distinct. ∗ Exercises 6.6.6. CHAPTER 6. CATEGORIES IN GROUP THEORY 189 1. Show that if f : G → F (X) is an extension of a group H by the free group on X, then f is a split extension. 2. Show that there exist group extensions f : G → F A(X) which cannot be split, where F A(X) is a free abelian group. 3. Let f : F (X) → G be a homomorphism, with F (X) the free group on X. Show that f is onto if and only if G is generated by {f (x) | x ∈ X}. 4. Let X be a set and let F (X) and F A(X) be the free group and the free abelian group on X, respectively. Let f : F (X) → F A(X) be the unique homomorphism such that the following diagram commutes. X? ?? η η ?? ?? ? f / F (X) F A(X) Here, the maps η are the canonical inclusions of X in the free group and free abelian group, respectively. ab Show that f factors uniquely through an isomorphism f : F (X) ∼ = F A(X) from the abelianization of F (X) onto F A(X). 5. Let f : F A(X) → A be a homomorphism, with A an abelian group and with F A(X) the free abelian group on X. Show that f is onto if and only if G is generated by {f (x) | x ∈ X}. 6. What is the free G-set on a set X? 7. What is the free monoid on a set X? 8. What is the free abelian monoid on a set X? 6.7 Generators and Relations Let G be a group and suppose that the set X ⊂ G generates G. Then there is a unique homomorphism f : F (X) → G from the free group on X whose restriction to X is its inclusion in G. By Problem 3 of Exercises 6.6.6, f is surjective. In this way, we see that every group is a factor group of a free group. But that, by itself, isn’t very useful information. We need to be able to say something about the kernel, K, of f : F (X) → G before we can get any useful information out of it. Free groups are very complicated objects, and we shall not attempt to say anything about the case where X is inﬁnite. Thus, we shall assume that G is ﬁnitely generated, by the set X = {x1 , . . . , xk }. We shall write F (X) = F (x1 , . . . , xk ). It happens that subgroups of free groups are always free groups.2 But even if the free group is ﬁnitely generated, if it is free on more than one generator, then it turns out to have many subgroups that are not ﬁnitely generated. But some inﬁnitely generated subgroups of a free group turn out to be normally generated by a ﬁnite set. 2 Our favorite proof of this uses the topology of covering spaces of graphs. See Section 3.8 of E.H. Spanier’s Algebraic Topology, McGraw–Hill, 1966. CHAPTER 6. CATEGORIES IN GROUP THEORY 190 Deﬁnitions 6.7.1. A ﬁnite presentation for a group G consists of a ﬁnite subset X = {x1 , . . . , xk } ⊂ G which generates G, together with a ﬁnite subset R = {R1 , . . . , Rl } of the free group F (x1 , . . . , xk ) on x1 , . . . , xk , such that the kernel of the homomorphism f : F (x1 , . . . , xk ) → G that’s induced by {x1 , . . . , xn } ⊂ G is normally generated by R in F (x1 , . . . , xk ). Given the above presentation for G, we say that G has generators X and relations R. In general, given variables x1 , . . . , xn and elements R1 , . . . , Rm of the free group F (x1 , . . . , xn ), we write x1 , . . . , xn | R1 , . . . , Rm for the group with generators x1 , . . . , xn and relations R1 , . . . , Rm . Thus, x1 , . . . , xn | R1 , . . . , Rm = F (x1 , . . . , xn )/R1 , . . . , Rm . If a group G admits a ﬁnite presentation, then we say that G is ﬁnitely presented. Note that a ﬁnite presentation of G is the construction of a pushout diagram F (r1 , . . . , rm ) / F (x1 , . . . , xn ) e / G. It turns out that not every ﬁnitely generated group is ﬁnitely presented. Thus, subgroups of ﬁnitely generated free groups can be very complicated indeed. One can show, via the theory of covering spaces, that the commutator subgroup of the free group, F (x, y), on two letters is inﬁnitely generated. But recall that the ab abelianization, F (x, y) , is naturally isomorphic to F A(x, y), the free abelian group of x and y. Thus, the next example shows that the commutator subgroup of F (x, y) is normally generated by a single element, xyx−1 y −1 . Example 6.7.2. We show here that the free abelian group on two letters has the presentation F A(x, y) = x, y | xyx−1 y −1 . First note that since F A(x, y) is abelian, xyx−1 y −1 is in the kernel of the map π : F (x, y) → F A(x, y) which takes x to x and y to y. Thus, there is an induced map π : G → F A(x, y), where G = x, y | xyx−1 y −1 = F (x, y)/xyx−1 y −1 . Note that x and y commute in G. Thus, since x and y generate G, G is abelian. Thus, the universal property of the free abelian group provides a homomorphism ϕ : F A(x, y) → G with ϕ(x) = x and ϕ(y) = y. Now notice that the composites ϕ ◦ π and π ◦ ϕ restrict to the identity map on generators for the two groups. Thus, ϕ and π are inverse isomorphisms. The hard part in showing that a given group H is given by a particular presentation is generally in producing a homomorphism from H to the stated quotient of the free group on the generators. In the above example, the homomorphism is obtained via the universal property for the free abelian group. Exercises 6.7.3. 1. Give a ﬁnite presentation for the free abelian group on n generators. CHAPTER 6. CATEGORIES IN GROUP THEORY 191 2. Show that Zn = x | xn . 3. Show that D2n = a, b | a2 , bn , aba−1 b. 4. Show that Q4n = a, b | b2n , a2 bn , aba−1 b. 5. Find a presentation for A4 . 6. Show that x, y | x2 , y 2 , (xy)3 is a presentation for S3 . 7. Show that x, y | x2 , y 2 , (xy)n is a presentation for D2n for each n ≥ 2. 8. Show that x, y, z | x2 , y 2 , z 2 , (xy)3 , (yz)3 , xzx−1 z −1 is a presentation for S4 , via setting x = (1 2), y = (2 3), and z = (3 4). 9. Let G = x, y | xn , y m , xyx−1 y k for integers n, m, k with m, n > 0. Show that G is ﬁnite and that y G. Do x and y necessarily have order n and m, respectively? 10. Let G = x, y | x4 , y 4 , (xy)4 . Show that Q8 and Z4 × Z4 are both factor groups of G. What extra relations, when added to those of G, will give Q8 ? What extra relations, when added to those of G, will give Z4 × Z4 ? Do any of the relations originally in G become redundant when we do this? 11. Show that x, y | xyx−1 y −2 is a presentation for the group Z[ 12 ]μ2 Z of Problem 19 of Exercises 4.6.13. 12. Given ﬁnite presentations for H and K and a homomorphism α : K → Aut(H), derive a ﬁnite presentation for H α K. 6.8 Direct and Inverse Limits We consider yet another pair of examples of universal properties that replace one diagram with another. We treat inverse limits here and treat direct limits in a separate subsection. Inverse limits are related to pullbacks. We suppose given objects Xi for i ≥ 0 and morphisms pi : Xi → Xi−1 for i ≥ 1. (The indices i here are integers.) We call this data an inverse system. Pictorially, this gives / X 3 p3 / X 2 p2 / X 1 p1 / X 0 . ··· The inverse limit of this system can be thought of as a universal object to tack on at −∞. What this means is that the inverse limit is a universal diagram of the following sort. X JTTTT JJ TTTT JJ TTT TTTT JJ $ TT* / ··· X1 p / X0 . 1 Deﬁnition 6.8.1. Suppose given an inverse system, consisting of objects Xi for i ≥ 0 and morphisms pi : Xi → Xi−1 for i ≥ 1. An inverse limit for this system consists of an object X = ← lim −Xi , together with morphisms πi : X → Xi for i ≥ 0 such that pi ◦ πi = πi−1 for i ≥ 1, and the following universal property is satisﬁed: For each object Y and each collection of morphisms fi : Y → Xi such that pi ◦ fi = fi−1 for all i ≥ 1, there is a unique morphism f : Y → X such that πi ◦ f = fi for all i ≥ 0. CHAPTER 6. CATEGORIES IN GROUP THEORY 192 We give the construction of inverse limits in Sets as follows. Suppose given an inverse system of sets: ··· / X3 p3 / X2 p2 / X1 p1 / X0 . ! Let X be the subset of the inﬁnite product i≥0 Xi consisting of those tuples (xi | i ≥ 0) with the property that pi (xi ) = xi−1 for all i ≥ 1, and let πi : X → Xi be the restriction to X of the i-th projection map of the product (which we shall also write as ! πi : i≥0 Xi → Xi ). Proposition 6.8.2. Using the notations just given, the set X, together with the maps πi : X → Xi , constitutes the inverse limit of the stated inverse system in Sets. Proof Suppose given functions fi : Y → Xi , i ≥ 0, such that p!i ◦ fi = fi−1 for i ≥ 1. Then the universal property of products gives a function f : Y → i≥0 Xi with πi ◦f = fi for all i. As tuples, this says f (y) = (fi (y) | i ≥ 0) for all y ∈ Y . But since pi ◦ fi = fi−1 , we have f (y) ∈ lim ←− Xi for all y ∈ Y . In other words, f takes values in the inverse limit. By construction, πi ◦ f = fi . Moreover, f is the unique map with this property because X is a subset of the product, and a map into the product is determined by its coordinate functions. We now give a useful way to interpret the construction ! of inverse limits. We shall write (. . . , xi , . . . , x1 , x0 ) for an element of the product i≥0 Xi , reversing the usual order. ! ! Then there is a “shift” map s : i≥0 Xi → i≥0 Xi given by s(. . . , xi , . . . , x1 , x0 ) = (. . . , pi+1 (xi+1 ), . . . , p2 (x2 ), p1 (x1 )). It is easy to see that the inverse limit lim ←− Xi is the ! set of all x ∈ i≥0 Xi such that s(x) = x. In other words, the inverse limit is the set of ! ﬁxed points of the shift map on i≥0 Xi . As was the case for products and pullbacks, the construction of inverse limit that works in Sets also produces the inverse limit in most of the other categories we’ve been considering. Proposition 6.8.3. Suppose given an inverse system ··· / G3 h3 / G2 h2 / G1 h1 / G0 ! in the category of groups, and let G ⊂ i≥0 Gi be the inverse limit of the induced inverse system of sets (i.e., G = {(gi | i ≥ 0) | hi (gi ) = gi−1 for all i ≥ 0}). Then G is a subgroup of the direct product of the Gi , and ! with this multiplication, together with the restrictions to G of the projection maps πi : i≥0 Gi → Gi , forms the inverse limit of this system in the category of groups. In addition, if the groups Gi are abelian, so is G, and in this case, G, together with the projection maps is the inverse limit of the given system in the category of abelian groups. Proof That G is a subgroup of the direct product follows from the fact that the hi are group homomorphisms. Since the direct product is the product in the category of groups, G, together with the projection maps, satisﬁes the universal property for the inverse limit by exactly the same proof as that given for the category of sets. A similar argument works for abelian groups. CHAPTER 6. CATEGORIES IN GROUP THEORY 193 We shall see in Chapter 7 that the same construction gives the inverse limit in the p , in some category of rings. We shall study one such example, the p-adic integers, Z depth. Zp is the inverse limit of the unique system of ring homomorphisms ··· / Zp3 / Zp2 / Zp . Exercises 6.8.4. 1. Show that if each of the functions pi : Xi → Xi−1 in an inverse system of sets is lim surjective, then so is each function πi : ← − Xi → Xi . 2. Show that if each of the functions pi : Xi → Xi−1 in an inverse system of sets is bijective, then so is each function πi : ← lim − Xi → Xi . 3. Show that the construction given for Sets, Gp, and Ab also produces the inverse limit in G-sets, Mon, and Amon. 4. Show that if X = ← lim − Xi is the inverse limit for an inverse system consisting of objects Xi , i ≥ 0, and morphisms pi : Xi → Xi−1 in a category C, then for any object Y of C, MorC (Y, X) is the inverse limit in Sets of the MorC (Y, Xi ). Direct Limits Direct limit is the concept dual to inverse limit: Deﬁnition 6.8.5. A direct system in C consists of objects Xi for i ≥ 0, together with morphisms ji : Xi → Xi+1 for i ≥ 0. Pictorially, we get X0 j0 / X1 j1 / X2 j2 / X3 / ··· A direct limit for this system consists of an object X = − lim → Xi , together with morphisms ιi : Xi → X for i ≥ 0, such that ιi+1 ◦ ji = ιi for i ≥ 0 and such that the following universal property is satisﬁed. Given an object Y and morphisms fi : Xi → Y such that lim fi+1 ◦ji = fi for all i ≥ 0, there is a unique morphism f : − → Xi → Y such that f ◦ιi = fi for all i ≥ 0. Surprisingly (since the direct limit is related to coproducts and pushouts in the same way that the inverse limit is related to products and pullbacks), the construction of the direct limit is the same in all the categories we’ve been discussing. The intuition is that if the morphisms ji are all injective, then the direct limit is the union of the Xi . (See Problem 6 of Exercises 6.8.8.) If the ji are not injective, we need to work a little harder to construct the direct limit. Assume that we’re working in Sets. We shall construct − lim by passage to equiva→X i lence classes from an equivalence relation on the disjoint union i≥0 Xi . Here, we use the equivalence relation generated by the following notion of elementary equivalence: if x ∈ Xi , we set x equivalent to ji (x) ∈ Xi+1 . We write x ∼e ji (x). We then declare two elements x, x ∈ i≥0 Xi to be equivalent, written x ∼ x , if there is a sequence x = x0 , . . . , xn = x such that for i = 0, . . . , n − 1, either xi ∼e xi+1 or xi+1 ∼e xi . We deﬁne − lim → Xi to be the set of equivalence classes under ∼, and write π : i≥0 Xi → lim −→ Xi are deﬁned to be the −→ Xi for the quotient map. The structure maps ιi : Xi → lim CHAPTER 6. CATEGORIES IN GROUP THEORY 194 composites Xi where ιi : Xi → i≥0 ιi / + Xi i≥0 π / lim X , −→ i Xi is the natural inclusion of Xi in the coproduct. Proposition 6.8.6. The above construction gives the direct limit in Sets. Proof Given functions fi : Xi → Y with fi+1 ◦ ji = fi for all i ≥ 0, there is a unique function f from the coproduct i≥0 Xi to Y such that f ◦ ιi = fi for all i, where ιi is the natural inclusion of Xi in the coproduct. Write ≈ for the equivalence relation on i≥0 Xi given by setting x ≈ x if f (x) = f (x ). Since fi+1 ◦ ji = fi for all i ≥ 0, we see that x ∼e x implies that x ≈ x . By Proposition 1.5.3, if x ∼ x , then f (x) must equal f (x ). Thus, f factors (necessarily uniquely, since π is onto) through a function f : − lim → Xi → Y . But now f clearly has the desired properties. Example 6.8.7. Let p be a prime. We shall construct a group by taking direct limits of cyclic p-groups. Speciﬁcally, note the equivalence class of p mod pi has order pi−1 in the additive group Zpi . Thus, there is a group homomorphism ji−1 : Zpi−1 → Zpi which takes 1 to p: explicitly, ji−1 (n) = pn. Since p has order pi−1 , ji−1 is injective. Clearly, the maps {ji | i ≥ 1} form a direct system in Ab. We write Tp for the direct limit of this system, which by Problem 4 below is an abelian group. Exercises 6.8.8. 1. For any direct system of sets, show that any element of the direct limit lies in the image of ιi for some i. 2. In Sets, show that if x, y ∈ Xi are equivalent in − lim → Xi , then there exists k ≥ 0 such that ji+k ◦ · · · ◦ ji (x) = ji+k ◦ · · · ◦ ji (y). 3. Suppose given a direct system in Gp. Show that the direct limit of the underlying sets has a unique group structure that makes it the direct limit of this system in Gp. † 4. Repeat the preceding problem in Ab, G-sets, Mon, and Amon. 5. In Sets, suppose there is a set U such that Xi ⊂ U for all i ≥ 0. Suppose also that each composite ji Xi −→ Xi+1 ⊂ U coincides with the inclusion Xi ⊂ U . Show that we may identify the direct limit lim X with the union X i i ⊂ U. i≥0 −→ † 6. More generally, suppose given a direct system of sets such that each function ji : Xi → Xi+1 is injective. Show that the structure maps ιi : Xi → lim −→ Xi are also injective. In addition, show that − lim X is the union of the images of the Xi . → i 7. Let Tp , as deﬁned above, be the direct limit of the cyclic groups Zpi and let ιi : Zpi → Tp be the structure map for the direct limit. By Problem 6, ιi is an injection. Show that Tp is a p-torsion group and that the image of ιi the subgroup consisting of those elements of exponent pi . CHAPTER 6. CATEGORIES IN GROUP THEORY 195 8. Let p be a prime and let Z[ p1 ] ⊂ Q be the subgroup of the rational numbers consisting of all quotients of the form pmi , where m ∈ Z and i ≥ 0. Show that Z[ p1 ] is isomorphic to the direct limit of the system p p Z → Z → Z → ... where each group in the direct system is Z and each map is multiplication by p. 9. Show that the torsion group Tp above is isomorphic to the factor group Z[ p1 ]/Z. 10. Show that any countable abelian torsion group is a direct limit of ﬁnite abelian groups. 11. Show that if G is ﬁnite, then a countable G-set is a direct limit of ﬁnite G-sets. 12. Show that if X = − lim → Xi is the direct limit for a direct system consisting of objects Xi , i ≥ 0, and morphisms ji : Xi → Xi+1 in a category C, then for any object Y of C, MorC (X, Y ) is the inverse limit in Sets of the MorC (Xi , Y ). 6.9 Natural Transformations and Adjoints In our study of semidirect products, we found that it was interesting to be able to determine when two homomorphisms are related in certain ways (e.g., conjugation). Similarly, it is often of interest to relate the eﬀect of two diﬀerent functors. Deﬁnition 6.9.1. Let C and D be categories, and let F and G be two functors from C to D. A natural transformation η : F → G consists of a D-morphism ηX : F (X) → G(X) for each X ∈ ob(C) such that for each f ∈ MorC (X, Y ), the following diagram commutes. F (f ) / F (Y ) F (X) ηX ηY G(f ) / G(Y ) G(X) We say that η is a natural isomorphism if ηX is an isomorphism in D for each object X of C. With straightforward modiﬁcations, we obtain the deﬁnitions of natural transformations or isomorphisms between two contravariant functors. The value of naturality is not immediately apparent. However, it turns out to be a strong tool in certain kinds of mathematical argument. On the theoretical level, naturality permits us to deﬁne equivalence of categories. Deﬁnition 6.9.2. An equivalence of categories between C and D consists of functors F : C → D and G : D → C, together with natural isomorphisms between G ◦ F and the identity functor of C, and between F ◦ G and the identity functor of D. One place where equivalence of categories turns out to be useful is in the notion of skeletal subcategories. CHAPTER 6. CATEGORIES IN GROUP THEORY 196 Deﬁnition 6.9.3. Let D be a subcategory of C, and suppose that the inclusion functor i : D → C is an equivalence of categories. Then we say that D is a skeleton (or skeletal subcategory) of C. We shall illustrate a nice feature of skeletal subcategories presently. First, we have another deﬁnition. Deﬁnition 6.9.4. A small category is a category whose objects form a set. For instance, consider the category of ﬁnite groups. It is not small. In fact, the collection of all objects that are isomorphic to the trivial group is too big to be a set. But we can recover most of the beneﬁts of smallness if we can construct a skeleton which is small. Recall from Corollary 2.6.6 that there are countably many isomorphism classes of ﬁnite groups. In particular, the isomorphism classes of ﬁnite groups form a set. (This does not happen in Gp, as there is no bound on the cardinality of the underlying sets of the objects in Gp.) Proposition 6.9.5. Let D be a full subcategory of C, and suppose that we’re given, for each X ∈ ob(C) an isomorphism ηX : X ∼ = F (X) for some object F (X) of D. Then D is a skeleton for C. Proof We claim that there is a unique way to make F into a functor from C to D in such a way that η is a natural transformation. But if this is the case, we’re done, since η then provides a natural isomorphism from the identity functor of C to i ◦ F , as well as from the identity functor of D to F ◦ i, where i is the inclusion of D in C. Because the morphisms ηX are all isomorphisms, it is easy to give the functoriality −1 of F : For f : X → Y in C, we take F (f ) = ηY ◦ f ◦ ηX . This immediately gives the naturality diagram for η, and the functoriality of F follows by cancelling inverses. However, the fact that we have to choose an isomorphism ηX : X ∼ = F (X) for each object X of C can be problematical from a set-theoretic point of view, as it requires an extension from sets to classes of the Axiom of Choice. Thus, it is not clear that the category of ﬁnite groups has a small skeleton. For any group G, we could use Cayley’s Theorem to identify G with a subgroup of S(G), the group of symmetries on G, but there’s no natural ordering of the elements of G, and hence no natural identiﬁcation of S(G) with S|G| . However, this need not be a serious issue. Any real mathematical question that we cared about would probably either concern isomorphism classes of objects or concern information relating to particular sorts of diagrams of objects. The latter questions can generally be phrased in terms of functors from a small category into the category one wishes to study. Thus, for the questions we care about, the next result will show that we do not really need a small skeleton. The proof of the next proposition is based on the method of proof of Proposition 6.9.5. It is left to the reader. Proposition 6.9.6. Let D be a full subcategory of C such that D contains an object in every isomorphism class of objects of C. Let E be any small category. Then any functor from E into C is naturally isomorphic to a functor that takes value in D. Also, since D is full in C, any two functors from E into D that are naturally isomorphic as functors into C are also naturally isomorphic as functors into D. CHAPTER 6. CATEGORIES IN GROUP THEORY 197 Thus, the category of ﬁnite subgroups of S∞ , which is small, contains all the information of interest with regard to the category of ﬁnite groups. Just as group isomorphisms are special examples of homomorphisms, equivalences of categories are special examples of what are called adjoint functors. Deﬁnitions 6.9.7. We say that a pair of functors R : C → D and L : D → C is an adjoint pair if for each X ∈ ob(D) and Y ∈ ob(C), there is an isomorphism (of sets) ΦX,Y : MorC (L(X), Y ) ∼ = MorD (X, R(Y )) which is natural in the following sense: given morphisms f1 : X → X in D and f2 : Y → Y in C, then the following diagram commutes. ΦX,Y / MorD (X, R(Y )) MorC (L(X), Y ) L(f1 )∗ ◦ f2 ∗ f1∗ ◦ R(f2 )∗ ΦX ,Y / MorD (X , R(Y )) MorC (L(X ), Y ) Here, g∗ (h) = g ◦ h = h∗ (g). If R and L form an adjoint pair, we call R a right adjoint for L and call L a left adjoint for R. The pair itself can be called an adjunction. Another way of expressing the naturality property for adjoint functors is as follows. Fixing X ∈ ob(D), then ﬁlling in the blanks provides two functors, MorC (L(X), −) and MorD (X, R(−)), from C to Sets, and ΦX,− provides a natural isomorphism between them. At the same time, for each Y ∈ ob(C), we have contravariant functors MorC (L(−), Y ) and MorD (−, R(Y )) from D to Sets, and Φ−,Y provides a natural isomorphism between them. It turns out that we’ve already seen several examples of adjoint functors. For instance, if C comes equipped with a forgetful functor to Sets and if ηX : X → F (X) is a free C-object for each set X, then the universal property of free objects says precisely that ηX ∗ : MorC (F (X), Y ) ∼ = MorSets (X, Y ) for each object Y of C. The right adjoint here is the forgetful functor to Sets, which doesn’t really require notation. Naturality in the variable Y is immediate, while naturality in the ﬁrst variable is shown in Proposition 6.6.5. We’ve seen some other examples as well. Exercises 6.9.8. 1. Show that abelianization is a left adjoint to the inclusion of Ab in Gp. 2. Let H be a subgroup of the group G. Then there is a forgetful functor from G-sets to H-sets. Show that this forgetful functor has a left adjoint, given by L(X) = G×H X for any H-set X. 3. Let H be a subgroup of the group G and let R : G-sets → Sets be given by R(X) = X H , the H-ﬁxed point set of X. Deﬁne L : Sets → G-sets by L(X) = (G/H) × X. Show that these form a pair of adjoint functors. 4. Show that passage from a monoid to its group of invertible elements is a right adjoint to the forgetful functor from groups to monoids. CHAPTER 6. CATEGORIES IN GROUP THEORY 198 5. Suppose given an adjunction ΦX,Y : MorC (L(X), Y ) ∼ = MorD (X, R(Y )). For X ∈ ob(D), let ηX : X → R(L(X)) be given by ηX = ΦX,L(X) (1L(X) ). Show that η is a natural transformation from the identity functor to R ◦ L. It is customarily called the unit of the adjunction. Show also that this unit determines the isomorphism Φ as follows: for f ∈ MorC (L(X), Y ), ΦX,Y (f ) = R(f ) ◦ ηX . 6. Suppose given an adjunction ΦX,Y : MorC (L(X), Y ) ∼ = MorD (X, R(Y )). For Y ∈ ob(C), let εY : L(R(Y )) → Y be given by εY = Φ−1 R(Y ),Y (1R(Y ) ). Show that ε gives a natural transformation from L ◦ R to the identity functor. We call it the counit of the adjunction. Show that it determines the isomorphism Φ as follows: for g ∈ MorD (X, R(Y )), Φ−1 X,Y (g) = εY ◦ L(g). 7. Show that left adjoints preserve coproducts, pushouts, and direct limits. This means, for instance, that if we apply a left adjoint to a pushout diagram, the resulting diagram is also a pushout diagram. 8. Show that right adjoints preserve products, pullbacks, and inverse limits. The last two problems should explain some of the results in the preceding sections, as well as predict some others that we haven’t proven yet. That, indeed is one of the main points in favor of category theory. And in some cases, knowing about adjoints can help explain why a theorem is true. For instance, it may be instructive to reconsider the notion of adjoint when one looks at the Frobenius Reciprocity Theorem in representation theory. 6.10 General Limits and Colimits Products, coproducts, pushouts, pullbacks, direct and inverse limits are examples of a couple of much more general types of construction. In all these cases, we’re given some kind of diagram in C, and we’re asked to extend it to a larger diagram that satisﬁes a universal property. The key to generalizing is in what constitutes a diagram. Deﬁnition 6.10.1. Let I be a small category and let C be an arbitrary category. By an I-diagram in C, we mean a functor X : I → C. We shall also refer to X as a diagram in C with indexing category I. We’ll show in Exercises 6.10.7 how to interpret the universal constructions we’ve already seen in terms of functors. By analogy with the constructions we’ve seen, the limit of an I-diagram in C will be deﬁned to be a diagram deﬁned on a larger indexing category that satisﬁes a certain universal property. Deﬁnition 6.10.2. For a category I, let I + be the category deﬁned as follows. The objects of I + consist of those in I, together with one new object, denoted ∗. The morphisms in I + between objects that come from I are the same as those in I (i.e., I is a full subcategory of I + ). In addition, we declare that MorI + (∗, i) has a single element for all objects i of I + , while MorI + (i, ∗) = ∅ if i = ∗. There is, of course, a unique composition law which makes this a category. Clearly, ∗ is an initial object for I + . Indeed, I + is the result of adjoining an initial object to I. (If I already had an initial object, that object will no longer be initial in I + .) CHAPTER 6. CATEGORIES IN GROUP THEORY 199 Deﬁnition 6.10.3. Let I be a small category and let X : I → C be a functor. A limit + + for X, denoted ← lim − X, is a functor X : I → C whose restriction to I ⊂ I is X, and which is universal in the following sense: if Y : I + → C is another such extension of X to I + , then there is a unique natural transformation η : Y → X such that ηi is the identity map of Xi whenever i is an object of I. Thus, the limit of an I-diagram is an I + -diagram. Colimits are deﬁned in terms of the category obtained by adjoining a terminal object to I. Deﬁnition 6.10.4. For a category I, let I+ be the category deﬁned as follows. The objects of I+ consist of those in I, together with one new object, denoted ∗. The morphisms in I+ between objects that come from I are the same as those in I (i.e., I is a full subcategory of I+ ). In addition, we declare that MorI+ (i, ∗) has a single element for all objects i of I+ , while MorI+ (∗, i) = ∅ if i = ∗. Colimits are obtained as follows. Deﬁnition 6.10.5. Let I be a small category and let X : I → C be a functor. A colimit for X, denoted − lim → X, is a functor X : I+ → C whose restriction to I ⊂ I+ is X, and which is universal in the following sense: if Y : I+ → C is another such extension of X to I+ , then there is a unique natural transformation η : X → Y such that ηi is the identity map of Xi whenever i is an object of I. Limit and colimit constructions indexed by arbitrary small categories can be quite complicated. It is often desirable to restrict attention to diagrams indexed by the category associated to a partially ordered set. (Recall that a partial ordering on a set I is a relation, generally denoted ≤ by default, that is reﬂexive, transitive, and antisymmetric.) Deﬁnition 6.10.6. Let I be a partially ordered set. The category, I, associated to I is the category whose objects are the elements of I, and whose morphisms are deﬁned as follows: for i, j ∈ I, MorI (i, j) is empty unless i ≤ j, in which case it has one element, which we denote by i ≤ j. I is a category, as transitivity gives the composition of morphisms, while reﬂexivity gives the existence of identity morphisms. Since I is a set, I is small. Exercises 6.10.7. 1. Show that a small category I comes from a partially ordered set if and only if the following properties hold. (a) For each i, j ∈ ob(I), MorI (i, j) has at most one element. (b) If i = j and if MorI (i, j) = ∅, then MorI (j, i) = ∅. 2. Let I be a set. Then we may consider it to be a partially ordered set in which i ≤ j if and only if i = j. Let I be the induced category and let C be an arbitrary category. Note that an I-diagram, X, in C is simply a family {X(i)|i ∈ I} of Cobjects indexed by I. Show that specifying a limit for X is equivalent to specifying a product for {X(i)|i ∈ I} and that specifying a colimit for X is equivalent to specifying a coproduct for {X(i)|i ∈ I}. CHAPTER 6. CATEGORIES IN GROUP THEORY 200 3. Let I be a partially ordered set with three elements, i, j and k, where i ≤ j, i ≤ k, and there are no other nonidentity morphisms in the induced category I. Show that ordinary diagrams in a category C which have the form • / • • are in one-to-one correspondence with I-diagrams in C. Under this correspondence, show that specifying a colimit for a functor X : I → C is equivalent to specifying a pushout for the associated ordinary diagram in C. 4. Give the dual of the preceding exercise. 5. Let N be the non-negative integers with the usual ordering, and with associated category N . Show that direct systems in a category C are in one-to-one correspondence with functors from N into C. Under this correspondence, show that specifying a colimit for an N -diagram is equivalent to specifying a direct limit for the associated direct system. 6. Give the dual of the preceding exercise. 7. Let I be a small category and let C be any category. Show that there is a category, which we denote C I , whose objects are the I-diagrams in C and whose morphisms are the natural transformations between them. 8. Suppose we’re given a construction that produces a colimit for every I-diagram in C. Show that we get an induced functor from C I to C I+ . 9. Suppose we’re given a construction that produces a limit for every I-diagram in C. + Show that we get an induced functor from C I to C I . 10. Show that left adjoints preserve colimits of I-diagrams. 11. Show that right adjoints preserve limits of I-diagrams. Chapter 7 Rings and Modules The study of rings and their modules is vastly more complicated than that of groups. In this chapter, long as it is, we shall do little more than set the stage for this study. We shall give some examples that we shall study in greater depth in later chapters, and shall give a number of basic deﬁnitions, constructions, and questions for further study. Section 7.1 deﬁnes concepts such as zero-divisors, unit groups, and algebras over commutative rings. Examples are given from matrix rings, the rings Zn , the complex numbers, C, and its subrings, the division ring H of quaternions, and the p-adic integers, p. Z Section 7.2 deﬁnes left, right, and two-sided ideals. Quotient rings are deﬁned, and their universal property is given. Principal ideals and generators for ideals are discussed. Operations on ideals are considered, and their use is illustrated via the Chinese Remainder Theorem. Section 7.3 develops polynomials in one and several variables. Evaluation maps are studied, with the usual applications to the theory of roots. Finitely generated algebras are discussed. The cyclotomic extensions Z[ζn ] and Q[ζn ] are featured as examples. Section 7.4 studies the symmetric polynomials, showing that they form a polynomial algebra on the elementary symmetric functions. Applications are given to the theory of discriminants. Section 7.5 develops group rings and monoid rings. It is shown that polynomials form an example of the latter. Augmentation ideals are studied, and are generalized to the notion of augmented algebras. The relationship between Z[Zn ] and Z[ζn ] is studied in the exercises. Then, in Section 7.6, the theory of prime and maximal ideals is developed for commutative rings. Krull dimension is introduced. So are the ascending and descending chain conditions for ideals, in the context of the discussion of maximality principles for families of ideals. Section 7.7 introduces modules. Cyclic modules, generating sets, annihilators, and internal and external operations on modules are studied. The theories of free modules and of exact sequences are developed. The section closes with a discussion of rings without identity. Using the techniques developed for studying modules, the ascending and descending chain conditions are studied for modules and for left and right ideals in Section 7.8. The relationship between chain conditions and extensions of modules is studied, as is the relationship between the Noetherian property and ﬁnite generation. The section closes with the Hilbert Basis Theorem and its application to ﬁnitely generated algebras over a 201 CHAPTER 7. RINGS AND MODULES 202 Noetherian ring. Then, Section 7.9 classiﬁes the vector spaces over a division ring, and shows that the ﬁnitely generated free modules over a commutative ring have a well deﬁned rank. Section 7.10 gives the relationship between matrix rings and the endomorphism rings of ﬁnitely generated free modules. We close the chapter with a development of rings and modules of fractions. Local rings and localization are studied. The ﬁeld of fractions of an integral domain is constructed, showing that a ring is an integral domain if and only if it is a subring of a ﬁeld. 7.1 Rings We give some basic deﬁnitions and examples of rings. We then treat the complex numbers, the quaternions, and the p-adic integers in subsections. We shall follow the tradition that insists that a ring should have a multiplicative identity. We give a brief discussion of rings without identity in the last subsection of Section 7.7. Deﬁnitions 7.1.1. A ring A is a set with two binary operations, addition and multiplication.1 Addition gives A the structure of an abelian group, with identity element 0. Multiplication provides A with a (not necessarily abelian) monoid structure, with identity element 1. The two operations interrelate via the distributive laws: a(b + c) = ab + ac and (b + c)a = ba + ca for all a, b, c ∈ A. If the multiplication does satisfy the commutative law, we call A a commutative ring. Let A and B be rings. Then a ring homomorphism f : A → B from A to B is a function such that f (a + b) = f (a) + f (b) and f (ab) = f (a)f (b) for all a, b ∈ A, and f (1) = 1. As usual, a ring isomorphism is a ring homomorphism that is bijective. A subring of a ring A is a subgroup B of the additive group of A, such that B is closed under multiplication and contains the multiplicative identity element 1. Finally, the invertible elements in the multiplicative monoid of a ring A are called the units of A. They form a group under multiplication, denoted A× . Let f : A → B be a ring homomorphism. Then f gives a group homomorphism between the additive groups of A and B, so we can talk about its kernel. Note that the kernel of f cannot be a subring of A unless 1 = 0 in B, because f (1) = 1. Nevertheless, ker f is more than just a subgroup of A. It is a two-sided ideal, a concept we shall study below. Images are a diﬀerent story: Lemma 7.1.2. Let f : A → B be a ring homomorphism. Then the image of f is a subring of B. The next example is the exception, rather than the rule. Example 7.1.3. There is a ring with only one element. We denote the element by 0 and set 0 + 0 = 0 and 0 · 0 = 0. In other words, both addition and multiplication are given by the unique binary operation on a one-element set. The element 0 serves as identity for both operations. Thus, with the multiplicative identity set equal to 0, the properties of a ring are satisﬁed. We call this ring the 0 ring and denote it by 0. 1 The use of A for a ring derives from anneau, the French word for ring. CHAPTER 7. RINGS AND MODULES 203 The next lemma shows that the 0 ring is the only ring in which 0 = 1. Lemma 7.1.4. Let A be a ring. Then 0 · a = a · 0 = 0 for all a ∈ A. Proof We have 0a = (0 + 0)a = 0a + 0a. But then subtracting 0a from both sides gives 0 = 0a. The case of a · 0 is similar. The distributive laws now give a useful fact. Corollary 7.1.5. Let A be a ring and let −1 ∈ A be the additive inverse of 1. Then for any a ∈ A, (−1) · a = a · (−1) is the additive inverse of a. The next result is an important generalization of the preceding ones. The proof is left to the reader. Lemma 7.1.6. Let A be a ring. For m ∈ Z, write m for the m-th power of 1 in the additive group of A. (Thus, if m is positive, m is the sum of m copies of 1 in A.) Then for any a ∈ A, m · a = a · m is the m-th power of a in the additive group of A. Examples 7.1.7. The most obvious examples of rings are the integers, rational numbers, and real numbers, Z, Q, and R, with the usual addition and multiplication. The latter two examples turn out to have some interesting subrings, but Z does not: Since 1 generates Z as an abelian group, the only subring of Z is Z itself. We’ve also seen (in Proposition 2.4.7) that multiplication in Zn induces a commutative ring structure on Zn such that the canonical map π : Z → Zn is a ring homomorphism. (Since π is surjective, this is the only ring structure that makes π a ring homomorphism.) The integers play a special role in ring theory: Lemma 7.1.8. Let A be a ring. Then there is a unique ring homomorphism from Z to A. Moreover, the image of each m ∈ Z under this homomorphism commutes2 with every element of A. Proof Ring homomorphisms are required to preserve the multiplicative identity elements. But recall from Proposition 2.5.6 that for any group G, and any element x ∈ G, there is a unique group homomorphism from Z to G that carries 1 to x. Thus, there is at most one ring homomorphism from Z to A: as a homomorphism of additive groups, it must be the homomorphism f : Z → A for which f (1) = 1. It suﬃces to show that this group homomorphism is also a ring homomorphism for any ring A, and that f (m) commutes with every element of A. Note that since f is a homomorphism of additive groups and since f (1) = 1, f (n) must be the n-th power of 1 in the additive group structure on A, for all n ∈ Z. Thus, Lemma 7.1.6 says that for each n ∈ Z, f (n) commutes with every element of a and that f (n) · a is the n-th power of a in the additive group structure on A. Thus, f (m)f (n) is the m-th power of the n-th power of 1 with respect to the additive group operation in A. So f is a ring homomorphism by the power laws for a group. 