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MATH20212: Algebraic Structures 2 This follows on from Algebraic Structures 1 MATH20201, so some basic knowledge and understanding of groups and rings is assumed. This course concentrates rather more on rings than on groups. The central aims of the course are: • that you become more acquainted with the basic examples and the general ideas, to the point where you have internalised the main concepts and can use them in new situations; • to introduce the general notion of a homomorphism (a structure-preserving map) and its kernel; • to introduce the factor, or quotient, construction. It is not expected that the average student will master these two rather sophisticated new concepts (especially the factor/quotient construction) by the end of the course: this is just an introduction; you can renew your acquaintance with them, and have the opportunity to understand them better, in later algebra courses. Like Algebraic Structures 1, there is a great deal of emphasis on examples: subrings of Z, Q and C; polynomial rings; matrix rings; groups of permutations and symmetries; groups of units of rings. Polynomials are the central examples. Algebraic Structures 1 and 2 together give a foundation for further courses in algebra, as well as those (e.g. some in geometry and topology) which use algebraic structures. These notes cover the material of the whole course but not everything is here. Lectures will be used for explaining the concepts, giving the proofs (only a few proofs are given in these notes) and presenting details of examples. There are a lot(!) of exercises in these notes. Some will be done in class; you should try some of the rest, giving priority to those marked **, then those marked *. It is really important to keep up to date with your work on examples. Work on the examples relevant to what has been covered recently in lectures: don’t get behind (you can always come back and use those you skipped over for revision). The Friday 11.00 slot will be used for going over examples and should be regarded as being as important as the lectures. Prepare for this (try the examples beforehand) and be ready to participate (asking questions, suggesting solutions). Further reading is not necessary but might be interesting and/or useful. There is the course text, Fraleigh. I will maintain a list other good sources on the website for the course. Finally, please let me know of any errors/typos you find in the notes, examples, solutions. 1 Part I Rings 1 Definitions and Examples, including Review Idea: a ring is an “arithmetic system” with two operations, usually written + and ×, because the ur-example is the ring of integers. The ring of integers modulo n is another example, as is the ring of n×n square matrices with integer entries. There are many more examples. Definition, in brief. A ring is given by the following data: a set R and two binary operations, written + and ×, on R. The conditions to be satisfied are: that (R, +) is an abelian group; that × is associative and distributes over +. The identity element of the abelian group (R, +) is denoted 0. In this course “ring” will mean “ring with 1 6= 0”: that means one more datum is given, namely a specified element, 1 ∈ R, which is different from 0, and there is one more condition, namely that 1 is an identity for ×. Definition, in detail. You can find this in your notes for Algebraic Structures 1, in most of the books mentioned in this course, and on the web. I’ll remind you in the lecture. Notation: we write (R; +, ×, 0, 1) if we want to show all the data but usually we don’t want to do that, so write just R. Examples 1.1. Examples of rings: (1) Z; (2) Z[i] = {a + bi : a, b ∈ Z}, where i denotes a square root of −1. In order to check that these are rings there is no need to check all the details directly because both are subrings of the ring, C, of complex numbers. So it is, by the lemma below, enough to check that each of the above sets contains 0 and 1 and is closed under taking negatives and under the operations, + and ×. Lemma 1.2. If S is a ring and R ⊆ S is a subset then R is a ring, a subring of S, iff: • 0, 1 ∈ R; • r, s ∈ R implies r + s, r × s ∈ R; • r ∈ R implies −r ∈ R. That was loosely said: what we should have said was the following. “If (S; +, ×, 0, 1) is a ring and R ⊆ S then (R; +0 , ×0 , 0, 1) is a ring, where +0 is the restriction of