2 When we say that two elements of a ring commute, we mean this with respect to the operation of multiplication. The point is that the addition operation is abelian so that two elements always commute additively. CHAPTER 7. RINGS AND MODULES 204 In categorical language, this implies that Z is an initial object for the category of rings. Clearly, 0 is a terminal object. We now give some examples of unit groups. Examples 7.1.9. 1. It is easy to see, e.g., via the order inequalities for multiplication, that Z× = {±1}. 2. In Lemma 4.5.3, it is shown that m ∈ Z× n if and only if (m, n) = 1. Then, in Corollary 4.5.7, it is shown that if n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes, then × ∼ × Z× n = Zpr1 × · · · × Zprk . 1 k Finally, the unit groups Z× pr are explicitly calculated in Corollaries 4.5.20 and 4.5.21. We now give some language for describing properties of rings and their elements. Deﬁnitions 7.1.10. Suppose we have two elements a, b in a ring A, neither of which is 0, such that ab = 0. Then a and b are said to be zero-divisors in A. More speciﬁcally, if A is noncommutative, a is a left zero-divisor and b is a right zero-divisor. We say that an element a ∈ A is nilpotent if an = 0 for some integer n. Note that any nonzero nilpotent element is a zero-divisor. Generalizing the obvious facts about the integers, we say that an integral domain, or domain, for short, is a nonzero commutative ring with no zero-divisors. A nonzero ring is called a division ring if every nonzero element in it is a unit. If the ring is also commutative, we call it a ﬁeld. The next lemma connects some of these deﬁnitions. Lemma 7.1.11. A division ring has no zero-divisors. Thus, every ﬁeld is an integral domain. Proof Let A be a division ring and let a, b ∈ A such that a = 0 and ab = 0. It suﬃces to show that b = 0. Because A is a division ring and a = 0, a has a multiplicative inverse, a−1 . But then 0 = a−1 · 0 = a−1 · ab = (a−1 a)b = b. Of course, Z is an example of an integral domain, and Q and R are examples of ﬁelds. The following lemma could have been left as an exercise, but we shall make suﬃcient use of it that a proof has been given. The reader may wish to investigate alternative proofs. Lemma 7.1.12. Let n > 1. Then Zn is an integral domain if and only if n is prime. Moreover, for any prime number p, Zp is a ﬁeld. Proof Suppose that n is not prime. Then we can write n = kl, where k and l are both greater than 1. But then 1 < k, l < n, so k and l are nonzero in Zn . But k · l = kl = 0, and hence k and l are zero-divisors in Zn . On the other hand, if p is prime, then each nonzero element of Zp is shown to be a unit in Lemma 4.5.3. In a noncommutative ring, it is useful to keep track of the elements that commute with all other elements of the ring. CHAPTER 7. RINGS AND MODULES 205 Deﬁnition 7.1.13. Let A be a ring. Then the center of A is the set {a ∈ A | ab = ba for all b ∈ A} consisting of those elements of a that commute with every element of A. Lemma 7.1.14. Let A be a ring. Then the center of A is a subring of A. Our ﬁrst examples of noncommutative rings come from matrices. Deﬁnitions 7.1.15. Let A be a ring (not necessarily commutative). We write Mn (A) for the collection of n × n matrices with coeﬃcients in A. Given elements aij ∈ A with 1 ≤ i ≤ n and 1 ≤ j ≤ n, we write (aij ) ∈ Mn (A) for the matrix whose ij-th entry is aij . We deﬁne addition in Mn (A) so that the ij-th entry of (a ijn) + (bij ) is aij + bij and deﬁne multiplication so that the ij-th entry of (aij ) · (bij ) is k=1 aik bkj . Note that the order of the factors in the last expression is important if A is noncommutative. We write 0 for the matrix whose entries are all 0, and write In for the matrix 1 0 0 whose oﬀ-diagonal entries are all 0 and whose diagonal entries are all 1, e.g., I3 = 0 1 0 . 001 More generally, for a ∈ A, we write aIn for the matrix whose diagonal entries are all equal to a and whose oﬀ-diagonal entries are all 0. We write ι : A → Mn (A) for the map given by ι(a) = aIn for all a ∈ A. We summarize the most basic properties of matrix rings: Lemma 7.1.16. Let A be a ring. Then Mn (A) is a ring under the operations of matrix addition and matrix multiplication, and the map ι : A → Mn (A) is a ring homomorphism. Moreover, the elements ι(a) = aIn behave as follows: if M = (aij ) ∈ Mn (A), then aIn · M is the matrix whose ij-th entry is a · aij for all i, j, while M · aIn is the matrix whose ij-th entry is aij · a for all i, j. Thus, if B is the center of A, then ι(B) is contained in the center of Mn (A). The units in Mn (A) are those matrices M for which there exists an M ∈ Mn (A) such that M M = M M = In . Note that this precisely says that M is invertible in the usual sense from linear algebra. × Deﬁnitions 7.1.17. Let A be a ring. We write Gln (A) for Mn (A) , the group of invertible n × n matrices. We call it the n-th general linear group of A. The reader should be warned that if A is not commutative, then the usual sort of determinant theory fails for matrices over A, and hence invertibility cannot be detected by the familiar methods. When A is commutative, on the other hand, there is a good theory of determinants, which we shall develop in Chapter 10. We shall show that a matrix in Mn (A) is invertible if and only if its determinant is in A× . Another example relates to a generalization of matrix rings. Deﬁnition 7.1.18. Let G be an abelian group and let EndZ (G) be the set of all group homomorphisms from G to itself. We call it the endomorphism ring of G. Here, the ring operations are deﬁned as follows. For f, g ∈ EndZ (G), we let f + g be the homomorphism given by (f + g)(x) = f (x) + g(x), where additive notation is used for the group operation in G. The product, f · g, of f, g ∈ EndZ (G) is simply the composition f ◦ g. As usual, the justiﬁcation for this deﬁnition is left to the reader: CHAPTER 7. RINGS AND MODULES 206 Lemma 7.1.19. Under these operations, EndZ (G) is a ring, with additive identity the trivial homomorphism, and multiplicative identity the identity map. We shall generalize this when we study A-modules. If M is an A-module, then the A-module homomorphisms from M to itself will be shown to form a subring, EndA (M ), of EndZ (M ). If A is commutative, and if An is the free A-module of rank n, we shall see that EndA (An ) is isomorphic as a ring to Mn (A). Thus, endomorphism rings do generalize matrix rings when A is commutative. A useful concept in ring theory is that of an algebra over a commutative ring A. Deﬁnitions 7.1.20. An algebra over a commutative ring A consists of a ring B together with a ring homomorphism ν : A → B such that the image of ν is contained in the center of B. We call ν the structure map for B as an A-algebra. If B is a commutative ring, we call it a commutative A-algebra. If B and C are A-algebras, via ν : A → B and μ : A → C, then an A-algebra homomorphism from B to C is a ring homomorphism f : B → C such that the following diagram commutes. ν ooo7 B ooo f A OOO OOO ' μ C Examples 7.1.21. 1. Let B be any ring and let A be the center of B. Then the inclusion of A in B makes B an A-algebra. 2. Let B be any ring and let A be the center of B. Let ι : B → Mn (B) be the ring homomorphism given by ι(b) = bIn for b ∈ B. Then Lemma 7.1.16 shows that ι(A) is contained in the center of Mn (B).3 Thus, Mn (B) is an A-algebra via ι ◦ i, where i : A ⊂ B is the inclusion, and ι : B → Mn (B) is an A-algebra map. We now show that the notions of rings and of Z-algebras are equivalent. Lemma 7.1.22. Every ring A has a unique structure as a Z-algebra, and every ring homomorphism f : A → B is a Z-algebra homomorphism. Proof The fact that a ring has a unique Z-algebra structure is just a restatement of Lemma 7.1.8: For any ring A, there is a unique ring homomorphism from Z to A, and its image lies in the center of A. Now if ν : Z → A and μ : Z → B are the unique ring homomorphisms and if f : A → B is any ring homomorphism, we must have f ◦ ν = μ. Thus, f is a Z-algebra homomorphism. As in the case of groups, the direct product gives the simplest method of obtaining new rings from old. Deﬁnition 7.1.23. Let A and B be rings. Then the direct product A × B is the ring whose additive group is the direct product of the additive groups of A and B and whose multiplication is given coordinate-wise. In other words, the underlying set of A × B is the cartesian product of A and B, and the operations are given by (a, b) + (c, d) = (a + c, b + d) 3 Indeed, and (a, b) · (c, d) = (ac, bd). Problem 10 of Exercises 7.1.27 shows that ι(A) is equal to the center of Mn (B). CHAPTER 7. RINGS AND MODULES 207 The reader should supply the justiﬁcation that A × B is a ring: Lemma 7.1.24. Let A and B be rings. Then the direct product A × B is a ring and the projection maps from A × B to A and B are both ring homomorphisms. Moreover, if C is a ring and if f : C → A × B is a map, say f (c) = (f1 (c), f2 (c)), then f is a ring homomorphism if and only if f1 : C → A and f2 : C → B are ring homomorphisms. In category theoretic terms, Lemma 7.1.24 just says that the direct product is the product in the category of rings. The ﬁnite direct products A1 × · · · × An are deﬁned analogously, and satisfy a similar universal property. Indeed, the product of an inﬁnite family of rings is a ring via coordinate-wise multiplication: ! Deﬁnition 7.1.25. Suppose given a family {Ai | i ∈ I} of rings. We write i∈I Ai for the ring structure on the product of the Ai whose operations are coordinate-wise addition and multiplication. Thus, (ai | i ∈ I) + (bi | i ∈ I) = (ai + bi | i ∈ I) and (ai | i ∈ I) · (bi | i ∈ I) = (ai bi | i ∈ I). Again, the reader should justify the deﬁnition: ! Lemma 7.1.26. Let {Ai | i ∈ I} be a family of rings. Then i∈I Ai is a ring under coordinate-wise addition and multiplication. Here 0 is the I-tuple whose coordinates are ! all 0, and 1 is the I-tuple whose coordinates are all 1. The projection maps πj : for all j ∈ I. i∈I Ai → Aj are ring homomorphisms ! If B is a ring and if f : B → i∈I Ai is a map, say f (b) = (fi (b) | i ∈ I) for each b ∈ B, then f is a ring homomorphism if and only if fi is a ring homomorphism for all i ∈ I. Exercises 7.1.27. 1. Give the proof of Lemma 7.1.6. 2. Show that every subring of a ﬁeld is an integral domain.4 3. We say that n ∈ Z is square free if it is not divisible by the square of any prime number. Show that Zn contains no nonzero nilpotent elements if and only if n is square free. 4. Suppose given a ring structure on the additive group Zn . Show that the multiplicative identity element must generate Zn as an additive group. Deduce that any ring structure on Zn is isomorphic to the usual one. (Hint: Let e be the multiplicative identity for a given ring structure on Zn . Show that e fails to generate Zn additively if and only if there is an integer d > 1 such that d divides n and such that e is the additive d-th power of some element x ∈ Zn . Now show that no such ring structure could exist.) 5. Let f : A → B be a ring homomorphism. Show that f restricts to a group homomorphism f : A× → B × . (Note then that the passage from rings to their unit groups gives a functor from rings to groups.) 6. Give the proof of Lemma 7.1.16. 4 We shall see in Section 7.11 that a ring is an integral domain if and only if it is a subring of a ﬁeld. CHAPTER 7. RINGS AND MODULES 208 7. Let A and B be rings. Show that Mn (A × B) is isomorphic to Mn (A) × Mn (B). 8. Show that Mn (A) contains nonzero nilpotent elements, and hence zero-divisors, for any nonzero ring A, provided that n ≥ 2. 9. Let A and B be subrings of a ring C. Let a c R= a ∈ A, b ∈ B and c ∈ C . 0 b Show that R is a subring of M2 (C). † 10. Let B be the center of A. Show that ι : A → Mn (A) restricts to an isomorphism from B onto the center of Mn (A). 11. Let f : A → B be a ring homomorphism. Show that there is a ring homomorphism f∗ : Mn (A) → Mn (B) deﬁned by setting f∗ ((aij )) equal to the matrix whose ij-th entry is f (aij ). 12. Give the proof of Lemma 7.1.19. 13. Show that the unit group of EndZ (G) is the automorphism group Aut(G) of G. 14. Let G be the direct sum of a countably inﬁnite number of copies of Z. Find an element of EndZ (G) which has a left inverse, but is not a unit. 15. Show that the center of a direct product A × B is the direct product of the centers of A and B. 16. Show that the group of units of A × B is A× × B × . ‡ 17. Suppose given ring homomorphisms f : A → C and g : B → C. Recall that the pullback of f and g is the diagram f / B A ×C B g g A f / C where A ×C B = {(a, b) ∈ A × B | f (a) = g(b)}, and f and g are deﬁned by f (a, b) = b and g(a, b) = a. (a) Show that A ×C B is a subring of A × B and that f and g are ring homomorphisms. (b) Suppose given a ring R and ring homomorphisms f : R → B and g : R → A such that f ◦ g = g ◦ f . Show that there is a unique ring homomorphism h : R → A ×C B such that f ◦ h = f and g ◦ h = g . Show that h must satisfy h(r) = (g (r), f (r)) for all r ∈ R. (c) Show that the natural map (A ×C B)× → A× ×C × B × is an isomorphism. 18. Show that a direct product A × B of commutative rings is an integral domain if and only if one of A and B is an integral domain and the other is the zero ring. CHAPTER 7. RINGS AND MODULES 209 19. Show that a direct product A × B of rings is a division ring if and only if one of A and B is a division ring and the other is the zero ring. 20. Give the proof of Lemma 7.1.24. 21. Give the proof of Lemma 7.1.26. 22. Let A be a ring. Show that Mm (A) × Mn (A) to the subring of is isomorphic M 0 Mm+n (A) consisting of matrices of the form , where M is an m × m 0 N matrix, N is an n × n matrix, and the 0’s are zero-matrices of the appropriate size. The Complex Numbers The complex numbers, C, give yet another example of a ﬁeld. We review the deﬁnition and basic properties of C. Deﬁnitions 7.1.28. The elements of the complex numbers are formal sums a + bi, with a, b ∈ R. This presentation is unique in the sense that if a, b, c, d ∈ R, then a+bi = c+di if and only if a = c and b = d. Thus, the elements of C are in one-to-one correspondence with ordered pairs (a, b) of real numbers. So we may identify C with the plane R2 , which gives us a geometric picture of the complex numbers. We identify a ∈ R with the complex number a + 0i. We say that a complex number is in R if it has this form. Similarly, we write i for the complex number 0 + 1i, and refer to the numbers ai = 0 + ai, with a ∈ R as the pure imaginary numbers. If z = a + bi and w = c + di are complex numbers, with a, b, c, d ∈ R, we deﬁne their sum by z + w = (a + c) + (b + d)i, and deﬁne their product by zw = (ac − bd) + (ad + bc)i. For z = a + bi with a, b ∈ R, we deﬁne the complex conjugate, z, of z by z = a − bi. Again, there are veriﬁcations left to the reader. Lemma 7.1.29. The above operations of addition and multiplication give C the structure of a commutative ring. Moreover, the standard inclusion R ⊂ C is a ring homomorphism, and hence the additive and multiplicative identities of C are given by 0 = 0 + 0i and 1 = 1 + 0i, respectively. The function C → C that carries each z ∈ C to its complex conjugate, z, is a ring homomorphism. For z = a + bi with a, b ∈ R, we have the equality zz = a2 + b2 ∈ R ⊂ C. In particular, zz ≥ 0 for all z, and zz = 0 if and only if z = 0. We can now ﬁnd a multiplicative inverse for every nonzero element of C. Corollary 7.1.30. The complex numbers, C, is a ﬁeld. Explicitly, if 0 = z ∈ C, then z is a unit in C with inverse z −1 = 1 · z. zz Here, 1/(zz) is the inverse of the nonzero real number zz. CHAPTER 7. RINGS AND MODULES 210 Proof Since zz = 0, the result follows from the facts that C is commutative and that R is a subring of C. The notion of planar distance from analytic geometry may be derived from the complex product zz. Deﬁnition 7.1.31. We deﬁne the absolute value | | : C → R by √ |z| = zz = a2 + b2 for z = a + bi with a, b ∈ R. A very useful tool in studying the complex numbers is the complex exponential function. Deﬁnition 7.1.32. Let x ∈ R. We deﬁne the complex exponential eix ∈ C by eix = cos x + i sin x. More generally, if z = x + iy ∈ C, with x, y ∈ R, we deﬁne ez ∈ C by ez = ex eiy = ex cos y + iex sin y, where ex is the ordinary real exponential function applied to x. We shall be primarily concerned with the exponentials eix with x ∈ R. Lemma 7.1.33. Let x, y ∈ R, then ei(x+y) = eix · eiy for x, y ∈ R, where the product on the right is complex multiplication. Moreover, ei·0 = 1. Thus, there is a group homomorphism exp : R → C× deﬁned by exp(x) = eix . Proof Complex multiplication gives eix · eiy = (cos x + i sin x) · (cos y + i sin y) = (cos x cos y − sin x sin y) + i(cos x sin y + sin x cos y) = cos(x + y) + i sin(x + y), by the formulæ for the sine and cosine of a sum. But the last line is precisely ei(x+y) . From the known values of the sine and cosine of 0, we see that ei·0 = 1, as claimed, so that setting exp(x) = eix gives a monoid homomorphism from R to the multiplicative monoid of C. But any monoid homomorphism from a group to a monoid must take value in the group of invertible elements of the monoid, which in this case is C× . The exponential allows us to deﬁne some important elements of C, the standard primitive n-th roots of unity. Deﬁnition 7.1.34. Let n ≥ 1. The standard primitive n-th root of unity, ζn ∈ C, is deﬁned by 2π 2π ζn = ei·2π/n = cos + i sin . n n The fact that 2π is the smallest positive real number whose sine is 0 and whose cosine is 1 gives us the following lemma. CHAPTER 7. RINGS AND MODULES 211 Lemma 7.1.35. Let n ≥ 1. Then ζn has order n in C× . There is a noncommutative division ring, called the quaternions, H, that is given by a construction similar to that of the complex numbers. We shall treat it in the next set of exercises. Both C and H are examples of what are called Cliﬀord algebras over R. We shall treat the general case of Cliﬀord algebras in Chapter 9. We shall not prove it here, but C and H are the only division rings that are algebras over R in such a way that the induced real vector space structure is ﬁnite dimensional. Exercises 7.1.36. 1. Write Z2 = T = {1, T }. Show that Z2 acts on C via T · z = z for all z ∈ C. Show that the ﬁxed point set CZ2 is equal to R. 2. Write Zn = T = {1, T, . . . , T n−1 } for n ≥ 2. Show that Zn acts on C via T k · z = ζnk · z for all k, where ζn = ei·2π/n is the standard primitive n-th root of unity in C. Show that 1 + ζn + · · · + ζnn−1 is a ﬁxed point under this action. Deduce that 1 + ζn + · · · + ζnn−1 = 0. 3. Show that |zw| = |z| · |w| for all z, w ∈ C, that |z| ≥ 0 for all z ∈ C, and that |z| = 0 if and only if z = 0. 4. Show that |z + w| ≤ |z| + |w| for all z, w ∈ C. 5. Show that i is a square root of −1 in C. 6. Let S 1 = {z ∈ C | |z| = 1}. Show that S 1 is a subgroup of C× . 7. Show that eix ∈ S 1 for all x ∈ R, and hence exp gives a homomorphism from R to S 1 . What is the kernel of exp? 8. Show that exp : R → S 1 is onto. Deduce that S 1 is isomorphic as a group to R/Z. 9. Show that there is a homomorphism expC : C → C× given by expC (z) = ez . 10. Show that eiπ = −1 ∈ C. 11. Show that ζ1 = 1, ζ2 = −1, and ζ4 = i. 12. Let n = mk with m and k positive integers. Show that ζnk = ζm . 13. Show that if n = 2m with m a positive integer, then ζnk+m = −ζnk . 14. Let m > 0 be an odd integer. Show that the subgroup of C× generated by −ζm coincides with the subgroup generated by ζ2m . when a, b ∈ R. Show that φ is an 15. Deﬁne φ : C → M2 (R) by φ(a + bi) = ab −b a injective ring homomorphism. Show that φ(S 1 ) = SO(2) and that φ(eiθ ) = Rθ , the matrix that rotates R2 through the angle θ, for all θ ∈ R. 16. We construct the ring of quaternions. The quaternionic groups Q4n , which we analyzed in the material on group theory, will be seen to embed naturally in the group of units of the quaternions. An alternative name for the quaternions is the Hamiltonians. Because of this, and the fact that Q is already taken, we shall write H for the quaternions. The elements of H are formal sums a + bi + cj + dk, where a, b, c, and d are real CHAPTER 7. RINGS AND MODULES 212 numbers. Here, a + bi + cj + dk = a + b i + c j + d k if and only if a = a , b = b , c = c , and d = d , and hence we may identify H with the set R4 of 4tuples (a, b, c, d) of real numbers. We deﬁne the addition in H accordingly, so that (a + bi + cj + dk) + (a + b i + c j + d k) = (a + a ) + (b + b )i + (c + c )j + (d + d )k. We write 1 = 1 + 0i + 0j + 0k, i = 0 + 1i + 0j + 0k, j = 0 + 0i + 1j + 0k and k = 0 + 0i + 0j + 1k. For a ∈ R, we write a = a · 1 = a + 0i + 0j + 0k. Thus, we identify the real multiples of 1 with R. The multiplication in H is deﬁned so that 1 is the identity element, i2 = j 2 = k 2 = −1, ij = −ji = k, jk = −kj = i and ki = −ik = j. Assembling this information, we see that if α = a + bi + cj + dk and α = a + b i + c j + d k, then αα = (aa − bb − cc − dd ) + (ab + ba + cd − dc )i + (ac + ca + db − bd )j + (ad + da + bc − cb )k. Unlike the complex numbers, we see that multiplication in H is noncommutative. (a) Show that H is a ring and that the real numbers R ⊂ H form a subring. (b) Show that the center of H is R, so that H is an R-algebra. (c) Show that if we set x equal to either i, j, or k, then the set of all elements of H of the form a + bx with a, b ∈ R forms a subring of H that is isomorphic to C. (d) Deﬁne conjugation in H as follows: for α = a + bi + cj + dk, we set α = a − bi − cj − dk. Show that the following equalities hold for all α, α ∈ H. i. α + α = α + α ii. α · α = α · α iii. 1 = 1. iv. α = α. (According to deﬁnitions we shall give below, the ﬁrst three of the above properties say that conjugation is an antiautomorphism of H, and hence H is a self-opposite ring. This is useful in the study of matrices over H.) (e) Show that α and α commute for all α ∈ H, and that if α = a + bi + cj + dk, then αα is the real number a2 + b2 + c2 + d2 . (f) Deﬁne the absolute value | | : H → R by √ |α| = αα = a2 + b2 + c2 + d2 for α = a + bi + cj + dk with a, b, c, d ∈ R. Show that |α · α | = |α| · |α | for all α, α ∈ H, that |α| ≥ 0 for all α ∈ H, and that |α| = 0 if and only if α = 0. (g) Show that every nonzero element α ∈ H is a unit in H (and hence H is a division ring). Here, if α = 0, show that α−1 = 1 α. αα (h) Show that if we set b = cos(π/n) + i sin(π/n) and set a = j, then the subgroup of H× generated by a and b is isomorphic to Q4n , the quaternionic group of order 4n. (In particular, note that Q8 is isomorphic to the subgroup {±1, ±i, ±j, ±k} of H× . When studying Q8 alone, rather than the quaternionic groups in general, it is customary to use the notation ±1, ±i, ±j, ±k for its elements.) CHAPTER 7. RINGS AND MODULES 213 (i) Let S 3 = {α ∈ H | |α| = 1}. Show that S 3 is a subgroup of H × containing the above copies of Q4n . Show also that S 1 embeds as a subgroup of S 3 . 17. Let S = {α = ai + bj + ck ∈ H | a, b, c ∈ R and |α| = 1}.5 Show that α2 = −1 for all α ∈ S. Deduce that H has inﬁnitely many subrings isomorphic to C. The p-adic Integers p . For k ≥ 2, let εk : Zpk → Zpk−1 be the unique ring We construct the p-adic integers, Z homomorphism (in fact, the unique group homomorphism that carries 1 to 1) from Zpk to Zpk−1 . The p-adic integers are the inverse limit of the maps εk . To avoid a dependence on category theory here, we shall give the construction explic p is a subring of the inﬁnite product ! itly. Z k≥1 Zpk . It is customary to write the tuples in the inﬁnite product in the reverse order from the usual one. Thus, a typical element looks like (. . . , mk , . . . , m2 , m1 ), where each mk lies in Zpk . Under this convention, we have p = {(. . . , mk , . . . , m2 , m1 ) | εk (mk ) = mk−1 for all k ≥ 2}. Z Recall that the ring operations in the inﬁnite product are deﬁned via coordinate-wise addition and multiplication, so the fact that the εk are ring homomorphisms veriﬁes the assertion above: p is a subring of ! Lemma 7.1.37. Z k≥1 Zpk . The additive and multiplicative identities, of course, are 0 = (. . . , 0, . . . , 0, 0) and 1 = (. . . , 1, . . . , 1, 1), respectively. The next lemma is elementary, but important. p → Zpk be induced by the projection onto the k-th factor. Lemma 7.1.38. Let πk : Z Then πk is a ring homomorphism for all k ≥ 1. The p-adic integers are important in number theory. Despite the fact that they are made up from quotient rings of Z, the next lemma will show that the p-adic integers themselves form a ring of characteristic 0. (See Deﬁnition 7.2.12.) p is an embedding. Lemma 7.1.39. The unique ring homomorphism ι : Z → Z Proof We have ι(m) = (. . . , m, . . . , m, m), so m ∈ ker ι if and only if m is divisible by pk for all k ≥ 0. p . Recall that a The structure of the unit group is one of the interesting features of Z splitting map, or section, for a homomorphism f : G → K is a homomorphism s : K → G such that f ◦ s is the identity map of K. A map f that admits a section is called a split surjection. Recall from Corollary 4.7.6 that if f : G → K is a split surjection with G abelian, then G ∼ = ker f × K. p is a unit if and only Proposition 7.1.40. An element a = (. . . , mk , . . . , m2 , m1 ) of Z if m1 is a unit in Zp . The map × → Z× π1 : Z p p is a split surjection. 5 Geometrically, S is a 2-dimensional subsphere of S 3 . CHAPTER 7. RINGS AND MODULES 214 p . Since the multiplication in Z p is coordinateProof Let a = (. . . , mk , . . . , m2 , m1 ) ∈ Z wise, if a is a unit in Zp , then each mk must be a unit in Zpk . Conversely, suppose that m1 is a unit. According to Lemma 4.5.3, an element m is a unit in Zpk if and only if (m, p) = 1. In particular, m is a unit in Zpk if and only if εk (m) = m is a unit in Zpk−1 . Thus, if m1 is a unit, then each mk must be a unit in Zpk . Let nk be the inverse of mk in Zpk . Since each εk is a ring homomorphism, we must p have εk (nk ) = nk−1 . Thus, setting b = (. . . , nk , . . . , n2 , n1 ) speciﬁes an element of Z that is easily seen to be a multiplicative inverse for a. × → Z× . For p = 2, there’s nothing It remains to construct a splitting map for π1 : Z p p to show, as Z× is the trivial group. For p > 2, we consider the proof of Corollary 4.5.12. 2 Let ηk : Zpk → Zp be the unique ring homomorphism between these rings. Then the × kernel of the induced map ηk : Z× → Z× p has order relatively prime to |Zp | = p − 1. pk consisting Thus, the proof of Corollary 4.4.15 shows that if K is the subgroup of Z× pk ∼ = of those elements of exponent p − 1, then ηk : K −→ Z× p . But this says that there is a × unique section sk : Z× → Z of η . k p pk But then εk ◦ sk is a section for ηk−1 : Z× → Z× p , so εk ◦ sk = sk−1 . Thus, setting pk−1 s(m) = (. . . , sk (m), . . . , s2 (m), m) gives a well deﬁned function s : Z× p → Zp , that is easily seen to give a section for × × →Z . π1 : Z p p Exercises 7.1.41. p is an integral domain. (Hint: Suppose that ab = 0 in Z p , with † 1. Show that Z a = (. . . , mk , . . . ) and b = (. . . , nk , . . . ). Suppose that mk = 0 in Zpk . What does this say about the p-divisibility of mk+l and nk+l ?) × → Z× is isomorphic to the 2. Let p be an odd prime. Show that the kernel of π1 : Z p p × additive group of Zp . Deduce that the torsion subgroup of Zp maps isomorphically to Z× p under π1 . × → Z× is isomorphic to the additive group of Z 2. 3. Show that the kernel of π2 : Z 2 4 × maps isomorphically to Z× under π2 . Deduce that the torsion subgroup of Z 2 4 p → Zp . Show that † 4. Let a be in the kernel of the additive homomorphism π1 : Z p . (Hint: Consider the diagram a = pb for some b ∈ Z Zpk εk Zpk−1 ⊂ / Zpk+1 εk+1 ⊂ / Zpk where the inclusion maps send 1 to p.) p . Suppose that mk = 0 in Zpk for all k ≤ r and † 5. Let a = (. . . , mk , . . . , m2 , m1 ) ∈ Z ×. that mr+1 = 0 in Zpr+1 . Show that a = pr u, where u ∈ Z p CHAPTER 7. RINGS AND MODULES 215 6. Let fk : Ak → Ak−1 be a ring homomorphism for k ≥ 2. Give an isomorphism × × ∼ lim A = lim ←− k ←−(Ak ). 7. Give Zpk the discrete topology for ! all k. Show that Zp is a closed subspace of the product (Tikhonov) topology on k≥1 Zpk , and hence is compact by the Tikhonov p . For this reason, Z p may Theorem. Show that ι(Z) is dense in this topology on Z p. be thought of as a completion of Z in the topology that Z inherits from Z 7.2 Ideals The ideals in a ring are a very important part of its structure. We give the most basic results on ideals in this section, and shall continue to study them throughout the material on rings. Deﬁnitions 7.2.1. Let A be a ring. A left ideal of A is a subgroup a ⊂ A of the additive group of A with the additional property that if x ∈ a, then for any a ∈ A, we have ax ∈ a. (In other words, a is closed on the operation of multiplying its elements on the left by any element of A.) From the other side, a right ideal of A is a subgroup a ⊂ A of the additive group of A with the additional property that if x ∈ a, then for any a ∈ A, we have xa ∈ a. A two-sided ideal of A is a subgroup of the additive group of A which is simultaneously a left ideal and a right ideal. If A is commutative, all ideals are two-sided, and we refer to them simply as ideals. Examples 7.2.2. 1. Let A be a ring. Then A is a two-sided ideal of itself. 2. Let A be a ring and let 0 ⊂ A be the trivial subgroup. Since a · 0 = 0 · a = 0 for all a ∈ A, 0 is a two-sided ideal of A. Another set of examples comes from the principal ideals. Deﬁnition 7.2.3. Let A be a ring and let x ∈ A. The principal left ideal, Ax, generated by x is deﬁned by Ax = {ax | a ∈ A}. Similarly, the principal right ideal generated by x is xA = {xa | a ∈ A}. If A is commutative, then Ax = xA, and we call it simply the principal ideal generated by x, and denote it by (x). Principal ideals play a role in ideal theory similar to the role of cyclic subgroups in group theory: Lemma 7.2.4. Let A be a ring and let x ∈ A. Then Ax and xA are the smallest left and right ideals, respectively, that contain x. Some commutative rings have no ideals other than the principal ones. Deﬁnitions 7.2.5. A principal ideal ring is a commutative ring in which every ideal is principal. A principal ideal domain, or P.I.D., is an integral domain in which every ideal is principal. CHAPTER 7. RINGS AND MODULES 216 Indeed, the integers are better behaved than the general integral domain: Lemma 7.2.6. Every subgroup of Z is a principal ideal. In particular, Z is a P.I.D. Proof We know (Proposition 2.3.2) that the subgroups of Z all have the form n = {nk | k ∈ Z} for some n ∈ Z. But n is precisely (n), the principal ideal generated by n. We shall see below that if K is a ﬁeld,6 then the polynomial ring K[X] is a principal ideal domain. Principal ideals are useful in recognizing division rings and ﬁelds. Lemma 7.2.7. A nonzero ring A is a division ring if and only if the only left ideals of A are 0 and A. Proof Let A be a division ring and let a be a nonzero left ideal of A. Thus, there is an x ∈ a with x = 0. But then x is a unit in A. Now 1 = x−1 x ∈ a, as a is a left ideal. But then a = a · 1 ∈ a for the same reason, for all a ∈ A. Thus, a = A. Conversely, suppose that A is a ring with no left ideals other than 0 and A. Let 0 = x ∈ A. Then Ax is a nonzero left ideal, and hence Ax = A. Thus, there is an element y ∈ A with yx = 1. Thus, every nonzero element in the multiplicative monoid of A has a left inverse. As in Problem 13 of Exercises 2.1.16, it is easy to see that every nonzero element of A is a unit. Two-sided ideals play a role in ring theory somewhat analogous to that played by normal subgroups in the theory of groups. Proposition 7.2.8. Let a be any subgroup of the additive group of a ring A, and let π : A → A/a be the canonical map. Then A/a has a ring structure that makes π a ring homomorphism if and only if a is a two-sided ideal of A. Such a ring structure, if it exists, is unique. Proof Suppose given a ring structure on A/a such that π is a ring homomorphism. Then writing a ∈ A/a for π(a), we must have a · b = ab, so the ring structure is indeed unique. Moreover, for x ∈ a, 1 = 1 + x. Thus, for a ∈ A, a = a · 1 = a · 1 + x = a + ax. In other words, a + ax represents the same element of A/a as a does. But this says a + ax − a ∈ a, and hence ax ∈ a for a ∈ A and x ∈ a. Thus, a has to be a left ideal. But a similar argument, obtained by multiplying 1 and 1 + x on the right by a, shows that a must also be a right ideal, so that a is in fact two-sided. For the converse, suppose that a is a two-sided ideal. We wish to show that the multiplication a · b = ab gives a well deﬁned binary operation on A/a. This will be suﬃcient to complete the proof, as then π is a homomorphism with respect to this operation, so that the veriﬁcations that A/a is a ring under this multiplication will follow from the fact that A is a ring. Thus, for a, b ∈ A and x, y ∈ a, we must show that ab and (a + x)(b + y) represent the same element of A/a, and hence that (a + x)(b + y) − ab ∈ a. But (a + x)(b + y) − ab = ay + xb + xy, which is in a because a is a two-sided ideal. The rings A/a satisfy a universal property similar to that of factor groups in group theory. We call them quotient rings of A. 6 The use of K for a ﬁeld derives from the German Körper , which means ﬁeld in the mathematical sense. Amusingly, the German word for a farmer’s ﬁeld is a cognate of ours: Feld. CHAPTER 7. RINGS AND MODULES 217 Proposition 7.2.9. Let f : A → B be a ring homomorphism. Then ker f is a two-sided ideal of A. If a is another two-sided ideal of A, then a ⊂ ker f if and only if there is a ring homomorphism f : A/a → B which makes the following diagram commute, where π : A → A/a is the canonical map: /B AB = BB || BB | | B || π BB || f ! A/a f Moreover, if a ⊂ ker f , then the homomorphism f : A/a → B making the diagram commute is unique. When a = ker f , f induces a ring isomorphism between A/a and the image of f . Proof By Proposition 4.1.7, there is a unique homomorphism of additive groups, f : A/a → B making the diagram commute if and only if a ⊂ ker f . And if a = ker f , this homomorphism f gives a group isomorphism onto the image of f by the First Noether Isomorphism Theorem. Thus, it suﬃces to show that ker f is a two-sided ideal and that f is a ring homomorphism whenever a is a two-sided ideal contained in ker f . We leave these veriﬁcations to the reader. Since π : Z → Zn is a surjective ring homomorphism and has kernel (n), we obtain the following corollary. Corollary 7.2.10. There is an isomorphism of rings between Zn and the quotient ring Z/(n). The next corollary is another consequence of Proposition 7.2.9. Corollary 7.2.11. Let f : D → B be a ring homomorphism, where D is a division ring and B is any nonzero ring. Then f is injective. Proof The kernel of f is a two-sided ideal, and hence must be either 0 or D. As B is nonzero, ker f = D. So ker f = 0, and f is injective. Recall that for any ring B, there is a unique ring homomorphism f : Z → B. We may obtain useful information about B by applying Proposition 7.2.9 with a = ker f . Deﬁnition 7.2.12. Let A be a ring and let f : Z → A be the unique ring homomorphism from Z to A. Let ker f = (n), with n ≥ 0. Then we say that A has characteristic n. In particular, A has characteristic 0 if and only if f is injective. Examples 7.2.13. Since the kernel of the canonical map π : Z → Zn is (n), Zn has characteristic n. On the other hand, Z is a subring of Q, R, C, and H, so these have characteristic 0. Proposition 7.2.14. Let n > 0. Then a ring A has characteristic n if and only if it has a subring isomorphic as a ring to Zn . If A does have characteristic n, then the subring in question is the image of the unique ring homomorphism f : Z → A. CHAPTER 7. RINGS AND MODULES 218 Proof If A has characteristic n with n > 0, then the image of f is isomorphic to Z/(n) by Proposition 7.2.9. Conversely, suppose that A has a subring B that is isomorphic as a ring to Zn . Then if f : Z → B is the unique ring homomorphism, we may identify f with the canonical map π : Z → Z/(n). In particular, the kernel of f is (n). But if i : B → A is the inclusion, then i ◦ f is the unique ring homomorphism from Z to A. Since i is injective, the kernel of i ◦ f , whose generator is by deﬁnition the characteristic of A, is precisely the kernel of f . Let K be a ﬁeld. Then the image of the unique ring homomorphism f : Z → K, as a subring of K, must be an integral domain. Thus, Lemma 7.1.12 shows that if f is not injective (i.e., if the image of f is ﬁnite), then f must have image Zp for some prime p. Corollary 7.2.15. Let K be a ﬁeld of nonzero characteristic. Then the characteristic of K is a prime number. We can generalize the notion of a principal ideal as follows. Deﬁnitions 7.2.16. Let A be a ring and let S be any subset of A. Then the left ideal generated by S, written AS, is given as follows: AS = {a1 x1 + · · · + ak xk | k ≥ 1, ai ∈ A and xi ∈ S for 1 ≤ i ≤ k}. Similarly, the right ideal generated by S is given by SA = {x1 a1 + · · · + xk ak | k ≥ 1, ai ∈ A and xi ∈ S for 1 ≤ i ≤ k}. When A is commutative, of course, AS = SA, and we denote it by (S). If S is ﬁnite, say S = {x1 , . . . , xn }, we write (S) = (x1 , . . . , xn ). For A noncommutative, the two-sided ideal generated by S is ASA = {a1 x1 b1 + · · · + ak xk bk | k ≥ 1, ai , bi ∈ A and xi ∈ S for 1 ≤ i ≤ k}. The next lemma is immediate from the deﬁnitions. Lemma 7.2.17. Let A be a ring and let S be a subset of the center of A. Then the left, right, and two-sided ideals generated by A all coincide, i.e., AS = SA = ASA. In particular, if A is commutative, then ASA = (S). Warning: In the case of left ideals, we may apply the distributive law to reduce the sums a1 x1 + · · · + ak xk until the elements xi are all distinct. Thus, for instance, A{x} = Ax = {ax | a ∈ A}, the principal left ideal generated by x. The analogous result holds for the right ideals. But in the case of two-sided ideals in noncommutative rings, we cannot make this kind of reduction in general. Thus, in A{x}A, the principal two-sided ideal generated by x, we must consider all elements of the form a1 xb1 + · · · + ak xbk with k ≥ 1 and with ai , bi ∈ A for 1 ≤ i ≤ k. Lemma 7.2.18. Let A be a ring and let S ⊂ A. Then AS, SA, and ASA are the smallest left, right, and two-sided ideals, respectively, that contain S. This now gives the following universal property. CHAPTER 7. RINGS AND MODULES 219 Corollary 7.2.19. Let A be a ring and let S ⊂ A, and let a ⊂ A be the two-sided ideal generated by S. Let f : A → B be a ring homomorphism whose kernel contains S. Then there is a unique homomorphism f : A/a → B such that the following diagram commutes /B, AB = BB || BB | | B || π BB || f ! A/a f where π : A → A/a is the canonical map. Proof The kernel of f is a two-sided ideal of A containing S, so we must have a ⊂ ker f by Lemma 7.2.18. The result now follows from Proposition 7.2.9. When a ring homomorphism f : B → A is surjective, we can use information about A and about ker f to deduce information about B. An important topic in ideal theory is a discussion of the various ways in which we can obtain new ideals from old. Recall ﬁrst (from Lemma 4.1.18) that if G is a group and if H and K are subgroups of G with K ⊂ NG (H), then KH = {kh | k ∈ K, h ∈ H} is a subgroup of G. Thus, if G is abelian, written in additive notation, and if H and K are any subgroups of G, we have a subgroup H + K given by H + K = {h + k | h ∈ H, k ∈ K}. The next lemma is elementary. Lemma 7.2.20. Let a and b be left (resp. right) ideals in the ring A. Then the subgroups a + b and a ∩ b are also left (resp. right) ideals of A. We have already seen an important use of sums of ideals. Indeed, the greatest common divisor of two integers, m and n, was deﬁned to be the generator of the ideal (m) + (n). We can generalize the notion of the ideals generated by a set. Deﬁnitions 7.2.21. Let A be any ring. Let a be a left ideal of A, let b be a right ideal of A, and let S be any subset of A. We write aS = {a1 x1 + · · · + ak xk | k ≥ 1, ai ∈ a and xi ∈ S for 1 ≤ i ≤ k} Sb = {x1 b1 + · · · + xk bk | k ≥ 1, xi ∈ S and bi ∈ b for 1 ≤ i ≤ k} Thus, under either of the above deﬁnitions, ab = {a1 b1 + · · · + ak bk | k ≥ 1, ai ∈ a and bi ∈ b for 1 ≤ i ≤ k}. These constructions behave as follows. Lemma 7.2.22. Let A be any ring. Let a be a left ideal of A, let b be a right ideal of A, and let S be any subset of A. Then aS is a left ideal of A and Sb is a right ideal of A. Combining these two facts, we see that ab is a two-sided ideal of A. Finally, if a and b are both two-sided ideals of A, then we have an inclusion of twosided ideals ab ⊂ a ∩ b. CHAPTER 7. RINGS AND MODULES 220 Proof We leave the assertions of the ﬁrst paragraph to the reader. For the ﬁnal statement, consider an element of the form a1 b1 + · · · + ak bk with ai ∈ a and bi ∈ b for all i. Since b is a left ideal, each term ai bi lies in b. Since a is a right ideal, each term ai bi lies in a. So their sum lies in a ∩ b. We shall explore the eﬀects of taking products and intersections of ideals in Z in the next set of exercises. The following use of sums and intersections can be very useful in the study of rings, as well as for the module theory we shall introduce in Section 7.7. Proposition 7.2.23. (Chinese Remainder Theorem) Suppose given two-sided ideals a1 , . . . , ak of the ring A such that ai + aj = A whenever i = j. Let f : A → A/a1 × · · · × A/ak be given by f (a) = (a, . . . , a). Then f induces a ring isomorphism ∼ = f : A/b −→ A/a1 × · · · × A/ak , where b = k i=1 ai . Proof By the obvious extension of Lemma 7.1.24 to k-fold products, kf is a ring homomorphism. Thus, it suﬃces to show that f is surjective, with kernel i=1 ai . The latter statement is immediate, as the kernel of a map into a product is always the intersection of the kernels of the associated maps into the factors. Thus, we need only show that f is surjective. But the product A/a1 × · · · × A/ak is generated as an abelian group by the images of the A/ai under the canonical inclusions. Thus, it suﬃces to show that these images are contained in the image of f . We show this for i = 1. Now, we claim that it suﬃces to show that (1, 0, . . . , 0) is in the image of f . To see this, suppose that f (x) = (1, 0, . . . , 0). Since f is a ring homomorphism f (ax) = (a, . . . , a) · (1, 0, . . . , 0) = (a, 0, . . . , 0) for all a ∈ A, and hence A/a1 × 0 × · · · × 0 is indeed contained in the image of f . We now make use of the fact that a1 + ai = A for all i = 1. Thus, for 2 ≤ i ≤ k, there are elements ai ∈ a1 and bi ∈ ai , with ai + bi = 1. Then if πi : A → A/ai is the canonical map for 1 ≤ i ≤ k, we have π1 (bi ) = 1 and πi (bi ) = 0 for 2 ≤ i ≤ k. Let b = b2 · · · bk . Since the canonical maps πi are ring homomorphisms, π1 (b) = 1, while πi (b) = 0 for 2 ≤ i ≤ k. But this just says that f (b) = (1, 0, . . . , 0) as desired. Exercises 7.2.24. 1. Let A and B be rings. Show that the left ideals of A × B all have the form a × b, where a is a left ideal of A and b is a left ideal of B. 2. Let A be an integral domain and let a, b ∈ A. Show that (a) = (b) if and only if there is a unit x ∈ A× such that ax = b. 3. Let A be a ring. Show that there is a ring homomorphism from Zn to A if and only if the characteristic of A divides n. Show that this ring homomorphism, if it exists, is unique. 4. Let f : A → B be a ring homomorphism. Show that f −1 (a) is a left (resp. right or two-sided) ideal of A for each left (resp. right or two-sided) ideal a of B. CHAPTER 7. RINGS AND MODULES 221 † 5. Let f : A → B be a surjective ring homomorphism. Show that f (a) is a left (resp. right or two-sided) ideal of B whenever a is a left (resp. right or two-sided) ideal of A. Deduce that there is a one-to-one correspondence between the ideals of B and the ideals of A containing ker f . 6. Show that every subgroup of Zn is a principal ideal, and hence that Zn is a principal ideal ring. 7. Find a ring homomorphism f : A → B that is not surjective and an ideal a ⊂ A such that f (a) is not an ideal of B. 8. Let a ⊂ A be a left ideal. Show that a = A if and only if a contains an element which has a left inverse. 9. Show that a principal left ideal Ax = A if and only if x has a left inverse. 10. A division ring has been shown to have no two-sided ideals other than 0 and the ring itself. Show that the paucity of two-sided ideals does not characterize division rings: Show that if K is a ﬁeld and n ≥ 2, then Mn (K) has no two-sided ideals other than 0 and itself. (Hint: The cleanest way to show this uses linear algebra. The reader may wish to come back to this problem after reading Sections 7.9 and 7.10.) † 11. Repeat the preceding problem, replacing K by a division ring D. † 12. Let A be a ring. Deﬁne the opposite ring Aop as follows. As an abelian group, Aop is just A. For a ∈ A, we write ) a for the associated element of Aop . We deﬁne the ) Show that Aop is a ring. Show that the left multiplication in Aop by ) a · )b = ba. ideals of A are in one-to-one correspondence with the right ideals of Aop , and that the right ideals of A are in one-to-one correspondence with the left ideals of Aop . ‡ 13. Let A be a ring. An anti-automorphism of A is an automorphism f : A → A of the additive group of A, such that f (ab) = f (b)f (a) for all a, b ∈ A and f (1) = 1. Show that A admits an anti-automorphism if and only if A and Aop are isomorphic. 14. We say that a ring is self-opposite if it admits an anti-automorphism. Show that commutative rings are self-opposite. 15. Let A be a self-opposite ring. Show that Mn (A) is self-opposite for all n ≥ 1. (Hint: See Problem 5 of Exercises 7.10.23.) 16. Let m, n ∈ Z. Calculate the generator of the product ideal (m)(n). 17. Let m, n ∈ Z. Calculate the generator of (m) ∩ (n). 18. Let n = pr11 . . . prkk , where p1 , . . . , pk are distinct primes. Show that Zn is isomorphic as a ring to the direct product Zpr11 × · · · × Zprk . k 19. Use the preceding problem to give a quick proof that if n = pr11 . . . prkk , then Aut(Zn ) is isomorphic to Aut(Zpr11 ) × · · · × Aut(Zprk ). k 20. Suppose given an inﬁnite family {ai | i ∈ I} of left ideals of A. Deﬁne the sum of this inﬁnite family by ai = {ai1 + · · · + aik | k ≥ 1, i1 , . . . , ik ∈ I, and aij ∈ aij for 1 ≤ j ≤ k}. i∈I CHAPTER 7. RINGS AND MODULES 222 (a) Show that i∈I ai is the smallest left ideal that contains each ai . (b) If S is any subset of A, show that AS = x∈S Ax. 7.3 Polynomials One of the most important constructions in ring theory is that of polynomial rings. We study the polynomials in one variable now and study the polynomials in several variables in a subsection (page 228). Deﬁnitions 7.3.1. Let A be a ring (not necessarily commutative) and let X be a variable. We deﬁne a ring A[X] as follows. The elements of A[X] are formal sums n i a X = an X n + · · · + a1 X + a0 , where n is a nonnegative integer and ai ∈ A for all i i=0 n i. We often use functional notation for these sums, writing, say, f (X) = i=0 ai X i for the element above. If an = 0, we say that f (X) has degree n and that an is its leading coeﬃcient. Otherwise, m if m is the largest integer for which am = 0, we identify f (X) with the polynomial i=0 ai X i = am X m + · · · + a1 X + a0 , and hence it has degree m. (In the case of 0 = 0X 0 , there are no possible identiﬁcations with lower degree polynomials, and we say that 0 has degree −∞.) We say that a polynomial f (X) is monic if its leading coeﬃcient is 1. As is implicit in the notations above, we write either a or aX 0 for the degree 0 polynomial whose 0-th coeﬃcient is a. We write ι : A → A[X] for the map given by ι(a) = aX 0 for all a ∈ A. We also write X n for 1X n and write X for X 1 . Note that ntwo polynomials are equal if and only if their coeﬃcients are equal. (Here, if f (X) = i=0 ai X i , then the coeﬃcients of the X m for m > n are all implicitly n 0.) Addition of polynomials is obtained by adding the coeﬃcients: If f (X) = i=0 ai X i n and g(X) = i=0 bi X i (where n is the larger of the degrees of f and g), then f (X) + g(X) = n (ai + bi )X i . i=0 of polynomials is given as follows: if f (X) = mMultiplication i b X , then their product is given by i i=0 ⎛ ⎞ n+m i ⎝ aj bi−j ⎠ X i . f (X)g(X) = i=0 n i=0 ai X i and g(X) = j=0 Note that if A is noncommutative, then the order of the products ai bi−j is important. Finally, we give a special name to some of the simplest polynomials: We shall refer to a polynomial of the form aX i for some a ∈ A and i ≥ 0 as a monomial. Note that every polynomial is a sum of monomials, so that the monomials generate A[X] as an abelian group. We will occasionally use other letters, such as T or Y for the variable of a polynomial ring. We shall refer to the variable as an indeterminate element over the ring A. Again, the reader should justify the deﬁnitions: Lemma 7.3.2. Let A be a ring. Then A[X] is a ring under addition and multiplication of polynomials, and ι : A → A[X] is a ring homomorphism. The element X ∈ A[X] commutes with every element of ι(A), and if A is commutative, so is A[X]. CHAPTER 7. RINGS AND MODULES 223 Note that since every polynomial is a sum of monomials, the distributive law shows that the multiplication in A[X] is determined by the rule aX i · bX j = abX i+j . In particular, we see that if A is a subring of the real numbers R, then the multiplication formula is exactly the one we get by multiplying polynomial functions over R. So polynomial rings are familiar objects after all. Polynomial rings have a very important universal property. Proposition 7.3.3. Let g : A → B be a ring homomorphism and let b ∈ B be an element that commutes with every element of g(A). Then there is a unique ring homomorphism g b : A[X] → B such that g b (X) = b and the following diagram commutes. g /B AD DD z= z DD zz D zzgb ι DD z z ! A[X] Moreover, g b satisﬁes the formula " n # n i gb ai X = g(ai )bi . i=0 i=0 0 In this formula, we adopt the convention that b = 1 for any b. Proof For the uniqueness statement, note that any ring homomorphism g b that carries X to b and makes the diagram commute must satisfy the stated formula: " n # n i = gb ai X g b (ι(ai ))(g b (X))i i=0 i=0 = n g(ai )bi . i=0 Also, the stated formula does make the diagram commute and carry X to b. Thus, it suﬃces to show that g b , as deﬁned by the formula, gives a ring homomorphism. That it is a homomorphism of additive groups follows from the distributive law in B. And g b (1) = ι(1) = 1, so it suﬃces to show that g b (f1 (X)f2 (X)) = g b (f1 (X))g b (f2 (X)) for all f1 (X), f2 (X) ∈ A[X]. Since the monomials generate A[X] additively, distributivity shows that it is suﬃcient to check this for the case that f1 (X) and f2 (X) are monomials. Thus, we must check that for a, a ∈ A and i, j ≥ 0, we have g(aa )bi+j = g(a)bi g(a )bj . Since b commutes with any element of g(A), so does bi by induction. Thus, g(a)bi g(a )bj = g(a)g(a )bi bj . That bi bj = bi+j follows from the power laws for monoids (Problem 5 of Exercises 2.2.18). Thus, the result follows from the fact that g is a ring homomorphism. If A is commutative, then so is A[X], and A[X] is an A-algebra via ι : A → A[X]. As such, it satisﬁes an important universal property. Corollary 7.3.4. Let A be a commutative ring and let B be an A-algebra, via ν : A → B. Then for each b ∈ B, there is a unique A-algebra homomorphism εb : A[X] → B with the property that εb (X) = b. It is given by the formula n n εb ( ai X i ) = ν(ai )bi , i=0 i=0 CHAPTER 7. RINGS AND MODULES 224 where we set b0 = 1 for any b. Proof Since ν : A → B is the structure map for B as an A-algebra, each element of ν(A) commutes with each b ∈ B. Thus, Proposition 7.3.3 provides a unique ring homomorphism ν b : A[X] → B with ν b ◦ ι = ν (i.e., ν b is an A-algebra map) and ν b (X) = b. So we set εb = ν b . The formula comes from Proposition 7.3.3. Deﬁnitions 7.3.5. Let B be an A-algebra and let b ∈ B. We shall refer to the unique A-algebra homomorphism, εb : A[X] → B, that carries X to b as the evaluation map obtained by evaluating X at b. In consonance with this language, for f (X) ∈ A[X], we sometimes write εb (f (X)) = f (b), as it is obtained by evaluating f (X) at X = b. We shall write A[b] for the image of εb . We call it the A-subalgebra of B generated by b, or the algebra obtained by adjoining b to A. Note that since εb : A[X] → A[b] is onto, A[b] must be commutative even if B is not. Examples 7.3.6. Recall that every ring is a Z-algebra in a unique way. Consider the complex numbers, C, as a Z-algebra, and form the Z-subalgebra, Z[i], obtained by adjoining i to Z. The ring Z[i] is known as the Gaussian integers. Note that i has order 4 in the unit group C× . Thus, the powers of i consist of i, −1, −i, and 1. Collecting terms, we see that every element of Z[i] has the form m + ni, with m, n ∈ Z, and hence Z[i] = {m + ni ∈ C | m, n ∈ Z}. As the addition and multiplication operations are inherited from those in C, we have (m + ni) + (k + li) = (m + k) + (n + l)i and (m + ni) · (k + li) = (mk − nl) + (ml + nk)i for m, n, k, l ∈ Z. Note also that since C is a Q-algebra, the same arguments may be used if we replace Z by Q: Adjoining i to Q, we obtain Q[i] = {a + bi ∈ C | a, b ∈ Q}. Recall from Lemma 7.1.35 that for n ≥ 1, the complex number ζn = ei2π/n has order n in C× . Also, ζ4 = i. Thus, the next example is a generalization of the previous one. Examples 7.3.7. The non-negative powers of ζn are 1, ζn , . . . , ζnn−1 . Thus, if A = Z or Q, the elements of A[ζn ] may all be written as sums a0 + a1 ζn + · · · + an−1 ζnn−1 with a0 , . . . , an−1 ∈ A. We shall refer to Z[ζn ] as the cyclotomic integers obtained by adjoining a primitive n-th root of unity and to Q[ζn ] as the cyclotomic extension of Q obtained by adjoining a primitive n-th root of unity. As we saw in the case of ζ4 = i, elements of A[ζn ], with A = Z or Q, do not have unique representations as sums of the form a0 + a1 ζn + · · · + an−1 ζnn−1 with a0 , . . . , an−1 ∈ A. We shall defer the determination of a unique representation for these elements until Section 8.8. We shall show in Proposition 8.2.5 that the rings Q[ζn ] are all subﬁelds of C. Remarks 7.3.8. Let A be a commutative ring. Note that if A is a subring of the center of B and if we view B as an A-algebra whose structure map is the inclusion of A in B, then for b ∈ B, we have n n εb ( ai X i ) = ai bi , i=0 n A[b] = { i=0 and hence i=0 ai bi | n ≥ 0 and ai ∈ A for 0 ≤ i ≤ n}. CHAPTER 7. RINGS AND MODULES 225 We now give an important property of A[b]. Lemma 7.3.9. Let B be an A-algebra and let b ∈ B. Then A[b] is the smallest Asubalgebra of B that contains b. Proof An A-subalgebra of B is any subring of B containing ν(A), where ν : A → B is the structure map of B as an A-algebra. Thus, if C is an A-subalgebra of B that n contains b, it must contain every sum i=0 ν(ai )bi , and hence contains A[b]. Since A[b] is an A-subalgebra of B containing b, the result follows. We shall now compute the kernels of the evaluation maps εa : A[X] → A, for a ∈ A and give some applications. The next result is a version of the Euclidean algorithm. n i Proposition 7.3.10. Let A be any ring and let g(X) = i=0 ai X be a polynomial whose leading coeﬃcient, an , is a unit in A. Then for any polynomial f (X) ∈ A[X], we may ﬁnd polynomials q(X), r(X) ∈ A[X] such that f (X) = q(X)g(X) + r(X) and such that the degree of r(X) is strictly less than that of g(X). Proof We argue by induction on the degree of f (X). If f (X) has degree less than n, then we may simply take q(X) = 0 and take r(X) = f (X), and the statement is satisﬁed. m Thus, suppose that f (X) = i=0 bi X i with bm = 0 and that the result is true for all polynomials of degree less than m. In particular, we may assume that m ≥ n. But m−n then (bm a−1 )g(X) is a polynomial of degree m whose leading coeﬃcient is the n X m−n )g(X) has degree less than m. By same as that of f (X), and hence f (X) − (bm a−1 n X m−n induction, we may write f (X) − (bm a−1 X )g(X) = q1 (X)g(X) + r(X), where r(X) n m−n ) and we has degree less than n. But then we may take q(X) = q1 (X) + (bm a−1 n X have f (X) = q(X)g(X) + r(X), as desired. As the reader may recall from calculus, the simplest application for division of polynomials is to the theory of roots. Deﬁnition 7.3.11. Let A be a commutative ring and let f (X) ∈ A[X]. Let B be an A-algebra. We say that b ∈ B is a root of f if f (X) lies in the kernel of the evaluation map εb : A[X] → B obtained by evaluating X at b. In particular, a ∈ A is a root of f if and only if f (a) = 0. Corollary 7.3.12. Let A be a commutative ring and let a ∈ A. Then the kernel of the evaluation map εa : A[X] → A is (X − a), the principal ideal generated by X − a. In particular, a ∈ A is a root of f (X) ∈ A[X] if and only if f (X) = (X − a)q(X) for some q(X) ∈ A[X]. Proof Since X − a has degree 1 and leading coeﬃcient 1, we may write any polynomial f (X) ∈ A[X] as f (X) = q(X)(X − a) + r(X) where r(X) has degree ≤ 0. In other words, f (X) = q(X)(X − a) + r, where r ∈ A. But the evaluation map is a ring homomorphism, so that f (a) = q(a)(a − a) + r = r. Thus, f (X) ∈ ker εa if and only if r = 0. Thus, f (X) ∈ ker εa if and only if f (X) = q(X)(X − a) for some q(X) ∈ A[X]. But f (X) = q(X)(X − a) just says that f (X) is in the principal ideal generated by X − a. Corollary 7.3.13. Let A be an integral domain and let f (X) ∈ A[X] be a polynomial of degree n. Then f has at most n distinct roots in A. CHAPTER 7. RINGS AND MODULES 226 Proof Suppose that a ∈ A is a root of f . Then f (X) = (X − a)g(X), where g(X) has degree n − 1. Thus, by induction, there are at most n − 1 roots of g. Thus, it suﬃces to show that if b = a is a root of f , then b is a root of g. But 0 = f (b) = (b − a)g(b). Since b − a = 0 and since A is an integral domain, b must be a root of g. We obtain a very useful consequence for ﬁeld theory. Corollary 7.3.14. Let K be a ﬁeld. Then any ﬁnite subgroup of K × is cyclic. Proof Recall from Problem 4 of Exercises 4.4.24 that a ﬁnite abelian group is cyclic if and only if it has at most d elements of exponent d for each d which divides its order. We claim that K × itself has at most d elements of exponent d for any integer d. To see this, note that 1 is the identity element of K × , so that if a ∈ K × has exponent d, we have ad = 1. But this says that a is a root of the polynomial X d − 1. Since X d − 1 has degree d, it can have at most d distinct roots, and hence K × can have at most d distinct elements of exponent d. As promised during our discussion of group theory, we obtain the following corollary. Corollary 7.3.15. Let K be a ﬁnite ﬁeld. Then K × is a cyclic group. Corollary 7.3.15 is a useful consequence of the determination of the kernel of the evaluation map εa : A[X] → A. Later on, we shall derive useful consequences from the determinations of the kernels of the evaluation maps εb : A[X] → B for particular A-algebras B, and elements b ∈ B. Thus, Corollary 7.3.12 should be thought of as a ﬁrst step in an important study. Exercises 7.3.16. 1. Give the proof of Lemma 7.3.2. 2. Suppose that A is noncommutative. What are the elements of the center of A[X]? 3. Let f (X), g(X) ∈ A[X]. Show that the degree of f (X) + g(X) is less than or equal to the larger of the degrees of f and g. 4. Let f (X), g(X) ∈ A[X]. Show that the degree of f (X) · g(X) is less than or equal to the sum of the degrees of f and g. 5. Show that A[X] is an integral domain if and only if A is. 6. Show that the units of A[X] are the image under ι of the units of A. 7. Let M= 1 0 1 1 ∈ M2 (R). Consider the evaluation map εM : R[X] → M2 (R). Show that a polynomial f (X) ∈ R[X] lies in the kernel of εM if and only if f (1) = 0 and f (1) = 0, where f (X) is the derivative of f in the sense of calculus. CHAPTER 7. RINGS AND MODULES 8. Let M= 227 a 1 0 a ∈ M2 (R) for some real number a. Consider the evaluation map εM : R[X] → M2 (R). Show that a polynomial f (X) ∈ R[X] lies in the kernel of εM if and only if f (a) = 0 and f (a) = 0, where f (X) is the derivative of f in the sense of calculus. 9. Let p be a prime. What are the elements of Z[1/p] ⊂ Q? 10. What are the elements of A[X 2 ] ⊂ A[X]? 11. What are the elements of A[X n ] ⊂ A[X]? 12. What are the elements of Z[2X] ⊂ Z[X]? 13. Show that C = R[i]. ‡ 14. Show that Q[i] is a subﬁeld of C. √ 15. What are the elements of Z[ 2] ⊂ R? √ 16. What are the elements of Q[ 2] ⊂ R? 17. Let m be a positive, odd integer. Show that Z[ζm ] = Z[ζ2m ] as a subring of C. 18. Show that every element of Z[ζ8 ] may be written in the form a + bζ8 + cζ82 + dζ83 with a, b, c, d ∈ Z. Write out the formula for the product (a + bζ8 + cζ82 + dζ83 ) · (a + b ζ8 + c ζ82 + d ζ83 ). 19. Let A be a ring and let 1 = a ∈ A. Show that ak − 1 = (a − 1)(1 + a + · · · + ak−1 ) for k ≥ 1. ‡ 20. Cyclotomic units Let m and k be relatively prime to n. Show that the element (ζnm − 1)/(ζnk − 1) ∈ Q[ζn ] actually lies in Z[ζn ]. Deduce that (ζnm − 1)/(ζnk − 1) is a unit in Z[ζn ]. More generally, if (m, n) = (k, n) = n, show that (ζnm − 1)/(ζnk − 1) is a unit in Z[ζn ]. The elements (ζnm − 1)/(ζnk − 1) are often called the cyclotomic units. 21. Let n > 3 be odd. Show that the cyclotomic unit (ζn2 − 1)/(ζn − 1) has inﬁnite order in Z[ζn ]× . 22. If (m, n) = 1 and m ≡ ±1 mod n, show that the cyclotomic unit (ζnm − 1)/(ζn − 1) has inﬁnite order in Z[ζn ]× . (Hint: If 0 < φ < π/2, show that ei(θ−φ) + ei(θ+φ) = reiθ for some positive real number r. Show also that r > 1 if φ < π/3.) 23. Let n be an even number greater than 6. Show that Z[ζn ]× is inﬁnite. 24. Show that in the category of rings, Z[X] is the free object on a one-element set. CHAPTER 7. RINGS AND MODULES 228 25. Show that every element of Z6 is a root of X 3 − X ∈ Z6 [X]. Deduce that the hypothesis that A be an integral domain in Corollary 7.3.13 is necessary. 26. Show that X 2 + 1 has inﬁnitely many roots in the quaternion ring, H. (The issue here is not zero-divisors, as H is a division ring. The point here is that evaluating polynomials at an element α ∈ H does not give a ring homomorphism from H[X] to H unless α lies in the center of H.) !∞ 27. Show that X p − X has inﬁnitely many roots in i=1 Zp . 28. Let p1 , . . . , pk be distinct primes and let m be the least common multiple of p1 − 1, . . . , pk − 1. Show that every element of Zn is a root of X m+1 − X, where n = p1 . . . pk . 29. Let n > 1 be an integer and let (n, X) be the ideal of Z[X] generated by n and X. Show that Z[X]/(n, X) is isomorphic to Zn . 30. Let B be any ring and let A be the center of B. Let a ∈ A. Show that there is a unique homomorphism εa : B[X] → B such that εa (X) = a and εa ◦ ι = 1B , where ι : B → B[X] is the standard inclusion. Show that the kernel of εa is the two-sided ideal of B[X] generated by X − a. 31. Let A be a ring and X a variable. We deﬁne the ring of formal power series, A[[X]], with coeﬃcients in A. ∞ The elements of A[[X]] are formal sums i=0 ai X i , with ai ∈ A for all i ≥ 0. We shall refer to them as formal power series with coeﬃcients in A. Unlike the case of polynomial rings, we do not assume that only ﬁnitely many of the coeﬃcients are nonzero; indeed, all of the coeﬃcients may be nonzero. Here, the sum and product operations are given by ∞ "∞ i=0 i=0 ai X i + ∞ bi X i i=0 = ∞ (ai + bi )X i , and i=0 ⎛ ⎞ # "∞ # ∞ i ⎝ = ai X i · bi X i aj bi−j ⎠ X i . i=0 i=0 j=0 We have an inclusion map η : A[X] → A[[X]], that identiﬁes a polynomial, f (X), of degree n with the formal power series whose coeﬃcients of index ≤ n agree with those of f , and whose higher coeﬃcients are all 0. (a) Show that A[[X]] is a ring and that η is a ring homomorphism. (b) Show that ∞the polynomial 1 − X is a unit in A[[X]] (but not in A[X]) with inverse i=0 X i = 1 + X + X 2 + · · · + X n + · · · . ∞ (c) Let A be commutative. Show that a formal power series i=0 ai X i is a unit in A[[X]] if and only if a0 ∈ A× . Polynomials in Several Variables We can also take polynomials in more than one variable. When considering polynomials in several variables, the most common case, which is the default if nothing more is said, is to assume that the variables commute with each other. In fact, we shall assume that CHAPTER 7. RINGS AND MODULES 229 this is the case whenever we use the word polynomial, but some people would explicitly call these “polynomials in several commuting variables.” Deﬁnitions 7.3.17. Suppose given n variables X1 , . . . , Xn . A primitive7 monomial in these variables is a product X1i1 · · · Xnin , with ij ≥ 0 for 1 ≤ j ≤ n. We abbreviate X1i1 · · · Xnin = X I for I = (i1 , . . . , in ) and write I = {(i1 , . . . , in ) | i1 , . . . , in ∈ Z, i1 , . . . , in ≥ 0} for the set of all n-tuples of non-negative integers. (We call the elements of I multiindices.) For any ring A, we deﬁne a monomial in X1 , . . . , Xn with coeﬃcient in A to be a product aX1i1 · · · Xnin where a ∈ A and X1i1 · · · Xnin is a primitive monomial in X1 , . . . , Xn . A polynomial in X1 , . . . , Xn with coeﬃcients in A is a ﬁnite sum of monomials in X1 , . . . , Xn with coeﬃcients in A. The elements of the polynomial ring A[X1 , . . . , Xn ] are the polynomials in X1 , . . . , Xn with coeﬃcients in A. Since there’s no preferred order for the multi-indices of the monomials, we shall write the generic polynomial in the form f (X1 , . . . , Xn ) = aI X I , I∈I where all but ﬁnitely many of the coeﬃcients aI ∈ A are 0. So f is the sum of the ﬁnitely many monomials whose coeﬃcients are nonzero. Addition of polynomials is obtained as in the one-variable case, by adding the co I eﬃcients of the monomials: if f (X , . . . , X ) = a X and g(X1 , . . . , Xn ) = 1 n I I∈I I b X , then I I∈I f (X1 , . . . , Xn ) + g(X1 , . . . , Xn ) = (aI + bI )X I . I∈I The multiplication in A[X1 , . . . , Xn ] is set up to have the following eﬀect on monomials: If I = (i1 , . . . , in ) and J = (j1 , . . . , jn ), then aX I · bX J = abX I+J , where I + J = (i1 + j1 , . . . , in + jn ). For generic polynomials, this is expressed as follows: If f (X1 , . . . , Xn ) = I∈I aI X I and g(X1 , . . . , Xn ) = I∈I bI X I , then ⎛ ⎞ ⎝ aJ bI−J ⎠ X I . f (X1 , . . . , Xn ) · g(X1 , . . . , Xn ) = I∈I J≤I Here, for I = (i1 , . . . , in ) and J = (j1 , . . . , jn ) in I, J ≤ I means that jk ≤ ik for all indices k (so that ≤ is only a partial order on I), and we deﬁne I − J to be the expected multi-index (i1 − j1 , . . . , in − jn ). We write 0 for the multi-index whose coordinates are all 0, and identify a ∈ A with a · X 0 . We write ι : A → A[X1 , . . . , Xn ] for the map that takes each a ∈ A to a = aX 0 ∈ A[X1 , . . . , Xn ]. We identify each primitive monomial X I with 1X I and identify each variable Xi with the primitive monomial X (0,...,0,1,0,...,0) , where the 1 occurs in the i-th place. When n = 2, we shall customarily write A[X, Y ] in place of A[X1 , X2 ]. We require the usual justiﬁcations. 7 The use of the word primitive here has no connection whatever to the notion of a primitive element in a Hopf algebra. CHAPTER 7. RINGS AND MODULES 230 Lemma 7.3.18. Suppose given a ring A and variables X1 , . . . , Xn . Then the polynomials A[X1 , . . . , Xn ] form a ring under the above addition and multiplication formulæ, and ι : A → A[X1 , . . . , Xn ] is a ring homomorphism. Each variable Xi commutes with every element of ι(A), and the variables Xi and Xj commute with each other for all i, j. If A is commutative, so is A[X1 , . . . , Xn ]. More generally, a polynomial f (X1 , . . . , Xn ) lies in the center of A[X1 , . . . , Xn ] if and only if its coeﬃcients lie in the center of A. Polynomials in several variables satisfy universal properties similar to those of polynomials in one variable. We leave the proofs to Exercises 7.3.27. Proposition 7.3.19. Let g : A → B be a ring homomorphism and suppose given elements b1 , . . . , bn ∈ B such that the following two conditions hold. 1. Each bi commutes with every element of g(A). 2. For i, j ∈ {1, . . . , n}, bi commutes with bj . Then there is a unique ring homomorphism g b1 ,...,bn : A[X1 , . . . , Xn ] → B such that g b1 ,...,bn (Xi ) = bi for 1 ≤ i ≤ n and the following diagram commutes. g /B A LL LLL rr8 r r LLL r rrr ι LLL rrr gb1 ,...,bn % A[X1 , . . . , Xn ] Moreover, g b1 ,...,bn satisﬁes the formula " # I = g b1 ,...,bn aI X g(aI )bI . I∈I I Here, if I = (i1 , . . . , in ), we set b = bi11 I∈I . . . binn , where b0i is set equal to 1 for all i. As in the case of a single variable, Proposition 7.3.19 admits a useful restatement in the context of algebras over a commutative ring. Corollary 7.3.20. Let A be a commutative ring and let B be an A-algebra, via ν : A → B. Suppose given b1 , . . . , bn ∈ B such that bi commutes with bj whenever i, j ∈ {1, . . . , n}. Then there is a unique A-algebra homomorphism εb1 ,...,bn : A[X1 , . . . , Xn ] → B with the property that εb (Xi ) = bi for 1 ≤ i ≤ n. It is given by the formula " # I = εb1 ,...,bn aI X ν(aI )bI . I∈I I∈I Here, if I = (i1 , . . . , in ), we set bI = bi11 . . . binn , where b0i is set equal to 1 for all i. We now codify some terms. Deﬁnitions 7.3.21. Let A be a commutative ring. Let B be an A-algebra and let b1 , . . . , bn ∈ B such that bi commutes with bj whenever i, j ∈ {1, . . . , n}. We shall refer to the A-algebra homomorphism εb1 ,...,bn as the evaluation map obtained by evaluating Xi at bi for 1 ≤ i ≤ n. We write A[b1 , . . . , bn ] for the image of εb1 ,...,bn . We call A[b1 , . . . , bn ] the A-subalgebra of B generated by b1 , . . . , bn . Suppose now that B is a commutative A-algebra. We say that B is ﬁnitely generated as an A-algebra (or has ﬁnite type as an A-algebra) if B = A[b1 , . . . , bn ] for some collection of elements b1 , . . . , bn ∈ B. CHAPTER 7. RINGS AND MODULES 231 A special case of the evaluation maps is of fundamental importance in algebraic geometry. Proposition 7.3.22. Let A be a commutative ring and let a1 , . . . , an ∈ A. Let a = (X1 − a1 , . . . , Xn − an ), the ideal of A[X1 , . . . , Xn ] generated by X1 − a1 , . . . , Xn − an . Then a is the kernel of the evaluation map εa1 ,...,an : A[X1 , . . . , Xn ] → A. Proof Since each Xi −ai lies in the kernel of εa1 ,...,an , Corollary 7.2.19 shows that there is a unique ring homomorphism εa1 ,...,an : A[X1 , . . . , Xn ]/a → A such that the following diagram commutes. εa1 ,...,an / A[X1 , . . . , Xn ] q8 A RRR q q RRR qq RR qqq q π RRRR q ε a ,...,a n R( qq 1 A[X1 , . . . , Xn ]/a Here, π : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn ]/a is the canonical map. It suﬃces to show that εa1 ,...,an is an isomorphism. Let ι = π ◦ ι : A → A[X1 , . . . , Xn ]/a, where ι is the A-algebra structure map of A[X1 , . . . , Xn ]. Then εa1 ,...,an ◦ ι = εa1 ,...,an ◦ π ◦ ι = εa1 ,...,an ◦ ι = 1A , where the last equality comes from the fact that εa1 ,...,an is an A-algebra homomorphism. Thus, it suﬃces to show that ι is onto. Since every element of A[X1 , . . . , Xn ] is a sum of monomials and since π is onto, it suﬃces to show that π(aX1i1 . . . Xnin ) lies in the image of ι for all a ∈ A and all n-tuples (i1 , . . . , in ) of nonnegative integers. Since π(Xi − ai ) = 0, we have π(Xi ) = ι(ai ) for all i. Thus, π(aX1i1 . . . Xnin ) = ι(aai11 . . . ainn ), and the result follows. The next result could have been left as an exercise in universal properties, but it is suﬃciently important that we give it here. Proposition 7.3.23. Let X1 , . . . , Xn be indeterminates over the ring A. Then there is a unique ring homomorphism α : (A[X1 , . . . , Xn−1 ])[Xn ] → A[X1 , . . . , Xn ] that carries Xi to Xi for 1 ≤ i ≤ n, and commutes with the natural inclusions of A in the two sides. Moreover, α is an isomorphism whose inverse is the unique ring homomorphism β : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn−1 ][Xn ] which carries Xi to Xi for 1 ≤ i ≤ n, and commutes with the natural inclusions of A in the two sides. Thus, A[X1 , . . . , Xn ] may be considered to be the polynomial ring in the variable Xn with coeﬃcients in A[X1 , . . . , Xn−1 ]. In particular, every polynomial f (X1 , . . . , Xn ) in n variables over A may be written uniquely in the form f (X1 , . . . , Xn ) = fk (X1 , . . . , Xn−1 )Xnk + · · · + f0 (X1 , . . . , Xn−1 ) for some k ≥ 0, where fi (X1 , . . . , Xn−1 ) ∈ A[X1 , . . . , Xn−1 ] for all i. CHAPTER 7. RINGS AND MODULES 232 Proof Here, the natural inclusion of A in A[X1 , . . . , Xn−1 ][Xn ] is the composite ι ι A− → A[X1 , . . . , Xn−1 ] − → A[X1 , . . . , Xn−1 ][Xn ], where ι is the standard inclusion of A in the polynomial ring in n − 1 variables and ι is the standard inclusion of A[X1 , . . . , Xn−1 ] in its polynomial ring in the variable Xn . Notice that the variables Xi lie in the centers of each of the rings under consideration. Thus, the existence and uniqueness of β is immediate from Proposition 7.3.19. For α, note that Proposition 7.3.19 provides a unique homomorphism α0 : A[X1 , . . . , Xn−1 ] → A[X1 , . . . , Xn ] that carries each Xi to Xi and commutes with the natural inclusions of A in both sides. But now the universal property of a polynomial ring in a single variable allows us to extend α0 to a map α carrying Xn to Xn . For the uniqueness of α, we claim that any two ring homomorphisms f, g : A[X1 , . . . , Xn−1 ][Xn ] → B that agree on A and agree on each of the variables Xi must be equal. To see this, note ﬁrst that f and g must agree on A[X1 , . . . , Xn−1 ] by the uniqueness property of Proposition 7.3.19. But then, since f and g also agree on Xn , they must be equal by the uniqueness property of Proposition 7.3.3. But now notice that each of the composites α ◦ β and β ◦ α carries each Xi to Xi and commutes with the standard inclusions of A. Thus, α ◦ β is the identity by the uniqueness property of Proposition 7.3.19, while β ◦ α is the identity by the uniqueness property derived in the preceding paragraph. If A is commutative and f (X1 , . . . , Xn ) ∈ A[X1 , . . . , Xn ], then there is a function f : An → A which takes (a1 , . . . , an ) to f (a1 , . . . , an ), the eﬀect of evaluating f at (a1 , . . . , an ). Indeed, much of our intuition about polynomials will have come from calculus, where polynomials are treated as functions. Of course, if A is ﬁnite, then there are only ﬁnitely many functions from An to A, but inﬁnitely many polynomials in A[X1 , . . . , Xn ]. Thus, in this case, a polynomial is not determined by its eﬀect as a function on An . Finiteness is not !∞the only diﬃculty in trying to treat polynomials as functions. For instance, if A = i=1 Z2 , then the polynomial X 2 is easily seen to induce the identity map of A, which is also induced by X. This suggests that if we wish to interpret polynomials as functions, then we might think about restricting attention to integral domains with inﬁnitely many elements. First, we shall formalize the situation. Deﬁnition 7.3.24. Let X be any set and let A be a ring. The ring of functions from X to A, written Map(X, A), is the ring whose elements are the functions from X to A and whose operations are given by (f + g)(x) = f (x) + g(x) and (f · g)(x) = f (x)g(x) for all f, g ∈ Map(X, A) and x ∈ X. The additive and multiplicative identities are the constant functions to 0 and 1, respectively.8 The next lemma follows from the fact that the evaluation maps are ring homomorphisms. 8 If X is a compact topological space, then the ring C(X) of continuous functions from X to C plays an important role in the theory of C ∗ algebras. Of course, C(X) is a subring of Map(X, C). CHAPTER 7. RINGS AND MODULES 233 Lemma 7.3.25. Let A be a commutative ring and let φ : A[X1 , . . . , Xn ] → Map(An , A) be the map which takes each polynomial f to the function obtained by evaluating f on the elements of An . Then φ is a ring homomorphism. Thus, two polynomials, f and g, induce the same function from An to A if and only if f − g ∈ ker φ. Thus, the elements of A[X1 , . . . , Xn ] are determined by their eﬀect as functions if and only if ker φ = 0. The next proposition gives a slightly stronger result when A is an inﬁnite domain. Proposition 7.3.26. Let S be an inﬁnite subset of the integral domain A. Suppose given a polynomial f (X1 , . . . , Xn ) in n variables over A with the property that f (a1 , . . . , an ) = 0 for all a1 , . . . , an ∈ S. Then f = 0. Thus, an element of A[X1 , . . . , Xn ] is determined by its eﬀect on S × · · · × S ⊂ An for any inﬁnite subset S of A. Proof We argue by induction on n, with the result being true for n = 1, as f would have inﬁnitely many roots. Suppose, inductively, that the result is true for polynomials of n − 1 variables, and let a ∈ S. Then evaluating Xn at a gives a polynomial f (X1 , . . . , Xn−1 , a) in n − 1 variables over A, which vanishes on all (n − 1)-tuples, a1 , . . . , an−1 of elements of S. By the inductive assumption, f (X1 , . . . , Xn−1 , a) = 0. Write B = A[X1 , . . . , Xn−1 ]. By Proposition 7.3.23, we may identify A[X1 , . . . , Xn ] with B[Xn ]. Under this identiﬁcation, f (X1 , . . . , Xn−1 , a) is the image of f under the evaluation map εa : B[Xn ] → B obtained by evaluating Xn at a. Thus, when f is thought of as an element of B[Xn ], each a ∈ S is a root of f . Since B is a domain and since f has inﬁnitely many roots in B, f = 0. Exercises 7.3.27. 1. Give the proof of Lemma 7.3.18. 2. Give the proof of Proposition 7.3.19. 