the operation + to R × R and ×0 is the restriction of × to R × R, iff (list of conditions as in the lemma)” or, more briefly, “If (S; +, ×, 0, 1) is a ring and R ⊆ S then R, equipped with the operations inherited from S, is a ring iff (list of conditions as in the lemma)”. Anyway, the point is, to check that a subset of a ring S is a subring you don’t have to check everything: many of the conditions come for free because they hold everywhere in the ring S. By the way, don’t get confused by some notation above: × was used both for the “multiplication” operation in the ring and for the cartesian product of sets (as in “R × R”). In practice we usually write just rs, or sometimes, r.s for the multiplication r × s. 2 Examples 1.3. More examples of rings. Small rings can be presented by giving their “addition and multiplication tables”. Here are three examples. Can you identify the rings? + 0 1 2 3 0 0 1 2 3 1 1 2 3 0 2 2 3 0 1 3 3 0 1 2 × 0 1 2 3 0 0 0 0 0 1 0 1 2 3 + 0 1 x 1+x 0 0 1 x 1+x 1 1 0 1+x x x x 1+x 0 1 1+x 1+x x 1 0 + 0 1 α 1+α 0 0 1 α 1+α 1 1 0 1+α α α α 1+α 0 1 1+α 1+α α 1 0 2 0 2 0 2 3 0 3 2 1 × 0 1 x 1+x × 0 1 α 1+α 0 0 0 0 0 0 0 0 0 0 1 0 1 x 1+x 1 0 1 α 1+α x 0 x 0 x 1+x 0 1+x x 1 α 0 α 1+α 1 There is an issue here: these tables certainly specify two operations on each relevant set and it would be easy to pick out the 0 and 1 even if they had not been identified in name, but it could be rather tedious to check that all the conditions (associativity and distributivity in particular) for a ring are satisfied. Of course you could give the tables to a computer to check. We will come back to these examples and, by recognising what they are, we will be able to avoid the tedious case-by-case checking of the ring axioms. Some elementary consequences of the definition are recalled in Exercise 1.4 Exercise* 1.4. Some elementary properties. If last semester’s course is by now but a hazy memory then you should do these exercises to bring the basics back into focus. If not then you should be able to rattle them off. In any case do them. Let R be any ring. Show the following. (1) There is just one element z with the property that z + r = r for every r ∈ R. (2) There is just one element e with the property that er = r for every r ∈ R. (3) For each element r ∈ R there is just one element r0 such that r + r0 = 0. Also recall extended associativity and distributivity: r1P × r2 × · · · × rn is Pn n unambiguous (no parentheses are needed) and r× i=1 si = i=1 (r×si ). These are proved by induction (and the first is probably more difficult to formulate precisely than to prove). Induction is also used to define powers and to prove the expected things about them: for r ∈ R set r0 = 1 and, inductively, rn+1 = r×rn , also define r−n = (r−1 )n . The characteristic, char(R), of a ring R is the least positive integer n such that 1 + · · · + 1 = 0 (n 1s), that is, such that n × 1 = 0, that is, such that n = 0. If, as may well be the case, there is no such positive n, then the characteristic of R is defined to be 0. For example, Z has characteristic 0, as have the rings Q and C. On the other hand char(Zn ) = n. 3 1+α 0 1+α 1 α Lemma 1.5. Suppose that char(R) = n > 0. Then: (1) nr = 0 for every r ∈ R; (2) if m is an integer then m1 = 0 iff n|m. The ring R is a domain if rs = 0 implies r = 0 or s = 0. A non-zero element r ∈ R is a zero-divisor if there is a non-zero element s ∈ R with rs = 0 or sr = 0. So a domain is a ring without zero-divisors. (A commutative domain is also called an integral domain.) Exercise 1.6 asks you to prove that the characteristic of a domain is either 0 or a prime integer. Exercise** 1.6. Show that if R is a domain then char(R) = 0 or char(R) is a prime integer. Examples 1.7. (1) Z is a domain; (2) Z[i] is a domain - this can be proved directly or use that C is a domain then apply the next result. Lemma 1.8. If R is a subring of S and S is a domain then R is a domain. Examples 1.9. More examples: (3) Q is a domain; (4) the polynomial ring Q[X] is a domain - to prove this consider leading terms of polynomials. The next result is the general case (this result and its corollary were proved in Algebraic Structures I). Proposition 1.10. Suppose that R is a domain. Then the polynomial ring R[X] is a domain. Corollary 1.11. Suppose that R is a domain. Then the ring, R[X1 , . . . , Xn ], of polynomials in n indeterminates and with coefficients