3. Give the proof of Corollary 7.3.20. ‡ 4. Let B be an A-algebra and let b1 , . . . , bn ∈ B such that bi commutes with bj whenever i, j ∈ {1, . . . , n}. Show that A[b1 , . . . , bn ] is the smallest A-subalgebra of B containing b1 , . . . , bn . 5. Let A be a commutative ring. What are the elements of A[X 2 , X 3 ] ⊂ A[X]? 6. Let A be a commutative ring. What are the elements of A[X 3 , X 4 ] ⊂ A[X]? 7. Let A be a commutative ring and let X and Y be indeterminates. Consider the A-algebra structure on A[X] × A[Y ] given by the structure map Δ ι×ι A −→ A × A −−→ A[X] × A[Y ] where Δ is the diagonal map Δ(a) = (a, a) and the maps ι are the standard structure maps for the polynomial rings A[X] and A[Y ]. What are the elements of the A-subalgebra of A[X] × A[Y ] generated by (X, 0) and (0, Y )? CHAPTER 7. RINGS AND MODULES 234 8. Show that Z[1/2, 1/3] = Z[1/6] as subrings of Q. (Hint: By Problem 4, it suﬃces to show that 1/2, 1/3 ∈ Z[1/6] and that 1/6 ∈ Z[1/2, 1/3].) 9. Show that Q is not ﬁnitely generated as a Z-algebra. 10. Let m and n be relatively prime positive integers. Show that Z[ζm , ζn ] = Z[ζmn ] 2π as subrings of C. Here, ζk = ei· k is the standard primitive k-th root of unity. 11. Let A be a commutative ring. Show that A[X1 , . . . , Xn ] is the free commutative A-algebra on the set {X1 , . . . , Xn }. 7.4 Symmetry of Polynomials We consider some issues relating to permutations of the variables in a polynomial ring. Throughout this section, A is a commutative ring. Suppose given n variables, X1 , . . . , Xn . Then the symmetric group on n letters, Sn , acts on the set of variables in the obvious way: σ · Xi = Xσ(i) . By Corollary 7.3.20, the induced map σ : {X1 , . . . , Xn } → {X1 , . . . , Xn } ⊂ A[X1 , . . . , Xn ] extends uniquely to an A-algebra homomorphism σ∗ : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn ] via σ∗ (f (X1 , . . . , Xn )) = f (Xσ(1) , . . . , Xσ(n) ). Since an A-algebra homomorphism out of A[X1 , . . . , Xn ] is determined by its eﬀect on the variables, σ ·f = σ∗ (f ) is easily seen to give an action of Sn on A[X1 , . . . , Xn ] through A-algebra homomorphisms. We ﬁrst consider the ﬁxed points of this action. Deﬁnition 7.4.1. We write A[X1 , . . . , Xn ]Sn for the ﬁxed points of the action of Sn on A[X1 , . . . , Xn ]. Thus, a polynomial f (X1 , . . . , Xn ) is in A[X1 , . . . , Xn ]Sn if and only if σ∗ (f ) = f for all σ ∈ Sn . We shall refer to A[X1 , . . . , Xn ]Sn as the symmetric polynomials in n variables over A. The reader may easily check that the symmetric polynomials form an A-subalgebra of A[X1 , . . . , Xn ]. We shall show that it is a polynomial ring on certain symmetric polynomials. We shall argue by an induction in which the next lemma will play a part. Lemma 7.4.2. Let π : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn−1 ] be the A-algebra map that takes Xi to Xi if i < n and takes Xn to 0. Then π carries the symmetric polynomials in n variables into the symmetric polynomials in n − 1 variables, inducing an A-algebra homomorphism π : A[X1 , . . . , Xn ]Sn → A[X1 , . . . , Xn−1 ]Sn−1 . Proof Recall from Proposition 7.3.23 that A[X1 , . . . , Xn ] may be identiﬁed with the polynomial ring on A[X1 , . . . , Xn−1 ] in the variable Xn . Thus, any nonzero element of A[X1 , . . . , Xn ] may be written uniquely in the form f (X1 , . . . , Xn ) = m i=0 fi (X1 , . . . , Xn−1 )Xni CHAPTER 7. RINGS AND MODULES 235 for some m ≥ 0, with f0 , . . . , fm ∈ A[X1 , . . . , Xn−1 ], and fm = 0. Now Sn−1 acts as permutations on the ﬁrst n − 1 variables of the polynomials in n variables, and if σ ∈ Sn−1 , then σ∗ (f ) = m σ∗ (fi )Xni . i=0 But by the uniqueness of the representation of polynomials in n variables as polynomials in Xn with coeﬃcients in A[X1 , . . . , Xn−1 ] we see that if f ∈ A[X1 , . . . , Xn ]Sn , then each fi must be ﬁxed by each σ ∈ Sn−1 . Thus, if f is a symmetric polynomial in n variables, then each fi is a symmetric polynomial in n − 1 variables. But since π(Xn ) = 0, we see that π(f ) = f0 , and the result follows. We now deﬁne the elements we will show to be the polynomial generators of the symmetric polynomials. Deﬁnition 7.4.3. The elementary symmetric polynomials, or elementary symmetric functions, on n variables are deﬁned by ⎧ 1 if i = 0 ⎪ ⎨ X . . . X if 1 ≤ i ≤ n j j 1 i si (X1 , . . . , Xn ) = ⎪ ⎩ j1 <···<ji 0 if i > n. More generally, if B is a commutative ring and if b1 , . . . , bn ∈ B, then we write si (b1 , . . . , bn ) ∈ B for the element obtained by evaluating si (X1 , . . . , Xn ) at b1 , . . . , bn . Note that we may do this because the elementary symmetric functions have integer coeﬃcients. Thus, for instance, s1 (X1 , . . . , Xn ) = X1 + · · · + Xn , or at the other extreme, sn (X1 , . . . , Xn ) = X1 · · · Xn . Note that the summands of si (X1 , . . . , Xn ) are the products of i distinct variables, and that each such product occurs exactly once. Thus, if σ ∈ Sn , then σ∗ permutes the summands of each si (X1 , . . . , Xn ), and hence the elementary symmetric functions are indeed symmetric polynomials. The elementary symmetric functions are ubiquitous in mathematics. One of the reasons for this is that they give the formula for computing the coeﬃcients of a polynomial with a predetermined collection of roots. Lemma 7.4.4. Suppose given n elements, a1 , . . . , an ∈ A, not necessarily distinct. Then the following equality holds in A[X]. (X − a1 ) . . . (X − an ) = X n − s1 X n−1 + · · · + (−1)n sn , where si = si (a1 , . . . , an ) for i = 1, . . . , n. Proof In the expansion of (X − a1 ) . . . (X − an ) from the distributive law, the terms of degree k in X are those obtained by choosing k of the factors from which to choose an X, leaving n − k factors of the form (−ai ). Adding these up, we get ⎞ ⎛ ⎝ (−aj1 ) . . . (−ajn−k )⎠ · X k = (−1)n−k sn−k (a1 , . . . , an ) · X k j1 <···<jn−k for the monomial of degree k in the resulting polynomial in X. CHAPTER 7. RINGS AND MODULES 236 The reader may easily verify the following lemma. Lemma 7.4.5. Let π : A[X1 , . . . , Xn ]Sn → A[X1 , . . . , Xn−1 ]Sn−1 be obtained by setting Xn = 0, as above. Then si (X1 , . . . , Xn−1 ) if 0 ≤ i < n π(si (X1 , . . . , Xn )) = 0 if i ≥ n There are various deﬁnitions of degree for a polynomial in several variables. The notion of total degree is one of them. Deﬁnition 7.4.6. The total degree of a monomial aX1r1 . . . Xnrn is r1 + · · · + rn . The total degree of a polynomial f ∈ A[X1 , . . . , Xn ] is the maximum of the total degrees of the nonzero monomials in f . We are now ready to show that A[X1 , . . . , Xn ]Sn is a polynomial algebra on the generators s1 (X1 , . . . , Xn ), . . . , sn (X1 , . . . , Xn ). Proposition 7.4.7. Let αn : A[Y1 , . . . , Yn ] → A[X1 , . . . , Xn ]Sn be the A-algebra homomorphism that carries Yi to si (X1 , . . . , Xn ), where Y1 , . . . , Yn are indeterminates over A. Then α is an isomorphism. Proof We argue by induction on n. Since s1 (X) = X, the result is trivial for n = 1. Thus, we shall assume, inductively, that αn−1 is an isomorphism. We ﬁrst show that αn is surjective. Suppose given a symmetric polynomial f (X1 , . . . , Xn ). We shall argue by induction on the total degree of f that f is in the image of α. The result is trivial in total degree 0. Since αn−1 is surjective, there is a polynomial, g, of n − 1 variables such that π(f ) = g(s1 (X1 , . . . , Xn−1 ), . . . , sn−1 (X1 , . . . , Xn−1 )) Thus, our calculation of the eﬀect of π on the elementary symmetric functions shows that if we set h(X1 , . . . , Xn ) = f (X1 , . . . , Xn ) − g(s1 (X1 , . . . , Xn ), . . . , sn−1 (X1 , . . . , Xn )), then h lies in the kernel of π. m i Thus, if we write h(X1 , . . . , Xn ) = i=0 hi (X1 , . . . , Xn−1 )Xn , we see that h0 = π(h) = 0, and hence h(X1 , . . . , Xn ) = Xn · k(X1 , . . . , Xn ) for some polynomial k. Let πi : A[X1 , . . . , Xn ] → A[X1 , . . . , Xn−1 ] be the A-algebra map obtained by setting ⎧ if j < i ⎨ Xj 0 if j = i πi (Xj ) = ⎩ Xj−1 if j > i, and write πi : A[X1 , . . . , Xn ]Sn → A[X1 , . . . , Xn−1 ]Sn−1 for the induced map. Thus, π = πn . Since h is in the kernel of π, symmetry shows that it must also be in the kernel of πi for 1 ≤ i ≤ n − 1. Since h(X1 , . . . , Xn ) = Xn · k(X1 , . . . , Xn ), we have 0 = πn−1 (h(X1 , . . . , Xn )) = πn−1 (Xn )πn−1 (k(X1 , . . . , Xn )). Since πn−1 (Xn ) = Xn−1 is not a zero divisor in A[X1 , . . . , Xn−1 ], k must lie in the kernel of πn−1 , and hence k(X1 , . . . , Xn ) = Xn−1 · p(X1 , . . . , Xn ) for a suitable polynomial p, by the argument above. CHAPTER 7. RINGS AND MODULES 237 An easy induction now shows that h(X1 , . . . , Xn ) = X1 · · · Xn · q(X1 , . . . , Xn ) for a suitable polynomial q. Since both h and X1 · · · Xn = sn (X1 , . . . , Xn ) are symmetric, and since sn (X1 , . . . , Xn ) is not a zero divisor, we see that q must be a symmetric polynomial. Note that by the known properties of π, h must have total degree less than or equal to that of f , while q has total degree less than that of h. Thus, q(X1 , . . . , Xn ) is in the image of αn by induction. But then f must also be in the image of αn . To see that αn is injective, we argue by contradiction. Let f (Y1 , . . . , Yn ) be an element of the kernel of αn with the lowest possible total degree. Write f (Y1 , . . . , Yn ) = m i f (Y i 1 , . . . , Yn−1 )Yn . Note that i=0 0 = (π ◦ αn )(f ) = f0 (s1 (X1 , . . . , Xn−1 ), . . . , sn−1 (X1 , . . . , Xn−1 )). Since αn−1 is injective, this forces f0 (Y1 , . . . , Yn−1 ) to be 0. But then f (Y1 , . . . , Yn ) = Yn · g(Y1 , . . . , Yn ), and hence αn (f ) = sn (X1 , . . . , Xn ) · αn (g). Thus, g lies in the kernel of αn . Since its total degree is less than that of f , this contradicts the minimality of the degree of f . Another important topic regarding symmetries of polynomials concerns discriminants. Proposition 7.4.8. Consider the following element of A[X1 , . . . , Xn ]: . (Xj − Xi ). δ(X1 , . . . , Xn ) = i<j Then for σ ∈ Sn , we have σ∗ (δ) = ε(σ) · δ, where ε(σ) is the sign of σ. Proof We may may identify the set of indices i < j in the above product with the set, S, of all two-element subsets of {1, . . . , n}. Of course, Sn acts on S, via σ({i, j}) = {σ(i), σ(j)}. Using this action to re-index the factors of δ, we see that . (Xmax(σ(i),σ(j)) − Xmin(σ(i),σ(j)) ). δ(X1 , . . . , Xn ) = i<j For i < j, σ∗ (Xj − Xi ) = Xσ(j) − Xσ(i) = ±(Xmax(σ(i),σ(j)) − Xmin(σ(i),σ(j)) ), where the sign is positive if σ preserves the order of i and j and is negative otherwise. Thus, σ∗ (δ) = (−1)k δ, where k is the number of pairs i < j for which σ(j) < σ(i). Deﬁne ψ : Sn → ±1 by ψ(σ) = (−1)k where (−1)k is deﬁned by the equation σ∗ (δ) = (−1)k δ. Then it follows immediately from the fact that the homomorphisms σ∗ deﬁne an action of Sn on A[X1 , . . . , Xn ] that ψ is a homomorphism. It suﬃces to show that ψ = ε. Since Sn is generated by transpositions, it suﬃces to show that if τ is a transposition, then ψ(τ ) = −1. Thus, let τ = (i j), with i < j. Then the pairs on which τ reverses order consist of the pairs i < k and k < j, where i < k < j, together with the pair i < j itself. The signs associated with i < k < j cancel each other out, leaving a single −1 from the pair i < j. Deﬁnition 7.4.9. The discriminant, Δ(X1 , . . . , Xn ), is deﬁned by Δ = δ 2 , where δ = ! i<j (Xj − Xi ), as above. CHAPTER 7. RINGS AND MODULES 238 Since each σ∗ (δ) = ±δ, we obtain the following corollary. Corollary 7.4.10. The discriminant, Δ, is a symmetric polynomial in the variables X1 , . . . , Xn . Thus, there is a unique polynomial Fn (X1 , . . . , Xn ), such that Δ(X1 , . . . , Xn ) = Fn (s1 (X1 , . . . , Xn ), . . . , sn (X1 , . . . , Xn )). Example 7.4.11. The discriminant on two variables is given by Δ(X, Y ) = (Y − X)2 = X 2 + Y 2 − 2s2 (X, Y ). Now note that (s1 (X, Y ))2 = (X + Y )2 = X 2 + Y 2 + 2s2 (X, Y ). Thus, Δ = s21 − 4s2 , and hence F2 (X, Y ) = X 2 − 4Y . We shall calculate F3 (X, Y, Z) in the next set of exercises. It takes considerably more work than F2 . Since the polynomials Fn (X1 , . . . , Xn ) have integer coeﬃcients, one may save considerable time by using a computer algebra package to calculate them. Recall from Lemma 7.4.4 that if a polynomial f (X) factors as a product of monic polynomials of degree 1, then the coeﬃcients of f are, up to sign, the elementary symmetric functions on the roots. Deﬁnition 7.4.12. Let f (X) = a0 X n + a1 X n−1 + · · · + an be a polynomial of degree n in A[X] and let Fn (X1 , . . . , Xn ) be the polynomial determined in Corollary 7.4.10. Then the discriminant, Δ(f ), of f is the following element of A: Δ(f ) = Fn (−a1 /a0 , a2 /a0 , . . . , (−1)n an /a0 ). Of course, we don’t know much at this point about the polynomials Fn , so the following corollary expresses the discriminant in a more familiar setting. It follows immediately from Lemma 7.4.4 and Corollary 7.4.10. Corollary 7.4.13. Let f (X) be a monic polynomial in A[X], and suppose that f factors as f (X) = (X − b1 ) . . . (X − bn ) in B[X], where B is a commutative ring that contains A as a subring. Then the discriminant, Δ(f ), may be calculated by Δ(f ) = = Δ(b1 , . . . , bn ) . (bj − bi )2 . i<j Exercises 7.4.14. 1. Write X12 + · · · + Xn2 as a polynomial in the elementary symmetric functions on X1 , . . . , Xn . CHAPTER 7. RINGS AND MODULES 239 2. Show that πi : A[X1 , . . . , Xn ]Sn → A[X1 , . . . , Xn−1 ]Sn−1 is injective when restricted to the A-submodule consisting of the symmetric polynomials of total degree less than n. 3. Let f (X) = aX 2 + bX + c be a quadratic in A[X]. Show that Δ(f ) = (b2 − 4ac)/a2 . Deduce that a quadratic f (X) ∈ R[X] has a root in R if and only if Δ(f ) ≥ 0. 4. Calculate the polynomial F3 (X, Y, Z) that’s used to calculate the discriminant of a cubic. † 5. Show that if f (X) = X 3 + pX + q is a monic cubic with no X 2 term, then Δ(f ) = −4p3 − 27q 2 . 7.5 Group Rings and Monoid Rings Polynomial rings are a special case of a more general construction called monoid rings. Deﬁnitions 7.5.1. Let M be a monoid, and write the product operation of M in multiplicative notation, even if M is abelian. We deﬁne A[M ], the monoid ring of M over A, as follows. The elements of A[M ] are formal sums m∈M am m, such that the coefﬁcient am is zero for all but ﬁnitely many m in M . (We abbreviate this by saying that am = 0 almost polynomials, addition is given by adding everywhere.) As in the case of coeﬃcients: ( m∈M am m) + ( m∈M bm m) = m∈M (am + bm )m. The multiplication is deﬁned as follows. ⎛ ⎞ #" # " ⎝ am m bm m = ax by ⎠ m. m∈M m∈M m∈M {(x,y) | xy=m} The last sum ranges over all ordered pairs (x, y) of elements in M whose product is m. In the generic monoid, that collection of ordered pairs can be diﬃcult to determine. But in the examples we are concerned with, there will be no diﬃculty. We identify each m ∈ M with the element of A[M ] whose m-th coeﬃcient is 1 and whose other coeﬃcients are all 0. Thus, we have a natural inclusion M ⊂ A[M ]. Similarly, we identify a ∈ A with the element of A[M ] whose m-th coeﬃcient is 0 for all m = 1, and whose coeﬃcient of 1 is equal to a. Here, we write 1 for the identity element of M . We write ι : A → A[M ] for the map that gives this identiﬁcation. In the case where the monoid is a group G, we call A[G] the group ring (or group algebra, if A is commutative) of G over A. Example 7.5.2. Write Z2 = T = {1, T }. Then for any ring A, A[Z2 ] = {a + bT | a, b ∈ A}. Here, if a, b, c, d ∈ A, we have (a+bT )+(c+dT ) = (a+c)+(b+d)T and (a+bT )·(c+dT ) = (ac + bd) + (ad + bc)T . Lemma 7.5.3. Let A be a ring and let M be a monoid. Then A[M ] is a ring under the above operations, and ι : A → A[M ] is a ring homomorphism. The inclusion M ⊂ A[M ] is a monoid homomorphism from M to the multiplicative monoid of A[M ], and each m ∈ M commutes with every element of ι(A). If A is commutative, then A[M ] is an A-algebra, via ι : A → A[M ]. CHAPTER 7. RINGS AND MODULES 240 Proof Everything here is straightforward, with the possible exception of the associativity of multiplication in A[M ]. We shall treat that, and leave the rest to the reader. Thus, suppose given α, β, γ ∈ A[M ], with α= am m, β = bm m and γ = cm m. m∈M m∈M m∈M Then αβ = dm m, where dm = ax by . {(x,y) | xy=m} m∈M Thus, (αβ)γ = ⎛ ⎝ m∈M = ⎛ ⎛ ⎝ {(w,z) | wz=m} ⎝ m∈M dw cz ⎠ m {(w,z) | wz=m} ⎝ m∈M = ⎛ ⎞ ⎞ ⎞ ax by ⎠ cz ⎠ m {(x,y) | xy=w} ⎞ ax by cz ⎠ m. {(x,y,z) | xyz=m} Here, the last equality comes from distributing the terms cz through the sums {(x,y) | xy=w} ax by . For each ordered triple (x, y, z) with xyz = m, there is exactly one summand ax by cz thus obtained. As the reader may now check, the expansion of α(βγ) produces exactly the same result. Monoid rings have an important universal property. Proposition 7.5.4. Let f : A → B be a ring homomorphism. Let M be a monoid and let g : M → B be a homomorphism from M to the multiplicative monoid of B. Suppose that for each m ∈ M , g(m) commutes with every element of f (A). Then there is a unique ring homomorphism h : A[M ] → B such that the following diagram commutes: ⊃ ι / A[M ] o A? M ?? ?? ?? ?? h g f ?? B Explicitly, we have " h m∈M # am m = f (am )g(m). m∈M Proof The element m∈M am m is, in fact, a ﬁnite sum of terms am m. Moreover, if j : M ⊂ A[M ] is the inclusion, then am m = ι(am )j(m). Thus, if h : A[M ] → B is a ring CHAPTER 7. RINGS AND MODULES 241 homomorphism making the diagram commute, then we must have h m∈M am m = m∈M f (am )g(m), and hence h is unique as claimed. Thus, it suﬃces to show that if f and g satisfy the stated hypotheses, then h, as deﬁned by the formula above, is a ring homomorphism. By the distributive law in B and the fact that f is a homomorphism of abelian groups, we see that h is also a homomorphism of abelian groups. Also, h(1) = f (1)g(1) = 1, so it suﬃces to show that h(αβ) = h(α)h(β) for all α, β ∈ A[M ]. By the distributive laws in both A[M ] and B, it suﬃces to show that h(am · a m ) = h(am)h(a m ) for all a, a ∈ A and m, m ∈ M . But h(am · a m ) = h(aa mm ) = f (aa )g(mm ) = f (a)f (a )g(m)g(m ), while h(am)h(a m ) = f (a)g(m)f (a )g(m ), so the result follows from the fact that f (a ) commutes with g(m). In the case of algebras over a commutative ring, the statement becomes simpler. Corollary 7.5.5. Let A be a commutative ring and let B be an A-algebra. Let M be a monoid and let g : M → B be a monoid homomorphism from M to the multiplicative monoid of B. Then there is a unique A-algebra homomorphism h : A[M ] → B such that h the composite M ⊂ A[M ] − → B is equal to g. Explicitly, " # h am m = ν(am )g(m), m∈M m∈M where ν : A → B is the structure map of B as an A-algebra. Proof If B is an A-algebra with structure map ν, then ν(A) is contained in the center of B, and hence every element of ν(A) commutes with every element of g(M ). Thus, Proposition 7.5.4 provides a unique ring homomorphism h, given by the displayed formula, such that the diagram in the statement of Proposition 7.5.4 commutes, with f replaced by ν. But the left-hand triangle of the diagram commutes if and only if h is an A-algebra homomorphism, while the right-hand triangle commutes if and only if the h composite M ⊂ A[M ] − → B is equal to g. Note that if G is a group, then a monoid homomorphism g : G → B from G to the multiplicative monoid of B always takes value in B × , the group of invertible elements in the multiplicative monoid of B. Thus, g restricts to give a group homomorphism g : G → B × . We obtain the following corollary. Corollary 7.5.6. Let A be a commutative ring and let B be an A-algebra. Let G be a group and let f : G → B × be a homomorphism. Then there is a unique A-algebra h homomorphism h : A[G] → B such that the composite G ⊂ A[G] − → B is equal to f . Explicitly, ⎛ ⎞ h⎝ ag g ⎠ = ν(ag )f (g), g∈G g∈G where ν : A → B is the structure map of B as an A-algebra. There is an important homomorphism from a group ring A[G] to A. Deﬁnitions 7.5.7. Let A be a ring and let G be a group. Then the standard augmentation of the group ring A[G] is the unique ring homomorphism (which exists by CHAPTER 7. RINGS AND MODULES 242 Proposition 7.5.4) ε : A[G] → A with the property that ε(g) = 1 for all g ∈ G and such that ε◦ι = 1A , where ι is the standard inclusion of A in A[G]. Explicitly, Proposition 7.5.4 gives ⎛ ⎞ ε⎝ ag g ⎠ = ag . g∈G g∈G We deﬁne the augmentation ideal, I(G) (or IA (G), if the ring A is allowed to vary), to be the kernel of ε. Thus, we can obtain information about A[G] from information about A and information about I(G). We shall study augmentation ideals in the next set of exercises. In the case that A is commutative, augmentation ideals admit a nice generalization to the study of certain A-algebras. Deﬁnition 7.5.8. Let A be a commutative ring. An augmented A-algebra consists of an A-algebra, B, together with an A-algebra homomorphism ε : B → A. Here, since A is given the A-algebra structure map 1A , this simply means that ε is a ring homomorphism such that ε ◦ ν = 1A , where ν is that A-algebra structure map of B. The ideal of the augmentation ε is its kernel. Examples 7.5.9. 1. Let A be a commutative ring and let a ∈ A. Then the evaluation map εa : A[X] → A is an augmentation. By Corollary 7.3.12, the ideal of this augmentation is the principal ideal (X − a). 2. Similarly, if A is commutative and a1 , . . . , an ∈ A, then the evaluation map εa1 ,...,an : A[X1 , . . . , Xn ] → A is an augmentation. Here, Proposition 7.3.22 tells us the augmentation ideal is (X1 − a1 , . . . , Xn − an ). 3. Let M be any monoid and let A be a commutative ring. Then the trivial monoid homomorphism from M tothe multiplicative monoid of A induces an augmentation map ε : A[M ] → A via ε( m∈M am m) = m∈M am . Exercises 7.5.10. 1. Complete the proof of Lemma 7.5.3. 2. Show that in the group ring of a group G over A, we may write ax by = ax bx−1 g . {(x,y) | xy=g} x∈G 3. Let Z2 = T = {1, T }. For any ring A, show that (1 − T )(1 + T ) = 0 in A[Z2 ], and hence A[Z2 ] has zero-divisors. 4. Show that A[Zn ] has zero-divisors for any ring A and any integer n > 1. Deduce that if A is any ring and if G is a group with torsion elements of order > 1, then A[G] has zero-divisors. 5. Let A be a ring and let G be a ﬁnite group. Let Σ ∈ A[G] be the sum of all the elements of G: Σ = g∈G g. Show that for any x ∈ A[G], xΣ = (ι ◦ ε(x))Σ, where ε : A[G] → A is the standard augmentation and ι : A → A[G] is the standard inclusion. CHAPTER 7. RINGS AND MODULES 243 ‡ 6. Let ζn = ei·2π/n ∈ C be the standard primitive n-th root of unity. Let A be any subring of C and let Zn = T = {1, T, . . . , T n−1 }. Show that there is a unique A-algebra homomorphism f : A[Zn ] → A[ζn ] such that f (T ) = ζn . Show that f is onto. ‡ 7. Let Zn = T = {1, T, . . . , T n−1 }. Let f : Z[Zn ] → Z[ζn ] be the homomorphism of Problem 6. (a) Show that Σ = 1 + T + · · · + T n−1 lies in the kernel of f , and hence there’s an induced, surjective ring homomorphism f : Z[Zn ]/(Σ) → Z[ζn ], where (Σ) is the principal ideal generated by Σ. (b) Suppose given integers m, k which are relatively prime to n. Let l > 0 be an integer such that m ≡ kl mod n, and let umk = 1 + T k + · · · + (T k )l−1 , where the right-hand side is interpreted as lying in Z[Zn ]/(Σ). Show that if we chose a diﬀerent integer l with m ≡ kl mod n, then the above formula gives rise to precisely the same element of Z[Zn ]/(Σ). Show that umk is a unit in Z[Zn ]/(Σ), and that f (umk ) is the cyclotomic unit (ζnm − 1)/(ζnk − 1) from Problem 20 of Exercises 7.3.16. 8. Show that there is a commutative diagram of ring homomorphisms Z[Zn ] π / Z[Zn ]/(Σ) ε ε Z / Zn π where ε is the standard augmentation, the maps π are the canonical maps to the respective quotient rings, and Σ = 1 + T + · · · + T n−1 as in Problem 7. (a) Show that the above diagram induces an isomorphism ∼ = h : Z[Zn ] −→ Z ×Zn (Z[Zn ]/(Σ)) , via h(α) = (ε(α), π(α)) for α ∈ Z[Zn ]. (See Problem 17 of Exercises 7.1.27.) For this reason, we call the diagram a pullback square. Deduce that there is a pullback square: Z[Zn ]× π / × (Z[Zn ]/(Σ)) ε Z× ε π / Z× n CHAPTER 7. RINGS AND MODULES 244 (b) Let m1 , . . . , ms , k1 , . . . , ks be relatively prime to n and let r1 , . . . , rs > 0. Let u = urm11 k1 . . . urmss ks , the unit of Z[Zn ]/(Σ) constructed in Problem 7. Show that u lifts to a unit of Z[Zn ] if and only if mr11 . . . mrss ≡ ±k1r1 . . . ksrs mod n. 9. Let X = {X k | k ≥ 0}. Show that the operation X k · X l = X k+l turns X into a monoid isomorphic to the monoid, N, of non-negative integers under addition. Show that the monoid ring A[X] is isomorphic to the polynomial ring A[X] for any ring A. 10. Let X be as in the previous exercise and let Xn = X × · · · × X be the direct n times product of X with itself n times. For any ring A, show that the polynomial ring A[X1 , . . . , Xn ] is isomorphic to the monoid ring A[Xn ]. ‡ 11. Let A be a ring and let f : M → M be a monoid homomorphism. Show that there is a ring homomorphism f : A[M ] → A[M ] deﬁned by f ( ∗ ∗ m∈M am m) = ( a )m . −1 m m ∈M m∈f (m ) ‡ 12. Show that if f : A → B is a ring homomorphism and M is a monoid, then there is a ring homomorphism f∗ : A[M ] → B[M ] deﬁned by f∗ ( am m) = f (am )m. m∈M m∈M 13. Let G be a group and let A be a commutative ring. Let α = g∈G ag g be an element of A[G]. Show that α is in the center of A[G] if and only if axgx−1 = ag for all x, g ∈ G. The reader who has studied module theory should now show that the center of A[G] is free as an A-module, and give a basis for it. 14. Show that if A is a ring and if M1 and M2 are monoids, then there is an isomorphism between (A[M1 ])[M2 ] and A[M1 × M2 ]. (Hint: One could either argue directly, or via universal mapping properties. The latter method is quicker, and less cumbersome notationally.) 15. Let A and B be rings and let M be a monoid. Show that (A × B)[M ] is isomorphic to A[M ] × B[M ]. † 16. Let A be a ring and let a be the principal two-sided ideal in the polynomial ring A[X] generated by X n − 1. Show that the quotient ring A[X]/a is isomorphic to the group ring A[Zn ] of Zn over A. (Hint: There is a direct proof, that uses module theory, and there is an easier one using universal mapping properties.) 17. Let a be a two-sided ideal of A and let M be a monoid. Let π : A → A/a be the canonical map and let π∗ : A[M ] → (A/a)[M ] be the ring homomorphism given by π∗ ( m∈M am m) = m∈M π(am )m (see Problem 12). Show that π∗ is a surjection whose kernel is the two-sided ideal of A[M ] generated by a. (Here, we identify a with its image under ι : A → A[M ].) 18. Let A be any ring and let Z2 = T = {1, T }. Determine precisely which elements a+bT of A[Z2 ], a, b ∈ A, lie in the augmentation ideal IA (Z2 ). Deduce that IA (Z2 ) is the principal two-sided ideal generated by T − 1. CHAPTER 7. RINGS AND MODULES 245 19. Let A be a commutative ring and let G be a group. Let S ⊂ G be a generating set for G (i.e., G = S as in Deﬁnitions 2.2.11). Show that the augmentation ideal IA (G) is the two-sided ideal of A[G] generated by {x − 1 | x ∈ S}. (Hint: Consider the proof of Proposition 7.3.22.) 20. Let f : G → K be a split surjection of groups. Find a set of generators (as an ideal) for the kernel of the ring homomorphism f∗ : A[G] → A[K] given by f∗ ( g∈G ag g) = k∈K ( g∈f −1 k ag )k, where A is any commutative ring. (This map was shown to be a ring homomorphism in Problem 11.) 21. Show that a group ring A[G] over a commutative ring A is self-opposite. 22. Show that a group ring A[G] over a self-opposite ring A is self-opposite. 23. Show that if A is commutative, then the passage from monoids M to the monoid rings A[M ] gives a left adjoint to the forgetful functor from A-algebras to their multiplicative monoids. 24. Let A be commutative. Construct the free A-algebra on an arbitrary set X. 25. Let A be commutative. Construct the free commutative A-algebra on an arbitrary set X. (So far, we’ve only handled the case where X is ﬁnite.) 26. Show that if A is commutative, then the passage from groups G to the group rings A[G] provides a left adjoint to the functor that takes an A-algebra to its group of units. 7.6 Ideals in Commutative Rings We learned a lot about groups by studying their factor groups. Similarly, quotient rings will be useful for studying rings. The drawback here, of course, is that while the kernel of a surjective group homomorphism is again a group, the kernel of a surjective ring homomorphism is only an ideal and not a ring. Thus, there is no good analogue for a composition series for a ring. However, we can mimic the ﬁrst step by which the composition series for a ﬁnite group was constructed. In that step, we constructed a surjective homomorphism from our ﬁnite group to a simple group. The analogue for rings of a simple group is a ring with no two-sided ideals other than 0 and the ring itself, and hence has no useful quotient rings. Deﬁnition 7.6.1. A nonzero ring is Jacobson simple if it has no two-sided ideals other than 0 and the ring itself. Problem 11 of Exercises 7.2.24 shows that if D is a division ring, then Mn (D) is a Jacobson simple ring for all n ≥ 1. (We shall give a slicker argument for this in Chapter 12.) There are other examples of noncommutative Jacobson simple rings more complicated than these. But in the commutative case, Jacobson simple rings are as nice as we could ask. The next corollary is immediate from Lemma 7.2.7. Corollary 7.6.2. A commutative ring is Jacobson simple if and only if it is a ﬁeld. We wish to study those quotient rings A/a that are not the 0 ring, and hence, we are concerned with those ideals a that are not equal to A. We shall give them a name. CHAPTER 7. RINGS AND MODULES 246 Deﬁnition 7.6.3. An ideal a of A is proper if a ⊂ A is a proper inclusion (i.e., if a = A). We now introduce two important kinds of proper ideals. Deﬁnitions 7.6.4. Let A be a commutative ring. A proper ideal, p, of A is prime if ab ∈ p implies that either a or b must be in p. A proper ideal, m, of A is maximal if the only proper ideal containing m is m itself. Examples 7.6.5. Let A be a commutative ring. Then it is immediate from the deﬁnitions that the ideal 0 of A is prime if and only if A is an integral domain. Note that the ideal 0 is maximal if and only if A has no proper ideals other than 0. But we’ve just seen in Corollary 7.6.2 that this occurs if and only if A is a ﬁeld. These two examples give a characterization of prime and maximal ideals in terms of the quotient rings they determine. Lemma 7.6.6. Let A be a commutative ring. Then a proper ideal p of A is prime if and only if the quotient ring A/p is an integral domain. Also, a proper ideal m of A is maximal if and only if the quotient ring A/m is a ﬁeld. Thus, every maximal ideal is prime. Proof Let p be a prime ideal of A. We must show that A/p has no zero-divisors. Thus, let a and b be elements of A/p such that a · b = 0. But this says that ab = 0, and hence ab ∈ p. But then one of a or b is in p, and hence that element goes to 0 in A/p. Conversely, if A/p is an integral domain and if ab ∈ p, then ab = 0, so that one of a and b must be 0 in A/p. But then one of a and b must be in p. For the case of maximal ideals, the correspondence between the ideals in a quotient ring A/a and the ideals in A containing a (Problem 5 of Exercises 7.2.24) shows that a proper ideal m of A is maximal if and only if the 0 ideal is maximal in A/m. As shown in the preceding example, this occurs if and only if A/m is a ﬁeld. Examples 7.6.7. 1. For any prime p, the ideal (p) ⊂ Z is maximal, as Zp is a ﬁeld. In fact, Lemma 7.1.12 shows that if n > 1, then (n) is a prime ideal of Z if and only if n is a prime number. Thus, every nonzero prime ideal of Z is maximal. 2. Let K be a ﬁeld and let a ∈ K. Then the evaluation map εa : K[X] → K is surjective, so its kernel is a maximal ideal. By Corollary 7.3.12, ker εa is the principal ideal generated by X − a. p → Zp be the ring homomorphism from the p-adic integers onto Zp 3. Let π1 : Z that’s induced by the projection map. (See Lemma 7.1.38.) Then ker π1 , which by Problem 4 of Exercises 7.1.41 is the principal ideal generated by p, is maximal. The next lemma is an easy, but important observation. Lemma 7.6.8. Let f : A → B be a ring homomorphism between commutative rings and let p be a prime ideal of B. Then f −1 (p) is a prime ideal of A. Note that the analogue of the preceding lemma for maximal ideals is false: Let i : Z → Q be the inclusion. Then 0 is a maximal ideal of Q, while f −1 (0) = 0, which is prime in Z, but not maximal. Prime ideals satisfy an important primality principle with respect to the ideal product of Deﬁnitions 7.2.21. This would have made a good exercise, but its uses are ubiquitous enough that we give it here. CHAPTER 7. RINGS AND MODULES 247 Lemma 7.6.9. Let p be a prime ideal in the commutative ring A and let a and b be ideals with ab ⊂ p. Then at least one of a and b must be contained in p. Proof If ab ⊂ p and a ⊂ p, let a ∈ a with a ∈ p. Then ab ∈ p for all b ∈ b. Since a ∈ p, we must have b ∈ p for all b ∈ b. By Lemma 7.2.22, ab ⊂ a ∩ b for any ideals a, b of a commutative ring. Corollary 7.6.10. Let p be a prime ideal in the commutative ring A and let a and b be ideals such that a ∩ b ⊂ p. Then at least one of a and b must be contained in p. Prime ideals may be used to deﬁne a notion of dimension for a commutative ring. Deﬁnition 7.6.11. Let A be a commutative ring. We say that the Krull dimension of A is ≤ n if whenever we have a sequence p0 ⊂ p1 ⊂ · · · ⊂ pk of proper inclusions (i.e., inclusions that are not onto) of prime ideals of A, we have k ≤ n. Thus, the Krull dimension of A is ﬁnite if there is a maximum length for such sequences of proper inclusions of prime ideals in A, and if p 0 ⊂ p1 ⊂ · · · ⊂ pn is the longest such sequence that can occur in A, then the Krull dimension is n. Examples 7.6.12. 1. Since a ﬁeld has only one prime ideal, 0, every ﬁeld has Krull dimension 0. 2. The ﬁrst example in Examples 7.6.7 shows that the longest chains of proper inclusions of prime ideals in Z are those of the form 0 ⊂ (p), for p prime. Thus, Z has Krull dimension 1. Let us return to the analogy with group theory. As noted above, we’ve made use of the fact that any nontrivial ﬁnite group has a nontrivial simple group as a factor group. By analogy, we would like to show that any commutative ring has a Jacobson simple ring as a quotient ring. Since the only commutative rings that are Jacobson simple are ﬁelds, this is equivalent to showing that every commutative ring has a maximal ideal. To show this, we shall make use of Zorn’s Lemma, which is equivalent to the Axiom of Choice. The reader can ﬁnd the proof of this equivalence in any good book on set theory.9 First we shall need some setup. Recall that a partially ordered set is a set X together with a relation, ≤, which is reﬂexive (x ≤ x for all x ∈ X), transitive (if x ≤ y and y ≤ z, then x ≤ z), and antisymmetric (if x ≤ y and y ≤ x, then x = y). In a partially ordered set, it is not required that an arbitrary pair of elements in X be comparable by the relation. In other words, there may exist pairs x, y such that neither x ≤ y nor y ≤ x is true. On the other hand, a partially ordered set in which any pair of elements is comparable is called a totally ordered set. Let S be a subset of a partially ordered set. We say that x ∈ X is an upper bound for S if s ≤ x for all s ∈ S. Note that the upper bound x need not lie in S. Finally, a maximal element of X is an element y such that if x ∈ X with y ≤ x, then x and y must be equal. 9 There’s a good exposition of Zorn’s Lemma and the Well Ordering Principle in Topology, by James Dugundji, Allyn and Bacon, 1966. CHAPTER 7. RINGS AND MODULES 248 Lemma 7.6.13. (Zorn’s Lemma) Let X be a nonempty partially ordered set such that every totally ordered subset of X has an upper bound in X. Then X has at least one maximal element. Corollary 7.6.14. Let A be a commutative ring. Then every proper ideal of A is contained in a maximal ideal. Proof Let a be a proper ideal of A, and let X be the set of all proper ideals of A that contain a. Put a partial ordering on X by b ≤ b if b ⊂ b . Then a maximal element of X with respect to this ordering is precisely a maximal ideal containing a. Since a ∈ X, X is nonempty, so it suﬃces to show that any totally ordered subset of X has an upper bound in X. Let S be a totally ordered subset of X. We claim that the union b∈S b of the ideals in S is such an upper bound. Of course, unions of ideals are not generally ideals, but if we’re dealing with a totally ordered set of ideals, this problem disappears: For x, y ∈ b∈S b, we can ﬁnd b, b ∈ S such that x ∈ b and y ∈ b . But since any two elements of S are comparable, either b ≤ b or b ≤ b. But in either case, x+ y is in the larger of the two, as is ax for any a ∈ A. But then x + y and ax are in b∈S b, and hence the union is an ideal. Clearly, it is an upper bound for the ideals in S, so it suﬃces to show that it is proper. But if 1 ∈ b∈S b, then 1 must be an element of some b ∈ S. But this is impossible, since the ideals in X are all proper. If we make extra assumptions on our ring, we can obtain an even stronger sort of maximality principle for ideals. Deﬁnitions 7.6.15. Let A be a commutative ring. We say that A is Noetherian, or has the ascending chain condition (a.c.c.) if any ascending sequence of inclusions of ideals of A is eventually constant. In other words, given a sequence a 1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · · there is some n ≥ 1 such that an = ak for all k ≥ n. Alternatively, we say that A is Artinian, or has the descending chain condition (d.c.c.) if any decreasing sequence of ideals is eventually constant. This means that given a sequence a1 ⊃ a2 ⊃ · · · ⊃ ak ⊃ · · · there is some n ≥ 1 such that an = ak for all k ≥ n. It turns out that the Artinian condition (in a commutative ring) implies that the ring is Noetherian. The Artinian condition is a very strong one, and fails to hold in many rings of interest. For instance, as we shall see in the exercises below, the only Artinian integral domains are ﬁelds. The descending chain condition for left or right ideals turns out to be extremely useful for certain topics in noncommutative ring theory. For instance, descending chain conditions are important in the study of group rings K[G], where K is a ﬁeld and G is a ﬁnite group. Here is the strengthened maximality principle that holds in Noetherian rings. Proposition 7.6.16. Let A be a Noetherian ring and let X be the partially ordered set of ideals in A (ordered by inclusion). Let S be any nonempty subset of X. Then S, with the partial ordering inherited from X, has maximal elements. CHAPTER 7. RINGS AND MODULES 249 Proof We argue by contradiction. Suppose that S has no maximal elements, and let a1 be an element of S. Since a1 is not a maximal element of S, we can ﬁnd a proper inclusion a1 ⊂ a2 , with a2 in S. Since no element of S is maximal, induction now shows that we may choose a sequence a 1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · · of elements of S, such that each inclusion ak−1 ⊂ ak is proper.10 But this contradicts the hypothesis that A is Noetherian, and hence S must have maximal elements after all. We shall study Noetherian rings in much greater detail in the context of module theory. Exercises 7.6.17. 