in R, is a domain. A division ring is a ring in which every non-zero element has a right inverse and a left inverse: for every r ∈ R there is s ∈ R such that rs = 1 (r is right invertible) and there is t ∈ R such that tr = 1 (r is left invertible). In this case, see below, s = t and we write r−1 for this inverse of r and say just that r is invertible. A field is a commutative division ring. (Recall that a ring R is said to be commutative if the multiplication is commutative: rs = sr for all r, s ∈ R.) Lemma 1.12. If R is a ring and r ∈ R has both a right and a left inverse then these are equal. (Just bracket trs in two different ways.) Beware that an element of a ring can have, for instance, a left inverse without having a right inverse and, in such a case, there might be more than one left inverse. Example 1.13. Let K be your favourite field and let V be the (infinitedimensional) vector space over K with basis v1 , v2 , . . . , vn , . . . . Let R be the ring of all linear transformations from V to itself, the operations on R being (pointwise) addition of linear transformations and, for the “multiplication” on R, use composition of linear transformations (with the convention that sr means “do r then do s”). Note that an invertible element of R must be an isomorphism from V to itself. Let r ∈ R be “right-shift”: the linear map which is defined on the given basis by sending vn to vn+1 . Since this map is not surjective, it is 4 not an isomorphism, hence r is not invertible. But r is left invertible: if s is the linear map which sends vn+1 back to vn and sends v1 to 0, then sr = 1 (where 1 is, note, the identity map on V ). Let t be the linear map which sends vn+1 to vn and sends v1 to v1 : then also tr = 1, showing that a left inverse need not be unique. Proposition 1.14. Every division ring is a domain. Exercise 1.15 recalls the condition on n for Zn to be a domain, equivalently for Zn to be a field. Exercise 1.15. Show that the following conditions on the integer n ≥ 2 are equivalent: (i) Zn is a domain; (ii) Zn is a field; (iii) n is a prime (look at your notes for Algebraic Structures I if you get stuck). Examples 1.16. More, apart from the obvious ones like Q, R, C, Zp , examples of fields: (1) if K is a field then so is the ring, K(X), of rational functions in the variable/indeterminate X with √ coefficients in K, more generally so is √ K(X1 , . . . , Xn ); (2) Q[ 2] = {a + b 2 : a, b ∈ Q}. Some examples similar to (2) above are in Exercise 1.17. Exercise** 1.17. Show that the following also are fields: (3) Q[i] (where i2 = −1); (4) Q[ω] where ω is a primitive cube root of 1 (in this example you should first figure out the dimension of Z[ω] as a vector space over Q; the dimension is perhaps not as obvious as it might seem to be). Tidying up: recall that polynomials are really equivalence classes of polynomial expressions: we consider X 2 + 1 to be the same polynomial as 1 + 0X + X 2 for instance. A similar comment applies to rational functions: these are expressions of the form p/q where p and q are polynomials but, again, up to equivalence, for example we identify (X 2 + X)/(X 3 − X) with (X + 1)/(X 2 − 1). Actually, a more accurate term than “rational function” would be “rational expression”: the distinction between polynomials and polynomial functions also applies here. Example 1.18. There are division rings which are not fields, that is, which are not commutative. The best known is the ring of quaternions: H = {a + bi + cj + dk : a, b, c, d ∈ R} where i2 = j 2 = k 2 = −1 and ij = k, jk = i, ki = j. From these relations it follows, for example, that ji = −k 6= k = ij. Notice that this contains, as a subring, the field, C, of complex numbers. And, in case you were wondering, “H” stands for “Hamilton”. It was shown in Algebraic Structures 1 that every non-zero element has an inverse. A more subtle point is whether the above ring is really what we think it is: how, for instance, do we know that k 6= −k, even that 0 6= 1? For, certainly we can write down a presentation of the ring H above by giving generators (as an R-vectorspace) and relations between these but we should really prove that there are no unexpected consequences of the relations, more precisely we should prove that a + bi + cj + dk = a0 + b0 i + c0 j + d0 k iff a = a0 , b = b0 , c = c0 and d = d0 . One way to do this is to provide a “concrete” representation of the ring, that is, find some ring S and elements si , sj , sk in S which satisfy the defining relations of i, j, k and where, for some reason, we know that a + bsi + csj + dsk = a0 + b0 si + c0 sj + d0 sk iff a = a0 , b = b0 , c = c0 and d = d0 . That will be done later (Exercise 2.5). 