1. Let f : A → B be a surjective ring homomorphism. Show that there is a one-to-one correspondence between the prime ideals of B and those primes of A that contain ker f . Show also that there is a one-to-one correspondence between the maximal ideals of B and those maximal ideals of A containing ker f . 2. Let n > 1. Show that every prime ideal of Zn is maximal. Given the prime decomposition of n, list all the maximal ideals, m, of Zn and calculate the quotient ﬁelds Zn /m. 3. Show that an integral domain has Krull dimension 0 if and only if it is a ﬁeld. 4. Let p be a prime ideal in the commutative ring A and let A[X] · p be the ideal generated by p in A[X]. Show that A[X] · p is a prime ideal of A[X]. 5. Let A be an integral domain which is not a ﬁeld. Show that the principal ideal (X − a) of A[X] (here, a ∈ A) is a prime ideal that is not maximal. Deduce that A[X] has Krull dimension ≥ 2. 6. Let p be a prime and let n be any integer. Show that the ideal (p, X − n) of Z[X], generated by p and X − n, is maximal. 7. Let f : Q[X] → Q[i] ⊂ C be the unique Q-algebra map that carries X to i. Recall from Problem 14 of Exercises 7.3.16 that Q[i] is a ﬁeld. Show that ker f = (X 2 +1), and hence (X 2 + 1) is a maximal ideal in Q[X]. 8. Let Z4 = T = {1, T, T 2 , T 3 }. Recall from Problem 6 of Exercises 7.5.10 that there is a unique, surjective, Q-algebra homomorphism f : Q[Z4 ] → Q[i] with f (T ) = i. Show that ker f = (T 2 + 1), and hence (T 2 + 1) is a maximal ideal in Q[Z4 ]. 9. Let X1 , . . . , Xn be indeterminates over a ﬁeld K. Let a1 ∈ K and for 2 ≤ i ≤ n, let fi (X) ∈ K[X1 , . . . , Xi−1 ] ⊂ K[X1 , . . . , Xn ]. Show that (X1 − a1 , X2 − f2 (X), . . . , Xn − fn (X)) is a maximal ideal of K[X1 , . . . , Xn ]. 10. Let X1 , . . . , Xn be indeterminates over a ﬁeld K. Find a chain p0 ⊂ · · · ⊂ pn of proper inclusions of prime ideals of K[X1 , . . . , Xn ]. Deduce that K[X1 , . . . , Xn ] has Krull dimension ≥ n. 10 Note that the inﬁnite induction here makes use of the Axiom of Choice. Thus, while our conclusion is stronger than that of the Zorn’s Lemma argument for the existence of maximal ideals, the Noetherian property has not permitted us to dispense with the use of the choice axiom. CHAPTER 7. RINGS AND MODULES 250 11. Let A and B be commutative rings. What are the prime ideals of A × B? Suppose that A and B have ﬁnite Krull dimension. Calculate the Krull dimension of A × B in terms of the dimensions of A and B. 12. Show that Z is Noetherian but not Artinian. 13. Show that Zn is both Noetherian and Artinian. 14. Let A be an Artinian integral domain and let a be a nonzero element of A. Show that the descending chain condition on (a) ⊃ (a2 ) ⊃ · · · ⊃ (an ) ⊃ · · · implies that a is a unit. Deduce that A must be a ﬁeld. Deduce that a ﬁnite domain is a ﬁeld. 15. Let A be an Artinian commutative ring. Show that every prime ideal of A is maximal. 7.7 Modules Modules play a role in ring theory analogous to the role of G-sets in understanding groups. Among other things, module theory will enable us to obtain a better grasp of the theory of ideals, and hence of the structure of the rings themselves. We give some general concepts and constructions here, and study free modules and exact sequences in subsections. We then close the section with a subsection on the notion of rings without identity elements. Deﬁnitions 7.7.1. Let A be a ring. A left A-module is an abelian group M together with an operation that assigns to each a ∈ A and m ∈ M an element am ∈ M subject to the following rules: a(m + m ) = am + am (a + a )m = am + a m 1 · m = m, a(a m) = (aa )m, for all a, a ∈ A and all m, m ∈ M . If M is an abelian group, then an operation A × M → M making M an A-module is called an A-module structure on M . A submodule of M is a subgroup M ⊂ M of the additive group of M , such that am ∈ M for all a ∈ A and m ∈ M (i.e., M is closed under multiplication by elements of A). An A-module homomorphism between the left A-modules M and N is a homomorphism f : M → N of abelian groups such that f (am) = af (m) for all a ∈ A and all m ∈ M. Right modules and their homomorphisms are deﬁned by reversing the order of the a’s and m’s in the above deﬁnitions. CHAPTER 7. RINGS AND MODULES 251 Note that if K is a ﬁeld, then a K-module is nothing other than a vector space over K, and a K-module homomorphism between vector spaces is nothing other than a K-linear map. Another case where we understand what it means to be a module is that of Z-modules. The proof of the next lemma is similar to that of Lemma 7.1.6, which we left to the reader, so we shall leave this to the reader as well. Lemma 7.7.2. Every abelian group G is a Z-module. Moreover, the Z-module structure on G is unique: for n ∈ Z and g ∈ G, ng is the n-th power of g in the group structure of G. (Thus, if n > 0, ng = g + · · · + g, the sum of n copies of g.) Finally, every group homomorphism between abelian groups is a Z-module homomorphism. When A is commutative, any left A-module may be made into a right A-module by setting m · a = am. This process is clearly reversible, so if A is commutative, then there is a one-to-one correspondence between the left and right A-modules. Not only does the above correspondence fail for noncommutative rings, but the theories of left modules and of right modules can be quite diﬀerent for some rings. Example 7.7.3. Let A be a ring. Then A is a left A-module via the structure map a · a = aa . Indeed, the equalities that demonstrate that this is an A-module structure are all part of the deﬁnition of a ring. The submodules of A under this structure are precisely the left ideals of A. Note that A is also a right A-module, via a · a = a a. The submodules under this right module structure are the right ideals of A. In particular, module theory will help us study ideals, and hence will provide important information about the ring A. Here is another source of A-modules. Example 7.7.4. Let f : A → B be a ring homomorphism. Then B is a left A-module via a · b = f (a)b. Also, B is a right A-module via b · a = bf (a). For simplicity, we shall restrict attention to left modules for the rest of this section. Analogues of all the results will be true for right modules. Thus, for the duration, “Amodule” will mean left A-module. We shall, however, use the word “left” to emphasize certain one-sided properties. Given the modules from the examples above, we can construct yet more via factor groups. Lemma 7.7.5. Let N be a submodule of the A-module M . Then the factor group M/N is an A-module via a·m = am. Moreover, if f : M → M is an A-module homomorphism, then there is a factorization / M ME EE x< x EE xx E xx π EE " xx f M/N f (i.e., an A-module homomorphism f making the diagram commute) if and only if N ⊂ ker f . The factorization f , if it exists, is unique. Finally, the kernel of an A-module homomorphism is always a submodule, and hence ∼ = we obtain an isomorphism of A-modules f : M/(ker f ) −→ im f from any A-module homomorphism f : M → M . CHAPTER 7. RINGS AND MODULES 252 We now analyze the A-module homomorphisms out of A. Lemma 7.7.6. Let M be a left A-module. Then for each m ∈ M there is a unique A-module homomorphism fm : A → M with fm (1) = m. Proof We must have fm (a) = fm (a · 1) = afm (1) = am, so uniqueness follows, and it suﬃces that this formula deﬁnes an A-module homomorphism. But that follows immediately from the deﬁnition of module. In category theoretic terms, this says that A is the free A-module on one generator. There’s another way to express the result which introduces a useful piece of notation. Deﬁnition 7.7.7. Let M and N be A-modules. We write HomA (M, N ) for the set of A-module homomorphisms from M to N .11 Lemma 7.7.8. Let M and N be A-modules. Then HomA (M, N ) is an abelian group under the operation that sets (f + g)(m) = f (m) + g(m) for f, g ∈ HomA (M, N ). Under certain circumstances (e.g., if A is commutative), HomA (M, N ) is naturally an A-module. We shall discuss the details in Section 9.7. We now give an easy consequence of Lemma 7.7.6. Corollary 7.7.9. Let M be an A-module. Then there is an isomorphism of abelian ∼ = groups, ε : HomA (A, M ) −→ M , given by ε(f ) = f (1). ε is an example of what’s called an evaluation map, as it is obtained by evaluating the homomorphisms at 1. As shown in Proposition 9.7.4, ε is an A-module isomorphism under the appropriate conventions on the group of homomorphisms. We return to the study of the homomorphisms from A to M . Deﬁnitions 7.7.10. Let m ∈ M and let fm : A → M be the unique A-module homomorphism that carries 1 to m. Then we write Am for the image of fm and write Ann(m) (or AnnA (m), if there’s more than one ring over which M is a module) for the kernel. Thus, Am Ann(m) = {am | a ∈ A} and = {a ∈ A | am = 0}. We shall refer to Ann(m) as the annihilator of m, and say that a kills (or annihilates) m if am = 0. In general, we say that N is a cyclic module if N = An for some n ∈ N . Corollary 7.7.11. Let M be a left A-module and let m ∈ M . Then Ann(m) is a left ideal of A, and the cyclic module Am is isomorphic to A/Ann(m). There is also the notion of the annihilator of a module. Deﬁnition 7.7.12. Let M be an A-module. Then the annihilator of M , written as Ann(M ) or AnnA (M ), is given by Ann(M ) = {a ∈ A | am = 0 for all m ∈ M } = Ann(m). m∈M 11 We use the same notation for the set of A-module homomorphisms between a pair of right A-modules. This can, at times, require some speciﬁcity of language. CHAPTER 7. RINGS AND MODULES 253 Unlike the case of the annihilator of an element, we have the following. Lemma 7.7.13. The annihilator of an A-module is a two-sided ideal of A. Warning: If A is not commutative, then Ann(Am) may not be equal to Ann(m). In addition, Ann(m) may not be a two-sided ideal, while Ann(Am) always is. Deﬁnitions 7.7.14. Let f : A → B be a ring homomorphism. Let M be a B-module. The A-module structure on M induced by f is given by the structure map a · m = f (a)m for a ∈ A and m ∈ M . Let N be an A-module. We say that a B-module structure on N is compatible with the original A-module structure if the A-module structure induced by f from the Bmodule structure is the same as the one we started with (i.e., f (a) · m = am for all a ∈ A and m ∈ M ). We wish to study these relationships in the case where f is the canonical map π : A → A/a, where a is a two-sided ideal of A. Here, if M is an A/a module, then we shall implicitly consider M to be an A-module under the structure induced by π. Proposition 7.7.15. Let a be a two-sided ideal in the ring A. Then an A-module M admits an A/a-module structure compatible with the given A-module structure if and only if a ⊂ Ann(M ). Such an A/a-module structure is unique if it exists. Finally, if M and N are A/a-modules, then every A-module homomorphism from M to N is an A/a-module homomorphism as well. Proof If M is an A/a-module, then, in the induced A-module structure, if x ∈ a and m ∈ M , then xm = π(x)m = 0, because π(x) = 0. Thus, a ⊂ Ann(M ). Conversely, if a ⊂ Ann(M ), given a ∈ A, x ∈ a, and m ∈ M , (a + x)m = am. Thus, setting a · m = am gives a well deﬁned correspondence from A/a × M to M . And this correspondence is easily seen to satisfy the conditions required for an A/a-module structure on M . Two A/a-module structures that induce the same A-module structure must be equal, because π : A → A/a is surjective. Finally, given an A-module homomorphism f : M → N between A/a-modules, we have f (am) = f (am) = af (m) = af (m), since the A-module structures on both M and N are induced by the given A/a-module structures. For A = Z, this gives a characterization of Zn -modules. Corollary 7.7.16. An abelian group M admits a Zn -module structure if and only if M has exponent n as a group. The Zn -module structure on M , if it exists, is unique. Finally, if M and N are Zn -modules, then every group homomorphism f : M → N is a Zn -module homomorphism. Proof We use the fact that Zn = Z/(n). We know that every abelian group M has a unique Z-module structure, in which k · m is the k-th power of m in the abelian group structure on M for all k ∈ Z and m ∈ M . Thus, k annihilates M if and only if M has exponent k. But since (n) is the smallest ideal of Z containing n, we see that n annihilates M if and only if (n) ⊂ Ann(M ). The result now follows directly from Proposition 7.7.15. Now we consider operations on submodules. The constructions that work are similar to those that work for ideals. CHAPTER 7. RINGS AND MODULES 254 Deﬁnitions 7.7.17. Let N1 and N2 be submodules of M and let a be a left ideal of A. We deﬁne N1 + N2 and aN1 by N1 + N2 aN1 = {n + m | n ∈ N1 , m ∈ N2 } = {a1 n1 + · · · + ak nk | k ≥ 1, ai ∈ a, and ni ∈ N1 for i = 1, . . . , k} Lemma 7.7.18. Let N1 and N2 be submodules of the A-module M and let a be a left ideal of A. Then N1 + N2 , aN1 , and N1 ∩ N2 are all submodules of M . The modules aM have the following virtue. Lemma 7.7.19. Let M be an A-module and let a be a two-sided ideal of M . Then M/aM is an A/a-module. Moreover, if N is any A/a-module and if f : M → N is an A-module homomorphism, then there is a unique factorization /N M G x; GG x x GG xx G xx π GG xx f # M/aM f where f : M/aM → N is an A/a-module homomorphism. Proof Clearly, a ⊂ Ann(M/aM ), and hence M/aM is an A/a-module by Proposition 7.7.15. If f : M → N is an A-module homomorphism, with N an A/a-module, then a ⊂ Ann(N ), and hence aM must lie in ker f . Thus, there is a unique A-module homomorphism f : M/aM → N making the diagram commute, and f is an A/a-module homomorphism by Proposition 7.7.15. In categorical terms, this says that the passage from M to M/aM is a left adjoint to the forgetful functor from A/a-modules to A-modules. We can also consider external operations on modules. Recall that the direct sum M1 ⊕ · · · ⊕ Mk of a ﬁnite collection of abelian groups is an alternative notation for the direct product. Deﬁnition 7.7.20. Let M1 , . . . , Mk be A-modules. By the direct sum, M1 ⊕ · · · ⊕ Mk , of M1 , . . . , Mk , we mean their direct sum as abelian groups, endowed with the A-module structure obtained by setting a(m1 , . . . , mk ) = (am1 , . . . , amk ), for each a ∈ A and each k-tuple (m1 , . . . , mk ) ∈ M1 ⊕ · · · ⊕ Mk . Note that the canonical inclusions ιi : Mi → M1 ⊕ · · · ⊕ Mk are A-module maps, as are the projection maps πi : M1 ⊕ · · · ⊕ Mk → Mi . The proof of the following lemma is identical to that of its analogue for abelian groups (see Proposition 2.10.6). Lemma 7.7.21. Let fi : Mi → N be A-module homomorphisms for 1 ≤ i ≤ k. Then there is a unique A-module homomorphism f : M1 ⊕ · · · ⊕ Mk → N such that f ◦ ιi = fi for 1 ≤ i ≤ n. Explicitly, f (m1 , . . . , mk ) = f1 (m1 ) + · · · + fk (mk ). CHAPTER 7. RINGS AND MODULES 255 The next lemma will be useful, for instance, in our understanding of the properties of free modules. Lemma 7.7.22. Let a be a two-sided ideal in A and let M1 , . . . , Mk be A-modules. Let M = M1 ⊕ · · · ⊕ Mk . Then there is a natural isomorphism of A/a-modules between M/aM and (M1 /aM1 ) ⊕ · · · ⊕ (Mk /aMk ). Proof Recall from Proposition 7.7.15 that any A-module map between A/a-modules is an A/a-module map. Thus, it suﬃces to show that aM = (aM1 ) ⊕ · · · ⊕ (aMk ). Note that the inclusion aM ⊂ (aM1 ) ⊕ · · · ⊕ (aMk ) follows from the fact that if a ∈ a, then a(m1 , . . . , mk ) = (am1 , . . . , amm ) lies in (aM1 ) ⊕ · · · ⊕ (aMk ) for each k-tuple (m1 , . . . , mk ). Since (aM1 ) ⊕ · · · ⊕ (aMk ) is closed under addition, the stated inclusion follows. For the opposite inclusion, note that (m1 , . . . , mk ) = ι1 (m1 ) + · · · + ιk (mk ). Since each ιi is an A-module homomorphism, we have ιi (mi ) ∈ aM whenever mi ∈ aMi . Since aM is closed under addition, we see that (aM1 ) ⊕ · · · ⊕ (aMk ) ⊂ aM , as claimed. We shall also make use of* inﬁnite direct sums. Recall that if Mi is !an abelian group for i ∈ I, then the direct sum i∈I Mi is the subgroup of the product i∈I Mi consisting of those I-tuples (mi | i ∈ I) such that *all but ﬁnitely many of the mi are 0. The homomorphisms ιi : Mi → i∈I Mi are deﬁned by setting ιi (m) to be the Ituple whose i-th coordinate is m and whose other coordinates are all 0. Making use of m as an alternate notation for (m | i ∈ I), Lemma 6.5.9 gives i i i∈I mi = ιi (mi ), i∈I where the right-hand side is the sum in for which mi = 0. * i∈I i∈I Mi of the ﬁnite collection of terms ιi (mi ) Deﬁnition 7.7.23. Let Mi be an A-module for each i ∈ I. We deﬁne the direct sum * abelian group i∈I Mi to be the A-module whose underlying is the direct sum of the Mi and whose A-module structure is given by a i∈I mi = i∈I ami . * The maps ιi : Mi → i∈I Mi are easily seen to be A-module homomorphisms. The proof of the next result is identical to that of its analogue for Z-modules (Proposition 6.5.10). Proposition 7.7.24. Suppose given A-modules Mi for i ∈ I. Then for any collection of A-module homomorphisms {fi : Mi → N | i ∈ I}, there is a unique A-module homo* morphism f : i∈I Mi → N such that f ◦ ιi = fi for each i. Explicitly, # " mi = fi (mi ). f i∈I i∈I We also have an inﬁnite sum operation for submodules of a given module. Deﬁnition 7.7.25. Suppose given a set {Ni | i ∈ I} of submodules of an A-module M . Then the sum of the Ni is the submodule Ni = {ni1 + · · · + nik | k ≥ 1, ij ∈ I and nij ∈ Nij for j = 1, . . . , k}. i∈I Thus, i∈I Ni is the set of all ﬁnite sums of elements from the various submodules Ni . CHAPTER 7. RINGS AND MODULES 256 Sums of submodules satisfy the following properties. The proof is left as an exercise. Proposition 7.7.26. Let M be an A-module. If N1 and N2 are submodules of M , then N1 + N2 is the smallest submodule of M containing both N1 and N2 . Similarly, if Ni is a submodule of M for i ∈ I, then i∈I Ni is the smallest submodule of M that contains each of the Ni . If fi : Mi → M * is an A-module homomorphism for each i ∈ I (with I either ﬁnite or inﬁnite) and if f : i∈I Mi → M is the unique homomorphism such that f ◦ ιi = fi for all i, then im fi . im f = i∈I Exercises 7.7.27. 1. Give the proof of Lemma 7.7.13. 2. Give the proof of Lemma 7.7.18. 3. Give the proof of Corollary 7.7.26. 4. Let M be an abelian group, considered as a Z-module. Show that the annihilator Ann(M ) is nonzero if and only if M has some positive exponent. If Ann(M ) = (n), with n > 0, show that n is the smallest positive number which is an exponent for M. 5. Let N1 and N2 be submodules of M . Show that Ann(N1 + N2 ) = Ann(N1 ) ∩ Ann(N2 ). 6. Let N1 and N2 be submodules of M . Show that Ann(N1 ) + Ann(N2 ) ⊂ Ann(N1 ∩ N2 ). Can you give an example where this inclusion is proper? 7. Consider Z8 as a Z-module. What is Ann((2)Z8 )? 8. Let f : A → B be a ring homomorphism. In the left A-module structure on B that’s induced by f , what is Ann(B)? 9. Let A be a commutative ring. Let m be an A-module and let m ∈ M . Show that Ann(m) = Ann(Am). 10. Let A be any ring and let x ∈ Mn (A) be the matrix whose 11-entry is 1 and whose other entries are all 0. ⎛ ⎞ 1 0 ... 0 ⎜ 0 0 ... 0 ⎟ ⎜ ⎟ a=⎜ ⎟ .. ⎝ ⎠ . 0 0 ... (a) What are the elements of AnnMn (A) (x)? 0 CHAPTER 7. RINGS AND MODULES 257 (b) What are the elements of the principal left ideal Mn (A)x? (c) What are the elements of AnnMn (A) (Mn (A)x)? 11. Let A be a ring and let G be a ﬁnite group. Let Σ ∈ A[G] be the sum of all the elements of G: Σ = g∈G g. Show that AnnA[G] (Σ) = I(G), the augmentation ideal of G. 12. Let A be a ring. Let G be a ﬁnite group and let x ∈ G. Show that an element g∈G ag g ∈ A[G] lies in Ann(x − 1) if and only if for each left coset gx of x, the coeﬃcients of the elements in gx are all equal (i.e., agxi = agx j for all i, j ∈ Z). Deduce that if G = x, then Ann(x − 1) = A[G]Σ, where Σ = g∈G g. 13. Let (m) be an ideal in Z. Show that for n ≥ 1, Zn /(m)Zn is isomorphic to Z(m,n) , where (m, n) is the greatest common divisor of m and n. (Recall that Z1 is the trivial group.) 14. Let M be an abelian group and let r ≥ 0. Show that (pr )M/(pr+1 )M is a vector space over Zp . 15. Let M be a ﬁnite abelian group of exponent pr . Show that (pr−1 )M is a vector space over Zp . 16. Let M be a ﬁnite abelian group whose order is prime to n. Show that M/(n)M = 0. 17. Let M = Zpr1 × · · · × Zprk , where 1 ≤ r1 ≤ · · · ≤ rk . (a) Show that M/(p)M is the product of k copies of Zp . (b) Calculate, in similar terms, the Zp -vector spaces (pr )M/(pr+1 )M for r ≥ 1. (c) Calculate Ann(M ). 18. Suppose given a family of submodules {Ni | i ∈ I} of M . Show that the intersection i∈I Ni is a submodule of M . 19. Let N be a submodule of M . Show that N = x∈N Ax. 20. What is Ann(M1 ⊕ · · · ⊕ Mk )? † 21. Let A be a ring. Show that a left A-module structure on an abelian group M determines and is determined by a ring homomorphism from A to EndZ (M ). 22. Use the preceding problem to give a quick proof of Proposition 7.7.15. 23. Let A be a ring. Show that a right A-module structure on an abelian group M determines and is determined by a ring homomorphism from the opposite ring Aop to EndZ (M ). † 24. Let A be a ring and let M be either a right or a left A-module. Let EndA (M ) ⊂ EndZ (M ) be the collection of A-module homomorphisms from M to itself. Show that EndA (M ) is a subring of EndZ (M ). 25. Let A be a commutative ring and let M be an A-module. Show that EndA (M ) is an A-algebra via the map ι : A → EndA (M ) deﬁned by ι(a)(m) = am for all a ∈ A and m ∈ M . CHAPTER 7. RINGS AND MODULES 258 26. Let A be a commutative ring. Suppose given an A-algebra B and an A-module M . Show that there is a one-to-one correspondence between the A-algebra homomorphisms from B to EndA (M ) and the B-module structures on M that are compatible with the original A-module structure. 27. For any ring A and any A-module M , we write AutA (M ) for the group of units in EndA (M ). Suppose that A is commutative and that G is a group. Show that there is a one-to-one correspondence between the group homomorphisms from G to AutA (M ) and the A[G]-module structures on M that are compatible (with respect to the standard inclusion ι : A → A[G]) with the original A-module structure on M. Free Modules and Generators Finite generation is as useful a concept in module theory as it is in group theory. Deﬁnitions 7.7.28. Let m1 , . . . , mk be elements of the A-module M . By the submodule generated by m1 , . . . , mk , we mean the sum Am1 + · · · + Amk . Clearly, this is the smallest submodule of M containing the elements m1 , . . . , mk . The elements of Am1 + · · · + Amk all have the form a1 m1 + · · · + ak mk , for some a1 , . . . , ak ∈ A. Such a sum is called a linear combination of m1 , . . . , mk . Similarly, the submodule of M generated by an inﬁnite family {mi | i ∈ I} is the inﬁnite sum, i∈I Ami , of the submodules Ami . Here, a linear combination of the mi is a ﬁnite sum ai1 mi1 + · · · + aik mik , with i1 , . . . , ik ∈ I. If M is generated by a ﬁnite set m1 , . . . , mk , we say that M is a ﬁnitely generated A-module. Similarly, we say that a left ideal is ﬁnitely generated if it is ﬁnitely generated as an A-module. Clearly, A is generated by 1, while the cyclic module Ax is generated by x. A very important collection of ﬁnitely generated modules is given by the free modules An : Deﬁnitions 7.7.29. We write An for the direct sum of n copies of A, and write ei = (0, . . . , 0, 1, 0, . . . , 0). for the element of An whose i-th coordinate is 1 and whose other coordinates are all 0, with 1 ≤ i ≤ n. Note that (a1 , . . . , an ) = a1 e1 + · · · + an en , so that An is generated by e1 , . . . , en . We call e1 , . . . , en the canonical basis of An . (We adopt the convention that A0 = 0, the 0-module, whose canonical basis is taken to be the null set.) We call An the standard free A-module of rank n. The next lemma shows that An is the free A-module (in the sense of category theory) on the set {e1 , . . . , en }. Lemma 7.7.30. Let M be an A-module and let m1 , . . . , mn ∈ M . Then there is a unique A-module homomorphism f : An → M with f (ei ) = mi for i = 1, . . . , n. Explicitly, f (a1 , . . . , an ) = a1 m1 + · · · + an mn . Proof Let fi : A → M be given by fi (a) = ami . Thus, fi is the unique A-module homomorphism with fi (1) = mi . By Lemma 7.7.21 there is a unique A-module homomorphism from An to N with the property that f ◦ ιi = fi for 1 ≤ i ≤ n, where ιi : A → An CHAPTER 7. RINGS AND MODULES 259 is the inclusion of the i-th summand. Since ei = ιi (1), this gives f (ei ) = fi (1) = mi . The explicit formula of Lemma 7.7.21 gives f (a1 , . . . , an ) = f1 (a1 ) + · · · + fn (an ) = a1 m1 + · · · + an mn , as claimed. Note that the uniqueness statement, while embedded in the above, also follows from the fact that the e1 , . . . , en generate An : Since (a1 , . . . , an ) = a1 e1 + · · · + an en , we must have f (a1 , . . . , an ) = a1 f (e1 ) + · · · + an f (en ), and hence any A-module homomorphism on An is determined by its values on e1 , . . . , en . We can generalize this to the study of arbitrary direct sums of copies of A. * Deﬁnition 7.7.31. Let * I be any set, and let ei = ιi (1) ∈ i∈I A for i ∈ I. Thus, in the I-tuples notation for i∈I A, ei is the element whose i-th coordinate is 1 and whose other * coordinates are all 0. The elements ei , i ∈ I are called the canonical basis elements of i∈I A. * Note that an element of i∈I A may be written uniquely as a sum i∈I ai ei , where all but ﬁnitely many of the ai are equal to 0. The proof of the next lemma is identical to that of Lemma 7.7.30, using Proposition 7.7.24 in place of Lemma 7.7.21. Lemma 7.7.32. Let I be any set. Let M be an A-module and let mi ∈ M for all i ∈ I. * Then there is a unique A-module homomorphism f : i∈I A → M such * that f (ei ) = mi for all i ∈ I, where the elements ei are the canonical basis elements of i∈I A. Explicitly, # " f ai ei = ai mi . i∈I i∈I For some purposes it is useful to get away from the use of canonical bases. Deﬁnitions 7.7.33. An A-module M is a free module if it is isomorphic to a direct sum of copies of A. We say that M is free of ﬁnite rank if it is isomorphic to An for some integer n ≥ 0. We say that a ﬁnite sequence x1 , . . . , xn of elements of M is a basis for M if the unique A-module homomorphism f : An → M with f (ei ) = xi for i = 1, . . . , n is an isomorphism.12 If I is an inﬁnite set, we say * that a set {mi | i ∈ I} is a basis for M if the unique A-module homomorphism f : i∈I A → M with f (ei ) = mi for i ∈ I is an isomorphism. Note that in the case of inﬁnitely generated free modules, a basis comes equipped with an indexing set I. In the ﬁnitely generated case, we insist that a basis, if non-null, be indexed by one of the standard ordered sets {1, . . . , n}. Examples 7.7.34. 12 By abuse of language, we shall speak of a sequence x , . . . , x of elements of M as an ordered subset n 1 of M . In the case of a basis, the elements x1 , . . . , xn must all be distinct, so that the sequence in question does specify an ordering on an n-element subset of M . CHAPTER 7. RINGS AND MODULES 260 1. Let G be a ﬁnite group, and consider A[G] as a left A-module via the canonical inclusion ι : A → A[G]. Then the elements of G, in any order, form a basis for A[G]. The point is that every element of A[G] may be written uniquely as g∈G ag g, so the A-module homomorphism from A|G| to A[G] induced by an ordering of the elements of G is a bijection. 2. For the same reason, if G is an inﬁnite group, then {g | g ∈ G} forms a basis for the inﬁnitely generated free module A[G]. 3. Consider Mn (A) as a left A-module via the canonical inclusion ι : A → Mn (A). Let eij be the matrix whose ij-th coordinate is 1 and whose other coordinates are all 0, where 1 ≤ i, j ≤ n. Then the matrices eij , in any order, form a basis for Mn (A). Here, we make use of the obvious fact that (aij ) = aij eij 1≤i,j≤n for any matrix (aij ) ∈ Mn (A). The very same elements form bases for A[G] and Mn (A) when we consider them as right modules via the inclusions ι. Another very important example of free modules comes from quotients of polynomial algebras. Proposition 7.7.35. Let A be a commutative ring and let f (X) ∈ A[X] be a polynomial of degree n > 0 whose leading coeﬃcient is a unit. Then A[X]/(f (X)) is a free Amodule with basis given by the images of 1, X, . . . , X n−1 under the canonical map from A[X]. Here, (f (X)) is the principal ideal of A[X] generated by f (X), and the A-module structure on the quotient ring A[X]/(f (X)) comes from the A-algebra structure induced by the standard algebra structure on A[X]. Proof Write h : An → A[X]/(f (X)) for the A-module homomorphism that carries ei to X i−1 for i = 1, . . . , n. We show that h is an isomorphism. For g(X) ∈ A[X], the Euclidean algorithm for polynomials (Proposition 7.3.10) allows us to write g(X) = q(X)f (X) + r(X), where deg r(X) < deg f (X). But then the images of g(X) and r(X) under the canonical map π : A[X] → A[X]/(f (X)) are equal, and, if r(X) = an−1 X n−1 + · · · + a0 , then π(r(X)) = h(a0 , . . . , an−1 ). Thus, h is onto. Now suppose that (a0 , . . . , an−1 ) ∈ ker h and let g(X) = an−1 X n−1 + · · · + a0 . Then π(g(X)) = 0, and hence g(X) ∈ (f (X)). Thus, g(X) = q(X)f (X) for some q(X) ∈ A[X]. Since the leading coeﬃcient of f (X) is a unit, the degree of q(X)f (X) must be ≥ deg f unless q(X) = 0. Thus, q(X) = g(X) = 0, and hence ai = 0 for 0 ≤ i ≤ n − 1. Thus, h is injective. The next lemma is almost immediate from the deﬁnition, but is useful nonetheless, especially in understanding matrix groups. Lemma 7.7.36. Let f : M → N be an A-module homomorphism between the free modules M and N , and let x1 , . . . , xn be any basis of M . Then f is an isomorphism if and only if f (x1 ), . . . , f (xn ) is a basis for N . Proof Let g : An → M be the A-module homomorphism with g(ei ) = xi for i = 1, . . . , n. By the deﬁnition of basis, g is an isomorphism. The result now follows from the fact that f ◦ g is the unique A-module homomorphism from An to N which takes ei to f (xi ) for i = 1, . . . , n. CHAPTER 7. RINGS AND MODULES 261 Surprisingly, rank is not well deﬁned for free modules over arbitrary rings: There are rings for which An ∼ = Am for n = m. But such rings turn out to be exotic. We shall see shortly that rank is well deﬁned over commutative rings. The following corollary, which is immediate from Lemma 7.7.22 and the deﬁnition of An , will help us do so. Corollary 7.7.37. Let a be a two-sided ideal in the ring A. Then An /aAn is isomorphic to (A/a)n as an A/a-module. While free modules of ﬁnite rank may not always have a uniquely deﬁned rank, we can always tell the diﬀerence between the cases of ﬁnite and inﬁnite rank. * Lemma 7.7.38. Let I be a set. Suppose that the free module M = i∈I A is ﬁnitely generated. Then I must be a ﬁnite set. Thus, a ﬁnitely generated free module has ﬁnite rank. Proof Let x1 , . . . , xn be a generating set for M , and write xj = i∈I aij ei for j = 1, . . . , n. Then for each j, all but ﬁnitely many of the aij are 0. Thus, if S = {i ∈ I | aij = 0 for some j}, then S must be ﬁnite. But then, the submodule of M generated by x1 , . . . , xn must lie in the submodule generated by {ei | i ∈ S}. But this says that M itself must be generated by {ei | i ∈ S}. But clearly, if i is not in S, then ei is not in the submodule of M generated by {ei | i ∈ S}. Thus, S = I, and hence I is ﬁnite. It is useful to understand what it means to be a basis. Deﬁnitions 7.7.39. Let m1 , . . . , mn be elements of the A-module M . We say that m1 , . . . , mn are linearly dependent if there are elements a1 , . . . , an of A, not all 0, such that a1 m1 + · · · + an mn = 0. Similarly, we say that an inﬁnite set {mi | i ∈ I} is linearly dependent if there is a relation of the form ai1 mi1 + · · · + aik mik = 0, where k > 0, i1 , . . . , ik are distinct elements of I and not all of the aij are 0. We say that a collection of elements is linearly independent if it is not linearly dependent. Proposition 7.7.40. Let m1 , . . . , mn be elements of the A-module M and let f : An → M be the unique A-module homomorphism such that f (ei ) = mi for 1 ≤ i ≤ n. Then the following conditions hold. 1. The elements m1 , . . . , mn are linearly independent if and only if f is an injection. 2. The image of f is the submodule generated by m1 , . . . , mn . Thus, m1 , . . . , mn generate M if and only if f is a surjection. Thus, m1 , . . . , mk is a basis for M if and only if it is linearly independent and generates M. * Similarly, if {mi | i ∈ I} is a family of elements of M , and if f : i∈I A → M is the unique homomorphism such that f (ei ) = mi for all i, then f is injective if and only if {mi | i ∈ I} are linearly independent, and is surjective if and only if {mi | i ∈ I} generate M. CHAPTER 7. RINGS AND MODULES 262 Proof First, consider the case of a ﬁnite subset m1 , . . . , mk . Here, an element of the kernel of f is precisely an n-tuple (a1 , . . . , an ) such that a1 m1 + · · · + an mn = 0. So linear independence is equivalent to f being injective. Also, f (a1 , . . . , an ) = a1 m1 + · · · + an mn . Since this gives the generic element of the submodule generated by m1 , . . . , mn , the second condition also holds. In the inﬁnitely generated case, a nonzero element of i∈I A may be written uniquely in the form ai1 ei1 + · · · + aik eik , where k > 0, i1 , . . . , ik are distinct elements of I, and ai = 0 for all i. And f (ai1 ei1 + · · · + aik eik ) = ai1 mi1 + · · · + aik mik . Thus, there is a nonzero element of the kernel if and only if {mi | i ∈I} is linearly dependent. Proposition 7.7.26 shows that the image of f is i∈I Ami , so the result follows. The following corollary is immediate. Corollary 7.7.41. An A-module M is ﬁnitely generated if and only if it is a quotient module of An for some integer n. Exercises 7.7.42. 1. Both Z2 and Z3 are Z6 -modules, as they both have exponent 6. Show that Z2 ⊕ Z3 is a free Z6 -module, but that neither Z2 nor Z3 is free over Z6 . † 2. Show that every A-module is a quotient module of a free module. 3. Show that every quotient module of a ﬁnitely generated module is ﬁnitely generated. Exact Sequences Deﬁnition 7.7.43. Suppose given a “sequence,” either ﬁnite or inﬁnite, of A-modules and module homomorphisms: f g · · · → M − →M − → M → · · · We say that the sequence is exact at M if the kernel of g is equal to the image of f . We call the sequence itself exact if it is exact at every module of the sequence that sits between two other modules. One sort of exact sequence has special importance. Deﬁnition 7.7.44. A short exact sequence of A-modules is an exact sequence of the form f g 0 → M − →M − → M → 0. Note that exactness at M shows that g has to be onto, while exactness at M shows that f is an injection. Thus, M/g(M ) is isomorphic to M , and hence the short exact sequence presents M as an extension of M by M . Example 7.7.45. For any pair M, N of A-modules, the sequence ι π 0→M − →M ⊕N − →N →0 is exact, where ι is the standard inclusion map and π is the standard projection map. We shall make extensive use of exact sequences, both short and otherwise. For instance, we can use exactness to measure the deviation of a map from being an isomorphism. CHAPTER 7. RINGS AND MODULES 263 Deﬁnition 7.7.46. Let f : M → N be an A-module homomorphism. Then the cokernel of f is the module C = N/(im f ). In particular, f is onto if and only if the cokernel of f is 0. Lemma 7.7.47. Let f : M → N be an A-module homomorphism. Then there is an exact sequence f i π 0→K− →M − →N − → C → 0, where i is the inclusion of the kernel of f and π is the canonical map onto the cokernel of f . Conversely, given a six term exact sequence f 0 → K → M − → N → C → 0, f is injective if and only if K = 0 and is surjective if and only if C = 0. The next result can be very useful for recognizing isomorphisms. The technique used in the proof is called a diagram chase. Lemma 7.7.48. (Five Lemma) Suppose given a commutative diagram with exact rows M1 α1 / M2 f1 N1 β1 α2 / M3 α3 / M4 f2 f3 f4 / N2 / N3 / N4 β2 β3 α4 / M5 f5 β4 / N5 Then the following properties hold. 1. If f2 and f4 are injective and if f1 is surjective, then f3 is injective. 2. If f2 and f4 are surjective and if f5 is injective, then f3 is surjective. In particular, if f2 and f4 are isomorphisms, f1 is surjective, and f5 is injective, then f3 is an isomorphism. Proof Suppose ﬁrst that f2 and f4 are injective and f1 is surjective. Let m ∈ ker f3 . Then (f4 ◦ α3 )(m) = 0 by the commutativity of the diagram. In other words, α3 (m) is in the kernel of f4 . Since f4 is injective, m must be in the kernel of α3 . By exactness, m = α2 (m ) for some m ∈ M2 . Since the diagram commutes and m ∈ ker f3 , this shows that f2 (m ) ∈ ker β2 . By the exactness of the lower sequence, f2 (m ) = β1 (n) for some n ∈ N1 . Since f1 is surjective, n = f1 (m ) for some m ∈ M1 . But then f2 (α1 (m )) = β1 (n). But n was chosen so that β1 (n) = f2 (m ), so α1 (m ) has the same image as m under f2 . Since f2 is injective, m = α1 (m ). But m was chosen so that α2 (m ) = m. Since α2 ◦ α1 = 0, m must be 0, and hence f3 is injective as claimed. Now suppose that f2 and f4 are surjective and f5 is injective, and let n ∈ N3 . Since f4 is surjective, β3 (n) = f4 (m) for some m ∈ M4 . Since β4 ◦ β3 = 0, commutativity of the diagram shows that (f5 ◦ α4 )(m) = 0. But f5 is injective, and hence α4 (m) = 0. By the exactness of the upper sequence, m = α3 (m ) for some m ∈ M3 . Recall that m was chosen so that f4 (m) = β3 (n). By commutativity of the diagram, β3 carries n and f3 (m ) to the same element, so that n − f3 (m ) = β2 (n ) for some n ∈ N2 by exactness. But f2 is surjective, so that n = f2 (m ) for some m ∈ M2 . So f3 (m + α2 (m )) = n − f3 (m ) + f3 (m ) = n, and hence f3 is surjective. CHAPTER 7. RINGS AND MODULES 264 A useful illustration of the utility of the Five Lemma comes from an analysis of when a short exact sequence may be used to express the middle term as the direct sum of its surrounding terms. Recall that a section for a homomorphism g : M → N is a homomorphism s : N → M such that g ◦ s = 1N . Recall also that a retraction for a homomorphism f : N → M is a homomorphism r : M → N such that r ◦ f = 1N . We assume that all homomorphisms under discussion are A-module homomorphisms. Proposition 7.7.49. Suppose given a short exact sequence f g →M − → M → 0. 0 → M − Then the following additional conditions are equivalent. 1. f admits a retraction. 2. g admits a section. 3. There is an isomorphism h : M ⊕ M → M such that the following diagram commutes. 