5 √ Exercise* 1.19. Show that Z[ 2] and Z[ω] (ω a primitive cube root of 1) are rings (check that they are subrings of C). Are they domains? In each case identify the invertible elements. We give some more examples of non-commutative rings (i.e. rings which are not necessarily commutative) below. Example 1.20. The ring M2 (Z) of 2 × 2 matrices with integer entries is a ring which is not commutative (exercise: show that this is not commutative). More generally, if R is a ring, then the ring, Mn (R), of n×n matrices with entries in R is not commutative if n > 1 (exercise show this), even if R itself is commutative. (The ring operations in Mn (R) are matrix addition and multiplication; the 0 and 1 are the obvious candidates.) An element r of a ring R is nilpotent if there is some integer n ≥ 1 with rn = 0 (and the least such n is the index of nilpotence of r). An element r ∈ R is idempotent if r2 = r. For example 0 and 1 are idempotent. Exercises 1.21, 1.22, 1.23, 1.24, 1.26, 1.27 are mostly about nilpotent and idempotent elements. Exercise** 1.21. Find examples of non-zero nilpotent elements in M2 (Z). Find examples of idempotent elements in M2 (Z) apart from 0 and 1. Exercise** 1.22. Prove that if R is a domain then there are no nilpotent elements other than 0 and no idempotent elements other than 0 and 1. Exercise** 1.23. Find nilpotent elements of M3 (Z) with indices of nilpotence 2 and 3. Is there a nilpotent element of order 4? What is the highest index of nilpotence of an element of Mn (Z)? Can you prove it? Exercise* 1.24. When does the ring, Zn , of integers modulo n have non-zero nilpotent elements? (and what are they?) Exercise** 1.25. If R is a ring and r ∈ R the centraliser of r, denoted C(r), is the set of elements which commute with r: C(r) = {s ∈ R : rs = sr}. Show that C(r) is a subring of R. In the ring M2 (Z) find a description for the centraliser of a general element. Is it always the case that the centraliser of an element is a commutative ring? Here is a general construction: if R1 and R2 are rings then their product is their set-theoretic product R1 × R2 made into a ring by defining the operations + and × (don’t confuse “×” and “×”!) by (r1 , r2 ) + (s1 , s2 ) = (r1 + s1 , r2 + s2 ) and (r1 , r2 ) × (s1 , s2 ) = (r1 × s1 , r2 × s2 ) (to be more precise we could write (r1 +1 s1 , r2 +2 s2 ) to show that the additions refer to operations in different rings and similarly for multiplication). The element (0, 0) clearly is a, hence the, zero of this ring and the element (1, 1) is the identity element of R1 × R2 . Note that the elements (1, 0) and (0, 1) corresponding to the identity elements of the original rings are idempotents of the ring R1 × R2 and they are orthogonal in the sense that idempotents e1 and e2 are orthogonal if e1 e2 = 0 = e2 e1 . Exercise 1.26. Find a pair of (non-trivial) idempotents in the ring M2 (R). What are the idempotent elements of M2 (R): are there just a few or are they a-dimea-dozen? (If you think of the elements of M2 (R) as linear transformations of R2 to itself, with respect to a fixed basis, then one may see that, geometrically, an idempotent is a projection and that “orthogonal” means just that.) 6 Exercise* 1.27. Let e ∈ R be idempotent. Show that 1 − e is idempotent and that e and 1 − e are orthogonal. Example 1.28. Let R be the set of all functions f : [0, 1] −→ R from the unit interval to the set of reals. Make this a ring by defining + and × to be pointwise addition and multiplication respectively, for instance f + g is defined to be the function which takes a ∈ [0, 1] to f (a) + g(a). Exercise 1.29 asks you to check that this is indeed a ring. Exercise* 1.29. Let R be the set of all functions f : [0, 1] −→ R from the unit interval to the set of reals. Make this a ring by defining + and × to be pointwise addition and multiplication respectively, for instance f + g is defined to be the function which takes a ∈ [0, 1] to f (a) + g(a). Check that this gives a commutative ring and identify the 0 and 1 of this ring. (This example also works if we restrict to functions which are continuous, alternatively functions which are differentiable: why?) Exercise 1.30. Let R be the set of all functions f : R −→ R with pointwise addition for + and take the “multiplication” × to be composition of functions, with the convention that f g means do g then do f. Show that R is not a ring (although R satisfies some of the axioms for a ring it does not satisfy them all, so you should find out what goes wrong). Exercise 1.31. Could we do something like Exercise 1.29 above but using functions from [0, 1] to [0, 1]? Example 1.32. Consider the ring, R[X], of polynomials in X with coefficients from R. Now extend this by a new variable Y which does not commute with X but, rather, satisfies the relation Y X = XY + 1. We denote this ring, called the first Weyl algebra, by RhX, Y : Y X = XY + 1i (thus, given R, we are describing this ring in terms of generators and relations): its elements are “noncommutative polynomials” in X and Y. Every such polynomial can be brought into a “normal form”: a sum of monomials, each of which has the form X m Y n for some non-negative integers m, n. Example 1.33. Let R be any ring Pand let G be any group. The group ring RG is the set of formal finite sums ri gi (up to the usual identifications) where ri ∈ R and gi ∈ G. Addition is formal addition using the rule rg + sg = (r + s)g and multiplication is based on (rg) × (sh) = (rs)(gh). The identity element is 1e where 1 is the identity of R and e is the identity of G. The zero of R is the element 0e. (This was described in Algebraic Structures I.) 1.1 Extra exercises Exercise* 1.34. Is the binary operation on Z (the set of integers) which takes (a, b) to a − b associative? Is there an identity element for this operation? Exercise* 1.35. Is the binary operation on P (the set of positive integers) given by (a, b) 7→ ab associative? a b Exercise** 1.36. Let SL2 (Z) denote the set of all matrices of the form c d with a, b, c, d ∈ Z and ad − bc = 1. Is this set a ring under matrix addition and multiplication? Is this set a group under matrix multiplication? What about 7 a b the set c d multiplication? : a, b, c, d ∈ Z, ad − bc = −1 : is this a group under matrix Exercise 1.37. Find some subrings of M2 (Z). This exercise becomes more interesting if we don’t insist that subrings contain the identity element so, just for this exercise, look for subsets containing {0} which are closed under addition, subtraction and multiplication. Exercise** 1.38. Suppose that R is a ring. How many solutions in R can there be to the equation X 2 − 1 = 0? Exercise** 1.39. Suppose that R is a domain and let a, b ∈ R with a 6= 0. Prove that the equation ax = b has at most one solution in R. What if, instead of assuming a 6= 0, we assume b 6= 0: is the conclusion still correct? √ √ Exercise* 1.40. √ Let Q[ 2, 3] denote the smallest subring of R which contains √ 3. Show Q, 2√and √ √ that this consists exactly of the real numbers of the form a + b 2 + c 3 + d 6 where a, b, c, d ∈ Q. Choose some non-zero elements of √ this√ring and√find √ their inverses (this is, in fact, a field). Also show that Q[ 2, 3] = Q[ 2 + 3] (hint: think about what you actually have to do here - it’s not much). Exercise 1.41. Let R = Z[i] where i is a square root of −1. Prove that R is a domain. Give another proof. And another. Exercise* 1.42. If R1 is a ring with characteristic m and R2 is a ring with characteristic n what is the characteristic of R1 × R2 ? Exercise 1.43. Show directly that any ring with exactly three elements must be a domain. Exercise 1.44. Let R be a ring. Show that the identity a2 − b2 = (a + b)(a − b) is true in R iff R is commutative Exercise** 1.45. Find all idempotent and all nilpotent elements in the ring Z6 × Z12 . Exercise 1.46. Suppose that R is a ring such that a2 = a for every a ∈ R. Show that a + a = 0 for every a ∈ R. Also show that R is commutative. Exercise 1.47. Let V be a finite-dimensional vector space over a field K and let R be the ring of linear transformations from V to V (so addition is pointwise addition and multiplication is composition of transformations). Prove that if f : V −→ V is in R then f has a left inverse iff f has a right inverse. You may quote basic results from linear algebra in your proof. Exercise** 1.48. Let R be the polynomial ring Z8 [X]. Show that the polynomial 1 + 2X is invertible. Exercise 1.49. Suppose that K is a field with q elements and that G is a group with n elements. How many elements does the group ring KG have? Now let K = Z3 be the field with 3 elements and let G be a cyclic group of order 3, generated by an element a. Write down the general form of an element of R = KG. Simplify the following elements of R (i.e. write them in “standard form”): (i) (1e + 2a + a2 ) + (1e + 2a2 ); (ii) (1e + 2a + a2 ) × (1e + 2a2 ); (iii) (1e + 2a + a2 )3 . 