0 / M ι / M ⊕ M π / M / 0 / M / 0 h 0 / M f / M g Proof We ﬁrst show that the existence of a retraction for f implies the existence of a section for g. Thus, let r be a retraction for f . We claim then that the restriction, g , of g to the kernel of r is an isomorphism of ker r onto M . Note that this is suﬃcient to produce a section, s, for g by setting s to be the composite M (g )−1 / ker r ⊂ M. To prove the claim, let m ∈ ker r. If g(m) = 0, then m ∈ ker g = im f , and hence m = f (m ) for some m ∈ M . But since r ◦ f = 1M , r(m) = m . But m = 0, since m ∈ ker r, and hence m = 0 as well. Thus, the restriction of g to ker r is injective, and it suﬃces to show that g : ker r → M is onto. Let m ∈ M . Then m = g(m) for some m ∈ M . Consider the element x = m − f (r(m)). Then r(x) = r(m) − r(m), since r ◦ f = 1M . Thus, x ∈ ker r. But clearly, ∼ = g(x) = m , so g : ker r −→ M as claimed, and hence g admits a section. Now suppose given a section s for g. We deﬁne h : M ⊕ M → M by h(m , m ) = f (m ) + s(m ). This is easily seen to make the diagram that’s displayed in the third condition commute. But then h is an isomorphism by the Five Lemma, and hence the third condition holds. It suﬃces to show that the third condition implies the existence of a retraction for f . But this is immediate: just take r = π1 ◦ h−1 , where π1 is the projection of M ⊕ M onto its ﬁrst factor. Note that the speciﬁc choice of r, s, or h above determines the other two maps precisely. CHAPTER 7. RINGS AND MODULES 265 Deﬁnition 7.7.50. If a short exact sequence g f 0 → M − →M − → M → 0 satisﬁes the three properties listed in Proposition 7.7.49, then we say the sequence splits. A speciﬁcation of a speciﬁc retraction for f , a speciﬁc section of g, or a speciﬁc isomorphism h as in the third condition is called a splitting for the sequence. Note that the situation with regard to splitting extensions of A-modules is cleaner than that for nonabelian groups. Here, there is no analogue of a semidirect product, and every split extension is a direct sum. If the third term of a short exact sequence is free, then it always splits. Proposition 7.7.51. Suppose given a short exact sequence g f 0 → M − →M − → M → 0 with M free. Then the sequence splits. Proof Let {mi | i ∈ I} be a basis for M , and, for each i ∈ I, choose mi ∈ M such that π(mi ) = mi . We claim that there is an A-module map s : M → M such that s(mi ) = mi for all i ∈ I. Once we show this we’ll be done, as then g ◦ s is the identity on {mi | i ∈ I}. But any two A-module homomorphisms that agree on a generating set for a module must be equal, so g ◦ s = 1M , and hence s is a section for g. We now show that such an s exists. Recall from Lemma 7.7.32 that if N is any A-module and if* {ni | i ∈ I} are elements of N , then there’s a unique A-module homomorphism from i∈I A to N which carries the canonical basis vector ei to ni for each i ∈ I. * Let f : i∈I A → M be the unique A-module homomorphism that carries * ei to mi for each i ∈ I. Then f is an isomorphism by the deﬁnition of a basis. Let s : i∈I A → M be the unique A-module homomorphism that carries ei to mi for each i ∈ I. Then the desired homomorphism s is the composite s ◦ f −1 . Exercises 7.7.52. 1. Let A be any ring and let 0 → M → M → M → 0 be an exact sequence of A-modules. Show that if M and M are ﬁnitely generated, then so is M . † 2. Snake Lemma Suppose given a commutative diagram with exact rows: M i / M f 0 / N p / N / 0 f f j / M q / N Show that there is an exact sequence ker f → ker f → ker f − → coker f → coker f → coker f , δ CHAPTER 7. RINGS AND MODULES 266 where “coker” stands for cokernel, and δ is deﬁned as follows. For m ∈ ker f , let m = p(m). Then f (m) = j(n ) for some n ∈ N , and we let δ(m ) be the element of coker f represented by n . Show also that if i is injective, so is ker f → ker f , and if q is surjective, so is coker f → coker f . (Hint: The key here is showing that δ, as described, is a well deﬁned homomorphism.) ‡ 3. Let A be an integral domain and let (x) be the principal ideal generated by x ∈ A. Show that there is a short exact sequence fx π → A/(x) → 0 0 → A −→ A − of A-modules, where fx (a) = ax. 4. Let Zn = T and let A be any commutative ring. Show that there is an exact sequence of A[Zn ]-modules · · · → Ck → · · · → C0 → A → 0, where Ck = A[Zn ] for all k ≥ 0, and the maps are given as follows. The map C0 → A is the standard augmentation. Each map C2k+1 → C2k with k ≥ 0 is multiplication by T − 1, and each map C2k → C2k−1 with k ≥ 1 is multiplication by Σ = 1 + T + · · · + T n−1 . (In the parlance of homological algebra, the sequence · · · → Ck → · · · → C0 is called a resolution of A over A[Zn ]. Rings Without Identity Not all algebraists require rings to have multiplicative identity elements. Here, we discuss the relationship between such a theory and that of rings as we have deﬁned them. Deﬁnition 7.7.53. A ring without identity is an abelian group R together with an associative binary operation, to be thought of as multiplication, that satisﬁes the distributive laws: a(b + c) = ab + ac and (a + b)c = ac + bc. When we speak of a ring, we will continue to assume that it has a multiplicative identity. The new notion must be named in full: rings without identity. Example 7.7.54. Let a be a left or right ideal of a ring. Then, under the multiplication inherited from A, a is a ring without identity. We shall see presently that every ring without identity may be embedded as an ideal in a ring. But it may be possible to embed it as an ideal in many diﬀerent rings. We deﬁne homomorphisms, modules, and ideals for rings without identity in the obvious ways. Since rings with identity may also be considered in this broader context, we shall, at the risk of redundancy, refer to the modules as “nonunital modules.” Deﬁnitions 7.7.55. A homomorphism f : R → S of rings without identity is a homomorphism of additive groups with the additional property that f (rs) = f (r)f (s) for all r, s ∈ R. CHAPTER 7. RINGS AND MODULES 267 A nonunital module M over a ring R without identity is an abelian group together with a multiplication R × M → M such that the following conditions hold: r(m + m ) = rm + rm (r + r )m = rm + r m r(r m) = (rr )m, for all r, r ∈ R and all m, m ∈ M . A homomorphism f : M → N of nonunital modules over a ring R without identity is a group homomorphism such that f (rm) = rf (m) for all r ∈ R and m ∈ M . The basics of module theory, such as direct sums, exact sequences, quotient modules, etc. go over to this context without change. Free modules, on the other hand, are problematic, as R itself is not necessarily free in the categorical sense. There is no good notion of basis here. Left, right, and two-sided nonunital ideals may now be deﬁned in the obvious way, and the quotient by a two-sided nonunital ideal satisﬁes the analogue of Proposition 7.2.9. There is a natural way to adjoin an identity to a ring without identity. ) of R is Deﬁnition 7.7.56. Let R be a ring without identity. The associated ring, R, ) given by R = Z ⊕ R, as an abelian group, with the multiplication (m, r) · (n, s) = (mn, ms + nr + rs), where ms and nr denote the m-th power of s and the n-th power of r, respectively, in the additive group of R. ) for the map ι(r) = (0, r) for r ∈ R. Write ι : R → R The reader may check the details of the following proposition. ) is Proposition 7.7.57. Let R be a ring without identity. Then the associated ring R ) a ring with identity element (1, 0), and ι : R → R is a homomorphism of rings without ) identity, whose image is a two-sided ideal of R. Moreover, if A is a ring, then for each homomorphism f : R → A of rings without ) → A such that f) ◦ ι = f . identity, there is a unique ring homomorphism f) : R ) is a left adjoint to In categorical language, this says that the passage from R to R the forgetful functor that forgets that a ring has an identity element. Given the relationship between module structures and homomorphisms into endomorphism rings, the following proposition should not be a surprise. Proposition 7.7.58. Let M be a nonunital module over the ring R without identity. ) Then there is a unique R-module structure on M whose restriction to the action by elements in the image of ι gives the original nonunital R-module structure on M : (n, r) · m = nm + rm, where nm is the n-th power of m in the additive group of M . ) ) The nonunital R-submodules of M and the R-submodules of this induced R-module structure coincide. This correspondence is natural in M in that it gives a functor from the category of ) nonunital R-modules to the category of R-modules. CHAPTER 7. RINGS AND MODULES 268 Exercises 7.7.59. 1. Suppose the ring R without identity has exponent n as an additive group. Show that there is a ring structure on Zn ⊕R that extends the multiplication on R = 0⊕R. Show that the analogue of Proposition 7.7.57 holds for homomorphisms into rings of characteristic n. 2. Let A be a commutative ring and let R be a ring without identity which is also an A-module, such that r · as = ar · s = a(r · s) for r, s ∈ R, and a ∈ A. Construct an A-algebra associated to R and state and prove an analogue of Proposition 7.7.57 for appropriate homomorphisms into A-algebras. 7.8 Chain Conditions We return to the discussion of chain conditions that was begun in Section 7.6. The module theory we’ve now developed will be very useful in this regard. First, let us generalize the chain conditions to the study of modules and of left or right ideals. Deﬁnitions 7.8.1. A ring is left Noetherian if it has the ascending chain condition for left ideals. In other words, any ascending sequence a1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · · of inclusions of left ideals is eventually constant in the sense that there’s an n such that an = ak for all k ≥ n. Similarly, left Artinian rings are those with the descending chain condition for left ideals. Here, any descending sequence a1 ⊃ a2 ⊃ · · · ⊃ ak ⊃ · · · of inclusions of left ideals is eventually constant in the sense that there’s an n such that an = ak for all k ≥ n. Right Noetherian and right Artinian rings are deﬁned analogously via chain conditions on right ideals. An A-module M is a Noetherian A-module if it has the ascending chain condition for its submodules. It is an Artinian module if it has the descending chain condition for its submodules. Clearly, A is left Noetherian (resp. left Artinian) if and only if it is Noetherian (resp. Artinian) as a left A-module. We shall not make use of chain conditions on two-sided ideals. When we speak of a Noetherian or Artinian ring, without specifying left or right, we shall mean a commutative ring with the appropriate chain condition on its ideals. Examples 7.8.2. 1. Recall that a division ring has no left (or right) ideals other than 0 and the ring itself. Thus, division rings are left and right Noetherian and Artinian. 2. Recall that the left ideals of a product A × B all have the form a × b, with a a left ideal of A and b a left ideal of B. Thus, if A and B are both left Noetherian (resp. left Artinian), so is A × B. CHAPTER 7. RINGS AND MODULES 269 3. Regardless *∞ of any chain conditions that hold for a nonzero ring A, the inﬁnite direct sum i=1 A is neither Noetherian nor Artinian as an A-module. 4. The integers, Z, are not Artinian, but Zn is an Artinian Z-module for all n > 0. !∞ 5. The inﬁnite product i=1 Zp satisﬁes neither chain condition on its ideals, but the module structure on Zp obtained from the projection onto the ﬁrst factor satisﬁes both chain conditions. 6. Let A= a b 0 c a ∈ Q, b, c ∈ R . Let a ⊂ A be the set of all elements of the form ( 00 0b ) with b ∈ R. Then a is a two-sided ideal of A. Note that a b 0 d 0 ad = . 0 c 0 0 0 0 Thus, the left submodules of a may be identiﬁed with the Q-submodules of R. Since R is inﬁnitely generated as a vector space over Q, it satisﬁes neither chain condition by Example 3. (See Corollary 7.9.14 for a demonstration that R is free as a Q-module.) Thus, A itself is neither left Artinian nor left Noetherian. But 0 d a b 0 dc = , 0 0 0 c 0 0 so the right submodules of a may be identiﬁed with the R-submodules of R, which consist only of 0 and R itself. Thus, a is both right Noetherian and right Artinian as an A-module. There is a surjective ring homomorphism f : A → Q × R, given by a b f = (a, c). 0 c The kernel of f is a. Since Q × R is commutative, the left and right A-module structures on it coincide. Note that the only proper A-submodules of Q × R are Q × 0, 0 × R, and 0. Thus, Q × R satisﬁes both chain conditions as an A-module. Moreover, we have a short exact sequence 0→a→A→Q×R→0 of two-sided A-modules. As right modules, both a and Q × R are both Noetherian and Artinian. As we shall see in Lemma 7.8.7, this implies that A itself is both right Noetherian and right Artinian. Thus, A satisﬁes both chain conditions on one side, and satisﬁes neither chain condition on the other. The Noetherian property can be characterized in terms of ﬁnite generation. Proposition 7.8.3. An A-module M is a Noetherian module if and only if each of its submodules is ﬁnitely generated. CHAPTER 7. RINGS AND MODULES 270 Proof Suppose that every submodule is ﬁnitely generated and that we are given an ascending chain N1 ⊂ N2 ⊂ · · · ⊂ Nk ⊂ · · · of submodules of M . Let N = k≥0 Nk . Then it is easy to see that N is a submodule of M . Suppose that N is generated by n1 , . . . , nk . Then we may choose m to be large enough that n1 , . . . , nk ∈ Nm . But then the submodule of M generated by n1 , . . . , nk must be contained in Nm . Since n1 , . . . , nk generate N , we must have Nm = Nm+l = N for all l ≥ 0. Conversely, suppose that M is a Noetherian module and let N be one of its submodules. Suppose by contradiction that N is not ﬁnitely generated. Then by induction we may choose a sequence of elements ni ∈ N such that n1 = 0 and for each i > 1, ni is not in the submodule generated by n1 , . . . , ni−1 . Write Nk for the submodule generated by n1 , . . . , nk . Then each inclusion Nk−1 ⊂ Nk is proper, and hence the sequence N1 ⊂ N2 ⊂ · · · ⊂ Nk ⊂ · · · contradicts the hypothesis that M is a Noetherian module. Since a ring is left Noetherian if and only if it is Noetherian as a left module over itself, we obtain the following corollary. Corollary 7.8.4. A ring A is left Noetherian if and only if every left ideal is ﬁnitely generated. Note that every ideal in a principal ideal ring is a cyclic module, and hence is ﬁnitely generated. Corollary 7.8.5. Principal ideal rings are Noetherian. The next very useful lemma shows that chain conditions are inherited by submodules and quotient modules. Lemma 7.8.6. Let A be a ring and let M be a Noetherian (resp. Artinian) A-module. Then any submodule or quotient module of M is Noetherian (resp. Artinian) as well. Proof An ascending (resp. descending) chain in a submodule of M is a chain in M . But then an application of the appropriate chain condition in M shows that it must hold in N as well. For quotient modules M/N , the result follows from the usual one-to-one correspondence between submodules of the quotient and the submodules of M containing N . We wish to show that the free modules An share whatever chain conditions A has. The next lemma will help us to do so. Lemma 7.8.7. Suppose given a short exact sequence f g 0 → M − →M − → M → 0 of A-modules. Then M is Noetherian (resp. Artinian) if and only if both M and M are. CHAPTER 7. RINGS AND MODULES 271 Proof If M satisﬁes the stated chain condition, then so do M and M by Lemma 7.8.6. For the converse, we restrict attention to the Noetherian case. The Artinian case is similar. Thus, suppose that M and M are Noetherian modules, and suppose given an ascending chain N1 ⊂ N2 ⊂ · · · ⊂ N k ⊂ · · · of submodules of M . Then we obtain ascending chains g(N1 ) ⊂ g(N2 ) ⊂ · · · ⊂ g(Nk ) ⊂ · · · in M and f −1 (N1 ) ⊂ f −1 (N2 ) ⊂ · · · ⊂ f −1 (Nk ) ⊂ · · · in M . Thus, since M and M are Noetherian, for k large enough, we have g(Nk ) = g(Nk+l ) and f −1 (Nk ) = f −1 (Nk+l ) for all l ≥ 0. But we claim then that Nk = Nk+l . To see this, consider the following diagram, in which the vertical maps are the inclusion maps from the sequences above. Since f (f −1 (Ni )) = Ni ∩ ker g, the rows are exact. f g / f −1 (Nk ) / Nk / g(Nk ) / 0 0 0 / f −1 (Nk+1 ) f / Nk+1 g / g(Nk+1 ) / 0 Since the inclusions f −1 Nk ⊂ f −1 Nk+l and g(Nk ) ⊂ f (Nk+l ) are isomorphisms, the inclusion of Nk in Nk+l is also by the Five Lemma. Since sequences of the form 0→M →M ⊕N →N →0 are exact, a quick induction on n gives the next corollary. Corollary 7.8.8. If A is left Noetherian (resp. left Artinian), then An is Noetherian (resp. Artinian) as a left A-module for all n ≥ 0. Now, a ﬁnitely generated module is a quotient module of An for some n. Since passage to quotient modules preserves chain conditions, the next proposition is immediate. Proposition 7.8.9. Let A be a left Noetherian (resp. left Artinian) ring. Then any ﬁnitely generated left A-module is a Noetherian (resp. Artinian) module over A. This now allows us to characterize the Noetherian property. Corollary 7.8.10. A ring A is left Noetherian if and only if it has the property that every submodule of every ﬁnitely generated left A-module is ﬁnitely generated. Proof If every left submodule of A itself is ﬁnitely generated, then A is left Noetherian by Corollary 7.8.4. Conversely, if A is left Noetherian and M is a ﬁnitely generated A-module, then M is Noetherian, and hence each of its submodules is ﬁnitely generated by Proposition 7.8.3. CHAPTER 7. RINGS AND MODULES 272 Proposition 7.8.9 also has immediate applications to establishing chain conditions in some of the examples we’ve seen of rings. Corollary 7.8.11. Let f : A → B be a ring homomorphism and suppose that B is ﬁnitely generated as a left A-module. Then if A is left Noetherian or left Artinian, so is B. Proof Every B-module is an A-module via f . So a chain of ideals in B is a chain of A-submodules of the ﬁnitely generated A-module B. We obtain two immediate corollaries. Corollary 7.8.12. Let G be a ﬁnite group. Then if A is left or right Noetherian or Artinian, then so is the group ring A[G]. In particular the group ring D[G] over a division ring D satisﬁes all four chain conditions. Corollary 7.8.13. If A is left or right Noetherian or Artinian, then so are the matrix rings Mn (A) for n ≥ 1. In particular the matrix rings Mn (D) over a division ring D satisfy all four chain conditions. Of course, A[X] is inﬁnitely generated as an A-module, so we shall have to work harder to show that it inherits the Noetherian property from A. As a result, however, we shall obtain a considerable strengthening of the Noetherian case of Corollary 7.8.11 when A is commutative. Theorem 7.8.14. (Hilbert Basis Theorem) Let A be a left Noetherian ring. Then the polynomial ring A[X] is also left Noetherian. Proof We show that every left ideal of A[X] is ﬁnitely generated. Let a be a left ideal of A[X] and deﬁne b ⊂ A to be the set of all leading coeﬃcients of elements of a. Note that if a and b are the leading coeﬃcients of f and g, respectively, and if f and g have degrees m and n, respectively, with m ≥ n, then the leading coeﬃcient of f + X m−n g is a + b. Thus, b is a left ideal of A. Since A is left Noetherian, there is a ﬁnite generating set x1 , . . . , xk for b. But then we can ﬁnd polynomials f1 , . . . , fk whose leading coeﬃcients are x1 , . . . , xk , respectively. After stabilizing by multiplying by powers of X, if necessary, we may assume that the polynomials f1 , . . . , fk all have the same degree, say m. Let M be the A-submodule of A[X] generated by 1, X, X 2 , . . . , X m−1 . Thus, the elements of M are the polynomials of degree less than m. Now a ∩ M is a submodule of a ﬁnitely generated A-module. Since A is Noetherian, a ∩ M is a ﬁnitely generated A-module. We claim that every element of a may be written as the sum of an element of A[X]f1 + · · · + A[X]fk (i.e., the ideal generated by f1 , . . . , fk ) with an element of a ∩ M . This will be suﬃcient, as then a is generated as an ideal by f1 , . . . , fk together with any set of A-module generators for a ∩ M . But the claim is an easy induction on degree. Let f ∈ a have degree n. If n < m, then f ∈ a ∩ M , and there’s nothing to show. Otherwise, let a be the leading coeﬃcient of f and write a = a1 x1 + · · · + ak xk with ai ∈ A for all i. Then a is the leading coeﬃcient of a1 X n−m f1 + · · · + ak X n−m fk , which has degree n. But then f − (a1 X n−m f1 + · · · + ak X n−m fk ) has degree less than n, and hence the claim follows by induction. Recall that an A-algebra of ﬁnite type is one which is ﬁnitely generated as an Aalgebra. CHAPTER 7. RINGS AND MODULES 273 Corollary 7.8.15. Let A be a commutative Noetherian ring and let B be a commutative A-algebra of ﬁnite type. Then B is Noetherian. Proof The hypothesis on B is that there is a surjective ring homomorphism f : A[X1 , . . . , Xn ] → B for some n ≥ 0. By Corollary 7.8.11, it suﬃces to show that A[X1 , . . . , Xn ] is Noetherian. But Proposition 7.3.23 shows that A[X1 , . . . , Xn ] is a polynomial ring in one variable on A[X1 , . . . , Xn−1 ], so the result follows from the Hilbert Basis Theorem and induction. Exercises 7.8.16. 1. Let A be a nonzero ring. Show that A[X] is not left Artinian. 2. In multiplicative notation, we can write the abelian group Z as {X n | n ∈ Z}, with 1 = X 0 as the identity element. For a ring A, the group ring A[Z] is sometimes called the ring of Laurent polynomials in A, and is sometimes denoted by A[X, X −1 ]. (This is absolutely not to be confused with a polynomial ring in two variables.) Note that A[X, X −1 ] contains the polynomial ring A[X] as a subring. Show that for any f ∈ A[X, X −1 ] there is a unit u such that uf ∈ A[X]. Use this, together with either the statement or the proof of the Hilbert Basis Theorem, to show that if A is left Noetherian, so is A[X, X −1 ]. 3. Since an abelian group is precisely a Z-module, any ﬁnitely generated abelian group is a quotient group of Zn for some n. Show that if G is a ﬁnitely generated abelian group and if A is a left Noetherian ring, then A[G] is also left Noetherian. 4. Suppose there is a ﬁnitely generated group, say with n generators, such that Z[G] fails to be left Noetherian. Show, then, that if F is a free group on n or more generators, then Z[F ] must also fail to be left Noetherian. 7.9 Vector Spaces We give some of the basic properties of vector spaces, which we shall use in studying modules over more general rings. In particular, we shall show that if A is a commutative ring and if we have an Amodule isomorphism Am ∼ = An , then m = n. Thus, using vector spaces, we shall show that the rank of a ﬁnitely generated free module over a commutative ring is well deﬁned, independently of the choice of an isomorphism to one of the canonical free modules An (i.e., independently of the choice of a basis). For the sake of generality, we will work over a division ring, rather than a ﬁeld. (In particular, we shall not treat determinants here.) Thus, “vector space” here will mean a left module over a division ring D, and a D-linear map is a D-module homomorphism. The results, of course, will all extend to right modules as well. Deﬁnition 7.9.1. We say that a vector space V over D is ﬁnite dimensional if it is ﬁnitely generated as a D-module. Recall that a ﬁnite sequence, v1 , . . . , vk , of elements of a D-module V is a basis for V if the unique D-module homomorphism f : Dk → V for which f (ei ) = vi for i = 1, . . . , k is an isomorphism, where ei is the i-th canonical basis vector of Dk . Recall also (from Proposition 7.7.40) that v1 , . . . , vk is a basis if and only if v1 , . . . , vk are linearly independent and generate V . CHAPTER 7. RINGS AND MODULES 274 Proposition 7.9.2. Let V be a ﬁnite dimensional vector space over D, with generators v1 , . . . , vn . Then there is a subset vi1 , . . . , vik of v1 , . . . , vn which is a basis for V . In addition, if v1 , . . . , vm are linearly independent for some m ≤ n, then we may choose the subset vi1 , . . . , vik so that it contains v1 , . . . , vm . In particular, every ﬁnite dimensional D-vector space is free as a D-module. Proof If V = 0, then the null set is a basis. Otherwise, we argue by induction on n − m. If v1 , . . . , vn is linearly independent, there is nothing to prove. Otherwise, we can ﬁnd x1 , . . . , xn ∈ D, not all 0, such that x1 v1 + · · · + xn vn = 0. Now at least one of xm+1 , . . . , xn must be nonzero, as otherwise the elements v1 , . . . , vm would be linearly dependent. For simplicity, suppose that xn = 0. Then vn = (−x−1 n x1 )v1 + x )v . Thus, v is in the submodule generated by v , . . . , v · · · + (−x−1 n−1 n−1 n 1 n−1 . Since n v1 , . . . , vn generate V , V must in fact be generated by v1 , . . . , vn−1 . But the result now follows by induction. Corollary 7.9.3. Let V be a ﬁnite dimensional vector space and let v1 , . . . , vm be linearly independent in V . Then there is a basis of V containing v1 , . . . , vm . Proof Let v1 , . . . , vn be obtained from v1 , . . . , vm by adjoining any ﬁnite generating set for V , and apply Proposition 7.9.2. Proposition 7.9.4. Let v1 , . . . , vn be a basis for the vector space V and let w1 , . . . , wk be linearly independent in V . Then k ≤ n. Proof We shall show by induction on i ≥ 0 that we may relabel the v1 , . . . , vn in such a way that w1 , . . . , wi , v1 , . . . , vn−i generate V . If k were greater than n, this would say that w1 , . . . , wn is a generating set for V , and hence that wn+1 is a linear combination of w1 , . . . , wn , say wn+1 = x1 w1 +· · ·+xn wn , with x1 , . . . , xn ∈ D. But then x1 w1 + · · · + xn wn + (−1)wn+1 = 0, contradicting the linear independence of w1 , . . . , wk . Thus, for i ≥ 1, we suppose inductively that for some relabelling of the v’s, the set w1 , . . . , wi−1 , v1 , . . . , vn−i+1 generates V . Thus, we may write wi as a linear combination of these elements, say wi = x1 w1 + · · · + xi−1 wi−1 + xi v1 + · · · + xn vn−i+1 . Now it cannot be the case that xj = 0 for all j ≥ i, as then the elements w1 , . . . , wi would be linearly dependent. After relabelling, if necessary, we may assume that xn = 0, in which case vn−i+1 may be written as a linear combination of w1 , . . . , wi , v1 , . . . , vn−i , which then must generate V . Corollary 7.9.5. Let V be a ﬁnite dimensional vector space. Then any two bases of V have the same number of elements. Proof Since any basis of V is linearly independent, the number of elements in each basis must be less than or equal to the number in the other, by Proposition 7.9.4. Thus, we can make the following deﬁnition. Deﬁnition 7.9.6. Let V be a ﬁnite dimensional vector space. Then the dimension of V , written dim V , is the number of elements in any basis of V . By the deﬁnition of a basis, V has a basis with n elements (and hence has dimension n) if and only if V is isomorphic to Dn . Corollary 7.9.7. Two ﬁnite dimensional vector spaces over D have the same dimension if and only if they are isomorphic. In particular, Dn ∼ = Dm if and only if n = m. CHAPTER 7. RINGS AND MODULES 275 We can now deduce an important fact about free modules over commutative rings. Theorem 7.9.8. Let A be a commutative ring. Then An is isomorphic to Am if and only if m = n. Thus, ﬁnitely generated free modules over a commutative ring have a well deﬁned rank, independent of the choice of an isomorphism to some An . Proof Suppose that An is isomorphic to Am and let m be a maximal ideal of A. Then An /mAn is isomorphic as a vector space over the ﬁeld A/m to Am /mAm . But for k arbitrary, Corollary 7.7.37 provides an isomorphism of A/m-vector spaces between Ak /mAk and (A/m)k . Thus, An /mAn and Am /mAm have dimensions n and m, respectively, as A/m-vector spaces, and hence n = m. The following lemma is useful in detecting the diﬀerence between ﬁnite and inﬁnite dimensional vector spaces. Lemma 7.9.9. Let V be an inﬁnite dimensional vector space over D. Then there are linearly independent subsets of V of arbitrary length. By passage to the vector subspaces that they generate, we see that V has subspaces of every ﬁnite dimension. Proof We argue by contradiction. Suppose that the maximal length of a linearly independent subset of V is k, and let v1 , . . . , vk be linearly independent. Let W be the subspace of V generated by v1 , . . . , vk . Then W is ﬁnite dimensional, so there must be an element vk+1 of V that is not in W . We claim that v1 , . . . , vk+1 is linearly independent, contradicting the maximality of k. To see this, suppose that x1 v1 + · · · + xk+1 vk+1 = 0, with x1 , . . . , xk+1 ∈ D. If −1 xk+1 = 0, then vk+1 = (−x−1 k+1 x1 )v1 + · · · + (−xk+1 xk )vk . But this says that vk+1 ∈ W , contradicting the choice of vk+1 , so we must have xk+1 = 0. But this gives x1 v1 + · · · + xk vk = 0. Since v1 , . . . , vk are linearly independent, we must have x1 = · · · = xk = 0, and hence v1 , . . . , vk+1 are indeed linearly independent. This now gives a quick proof of the following fact, which we already knew from the fact that D is left Noetherian. Corollary 7.9.10. Let V be a ﬁnite dimensional vector space. Then any subspace of V is ﬁnite dimensional as well. Proof If V ⊂ V , then any linearly independent subset of V is a linearly independent subset of V . By Proposition 7.9.4, the maximal length of a linearly independent subset of V is dim V . So V must be ﬁnite dimensional by Lemma 7.9.9. We can now analyze the behavior of extensions with respect to dimension. Proposition 7.9.11. Suppose given a short exact sequence 0→V − →V − → V → 0 i π of vector spaces over D. Then V is ﬁnite dimensional if and only if both V and V are ﬁnite dimensional, in which case dim V = dim V + dim V . Proof Suppose V is ﬁnite dimensional. Since π : V → V is surjective, any set of generators of V is carried onto a set of generators of V by π. Since i(V ) is a subspace of V that is isomorphic to V , ﬁnite generation of V was just shown in Corollary 7.9.10. Now suppose that V and V are ﬁnite dimensional. We identify V with the image of i. Let v1 , . . . , vk be a basis for V and let z1 , . . . , zl be a basis for V . We claim that CHAPTER 7. RINGS AND MODULES 276 if we’re given any choice of w1 , . . . , wl ∈ V such that π(wi ) = zi for 1 ≤ i ≤ k, then v1 , . . . , vk , w1 , . . . , wl is a basis for V . First, we show that v1 , . . . , vk , w1 , . . . , wl is linearly independent. Suppose given x1 , . . . , xk+l ∈ D with x1 v1 + · · · + xk vk + xk+1 w1 + · · · + xk+l wl = 0. Then applying π, we see that xk+1 z1 + · · · + xk+l zl = 0. Since z1 , . . . , zl are linearly independent, we must have xk+1 = · · · = xk+l = 0. But then the original equation gives x1 v1 +· · ·+xk vk = 0. Since v1 , . . . , vk are linearly independent, we must have x1 = · · · = xk = 0 as well. The proof that v1 , . . . , vk , w1 , . . . , wl generate V works in an exact sequence of ﬁnitely generated modules over any ring. Let v ∈ V . Then π(v) = xk+1 z1 + · · · + xk+l zl for some l l xk+1 , . . . , xk+l ∈ D. But then π(v − i=1 xk+i wi ) = 0, and hence (v − i=1 xk+i wi ) ∈ ker π. The result now follows, since ker π = V is generated by v1 , . . . , vk . We obtain a characterization of isomorphisms of ﬁnite dimensional vector spaces. Corollary 7.9.12. Let V and W be ﬁnite dimensional vector spaces over D with the same dimension. Let f : V → W be a D-linear map. Then the following conditions are equivalent. 1. f is an isomorphism. 2. f is injective. 3. f is surjective. Proof Clearly, an isomorphism is both injective and surjective. We shall show that if f is either injective or surjective, then it is an isomorphism. If f is injective, then we get an exact sequence f π 0→V − →W − → C → 0, where C is the cokernel of f . But then dim C = dim W − dim V = 0. But then C = 0, as otherwise, C would have at least one linearly independent element in it. If f is surjective, then a similar argument shows that the kernel of f is 0. For inﬁnite dimensional vector spaces, we can give an analogue of Proposition 7.9.2. Proposition 7.9.13. Let V be an inﬁnite dimensional vector space over D. Let {vi | i ∈ I} be linearly independent in V , and suppose given a family {vi | i ∈ J} such that V is generated by {vi | i ∈ I ∪ J}. Then there is a subset, K, of I ∪ J containing I such that {vi | i ∈ K} is a basis for V . Proof Let S = {K ⊂ I ∪ J | I ⊂ K and {vi | i ∈ K} is linearly independent } . Then S is nonempty, and has a partial ordering given by inclusion of subsets. We claim that S satisﬁes the hypothesis of Zorn’s Lemma. = Thus, let T = {Kα | α ∈ A} be a totally ordered subset of S and let K α∈A Kα . We claim that K is linearly independent, and hence K is an upper bound for T in S. To see this, suppose that xi1 vi1 + · · · + xik vik = 0, where vi1 , . . . , vik are distinct Then for each j, vi ∈ Kα for some αj ∈ A. But since T is totally elements of K. j j ordered, there is an r ∈ {1, . . . , k} such that Kαj ⊂ Kαr for j = 1, . . . , k. But then CHAPTER 7. RINGS AND MODULES 277 vi1 , . . . , vik all lie in Kαr . Since Kαr is linearly independent, this forces xij = 0 for is linearly independent as well. j = 1, . . . , k. Thus, K Thus, by Zorn’s Lemma, there is a maximal element, K, in S. We claim that {vi | i ∈ K} is a basis for V . To see this, it suﬃces to show that {vi | i ∈ K} generates V , which in turn will follow if we show that each vi for i ∈ J is contained in the vector subspace generated by {vi | i ∈ K}. Thus, suppose that r is contained in the complement of K in I ∪ J, and let K = K ∪{r}. Then {vi | i ∈ K } must be linearly dependent by the maximality of K. Suppose given a relation of the form xi1 vi1 +· · ·+xik vik = 0, where vi1 , . . . , vik are distinct elements of K , and xij is nonzero for j = 1, . . . , k. Since {vi | i ∈ K} is linearly independent, then at least one of the ij must equal r, −1 say r = ik . But then vr = −x−1 r xi1 vi1 − · · · − xr xik−1 vik−1 lies in the vector subspace of V generated by {vi | i ∈ K}. Taking the set {vj | j ∈ J} to be the set of nonzero elements of V , if necessary, we see that every vector space over D has a basis. Corollary 7.9.14. Every module over a division ring is free. The following corollary is now immediate from Proposition 7.7.51. Corollary 7.9.15. Every short exact sequence 0 → V → V → V → 0 of modules over a division ring splits. 7.10 Matrices and Transformations Recall that an m × n matrix is one with m rows and n columns. We review the basic deﬁnitions and properties of nonsquare matrices. Deﬁnitions 7.10.1. Addition of nonsquare matrices is deﬁned coordinate-wise: If M1 = (aij ) and M2 = (bij ) are both m × n matrices with coeﬃcients in A, then M1 + M2 is the m × n matrix whose ij-th coeﬃcient is aij + bij . Multiplication of non-square matrices is deﬁned analogously to that of square matrices, except that we may form the product M1 M2 whenever the number of columns of M1 is equal to the number of rows of M2 . Thus, if M1 = (aij ) and M2 = (bij ) are m × n and n × l matrices, respectively, then we deﬁne M1 M2 to be the m × l matrix whose ij-th n entry is k=1 aik bkj for 1 ≤ i ≤ m and 1 ≤ j ≤ l. Nonsquare matrices behave analogously to square ones. The proof of the next lemma is left to the reader. Lemma 7.10.2. Let A be a ring and let M1 , M2 , and M3 be m × n, n × k, and k × l matrices over A, respectively. Then (M1 M2 )M3 = M1 (M2 M3 ). In addition, the distributive laws hold: with the Mi as above, if M1 and M2 are m × n and n × k matrices, respectively, then (M1 + M1 )M2 = M1 M2 + M1 M2 and M1 (M2 + M2 ) = M1 M2 + M1 M2 . Finally, for a ∈ A, let aIn ∈ Mn (A) be the matrix whose diagonal entries are all a and whose oﬀ-diagonal entries are all 0. Then if M1 = (bij ) and M2 = (cij ) for M1 and M2 above, then M1 · aIn is the matrix whose ij-th entry is bij a, and aIn · M2 is the matrix whose ij-th entry is acij . CHAPTER 7. RINGS AND MODULES 278 The following corollary is now immediate. Corollary 7.10.3. Right multiplication by the matrices aIn gives a right A-module structure to the set of m × n matrices over A, and left multiplication by aIm gives it a left A-module structure. In both cases the underlying abelian group structure of the module is given by matrix addition, but the two module structures have no direct relationship to each other if A is not commutative. (What we have instead is an example of an A-A-bimodule, as we shall encounter in our discussions of tensor products in Section 9.4. We shall not make use of bimodules here.) Let us ﬁrst establish notation: Deﬁnitions 7.10.4. For m, n ≥ 1, we write Mm,n (A) for the set of m × n matrices over A. We write eij ∈ Mm,n (A) for the matrix whose ij-th coordinate is 1 and whose other coordinates are all 0, for 1 ≤ i ≤ m and 1 ≤ j ≤ n. The next lemma generalizes a result we’ve seen for square matrices. Lemma 7.10.5. Let A be a ring. Then as either a right or a left A-module, Mm,n (A) is a free module with basis given by any ordering of {eij | 1 ≤ i ≤ m, 1 ≤ j ≤ n}. In the study of classical linear algebra over a ﬁeld, it is shown that if K is a ﬁeld, then the K-linear maps from K n to K m are in one-to-one correspondence with Mm,n (K) We shall generalize this here to show that if A is a commutative ring, then the A-module homomorphisms from An to Am are in one-to-one correspondence with Mm,n (A). Thus, matrices correspond to transformations of free modules. The situation for noncommutative rings is more complicated, but only by a very little. The (usually) standard situation, where an m × n matrix acts from the left on an n × 1 column vector, gives the right A-module homomorphisms from An to Am . Left module homomorphisms, on the other hand, are given by the right action of the n × m matrices on the 1 × n row vectors. We now give formal treatments of these correspondences, beginning with the case of right modules. Thus, for r ≥ 1, identify Ar with the right A-module of r × 1 column matrices (i.e., column vectors) for all r ≥ 1. A typical element is ⎞ ⎛ a1 ⎟ ⎜ x = ⎝ ... ⎠ . ar As shown above, the standard right module structure on Ar is given by setting xa equal to the matrix product x · aI1 for a ∈ A. We write ei for the basis element ei1 of Lemma 7.10.5, so that ⎛ ⎞ 0 ⎜ .. ⎟ ⎜ . ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎟ ei = ⎜ ⎜ 1 ⎟. ⎜ 0 ⎟ ⎜ ⎟ ⎜ . ⎟ ⎝ .. ⎠ 0 CHAPTER 7. RINGS AND MODULES 279 where nthe 1 is in the i-th place, with 1 ≤ i ≤ n. Thus, with x as above, we have x = i=1 ei ai . Regarding An and Am are right A-modules, recall that HomA (An , Am ) denotes the set of right A-module homomorphisms from An to Am . Recall also that HomA (An , Am ) is an abelian group under the operation that sets (f + g)(x) = f (x) + g(x) for f, g ∈ HomA (An , Am ) and x ∈ An . Proposition 7.10.6. Let m, n ≥ 1. Then there is an isomorphism of abelian groups ∼ = Φm,n : Mm,n (A) −→ HomA (An , Am ) deﬁned as follows. For an m × n matrix M , we deﬁne Φm,n (M ) : An → Am by setting (Φm,n (M ))(x) to be equal to the matrix product M x for all x ∈ An . Proof Φm,n (M ) is a group homomorphism from An to Am by the distributive law of matrix multiplication, and is an A-module homomorphism by associativity: M · (xa) = M (x · aI1 ) = (M x)aI1 = (M x)a. Thus, Φm,n : Mm,n (A) → HomA (An , Am ) is a well deﬁned function, and by the distributive law, it is a group homomorphism. It is easy to see, by direct calculation of matrix products, that M ei is the i-th column of M . Thus, if Φm,n (M ) and Φm,n (M ) have the same eﬀect on the canonical basis vectors, the matrices M and M must be equal. Thus, Φm,n is injective. Let f ∈ HomA (An , Am ), and let M be the matrix whose i-th column is f (ei ) for i = 1, . . . , n. Then Φm,n (M ) and f have the same eﬀect on the canonical basis elements. Since the canonical basis generates An as an A-module, Φm,n (M ) and f must have the same eﬀect on every element of An , and hence Φm,n (M ) = f . Thus, Φm,n is surjective. As we shall see in Section 9.7, there is a natural left A-module structure on HomA (An , Am ) coming from the left action of A on Am . Similarly, there is a natural right A-module structure on HomA (An , Am ) coming from the left action of A on An . As the interested reader may check upon reading Section 9.7, Φm,n gives an A-module isomorphism from the standard left module structure on matrices to the standard left module structure on homomorphisms, as well as from the standard right module structure on matrices to the standard right module structure on homomorphisms. Recall from Problem 24 of Exercises 7.7.27 that HomA (An , An ) forms a ring, otherwise denoted EndA (An ), called the endomorphism ring of An over A. The addition is the same one we’ve just considered: (f + g)(x) = f (x) + g(x). Multiplication is given by composition of functions: (f · g)(x) = f (g(x)). Proposition 7.10.7. For n ≥ 1, Φn,n induces an isomorphism of rings from the matrix ring Mn (A) to the endomorphism ring EndA (An ). (Here, An is considered as a free right A-module.) More generally, if M is an m × n matrix and M is an n × k matrix, then the A-module homomorphism induced by the matrix product M M is the composite of the homomorphism induced by M and the homomorphism induced by M : Φm,k (M M ) = (Φm,n (M )) ◦ (Φn,k (M )) . Proof We’ve seen that Φn,n is a group homomorphism. Moreover, the multiplicative identity element In of Mn (A) is known to induce the identity transformation of An . Thus, Φn,n is a ring homomorphism if and only if it respects multiplication, and hence the ﬁrst statement follows from the second. But the second statement is immediate from the associativity of matrix multiplication: (Φm,n (M ) ◦ Φn,k (M )) (x) = M (M x) = (M M )x = Φm,k (M M )(x). CHAPTER 7. RINGS AND MODULES 280 Recall that a matrix is invertible if it is a unit in Mn (A), and that the group of units there is Gln (A), the n-th general linear group of A. Note that M ∈ Gln (A) if and only if there is a matrix M ∈ Mn (A) with M M = M M = In . There is a module theoretic analogue of the general linear group. Deﬁnition 7.10.8. If N is an A-module, we shall write AutA (N ), the group of A-module automorphisms of N , for the unit group of EndA (N ). Since composition is the ring product in EndA (An ) and the identity function is the multiplicative identity element, an A-module automorphism is precisely an isomorphism f : N → N of A-modules. Since Φn,n is an isomorphism of rings, we obtain the following corollary. Corollary 7.10.9. An n × n matrix M over A is invertible if and only if the right Amodule homomorphism that it induces (i.e., Φn,n (M ) : An → An ) is an isomorphism. In particular, Φn,n restricts to an isomorphism ∼ = Φn,n : Gln (A) −→ AutA (An ). Recall from Lemma 7.7.36 that if f : N → N is an A-module homomorphism between free modules and if x1 , . . . , xn is a basis for N , then f is an isomorphism if and only if f (x1 ), . . . , f (xn ) is a basis for N . Since M ei is the i-th column of M , we obtain the following corollary. Corollary 7.10.10. An n × n matrix M over A is invertible if and only if its columns form a basis of An . It is useful to be able to do manipulations on the matrix level that match standard constructions in module theory. One such construction is the block sum (sometimes called Whitney sum13 ) of matrices: given an m × n matrix M and an m × n matrix M , the block sum M ⊕ M of M and M is the matrix M 0 , 0 M where the 0’s are 0-matrices of the appropriate size. In coordinates, the ij-th coordinate of M ⊕ M is given as follows. If 1 ≤ i ≤ m and 1 ≤ j ≤ n, then the ij-th coordinate of M ⊕ M is equal to the ij-th coordinate of M . If 1 ≤ i ≤ m and j > n, or if i > m and j ≤ n, then the ij-th coordinate of M ⊕ M is 0. If m + 1 ≤ i ≤ m + m and n + 1 ≤ j ≤ n + n , then the ij-th coordinate of M ⊕ M is equal to the (i − m)(j − n)-th coordinate of M . The use of the symbol “⊕” for block sum is meant to be suggestive. In moduletheoretic terms, we use it as follows: If f1 : N1 → N1 and g : N2 → N2 are A-module homomorphisms, we write f ⊕ g : N1 ⊕ N2 → N1 ⊕ N2 for the cartesian product of f and g. Thus, (f ⊕ g)(n1 , n2 ) = (f (n1 ), g(n2 )). 13 After the topologist Hassler Whitney. CHAPTER 7. RINGS AND MODULES 281 Proposition 7.10.11. Let M be an m × n matrix and let M be an m × n matrix over A. Then if we identify An ⊕ An with An+n in the standard way (i.e., identifying n n n n ( i=1 ei ai , i=1 ei ai ) with i=1 ei ai + i=1 ei+n ai ) and identify Am ⊕Am with Am+m similarly, then the A-module homomorphism represented by the block sum M ⊕ M is equal to the direct sum of the A-module homomorphisms represented by M and by M . Symbolically, we have Φm+m ,n+n (M ⊕ M ) = Φm,n (M ) ⊕ Φm n (M ). Proof Both sides agree on the canonical basis elements of An+n . So far we have only treated the module homomorphisms from An to Am , rather than treating the module homomorphisms between arbitrary ﬁnitely generated free modules. The reason is that there is no way to get from module homomorphisms to matrices that doesn’t depend on making explicit choices of bases for the two modules: a diﬀerent choice of basis results in a diﬀerent matrix to represent it. Thus, we should think of An or Am in this discussion as a ﬁnitely generated free module with a speciﬁc preferred basis. Of course, the point is that if N is a ﬁnitely generated free module, then an ordered subset B = n1 , . . . , nk of N is a basis if and only if the unique homomorphism fB : Ak → N with fB (ei ) = ni for 1 ≤ i ≤ k is an isomorphism. Thus, the bases of N with k elements are in one-to-one correspondence with the isomorphisms from Ak to N . Deﬁnition 7.10.12. Suppose given ﬁnitely generated free right A-modules N and N , and let B = n1 , . . . , nk and B = n1 , . . . , nm be bases for N and N , respectively. Write fB : Ak → N and fB : Am → N for the isomorphisms induced by B and B . Then if g : N → N is an A-module homomorphism, we deﬁne the matrix of g with respect to the bases B to be the m × k matrix deﬁned as follows. For 1 ≤ j ≤ k, write m B and g(nj ) = i=1 ni aij . Then the ij-th entry of the matrix of g is deﬁned to be aij . Note that the j-th column of the matrix of g with respect to the bases B and B is precisely fB−1 g(nj ). Thus, the following lemma is immediate. Lemma 7.10.13. The matrix of g : N → N with respect to the bases B and B is equal −1 to the matrix of (fB−1 ◦ g ◦ fB ) : Ak → Am , i.e., it is the matrix Φ−1 m,k (fB ◦ g ◦ fB ). Of particular interest is the use of matrices to study the A-module homomorphisms from a free right A-module N to itself (i.e., to study elements of EndA (N )). Since the source and target of these homomorphisms are the same, we need only one basis to deﬁne a matrix. Deﬁnition 7.10.14. Let B = x1 , . . . , xn be a basis for the free right A-module N and let fB : An → N be the A-module homomorphism that takes ei to xi for each i. Let g : N → N be an A-module homomorphism. Then the matrix of g with respect to the −1 −1 basis B is deﬁned to be Φ−1 n,n (fB ◦ g ◦ fB ), the matrix whose i-th column is fB (g(xi )). The next result makes it clearer what is going on. Lemma 7.10.15. Let B = x1 , . . . , xn be a basis for the free right A-module N and let fB : An → N be the A-module homomorphism that takes ei to xi for each i. Then there is a ring isomorphism fB∗ : EndA (N ) → EndA (An ) deﬁned by fB∗ (g) = fB−1 ◦ g ◦ fB . CHAPTER 7. RINGS AND MODULES 282 Moreover, the matrix of g with respect to the basis B is the image of g under the composite ring isomorphism f∗ Φ−1 n,n B EndA (An ) −−−→ Mn (A). EndA (N ) −→ In particular, the passage from a transformation to its matrix with respect to B gives an isomorphism of rings from EndA (N ) to Mn (A). Proof The function fB∗ is easily seen to be a ring homomorphism. It is an isomorphism because the correspondence that takes g ∈ EndA (An ) to fB ◦ g ◦ fB−1 gives an inverse for it. The rest follows from the deﬁnition of the matrix of g with respect to B. It is valuable to discern which properties of a transformation may be detected by its matrix regardless of the basis used. Thus, it is important to understand what happens to the matrix of g when we change the basis used to deﬁne it. Of course, as we’ve stated above, the rank of a free module is not well deﬁned over every ring A (though it is for the ones that we care about, as we’ll see below). We shall not consider the base-change problem between bases with a diﬀerent number of elements, but shall restrict attention to understanding what happens when we change between bases with the same number of elements. Deﬁnition 7.10.16. Suppose given bases B = x1 , . . . , xn and B = x1 , . . . , xn of the free right A-module N and let fB : An → N and fB : An → N be the A-module homomorphisms that take ei to xi and xi , respectively, for 1 ≤ i ≤ n. Then the base change matrix MBB is deﬁned to be the matrix of the identity map 1N : N → N , where B is used as the basis for the domain of 1N , and B is used as the basis for the target −1 copy of N . Notationally, we have MBB = Φ−1 n,n (fB fB ). Lemma 7.10.17. Suppose given bases B, B , and B for the free right A-module N , all of which have n elements. Then the base change matrices satisfy MB B · MBB = MBB . Additionally, MBB is the identity matrix, and hence MBB is invertible, with inverse given by MB B . Proof We ﬁrst show that the displayed formula holds. By deﬁnition, it suﬃces to show that Φn,n (MB B · MBB ) = fB−1 fB . But Φn,n is a ring homomorphism, so that Φn,n (MB B · MBB ) = Φn,n (MB B ) ◦ Φn,n (MBB ) = fB−1 ◦ fB ◦ fB−1 ◦ fB , and hence the formula holds. Now MBB = Φ−1 n,n (1) = In , and hence MB B MBB = In by setting B = B in the displayed equation. Exchanging the roles of B and B now gives the desired result. We may now express the diﬀerence between the matrices of a given homomorphism with respect to diﬀerent bases. Corollary 7.10.18. Suppose given bases B = x1 , . . . , xn and B = x1 , . . . , xn of the free right A-module N and let MBB be the base change matrix deﬁned above. Then if g ∈ EndA (N ) and if M is the matrix of g with respect to the basis B, then the matrix of −1 g with respect to B is the matrix product MBB M MBB . CHAPTER 7. RINGS AND MODULES 283 Proof Since Φn,n is a ring homomorphism, we have −1 −1 −1 Φn,n (MBB M MBB ◦ g ◦ fB ) ◦ (fB−1 ◦ fB ). ) = (fB ◦ fB ) ◦ (fB The result now follows. Deﬁnition 7.10.19. We say that two n × n matrices, M, M ∈ Mn (A) are similar if there is an invertible n × n matrix M such that M = M M (M )−1 . Corollary 7.10.20. Let N be a ﬁnitely generated free right A-module and let g ∈ EndA (N ). Suppose we’re given two bases of N with the same number of elements. Then the matrices of g with respect to these two bases are similar. Of course, we’d like to treat left modules as well. When A is commutative, the usual thing is to identify left modules with right modules, and to study homomorphisms of free left modules by the methods above. For noncommutative rings, this doesn’t work, and hence further argumentation is needed. In this case, we identify the free left module An with the 1 × n row matrices (i.e., row vectors). This has the advantage that the notation agrees with the notation we’re used to for free modules: A typical element is x = (a1 , . . . , an ). Again we write ei = (0, . . . , 0, 1, 0, . . . , 0), where the 1 is in the i-th place. Here, the key is that if M is an n × m matrix, then the matrix product ei M is precisely the i-th row of M . Given this, and the fact that the left A-module structure on An is obtained by setting ax equal to the matrix product (aI1 )x, the proof of the next proposition is entirely analogous to that of Proposition 7.10.6, and is left to the reader. Proposition 7.10.21. Write HomA (An , Am ) for the set of left A-module homomorphisms from An to Am . Then there is a group isomorphism Ψn,m : Mn,m (A) → HomA (An , Am ) deﬁned by setting (Ψn,m (M ))(x) equal to the matrix product xM . In this case, there is a reversal between composition of functions and matrix multiplication: if M is n × m and M is k × n, then (Ψn,m (M ) ◦ Ψk,n (M )) (x) = (xM )M = x(M M ) = Ψk,m (M M )(x). We obtain the following proposition. Proposition 7.10.22. For n ≥ 1, Ψn,n induces an isomorphism from the matrix ring Mn (A) to the opposite ring of the endomorphism ring of the free left A-module An . As in the case of right modules, the block sum of matrices represents the direct sum of the associated A-module homomorphisms. The treatment of matrices associated to diﬀerent bases is analogous to that given for right modules. We leave the details to the reader. Exercises 7.10.23. 1. Let N be a free right A-module with basis B = x1 , . . . , xn . Let g ∈ EndA (N ) and let M be the matrix of g with respect to the basis B. Let M be an n × n matrix that is similar to M . Show that M is the matrix of g with respect to some other basis of N . CHAPTER 7. RINGS AND MODULES 2. Suppose given matrices M1 M= M3 M2 M4 284 and M = M1 M3 M2 M4 , where M1 is m × n, M2 is m × s, M3 is r × n, and M4 is r × s, while M1 is n × k, M2 is n × t, M3 is s × k, and M4 is s × t. Show that M1 M1 + M2 M3 M1 M2 + M2 M4 . MM = M3 M1 + M4 M3 M3 M2 + M4 M4 3. Let M be an (n + k) × (n + k) matrix over A. Show that M has the form M1 M 2 M= 0 M3 with M1 n × n, M2 n × k, and M3 k × k if and only if (Φn,n (M ))(An ) ⊂ An (i.e., the standard right A-module homomorphism induced by M carries An into itself). Here An is the submodule of An+k generated by e1 , . . . , en . Show also that the matrices of the above form (where n is ﬁxed) give a subring of Mn+k (A). 4. Let B = Mn (A). Show that the matrix ring Mk (B) is isomorphic as a ring to Mnk (A). ‡ 5. If A is commutative, then, since left and right A-modules may be identiﬁed with each other, Ψ−1 n,n ◦ Φn,n provides an isomorphism from Mn (A) to its opposite ring for each n ≥ 1. Explicitly write down the eﬀect of Ψ−1 n,n ◦ Φn,n on a matrix (aij ). 7.11 Rings of Fractions Think for a moment about how our number system evolved. First, there were the nonnegative integers. Then negative numbers were adjoined, producing the integers. This process of adjoining inverses has since been generalized to a construction that produces an abelian group from any abelian monoid. It is called the Grothendieck construction, and will be studied below. The next stage in the development of our number system was the creation of the rational numbers, by adjoining multiplicative inverses to the nonzero integers. We shall generalize this last process in this section. One of the consequences will be that we can then embed any integral domain in a ﬁeld. Deﬁnition 7.11.1. Let A be a commutative ring. A multiplicative subset of A is a subset S ⊂ A such that 0 ∈ S and such that S is closed under multiplication: For s, t ∈ S, the product st is also in S.14 Note: A ring called “A” in this section will be implicitly assumed to be commutative. Other rings, however, will not. Let S be a multiplicative subset of the commutative ring A. We shall construct a new ring S −1 A, in which the elements of S are invertible. 14 Some people permit 0 to be an element of a multiplicative subset. Indeed, the entire theory of rings of fractions generalizes to that context, but, if 0 ∈ S, then the ring of fractions S −1 A is the 0 ring. So all we’re doing is excluding this rather boring case. CHAPTER 7. RINGS AND MODULES 285 The elements of S −1 A will be equivalence classes of ordered pairs (a, s) with a ∈ A and s ∈ S. We say that (a, s) is equivalent to (b, t) (written (a, s) ∼ (b, t)) if there is an s ∈ S such that s (ta − sb) = 0. Notice that if A is an integral domain, then this is just the familiar condition we use to test when two fractions a/s and b/t are equal in Q. Lemma 7.11.2. Let S be a multiplicative subset of the commutative ring A. Then the above relation ∼ is an equivalence relation. Proof The relation is obviously symmetric and reﬂexive, so it suﬃces to show transitivity. Suppose that (a, s) ∼ (b, t) and that (b, t) ∼ (c, u). Then there are elements s , s ∈ S with s (ta − sb) = 0 and s (ub − tc) = 0. We claim now that s s t(ua − sc) = 0, and hence (a, s) ∼ (c, u). Now s s tua = s s usb, because s (ta − sb) = 0. Similarly, s s usb = s s stc, because s (ub − tc) = 0. Putting this together, we see that s s t(ua − sc) = 0 as desired. Deﬁnition 7.11.3. Let S be a multiplicative subset of the commutative ring A. We deﬁne the ring of fractions S −1 A as the set of equivalence classes of pairs (a, s), and, to emphasize the analogy with fractions, we write a/s, or as , for the element of S −1 A represented by (a, s). The addition and multiplication operations in S −1 A are given by ta + sb a b + = s t st and a b ab · = . s t st We deﬁne the canonical map η : A → S −1 A by setting η(a) = as/s for any element s ∈ S. The proof of the following proposition is tedious but straightforward. We leave it to the reader. Proposition 7.11.4. Let S be a multiplicative subset of A. Then S −1 A is a commutative ring under the above operations. The additive identity is 0/s for any s ∈ S, and multiplicative identity is s/s for any s ∈ S. The canonical map η : A → S −1 A is a ring homomorphism whose kernel is the set of all a ∈ A that are annihilated by some s ∈ S. For s ∈ S, the element η(s) is a unit in S −1 A, with inverse 1/s. The generic element of S −1 A may then be characterized by a = η(a) · (η(s))−1 . s The reader may very well ask: Why don’t we insist that 1 ∈ S? The answer, it turns out, is that there isn’t a good reason other than the fact that it isn’t necessary. We’ll see in Exercises 7.11.27 that if 1 ∈ S, and if we take S = S ∪ {1}, then the natural inclusion −1 of S −1 A in S A is an isomorphism. The principal examples of rings of fractions are important enough to warrant names. Deﬁnitions 7.11.5. 1. Let A be an integral domain and let S be the set of all nonzero elements in A. The resulting ring S −1 A is called the ﬁeld of fractions, A(0) , of A. 2. Let A be any commutative ring and let p be a prime ideal of A. For S = {a ∈ A | a ∈ p} and call it the localization of A at p. we write S −1 A = Ap , CHAPTER 7. RINGS AND MODULES 286 3. Let A be any commutative ring and let a ∈ A be any element that is not nilpotent. Let S = {ak | k ≥ 0}. We write S −1 A = A [1/a] , and call it the ring obtained by adjoining an inverse to a. It is easy to see that Q is the ﬁeld of fractions of Z, so that we have, as promised, generalized the construction of the rationals from the integers. Note that the ﬁeld of fractions of an integral domain A really is a ﬁeld, as η(a)(η(s))−1 is a unit in A(0) whenever a = 0. We obtain the following corollary. Corollary 7.11.6. The ﬁeld of fractions, A(0) , of an integral domain A is a ﬁeld, and η : A → A(0) an embedding. Thus, a ring is an integral domain if and only if it is a subring of a ﬁeld. The following examples are important in ﬁeld theory and number theory, respectively. Examples 7.11.7. 1. Let K be a ﬁeld and let X1 , . . . , Xn be variables. We write K(X1 , . . . , Xn ) for the ﬁeld of fractions of the polynomial ring K[X1 , . . . , Xn ]. We call it the ﬁeld of rational functions over K in the variables X1 , . . . , Xn . p form an 2. Recall from Problem 1 of Exercises 7.1.41 that the p-adic integers Z integral domain. We write Qp for its ﬁeld of fractions, known as the ﬁeld of p-adic rationals. Note that a ﬁeld of fractions is a special case of localization, as the ideal (0) is prime in an integral domain. If p ∈ Z is prime, then the ring Z [1/p] is sometimes called the “localization of Z away from p.” As we shall see below, this is a deﬁnite abuse of the word localization. Lemma 7.11.8. Let B be a ring and let a and b be elements of B that commute with each other, where b ∈ B × . Then a commutes with b−1 . Proof Clearly, b annihilates ab−1 − b−1 a. Since b is a unit, the result follows. We return to the general case of a multiplicative subset of a commutative ring A. The ring of fractions S −1 A satisﬁes a useful universal property. Proposition 7.11.9. Let f : A → B be a ring homomorphism such that f (s) is a unit for all s ∈ S. Then there is a unique ring homomorphism f : S −1 A → B such that the following diagram commutes. f / B A? ? ?? ?? η ?? f S −1 A Moreover, if f gives an A-algebra structure on B, then f , if it exists, induces an S −1 A-algebra structure on B. We obtain a one-to-one correspondence between the S −1 Aalgebra structures on B and the A-algebra structures f : A → B with the property that f (S) ⊂ B × . And if g : B → C is an A-algebra homomorphism between S −1 A-algebras, then it is also an S −1 A-algebra homomorphism. CHAPTER 7. RINGS AND MODULES 287 Proof Suppose there is a ring homomorphism f that makes the diagram commute. Since η(s) is a unit in S −1 A for each s ∈ S, f (s) = f (η(s)) must be a unit in B. Moreover, since a/s = η(a)(η(s))−1 , f must satisfy the formula f (a/s) = f (a)f (s)−1 . Thus, f is uniquely deﬁned by the commutativity of the diagram. Conversely, if f (s) is a unit for each s ∈ S, we deﬁne f by the same formula: f (a/s) = f (a)f (s)−1 . This is well deﬁned, since if s (ta − sb) = 0, we have f (ta − sb) = 0 by the invertibility of f (s ). Making use of Lemma 7.11.8, if B is noncommutative, it is easy to see that f is a ring homomorphism. The statements about existence and uniqueness of algebra structures are straightforward. To see that every A-algebra homomorphism f : B → C between S −1 A-algebras is an S −1 A-algebra homomorphism, write ν and μ for the S −1 A-algebra structures on B and C, respectively. We want to show that f ◦ ν = μ. Our assumption is that f is an A-algebra homomorphism, which means that f ◦ ν agrees with μ if we precompose with η : A → S −1 A. But two ring homomorphisms out of S −1 A that agree when precomposed with η must be equal. Now Q is the ﬁeld of fractions of Z. Recall that for any ring B, there is a unique ring homomorphism f : Z → B, where f (n) = n · 1, the n-th “power” of 1 in the additive group structure on B. The homomorphism f gives B a unique Z-algebra structure, and B has characteristic 0 if and only if f is injective. We obtain the following corollary. Corollary 7.11.10. A ring B is a Q-algebra if and only if n · 1 is a unit in B for each 0 = n ∈ Z. The Q-algebra structure on B, if it exists, is unique, and any ring homomorphism between Q-algebras is a Q-algebra homomorphism. In particular, there is a unique embedding (as a ring) of Q in any ﬁeld of characteristic 0. We also have the following. Corollary 7.11.11. Let f : A → K be an embedding of an integral domain A into a ﬁeld K. Then there is a unique extension of f to an embedding f : A(0) → K, where A(0) is the ﬁeld of fractions of A. The image of f is the smallest subﬁeld of K containing the image of f . We now have a framework to classify the subrings of Q. Proposition 7.11.12. Let A be a subring of Q and let S = A× ∩ Z. Then A is isomorphic as a ring to S −1 Z. Proof The elements of S become units in A. Thus, Proposition 7.11.9 provides a unique ring homomorphism f : S −1 Z → A. We claim that f is an isomorphism. f First, note that the composite S −1 Z − → A ⊂ Q takes m/s to m/s. Since the equivalence relations deﬁning S −1 Z and Q are the same (because Z is an integral domain), the above composite is injective, and hence so is f . Now suppose given an element of A. Since A ⊂ Q, we may write it as m/n for m, n ∈ Z with n = 0. By reducing the fraction, we may assume that m and n are relatively prime. It suﬃces to show that n is a unit in A, i.e., that the element 1/n of Q lies in A. Since m and n are relatively prime, we can ﬁnd a, b ∈ Z with am + bn = 1. Dividing both sides by n, we see that a(m/n) + b = 1/n. Since a, m/n and b are in A, so is 1/n. CHAPTER 7. RINGS AND MODULES 288 The universal property for homomorphisms out of rings of fractions gives the simplest proof of the following characterization of modules over these rings. The situation is similar to that given by Proposition 7.7.15. Because S −1 A is an A-algebra, every S −1 A-module has an underlying A-module structure. So an immediate question is: Which A-modules admit a compatible S −1 Amodule structure? And if a compatible S −1 A-module structure exists, is it unique? (Compatibility here means that the A-module structure induced by the S −1 A-module structure is the same as the one we started with.) A good answer will give us a complete understanding of S −1 A-modules. Proposition 7.11.13. Let M be an A-module. Then M admits a compatible S −1 A module structure if and only if multiplication by s induces an isomorphism (of abelian groups) from M to M for each s ∈ S. Moreover, there is at most one S −1 A-module structure compatible with the original A-module structure on M . Finally, if the A-modules M and N admit compatible S −1 A-module structures, then every A-module homomorphism from M to N is an S −1 A-module homomorphism as well. Proof We use the correspondence given by Problem 21 of Exercises 7.7.27 between A-module structures on M and ring homomorphisms from A to EndZ (M ). Note that if f : A → EndZ (M ) and f : S −1 A → EndZ (M ) are ring homomorphisms, then the S −1 A-module structure on M induced by f is compatible with the A-module structure coming from f if and only if the diagram f / EndZ (M ) A JJ JJ t: JJ ttt J t J t η J$ tt f −1 S A commutes. For a given homomorphism f : A → EndZ (M ), the homomorphisms f : S −1 A → EndZ (M ) that make the diagram commute are characterized by Proposition 7.11.9: For a given f there is at most one f with this property, and such an f exists if and only if f (s) is a unit in EndZ (M ) (i.e., a group isomorphism of M ) for each s ∈ S. It remains to show that if f : M → N is an A-module homomorphism between the S −1 A-modules M and N , then f is in fact an S −1 A-module homomorphism. In particular, it suﬃces to show that f ((1/s)m) = (1/s)f (m) for s ∈ S and m ∈ M . Write s : M → M and s : N → N for the homomorphisms induced by multiplication by s. Then the statement that f ((1/s)m) = (1/s)f (m) for all m ∈ M amounts to the −1 statement that f ◦ −1 s = s ◦ f . Now, since f is an A-module homomorphism, we have f ◦ s = s ◦ f . So the result now follows as in the proof of Lemma 7.11.8. We’d like now to study ideals in rings of fractions. Indeed, the properties of the ideals in a localization form one of the primary motivations for studying rings of fractions. Given the relationship between ideals and modules, it seems reasonable to discuss modules of fractions ﬁrst. Let M be an A-module and let S be a multiplicative subset of A. We shall deﬁne an S −1 A-module S −1 M as follows. The elements of S −1 M are equivalence classes of ordered pairs (m, s) with m ∈ M and s ∈ S, where (m, s) is equivalent to (n, t) (written (m, s) ∼ (n, t)) if there is an s ∈ S with s (tm − sn) = 0. As expected, we write m/s for the element of S −1 M represented by the ordered pair (m, s). The proof of the following lemma is straightforward and is left to the reader. CHAPTER 7. RINGS AND MODULES 289 Lemma 7.11.14. The relation ∼ above is an equivalence relation. The resulting set S −1 M of equivalence classes is an abelian group and an S −1 A-module via m n tm + sn + = s t st a m am · = , s t st and respectively. There is a canonical A-module homomorphism η : M → S −1 M given by η(m) = ms/s for any s ∈ S. If N is an S −1 A-module and f : M → N is an A-module homomorphism, then there is a unique S −1 A-module homomorphism f : S −1 M → N that makes the following diagram commute. f / N M? ? ?? ?? η ??? f S −1 M We obtain an isomorphism η∗ HomS −1 A (S −1 M , N ) −→ HomA (M, N ) ∼ = via η ∗ (f ) = f ◦ η. Localization forms an important special case of modules of fractions: Deﬁnition 7.11.15. Let p be a prime ideal of A and let S be the complement of p in A. Then we write Mp for S −1 M . We call it the localization of M at p. Let S be an arbitrary multiplicative subset of A and let f : M → N be an A-module homomorphism. Let S −1 f : S −1 M → S −1 N be deﬁned by (S −1 f )(m/s) = f (m)/s. Then S −1 f is clearly the unique S −1 A-module homomorphism that makes the following diagram commute. f M η / N η S S −1 M −1 f/ S −1 N Clearly, passage from M to S −1 M and from f to S −1 f provides a functor from Amodules to S −1 A-modules (i.e., S −1 f ◦S −1 g = S −1 (f ◦g), and S −1 (1M ) = 1S −1 M , where 1N denotes the identity map of the module N ). Proposition 7.11.16. Suppose given an exact sequence f g · · · → M − →M − → M → · · · of A-modules. Then the induced sequence S −1 f S −1 g · · · → S −1 M −−−→ S −1 M −−−→ S −1 M → · · · of S −1 A-modules is also exact. CHAPTER 7. RINGS AND MODULES 290 Proof A sequence is exact if it is exact at every module in it. We show exactness at M . First, note that S −1 g ◦ S −1 f = S −1 (g ◦ f ) = S −1 0, since g ◦ f is the 0-homomorphism 0 (since im f ⊂ ker g). But S −1 0 = 0, so im(S −1 f ) ⊂ ker(S −1 g). Now let m/s ∈ ker S −1 g. Then a glance at the equivalence relation deﬁning S −1 M shows that tg(m) = 0 for some t ∈ S. But then g(tm) = tg(m) = 0, so tm ∈ ker g = im f . −1 f is onto. Let tm = f (m ). Then m/s = (S −1 f )( m st ). Thus, S This behavior may be abstracted as follows. Deﬁnition 7.11.17. A functor which takes exact sequences to exact sequences is called an exact functor. In particular, the passage from M to S −1 M is an exact functor. This has a number of uses, since certain kinds of relationships can be described by exactness properties. For ⊂ instance, N is a submodule of M if and only if the sequence 0 → N −→ M is exact. We immediately obtain the following corollary. Corollary 7.11.18. Let S be a multiplicative subset of A and let M be an A-module. Then if N is a submodule of M , the natural map from S −1 N to S −1 M is an inclusion of S −1 A-modules. Thus, if a is an ideal of A, then S −1 a is an ideal of S −1 A. We shall show that every ideal of S −1 A arises this way: Proposition 7.11.19. Let S be a multiplicative subset of A and let b be an ideal of S −1 A. Let η : A → S −1 A be the canonical homomorphism and let a = η −1 b. Then b = S −1 a. Proof Let x ∈ a. Then xs/s ∈ b for any s ∈ S, and hence so is x/t = (xs)/(ts) for any t ∈ S, since b is an ideal of S −1 A. Thus, S −1 a ⊂ b. Suppose given an arbitrary element x/s ∈ b with x ∈ A and s ∈ S. It suﬃces to show that x ∈ a. But this follows, since η(s) · (x/s) = η(x) (i.e., η(x) ∈ b). Corollary 7.11.20. Let S be a multiplicative subset of A. Then the prime ideals of S −1 A are those of the form S −1 p, where p is a prime ideal of A such that p ∩ S = ∅. Proof Since the preimage under a ring homomorphism of a prime ideal is prime, every prime ideal of S −1 A has the form S −1 p for some prime ideal p of A. Also, since S −1 p must be a proper ideal of S −1 A in order to be prime, S −1 p must not contain a unit of S −1 A, and hence p ∩ S must be empty. Thus, it suﬃces to show that if p is any prime ideal of A that doesn’t meet S, then S −1 p is prime in S −1 A. Suppose given such a p, and suppose that (a/s) · (b/t) ∈ S −1 p. Say (ab)/(st) = x/s , with x ∈ p and s ∈ S. This says there is an s ∈ S with s (s ab − stx) = 0. But then s s ab ∈ p. Since s and s are not in p, one of a and b must be. Thus, one of a/s and b/t is in S −1 p. Thus, it suﬃces to show that S −1 p is proper. Suppose that x/s = 1 in S −1 A, with x ∈ p and s ∈ S. Since 1 = s /s for any s ∈ S, we can ﬁnd s , s ∈ S with s (s x − s s) = 0. But this places s s s in p, which is impossible, and hence S −1 p cannot be all of S −1 A. Of course, if S −1 p and S −1 p are prime ideals of S −1 A, then S −1 p ⊂ S −1 p if and only if p ⊂ p, by Proposition 7.11.19. We now are in a position to explain the term “localization.” CHAPTER 7. RINGS AND MODULES 291 Deﬁnition 7.11.21. A local ring is a commutative ring that has only one maximal ideal. Fields, of course, are local rings, as are localizations, as we shall see shortly. Another p. example of a local ring that we’ve seen is Z Example 7.11.22. Recall from Proposition 7.1.40 that the ring homomorphism p → Zp π1 : Z p is a unit if and only if π1 (a) is nonzero in Zp . Thus, any has the property that a ∈ Z p p which doesn’t lie in ker π1 is a unit. Therefore, any proper ideal of Z element of Z must be contained in ker π1 , and hence ker π1 is the unique maximal ideal of Zp . Note that Problem 4 of Exercises 7.1.41 shows that ker π1 = (p), the principal ideal p is a local ring with maximal ideal (p). generated by p. Thus, Z Proposition 7.11.23. Let p be a prime ideal in the commutative ring A. Then the localization Ap is a local ring whose maximal ideal is pp . Proof Let S be the complement of p in A. Then p ∩ S = ∅, so that pp = S −1 p is a prime ideal of Ap . Since maximal ideals are prime, it suﬃces to show that any other prime ideal of Ap must be contained in pp . But the prime ideals of Ap all have the form pp , where p ∩ S = ∅. By the deﬁnition of S, this forces p ⊂ p, and the result follows. This, of course, begs the question of the relationship of the ﬁeld Ap /pp to A itself. Proposition 7.11.24. Let p be a prime ideal of A. Then Ap /pp is isomorphic to the ﬁeld of fractions of A/p. In particular, if m is a maximal ideal of A, then Am /mm is just A/m. Proof We have a commutative diagram A π A/p η η / Ap π / Ap /pp where π and π are the canonical maps. It is easy to check that p = η −1 (pp ), and hence η is an embedding. By Corollary 7.11.11, η extends uniquely to an embedding, η : K → Ap /pp of the ﬁeld of fractions, K, of A/p. But η is surjective, as π (a/s) = η(π(a)/π(s)). If p is maximal, then A/p is a ﬁeld, and hence A/p = K by Corollary 7.11.11. Localization turns out to be a very important technique in understanding A-modules. The point is that modules over a local ring are easier to classify than those over a non-local ring. This will become especially clear in our study of projective modules in Section 9.8. So it is valuable to be able to deduce information about a module M from information about the localizations Mp . Proposition 7.11.25. Let M be a module over the commutative ring A. Then the following conditions are equivalent. CHAPTER 7. RINGS AND MODULES 292 1. M = 0. 2. Mp = 0 for each prime ideal p of A. 3. Mm = 0 for each maximal ideal m of A. Proof Clearly, the ﬁrst condition implies the second, which implies the third. We shall show that the third implies the ﬁrst. Let M be an A-module such that Mm = 0 for each maximal ideal m of A. Let m ∈ M . It suﬃces to show that the annihilator of m is A. Thus, let m be a maximal ideal of A. Then m is in the kernel of the canonical map from M to Mm . Thus, m/1 = 0/1 in Mm , so there exists s in A which is not in m such that sm = 0. In other words, Ann(m) is not contained in m. But since Mm = 0 for every maximal ideal m of A, we see that Ann(m) is not contained in any of the maximal ideals of A. Since Ann(m) is an ideal, it must be all of A. Corollary 7.11.26. Let f : M → N be an A-module homomorphism. Then the following conditions are equivalent. 1. f is injective. 2. fp is injective for each prime ideal p of A. 3. fm is injective for each maximal ideal m of A. The same holds if we replace injective by surjective. Proof Consider the exact sequence i f π →M − →C→0 →N − 0→K− of Lemma 7.7.47. Since localization is an exact functor, we see that Kp is the kernel of fp and Cp is the cokernel of fp for each prime ideal p of A. Thus, the result follows immediately from Proposition 7.11.25. Exercises 7.11.27. 1. Show that Zn is a local ring if and only if n is a prime power. 2. Show that the localization, Z(p) , of Z at (p) embeds uniquely as a subring of the p. p-adic integers Z 3. Let a be any non-nilpotent element of the commutative ring A. Show that A[1/a] is isomorphic as an A-algebra to A[X]/(aX − 1), where (aX − 1) is the principal ideal generated by aX − 1 in the polynomial ring A[X]. 4. Let A be an integral domain and let K be its ﬁeld of fractions. Show that if 0 = a ∈ A, then the ring of fractions A[1/a] may be identiﬁed with the image of the evaluation map ε1/a : A[X] → K obtained by evaluating X at 1/a. Thus, the notation A[1/a] is consistent with that for adjoining an element to A. 5. Let A be a P.I.D. and let S be a multiplicative subset of A. Show that the ring of fractions S −1 A is also a P.I.D. CHAPTER 7. RINGS AND MODULES 293 ‡ 6. Let S be a multiplicative subset of A with 1 ∈ S. Suppose that S is generated as a monoid by a ﬁnite set s1 , . . . , sk , i.e., an element is in S if and only if it has the form sr11 . . . srkk where ri ≥ 0 for 1 ≤ i ≤ k. Show that S −1 A is isomorphic as an A-algebra to A[X1 , . . . , Xk ]/(s1 X1 − 1, . . . , sk Xk − 1). Here, X1 , . . . , Xk are indeterminates over A, and (s1 X1 − 1, . . . , sk Xk − 1) is the ideal generated by s1 X1 − 1, . . . , sk Xk − 1. It is common to write A[1/s1 , . . . , 1/sk ] for S −1 A in this context, a usage which may be of assistance in designing the proof. 7. Let A be a commutative ring and let S ⊂ A[X] be given by S = {X k | k ≥ 0}. Show that S −1 A[X] is isomorphic to the group ring A[Z] of Z over A. 8. We say that a multiplicative subset S ⊂ A is saturated if for any s ∈ S and any a ∈ A dividing s, we must have a ∈ S. Show that if S is any multiplicative subset, then T = {a ∈ A | a divides s for some s ∈ S} is a multiplicative subset. Show also that T is saturated. We call T the saturation of S. † 9. Let S be a multiplicative subset of A and let T be its saturation. Show that for −1 any multiplicative subset S with S ⊂ S ⊂ T , the natural maps S −1 A → S A → −1 T A are isomorphisms of rings. Show also that if M is an A-module, then the −1 natural maps S −1 M → S M → T −1 M are isomorphisms of S −1 A-modules. 10. To complete the classiﬁcation of the subrings of Q, it suﬃces to ﬁnd all of the saturated multiplicative subsets of Z. Show that there is a one-to-one correspondence between the saturated multiplicative subsets of Z and the subsets of the set of all primes in Z. Here, if T is a set of primes in Z, then the associated multiplicative subset is given by S = {±pr11 . . . prkk | k ≥ 0, p1 , . . . , pk ∈ T and ri ≥ 0 for i = 1, . . . , k}. 11. Let A be an integral domain and let B be any commutative ring. Show that the ideal p = (0 × B) of A × B is prime and that the localization (A × B)p is isomorphic to the ﬁeld of fractions of A. 12. Let p be a prime number and let M be a ﬁnite abelian group. Show that the localization M(p) is isomorphic to the subgroup Mp of M consisting of those elements whose order is a power of p. (Hint: Consider the restriction of the natural map η : M → M(p) to Mp .) 13. Let S be a multiplicative subset of Z and let M be a ﬁnite abelian group. Show that S −1 M is isomorphic to the subgroup of M consisting of those elements whose order is relatively prime to every element in S. 14. Let S be a multiplicative subset of A. Show that if S ⊂ A× , then the natural map η : A → S −1 A is an isomorphism. 15. Let S be a multiplicative subset of A and let M be an A-module that admits a compatible S −1 A-module structure. Show that the natural map η : M → S −1 M is an isomorphism. 16. Let F be a functor from A-modules to B-modules. Show that F is an exact functor if and only if for any short exact sequence f g →M − → M → 0 0 → M − CHAPTER 7. RINGS AND MODULES 294 of A-modules, the induced sequence F (f ) F (g) 0 → F (M ) −−−→ F (M ) −−−→ F (M ) → 0 of B-modules is also exact. Chapter 8 P.I.D.s and Field Extensions The theories of ﬁelds and of P.I.D.s are inextricably intertwined. If K is a ﬁeld, then the polynomial ring K[X] is a P.I.D. We shall show in Section 8.1 that the elements of a P.I.D. have prime decompositions similar to the ones in the integers. This is used to study the extension ﬁelds L of K, via the evaluation maps εα : K[X] → L obtained by evaluating X at an element α of L. Also, we give a classiﬁcation of the ﬁnitely generated modules over a P.I.D. in Section 8.9. We shall use this to classify the n × n matrices over a ﬁeld K up to similarity in Section 10.6. The point is that an n × n matrix induces a K[X]-module structure on K n in which X acts on K n by multiplication by the matrix in question. The theory of prime factorization in P.I.D.s also gives some elementary examples of the behavior of primes under ﬁnite extensions of the rational numbers, Q. The point is that if K is a ﬁnite extension ﬁeld of Q, then there is a subring O(K), called the ring of integers of K, that plays a role in K similar to the role of the integers in Q. An important topic in number theory is the way in which the prime numbers in Z factor in O(K). In the general case, O(K) is a Dedekind domain, but not a P.I.D. In that case, we study the factorizations of the principal ideal (p) of O(K) as a product of ideals. But in some nice cases, which we shall illustrate in the exercises to Section 8.1, the ring of integers is a P.I.D., and we may study this factorization in terms of factorizations of prime elements. The theory of prime factorization is developed in Section 8.1 in a generalization of P.I.D.s called unique factorization domains, or U.F.D.s. In Section 8.2, we study the algebraic extensions of a ﬁeld K. These are the extension ﬁelds with the property that the evaluation maps εα : K[X] → L have nontrivial kernels for each α ∈ L. They will be the primary objects of study in Chapter 11. We begin to analyze them here. Section 8.3 studies extension ﬁelds which are not algebraic. There is a notion of transcendence degree, which measures how far a given extension is from being algebraic. There is a theory of transcendence bases, which behave toward the transcendence degree in a manner analogous to the relationship between a basis for a vector space and its dimension. In Section 8.4, we construct an algebraic closure for a ﬁeld K. Uniqueness statements regarding algebraic closures are deferred to Section 11.1. We shall see that every algebraic extension of a ﬁeld K may be embedded in its algebraic closure. So algebraic closures will be an important tool in the material on Galois theory. Algebraic closures are also important in matrix theory, as the Jordan canonical form 295 CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 296 may be deﬁned for n × n matrices over an algebraically closed ﬁeld. Thus, we may study matrices over K by passage to the matrix ring of its algebraic closure. The various applications of prime factorization in the P.I.D. K[X] depend strongly on our ability to actually compute these factorizations. In particular, we need to be able to test polynomials for primality. In Section 8.5, we give criteria for primality in Q[X], or, more generally, in K[X], where K is the ﬁeld of fractions of a unique factorization domain. In the process, we show that if A is a U.F.D., then so is A[X]. In particular, the polynomial ring K[X1 , . . . , Xn ] is a U.F.D. for any ﬁeld K. In Section 9.3, we make use of unique factorization in K[X1 , . . . , Xn ], the study of transcendence bases, and the theory of Noetherian rings to prove Hilbert’s Nullstellensatz, which states that if the extension ﬁeld L of K is ﬁnitely generated as a K-algebra, then L is a ﬁnite extension of K. In consequence, we determine all of the maximal ideals of K[X1 , . . . , Xn ] when K is an algebraically closed ﬁeld. This is the starting point for the study of algebraic varieties. We proceed to deﬁne algebraic varieties and to describe the Zariski topology on aﬃne n-space, K n , as well as on the prime spectrum of a commutative ring. This material forms an introduction to the study of algebraic geometry. The next three sections study ﬁeld extensions, culminating in the determination of the degree of the cyclotomic extension Q[ζn ] over Q, as well as the determination of the minimal polynomial of ζn over Q. This establishes the most basic facts about a family of examples that has vital connections to a number of branches of mathematics. For instance, we shall see that the cyclotomic rationals play a vital role in computations of Galois theory. And their rings of integers, Z[ζn ], are of crucial importance in issues related to Fermat’s Last Theorem, as well as to calculations of the K-theory of integral group rings. The latter plays an important role in topology. Section 8.6 introduces the Frobenius homomorphism and discusses perfect ﬁelds. Section 8.7 introduces the notion of repeated roots, which forms the basis for the study of separable extensions. We shall return to these ideas in Chapter 11. 8.1 Euclidean Rings, P.I.D.s, and U.F.D.s Recall that the Euclidean algorithm was the main tool we used to analyze the structure of the integers. There is an analogue of it which holds in a variety of other rings, including the polynomial ring K[X] over a ﬁeld K. The consequences, for rings that satisfy this property, are very similar to the consequences that hold in Z. Deﬁnition 8.1.1. A Euclidean domain is an integral domain A that admits a function ϕ : A → Z with the following properties. 1. If b divides1 a and a = 0, then ϕ(b) ≤ ϕ(a). 2. For any a, b ∈ A with b = 0, there are elements q, r ∈ A such that a = qb + r and ϕ(r) < ϕ(b). We call the function ϕ : A → Z a Euclidean structure map for A. 1 I.e., b. a = bc for some c ∈ A. Once again, this just says that a ∈ (b), the principal ideal generated by CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 297 Examples 8.1.2. 1. Z is Euclidean, with the structure map given by the absolute value function | | : Z → Z. 2. Let K be a ﬁeld, and deﬁne ϕ : K[X] → Z by setting ϕ(f (X)) equal to the degree of f (X) if f (X) = 0, and setting ϕ(0) = −1. Then Proposition 7.3.10 shows that ϕ puts a Euclidean structure on K[X]. 3. In Problem 5 of Exercises 7.1.41, it is shown that every nonzero element of the p may be written as a product pk u with k ≥ 0 and u ∈ Z × . Deﬁne p-adic integers Z p p → Z by setting ϕ(pk u) = k and ϕ(0) = −1. Since p is not a unit in the ϕ:Z p , ϕ is a well deﬁned function. Note that if a and b are nonzero elements domain Z of Zp , then ϕ(b) ≤ ϕ(a) if and only if b divides a. It follows that ϕ is a Euclidean p. structure map for Z p is also a local ring, meaning Recall from the discussion in Example 7.11.22 that Z p is an that it has only one maximal ideal, in this case the principal ideal (p). Thus, Z example of the following notion. Deﬁnition 8.1.3. A discrete valuation ring, or D.V.R., is a Euclidean domain that is also a local ring. The Euclidean structure on Z was crucial in our analysis of the subgroups of Z. It will have similar value for the study of the ideals in other Euclidean domains. There is an additional property one could ask for from the function ϕ: Deﬁnition 8.1.4. We say that a Euclidean structure map ϕ : A → Z is positive deﬁnite if ϕ(a) > 0 for all 0 = a ∈ A and ϕ(0) = 0. Lemma 8.1.5. Let A be a Euclidean domain. Then there is a Euclidean structure map for A which is positive deﬁnite. Proof We claim that if ϕ : A → Z is any Euclidean structure map for A, then ϕ(0) < ϕ(a) for all a = 0. Given this, it will suﬃce to replace ϕ by a function ϕ given by ϕ (a) = ϕ(a) − ϕ(0). To prove the claim, let a = 0 and write 0 = qa + r with ϕ(r) < ϕ(a). But then r = (−q)a is divisible by a, which would force ϕ(a) ≤ ϕ(r) unless r = 0. But then r must equal 0, displaying ϕ(0) < ϕ(a) as desired. Recall that a principal ideal domain, or P.I.D., is an integral domain in which every ideal is principal. Proposition 8.1.6. A Euclidean domain is a principal ideal domain. Indeed, let ϕ : A → Z be a positive deﬁnite Euclidean structure map for A. Let a be a nonzero ideal of A and let a ∈ a be such that ϕ(a) is the smallest positive value taken on by the restriction of ϕ to a. Then a = (a). Proof Let 0 = b ∈ a and write b = qa + r with ϕ(r) < ϕ(a). But r = b − qa is in a, so by the minimality of ϕ(a), we must have ϕ(r) = 0. But then r = 0 and b ∈ (a). CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 298 Lemma 8.1.7. Let A be an integral domain and let a, b ∈ A. Then (a) = (b) if and only if a = ub for some unit u of A. Indeed, if a = 0 and (a) = (b), and if a = bc, then c must be a unit. Thus, if b = 0 and c is not a unit, then the inclusion (bc) ⊂ (b) is proper. Proof The case a = 0 is immediate, so we assume a = 0. Here, if (a) = (b), we can ﬁnd c, d such that a = bc and b = ad. Then a = adc. By the cancellation property in a domain, dc = 1. Since the units in K[X] (K a ﬁeld) are the nonzero elements of K, we obtain an immediate corollary. Corollary 8.1.8. Let K be a ﬁeld and let a be a nonzero ideal in K[X]. Let n be the smallest non-negative integer such that a contains an element of degree n. Then there is a unique monic polynomial, f (X), in a of degree n, and a is the principal ideal generated by f (X). Moreover, f (X) is the unique monic polynomial generating a. Deﬁnition 8.1.9. Let B be an algebra over the ﬁeld K and let b ∈ B. Let εb : K[X] → B be the K-algebra map obtained by evaluating X at b. If εb is injective, we say that the minimal polynomial of b over K is 0. Otherwise, the minimal polynomial of b over K is the monic polynomial of smallest degree in ker εb . We denote it by minb (X), if the context is obvious, or more formally by minb/K (X). Corollary 8.1.10. Let B be an algebra over the ﬁeld K and let b ∈ B. Then K[b] is isomorphic as a K-algebra to K[X]/(minb (X)). In particular, K[b] is ﬁnite dimensional as a K-vector space if and only if minb (X) = 0, and in this case, the dimension of K[b] over K is equal to the degree of minb (X). Proof If minb (X) = 0, then the determination of the dimension of K[b] follows from Proposition 7.7.35. Otherwise, K[b] is isomorphic to K[X], which is easily seen to be an inﬁnite dimensional vector space over K, with basis {X i | i ≥ 0}. It turns out that P.I.D.s have almost all the nice properties exhibited by the integers, including those shown in Sections 2.3 and 4.4. We shall treat the material from the former in this section and the latter in Section 8.9. Deﬁnitions 8.1.11. Let A be a commutative ring. Then p ∈ A is irreducible if p is not a unit, but whenever p = ab, then either a or b must be a unit. We say that p ∈ A is a prime element if p = 0 and the principal ideal (p) is a prime ideal. Note that if A is a domain, then (0) is a prime ideal, but 0 is neither a prime element nor an irreducible element. There is a basic ideal theoretic consequence of irreducibility. Lemma 8.1.12. Let p be an irreducible element in the commutative ring A and suppose that a divides p. Then either a is a unit, or p divides a. In ideal theoretic terms, if (p) ⊂ (a), then either (a) = (p) or (a) = A. Thus, the ideal generated by an irreducible element is a maximal element in the partially ordered set of proper principal ideals, ordered by inclusion. CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 299 Proof If a divides p, then p = ab for some b ∈ A. Since p is irreducible, either a or b is a unit. If b is a unit, then a = pb−1 , and hence p divides a. In terms of ideals, (p) ⊂ (a) if and only if p ∈ (a), which is equivalent to a dividing p. If a is a unit, (a) = A. Otherwise, p divides a, and hence (a) ⊂ (p). In a domain, Lemma 8.1.12 has a converse. Lemma 8.1.13. Let A be an integral domain. Then a nonzero nonunit p ∈ A is irreducible if and only if the principal ideal (p) satisﬁes the property that (p) ⊂ (a) implies that either (a) = (p) or (a) = A. Proof Suppose that (p) satisﬁes the stated condition and suppose that p = ab. Then (p) ⊂ (a). If (a) = A, then a is a unit. If (p) = (a), then Lemma 8.1.7 shows b to be a unit. Thus, p is irreducible. If every ideal is principal, then Lemma 8.1.13 may be restated as follows. Corollary 8.1.14. Let A be a P.I.D. Then a nonzero element p ∈ A is irreducible if and only if the principal ideal (p) is a maximal ideal of A. Thus, since maximal ideals are prime, every irreducible element in a P.I.D. is prime. But this does not hold in a general domain. Surprisingly, the reverse implication does hold. Lemma 8.1.15. Let p be a prime element in the integral domain A. Then p is irreducible. Proof Suppose that p = ab. Then ab ∈ (p), a prime ideal, so either a or b must lie in (p). Say a ∈ (p). But p is also in (a), since p = ab, so (a) = (p), and hence b is a unit by Lemma 8.1.7. Since maximal ideals are prime, the next corollary is immediate from Lemma 8.1.15 and Corollary 8.1.14. Corollary 8.1.16. Let A be a P.I.D. Then an element p ∈ A is irreducible if and only if it is a prime element. In particular, if p ∈ A is irreducible and if p divides ab, then it must divide either a or b. Moreover, since every nonzero prime ideal of A is generated by a prime element, every nonzero prime ideal of A is maximal, and has the form (p), where p is an irreducible element of A. Thus, the longest possible chain of proper inclusions of prime ideals in a P.I.D. has the form 0 ⊂ (p) where p is a prime element of A. This may only occur if A is not a ﬁeld. Corollary 8.1.17. A P.I.D. that is not a ﬁeld has Krull dimension 1. The statement in the ﬁrst paragraph of Corollary 8.1.16 has a more classical proof, which is modelled on the one that we used for the integers. We shall give it to illustrate that it really uses exactly the same ideas as the proof we just gave. CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 300 Deﬁnition 8.1.18. Let A be a P.I.D. Then the greatest common divisor of a and b is the ideal (c) = (a) + (b). We write (a, b) = (c). If (a) + (b) = A, then we say that a and b are relatively prime, and write (a, b) = 1. Note that the notation of (a, b) for the ideal (a) + (b) is consistent with the notation that (a, b) is the ideal generated by a and b. We will occasionally refer to c as the greatest common divisor of a and b when (a, b) = (c). But it is really more correct for g.c.d. to refer to the ideal (c), rather than the generator c. Lemma 8.1.19. Let A be a P.I.D. and let (a, b) = (c). Then x divides both a and b if and only if x divides c. Proof Since a and b are both in (c), c divides a and b. Thus, it suﬃces to show that an element dividing both a and b must divide c. But since (c) = (a) + (b), c = ra + sb for some r, s ∈ A. Thus, any element that divides a and b will divide c. The entire argument above reduces to the statement that (a) + (b) is the smallest ideal containing both a and b, in the context that every ideal is principal. Second Proof of Corollary 8.1.16 Let p be an irreducible element of A. We shall show that (p) is a prime ideal. Suppose that p divides ab but does not divide a. Since p is irreducible, Lemma 8.1.12 shows that any divisor of p is either a unit or is divisible by p. Thus, a and p cannot have any common divisors other than units. So a and p are relatively prime, and hence there are elements r, s ∈ A with ra + sp = 1. But then rab + spb = b. Since p divides ab, it divides the left-hand side, so it must divide b. Note the role that the maximality of (p) plays in the above argument. Now we would like to show that every nonzero element of a P.I.D. may be factored in some sense uniquely as a product of primes. This may be read as a statement that every P.I.D. is an example of a more general type of ring called a unique factorization domain, or U.F.D. Deﬁnition 8.1.20. A unique factorization domain, or U.F.D., is an integral domain in which every nonzero nonunit may be written as a product of prime elements. The next lemma has already been seen to hold in P.I.D.s. Lemma 8.1.21. Irreducible elements in a U.F.D. are prime. Thus, in a U.F.D., the prime elements and the irreducible elements coincide. Proof Let p be an irreducible element in the U.F.D. A. Then p is a product of prime elements, and hence is divisible by a prime element, q. Say p = qr. Since q is prime, it is not a unit. But p is irreducible, so r must be a unit. Thus, (p) = (q), a prime ideal, and hence p is prime. We now wish to show that every P.I.D. is a U.F.D. Since we’ve shown that every irreducible element in a P.I.D. is prime, it suﬃces to show that every nonzero nonunit is a product of irreducible elements. If our P.I.D. is actually a Euclidean domain, this may be shown by a quick induction on ϕ(a), much as the proof for the integers worked. For the general P.I.D., we need to work a little bit harder. The main tool comes from the Noetherian property. CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 301 Proposition 8.1.22. Every P.I.D. is a U.F.D. Proof Let A be a P.I.D. Since every ideal is principal, the ideals are certainly ﬁnitely generated, and hence A is Noetherian by Corollary 7.8.4. Let 0 = a ∈ A be a nonunit, and suppose by contradiction that a is not a product of primes. Since a is not a unit, (a) is proper, and hence (a) is contained in a maximal ideal (p1 ). Thus, a = p1 a1 for some a1 ∈ A. Now p1 is irreducible by Corollary 8.1.14, and hence prime, by Corollary 8.1.16. Since p1 is prime and a = p1 a1 is not a product of primes, a1 cannot be a product of primes. Also, the inclusion (a) ⊂ (a1 ) must be proper by Lemma 8.1.7, as p1 and a1 are nonzero nonunits. We may continue by induction, obtaining elements ak ∈ A for k ≥ 1, such that ak−1 = pk ak with pk prime, and hence ak cannot be a product of primes. Thus, we obtain an inﬁnite sequence of proper inclusions (a) ⊂ (a1 ) ⊂ · · · ⊂ (ak ) ⊂ · · · contradicting the fact that A is Noetherian. We shall show in Section 8.5 that if A is a U.F.D., so is A[X]. Taking A to be a P.I.D. that is not a ﬁeld, this will give a family of examples of U.F.D.s that are not P.I.D.s. We shall now discuss the uniqueness properties of the factorizations of the elements in a U.F.D. as products of primes. Deﬁnition 8.1.23. Let p and q be prime elements in the U.F.D. A. We say that p and q are equivalent if (p) = (q). In other words, p and q are equivalent if p = uq for some unit u. In the integers, this doesn’t say much, since the only units are ±1. Thus, it is easy to choose a representative for each equivalence class of primes in Z: Just pick the positive one. That’s what we did in Section 2.3. In the generic P.I.D. there can be inﬁnitely many units, so choosing a representative for each equivalence class of primes could get cumbersome. For this reason, we shall give a uniqueness statement ﬁrst that uses equivalence classes, and then follow it up with another in which actual representatives are chosen. We say that a collection p1 , . . . , pk of primes is pairwise inequivalent if pi and pj are inequivalent for i = j. Proposition 8.1.24. Let A be a U.F.D., and suppose given an equality upr11 . . . prkk = vq1s1 . . . qlsl where u and v are units, p1 , . . . , pk are pairwise inequivalent primes, q1 , . . . , ql are pairwise inequivalent primes, and the exponents ri and si are all positive. Then k = l, and after reordering (and relabelling) the q1 , . . . , ql if necessary, pi is equivalent to qi and ri = si for 1 ≤ i ≤ k. Proof We argue by induction on r = r1 + · · · + rk . Here, we take the empty product to be 1, so if r = 0, then the left side is just u. But then vq1s1 . . . qlsl is a unit, and hence q1s1 . . . qlsl must be the empty product, too. If r ≥ 1, then p1 divides the left-hand side and hence divides the right as well. In particular, it must divide qi for some i, since the ideal (p1 ) is prime. But since p1 is not a unit and qi is irreducible, p1 and qi must be equivalent, say qi = wp1 , with w a unit. CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 302 Reorder the q’s so that i = 1. Then p1 (up1r1 −1 . . . prkk − vwq1s1 −1 . . . qlsl ) = 0. Since A is an integral domain, we have upr11 −1 . . . prkk = vwq1s1 −1 . . . qlsl , and the result follows by induction. By a choice of representatives for the primes in A, we mean a collection {pi | i ∈ I} of primes in A such that for each prime p of A there is a unique i ∈ I such that p is equivalent to pi . Thus, for example, the positive primes form a choice of representatives for the primes in Z. Corollary 8.1.25. Let A be a U.F.D. and let {pi | i ∈ I} be a choice of representatives for the primes in A. Then any nonzero nonunit in A may be written uniquely in the form upri11 . . . prikk , where k > 0, u is a unit, i1 , . . . , ik are distinct elements of I, and ri > 0 for 1 ≤ i ≤ k. Proof Every element may be written in the above form because equivalent primes diﬀer by multiplication by a unit. Thus, the u term can absorb the diﬀerences between a given product of primes and a product of pi ’s. The uniqueness statement now follows from that in Proposition 8.1.24, since pi is not equivalent to pj unless i = j. Thus, if two elements of the form upri11 . . . prikk are equal, the primes in them, together with their exponents, must be identical. Thus, we’re reduced to the case where upri11 . . . prikk = vpri11 . . . prikk , with u and v units. But since we’re in an integral domain, we can cancel oﬀ the primes in the above equation, leaving u = v. Corollary 8.1.26. Let A be a U.F.D. and let a ∈ A. Suppose that a = upr11 . . . prkk , where u is a unit, p1 , . . . , pk are pairwise inequivalent primes, and ri ≥ 0 for i = 1, . . . , k. Then an element b divides a if and only if b = vps11 . . . pksk , where v is a unit and 0 ≤ si ≤ ri for 1 ≤ i ≤ k. Proof If b divides a, then any prime that divides b will also divide a. If a = bc, then c also divides a, and the same reasoning applies. Thus, b = vps11 . . . pskk , and c = wpt11 . . . ptkk , where v and w are units and the exponents are all non-negative. But then a = bc = vwps11 +t1 . . . pskk +tk . By the uniqueness of factorization, si + ti = ri for i = 1, . . . , k, and hence si ≤ ri as claimed. Conversely, if si ≤ ri for i = 1, . . . , k and if v is a unit, then vps11 . . . pskk is easily seen to divide a. We’d like to be able to talk about greatest common divisors in a U.F.D., but the deﬁnition that we used in P.I.D.s fails for U.F.D.s that are not P.I.D.s. The next corollary will help us. Corollary 8.1.27. Let A be a U.F.D. and let 0 = a, b ∈ A. Let p1 , . . . , pk be a set of representatives for the set of primes dividing either a or b. Thus, p1 , . . . , pk are pairwise inequivalent, and we may write a = upr11 . . . prkk and b = vps11 . . . pskk , where u and v are units, and the exponents are non-negative. Let ti = min(ri , si ) for 1 ≤ i ≤ k and let c = pt11 . . . ptkk . (Note that c = 1 if no prime element divides both a and b.) Then c divides both a and b. In addition, if d is any element that divides both a and b, then d divides c. CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 303 In the language of ideals, this says that (a) + (b) ⊂ (c), and if d is any other element such that (a) + (b) ⊂ (d), then (c) ⊂ (d). In particular, (c) is the smallest principal ideal that contains a and b. Proof This is almost immediate from Corollary 8.1.26: If d divides both a and b, then d = wpl11 . . . plkk , where, if 1 ≤ i ≤ k, then 0 ≤ li ≤ ri , because d divides a, and 0 ≤ li ≤ si , because d divides b. But then li ≤ ti for all i, so d divides c. Clearly, c divides a and b, so the result follows. Deﬁnitions 8.1.28. Let A be a U.F.D. and let a, b ∈ A. Then the greatest common divisor of a and b is the smallest principal ideal that contains (a) + (b). (Such an ideal exists by Corollary 8.1.27.) If the greatest common divisor of a and b is c, we write gcd(a, b) = (c), or sometimes, by abuse of notation, gcd(a, b) = c. We say that a and b are relatively prime if their greatest common divisor is (1). Notice that if A is a P.I.D., then these deﬁnitions agree with the old ones. Of course, Corollary 8.1.27 tells us exactly how to calculate greatest common divisors in a U.F.D. We state the result formally here: Corollary 8.1.29. Let A be a U.F.D. and let 0 = a, b ∈ A. Let p1 , . . . , pk be a set of representatives for the set of primes dividing either a or b. Thus, p1 , . . . , pk are pairwise inequivalent, and we may write a = upr11 . . . prkk and b = vps11 . . . pskk , where u and v are units, and the exponents are non-negative. Let ti = min(ri , si ) for 1 ≤ i ≤ k. Then the greatest common divisor of a and b is (c), where c = pt11 . . . ptkk . In particular, a and b are relatively prime if and only if no prime element of A divides both a and b. More generally, we can talk about the greatest common divisor of a set of elements in a U.F.D. Deﬁnition 8.1.30. Let a1 , . . . , an be elements of the U.F.D. A. Then the greatest common divisor, gcd(a1 , . . . , an ) of these elements is the smallest principal ideal containing (a1 ) + · · · + (an ). By abuse of notation, we also write gcd(a1 , . . . , an ) for any generator of this ideal. We say that a1 , . . . , an are relatively prime if gcd(a1 , . . . , an ) = 1. Notice the diﬀerence between saying that a1 , . . . , an are relatively prime and saying that they are pairwise relatively prime. The reader should supply the proof of the next corollary. Corollary 8.1.31. Let a1 , . . . , an be nonzero elements of the U.F.D. A. Let p1 , . . . , pk be a set of representatives for the set of all primes that divide the elements a1 , . . . , an , and write ai = ui pr1i1 . . . prkik for 1 ≤ i ≤ n, with ui a unit in A. Let tj = min(r1j , . . . , rnj ) for 1 ≤ j ≤ k. Then gcd(a1 , . . . , an ) = pt11 . . . ptkk . CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 304 Another construction which makes sense in a U.F.D. is the least common multiple. Once again we shall need a computation to set it up. Proposition 8.1.32. Let A be a U.F.D. and let a1 , . . . , an ∈ A. Deﬁne an element a ∈ A as follows. If any of the ai = 0, then set a = 0. Otherwise, let p1 , . . . , pk be a set of representatives for the primes that divide at least one of the ai , and let ai = ui pr1i1 . . . prkik for 1 ≤ i ≤ n, where ui is a unit and the exponents rij are all non-negative. Then deﬁne tj = max(r1j , . . . , rnj ), and set a = pt11 . . . pktk . Then the intersection of the principal ideals generated by the ai is the principal ideal generated by a: n (ai ) = (a). i=1 n Proof Here, i=1 (ai ) is the collection of elements divisible by each of the ai , so the result follows immediately from Corollary 8.1.26. Deﬁnition 8.1.33. Let A be a U.F.D. and let a 1 , . . . , an ∈ A. By the least common n multiple of a1 , . . . , an we mean the intersection, i=1 n(ai ), of the principal ideals they generate, or, by abuse of notation, any generator of i=1 (ai ). We also obtain a sharpening of the Chinese Remainder Theorem. Corollary 8.1.34. (Chinese Remainder Theorem, second form) Let A be a P.I.D. and let a1 , . . . , ak ∈ A, such that ai and aj are relatively prime for i = j. Let a = a1 . . . ak . Then there is an A-algebra isomorphism ∼ = f : A/(a) −→ A/(a1 ) × · · · × A/(ak ) given by f (x) = (x, . . . , x) for all x ∈ A. Proof Since ai and aj are relatively prime for i = j, (ai )+(aj ) = A for i = j. Thus, the ﬁrst form of the Chinese Remainder Theorem (Proposition 7.2.23) applies, showing that the A-algebra homomorphism f : A → A/(a1 ) × · · · × A/(ak ) given by f (a) = (a, . . . , a) k k is a surjection whose kernel is i=1 (ai ). But i=1 (ai ) = (a) by Proposition 8.1.32. If a U.F.D. A is not a P.I.D., then its prime elements generate prime ideals which will not, in general, be maximal ideals. In fact, they turn out to be minimal nonzero prime ideals of A. Deﬁnition 8.1.35. Let A be a domain. A minimal nonzero prime ideal of A is a nonzero prime ideal with the property that the only prime ideal which is properly contained in it is 0. Proposition 8.1.36. Let A be a U.F.D. Then a prime ideal p of A is a minimal nonzero prime ideal if and only if p = (p) for a prime element p of A. CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 305 Proof To show that every minimal nonzero prime ideal of A is principal, it suﬃces to show that every nonzero prime ideal q contains a prime element. But if 0 = a ∈ q, let a = p1 . . . pk , a product of (not necessarily distinct) prime elements of A. Because a ∈ q and q is prime, at least one of the pi must lie in q. Conversely, let p be a prime element and suppose that q is a prime ideal that’s properly contained in (p). Let K be the fraction ﬁeld of A and let a = (1/p)q = {(1/p)a | a ∈ q} ⊂ K. Since every element of q is divisible by p, a ⊂ A, and is easily seen to be an ideal of A. But now q = pa = (p)a. Since q is prime, either (p) ⊂ q or a ⊂ q by Lemma 7.6.9. Since q is properly contained in (p), we must have a ⊂ q. But the fact that q = (p)a shows that q ⊂ a, so q = a. In other words, q = (p)q = pq. But by induction, we see that q = pn q for all n, and hence each a ∈ q is divisible by pn for all n. By the uniqueness of factorization in A, no element but 0 can be divisible by all powers of a prime element, so q must be the 0 ideal. Thus, the only prime ideal that can be properly contained in a principal prime ideal is 0. Exercises 8.1.37. 1. Let A be a U.F.D. Suppose that a divides bc in A, where a and b are relatively prime. Show that a divides c. 2. Let p1 , . . . , pk be pairwise inequivalent primes in the U.F.D. A and let a = upr11 . . . prkk for some unit u. Show that a divides b if and only if pri i divides b for 1 ≤ i ≤ k. 3. Let K be a ﬁeld. Show that every degree 1 polynomial of K is irreducible in K[X]. If f (X) has degree 1, what is K[X]/(f (X))? 4. Let K be a ﬁeld and let f (X) ∈ K[X] be a polynomial whose degree is either 2 or 3. Show that f (X) is irreducible if and only if it has no roots. 5. Let K be the ﬁeld of fractions of the U.F.D. A. Let a/b ∈ K be a root of f (X) = am X m + · · · + a0 ∈ A[X], where am = 0 and where a, b ∈ A are relatively prime. Show that b must divide am and a must divide a0 . 6. The simplest procedure for ﬁnding primes in Z is to proceed inductively, testing each number by dividing it by all primes smaller than its square root. If it has no prime divisor less than or equal to its square root, it is prime. One could adopt a similar strategy in F [X] for a ﬁnite ﬁeld F . Here, a polynomial f (X) is irreducible if it has no irreducible factors of degree ≤ 12 deg f . Carry out this procedure for F = Z2 to ﬁnd all irreducible polynomials of degree ≤ 4. 7. Let A be a P.I.D. and let B be a subring of the ﬁeld of fractions of A with A ⊂ B. Show that B = S −1 A for a suitable multiplicative subset S of A. 8. Let A be a U.F.D. and let S be a saturated multiplicative subset of A. Let T be the set of prime elements of A that lie in S. Show that S = {up1r1 . . . prkk | k ≥ 0, u ∈ A× , p1 , . . . , pk ∈ T and ri ≥ 0 for i = 1, . . . , k}. 9. Let A be a U.F.D. with only one prime element, p, up to equivalence of primes. Show that there is a Euclidean structure map ϕ : A → Z obtained by setting ϕ(0) = −1 and setting ϕ(a) = k if pk is the highest power of p that divides the nonzero element a. Deduce that a ring is a discrete valuation ring if and only if it is a U.F.D. with at most one prime element up to equivalence of primes. CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 306 10. Let A be a U.F.D. and let S be a multiplicative subset of A. Show that S −1 A is a U.F.D. 11. Let A be a U.F.D. and let p be a prime element of A. Show that the localization A(p) of A at the prime ideal (p) is a discrete valuation ring. 12. Let A be a U.F.D. that has inﬁnitely many equivalence classes of prime elements. Show that the ﬁeld of fractions of A is inﬁnitely generated as an A-algebra. 13. Show that if A is not a ﬁeld, then A[X] is not a P.I.D. 14. Recall the ring of Gaussian integers, Z[i], introduced in Example 7.3.6: Z[i] is the subring of C consisting of the elements a + bi with a, b ∈ Z. (a) Show that Z[i] is Euclidean, via the function ϕ(a + bi) = a2 + b2 . (b) Show that complex conjugation restricts to an isomorphism of rings from Z[i] to itself. Show also that ϕ(z) = zz for z ∈ Z[i], where z is the complex conjugate of z. Deduce that ϕ(zw) = ϕ(z)ϕ(w) for z, w ∈ Z[i]. (c) Show that the only units in Z[i] are ±i and ±1. (d) Show that anelement z ∈ Z[i] is prime if and only if it has no divisors w with 1 < ϕ(w) ≤ ϕ(z). (e) Show that 1 + i is a prime and that 1 + i = −i · (1 + i). Show also that ϕ(1 + i) = 2, so that 2 = (1 + i) · (1 + i) = −i · (1 + i)2 . (In the language of algebraic number theory, this implies that 2 ramiﬁes in Z[i].) Deduce that 2 is not prime in Z[i]. (f) Show that if z = a+bi ∈ Z[i] with a, b ∈ Z is a prime element such that neither a nor b is zero and (z) = (1 + i), then z is a prime that is not equivalent to z. Deduce that every nonzero element n ∈ Z has a prime decomposition in Z[i] of the form n = u · (1 + i)2r · pr11 . . . prkk · z1s1 z s11 . . . zlsl z sl l , where u is a unit, the pi are primes in Z that remain prime in Z[i], and the zi are primes in Z[i] of the form a + bi, where a and b are positive integers with b < a. (g) Let z = a + bi where a and b are nonzero integers. Show that z is prime in Z[i] if and only if ϕ(z) is prime in Z. (h) Let p be a prime in Z that is not prime in Z[i]. Show that p = ϕ(z) for some z ∈ Z[i]. Deduce that there are integers a and b with p = a2 + b2 . (i) Show that a prime in Z that is congruent to 3 mod 4 remains prime in Z[i]. (j) Suppose that a prime p in Z remains prime in Z[i]. Deduce that Z[i]/(p) is a ﬁeld of order p2 , which is additively a vector space over Zp of dimension 2, with a basis given by the images of 1 and i under the canonical map from Z[i]. Deduce that the only elements of order 4 in the unit group (Z[i]/(p))× must lie outside the image of the unique ring homomorphism from Zp to Z[i]/(p). Deduce that Zp × has no elements of order 4, and hence that p must be congruent to 3 mod 4. CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 307 (k) Let p be a prime in Z that is congruent to 1 mod 4. Deduce from the results above that p = zz for some prime z of Z[i]. Show that the ﬁeld Z[i]/(z) is of characteristic p, and is generated additively by the images of 1 and i under the canonical map from Z[i]. Deduce from the fact that Zp has units of order 4 that the image of i under the canonical map must lie in the image of the unique ring homomorphism from Zp to Z[i]/(z). Deduce that Z[i]/(z) is isomorphic to Zp . 15. Characterize the collection of integers that may be written as the sum of two squares. 16. We study the subring Z[ζ3 ] of C. (a) Show that the kernel of the evaluation map εζ3 : Z[X] → Z[ζ3 ] is the principal ideal of Z[X] generated by X 2 + X + 1. Deduce from Proposition 7.7.35 that the elements of Z[ζ3 ] may be written uniquely in the form m + nζ3 with m, n ∈ Z. Write down the multiplication formula for the product of two such elements. (b) Using the fact that ζ 3 = ζ3−1 = ζ32 , where ζ 3 is the complex conjugate of ζ3 , show that setting ϕ(α) = αα deﬁnes a Euclidean structure map ϕ : Z[ζ3 ] → Z. (c) Show that Z[ζ3 ]× = {±1, ±ζ3 , ±ζ32 }, where ζ32 = −1−ζ3 and −ζ32 = ζ6 = 1+ζ3 . (d) Show that 3 ramiﬁes in Z[ζ3 ] in the sense that 3 = up2 , where u ∈ Z[ζ3 ]× and p is a prime of Z[ζ3 ]. (e) Show that an element of the form m + nζ3 , where m and n are both nonzero elements of Z, is prime in Z[ζ3 ] if and only if its image under ϕ is a prime in Z. (f) Show that a prime in Z remains prime in Z[ζ3 ] if and only if it is not in the image of ϕ : Z[ζ3 ] → Z. (g) Show that a prime p = 3 in Z remains prime in Z[ζ3 ] if and only if p is not congruent to 1 mod 3. CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 308 √ 17. Show that Z[i 2] is a Euclidean√domain via the square of the complex norm, as above. What are the units of Z[i√ 2]? Find congruences that will guarantee that a prime in Z remains prime in Z[i 2]. √ 18. Show that Z[ 2] admits a Euclidean structure map. 8.2 Algebraic Extensions Deﬁnitions 8.2.1. Let K be a ﬁeld. By an extension of K, or, more formally, an extension ﬁeld of K, we mean a ﬁeld L that contains K as a subﬁeld. We say that L is a ﬁnite extension of K if it is ﬁnite dimensional as a vector space over K. If L is a ﬁnite extension of K, then the degree of L over K, written [L : K], is the dimension of L as a K-vector space. If L is an inﬁnite extension of K, we write [L : K] = ∞. Throughout this section, K is a ﬁeld. Lemma 8.2.2. Let L be an extension of K and let L be an extension of L. Then L is ﬁnite over K if and only if both L is ﬁnite over L and L is ﬁnite over K, in which case [L : K] = [L : L] · [L : K]. Proof If L is ﬁnite over K, then the other two extensions are clearly ﬁnite as well. Conversely, if y1 , . . . , yn is a basis for L as a vector space over L, and x1 , . . . , xm is a basis for L as a vector space over K, then any ordering of {xi yj | 1 ≤ i ≤ m, 1 ≤ j ≤ n} will give a basis for L over K. We now consider the elements of an extension ﬁeld L of K. For each α ∈ L, there is a unique K-algebra homomorphism εα : K[X] → L that takes X to α. Here, εα evaluates polynomials at α: " n # n i εα ai X = ai αi . i=0 i=0 Thus, f (X) ∈ ker εα if and only if α is a root of f . In particular, ker εα = 0 if and only if α is a root of some nonzero polynomial over K. Recall from Corollary 8.1.10 that the kernel of εα is the principal ideal generated by minα (X), the minimal polynomial of α. Here, minα (X) = 0 if εα is injective, and is the monic polynomial of lowest degree in ker εα otherwise. In particular, minα (X) divides any polynomial that has α as a root. The image of εα , of course, is K[α], the K-subalgebra of L generated by α. Deﬁnition 8.2.3. Let L be an extension ﬁeld of K and let α ∈ L. We say that α is algebraic over K if α is a root of a nonzero polynomial over K, and hence minα (X) = 0. If α is not a root for any nonzero pol