8 2 Isomorphisms, Homomorphisms and Ideals Isomorphism: the idea is that two rings are isomorphic if they are the same abstract structure: their addition and multiplication tables “are the same up to re-arrangement and re-naming of elements”, and an isomorphism is a map which shows this. For instance, the polynomial rings R[X] and R[Y ] are isomorphic: the map from the first ring to the second which takes a polynomial in X and replaces every occurrence of X by Y is an isomorphism. For, clearly this map preserves (“commutes with”) addition and multiplication and it takes the 0, respectively 1, of the first ring to the 0, resp. 1, of the second. Isomorphism: definition. If R and S are rings then an isomorphism from R to S is a bijection θ : R −→ S such that, for all r, r0 ∈ R we have θ(r + r0 ) = θ(r) + θ(r0 ) and θ(r × r0 ) = θ(r) × θ(r0 ) (in each case the operation on the left hand side of the equation is an operation on elements of R and that on the right hand side is an operation on elements of S). Various things are preserved by θ: see the next lemma. If θ is an isomorphism from R to S then we write θ : R ' S. We say that R and S are isomorphic, and write R ' S, if there is an isomorphism from R to S. The relation of being isomorphic is an equivalence relation on rings (Exercise 2.1). Exercise 2.1. Show that isomorphism is an equivalence relation on rings. (Of course, first you’ll need to recall what is meant by an equivalence relation.) 2 Example 2.2. R[X] ] since the map θ : R[X] −→ R[X 2 ] defined by Pn Pn ' R[X i 2i θ( i=0 ai X ) = i=0 ai X (i.e. the map which substitutes X 2 for X in each polynomial) is easily seen to be an isomorphism. Here R is any ring. (On the other hand this operation, if it were regarded as a map from R[X] to R[X], would not be an isomorphism: why?) Lemma 2.3. Suppose that θ : R −→ S is an isomorphism. Then: (1) θ(1) = 1; (2) θ(0) = 0; (3) θ(−r) = −θ(r) for every r = R; (4) r ∈ R is invertible iff θ(r) ∈ S is invertible and, in that case, (θ(r))−1 = θ(r−1 ); (5) r ∈ R is nilpotent iff θ(r) is nilpotent (and then they have the same index of nilpotence). Example 2.4. Let R = C be the field of complex numbers and let S be the ring 1 0 0 −1 of all 2 × 2 matrices of the form a +b where a and b are 0 1 1 0 real numbers. Then the map θ from R to S which takes the complex number 1 0 0 −1 a + bi (a, b ∈ R) to the matrix a +b is, one may (and you 0 1 1 0 should) check, an isomorphism. There is a similar representation of quaternions as matrices with real coefficients, see Exercise 2.5. 1 0 Exercise* 2.5. Show that the ring of 2 × 2 matrices of the form a + 0 1 9 0 −1 i 0 0 i b +c +d with a, b, c, d ∈ R (and i a square 1 0 0 −i i 0 root of −1) is isomorphic to the ring of quaternions. Example 2.6. All rings with two elements are isomorphic, because the only elements of such a ring are 0 6= 1 and the map from one two-element ring R to another two-element ring S which takes 0R to 0S and 1R to 1S is quickly checked to be an isomorphism. Note that there is a ring with two elements take the ring, Z2 , of integers modulo 2. It is also the case that, up to isomorphism, there is just one ring with three elements (Exercise 2.7). Exercise** 2.7. Show that all rings with three elements are isomorphic. Example 2.8. There are rings with four elements which are not isomorphic. In fact, the tables in Examples 1.3 do all give different rings (indeed, there is even one more). Homomorphism: this is a weakening of the idea of isomorphism. The requirement that the map commute with the ring structure is kept but the map is not required to be an isomorphism. That is, the map might collapse some elements (it need not be injective) and its image need not be the whole of the second ring (it need not be surjective). If R and S are rings then a homomorphism from R to S is a map θ : R −→ S such that, for all r, r0 ∈ R we have θ(r + r0 ) = θ(r) + θ(r0 ) and θ(r × r0 ) = θ(r) × θ(r0 ). We also require that θ(1) = 1, that is, θ(1R ) = 1S . Compare the next lemma with that, 2.3, for isomorphisms. Lemma 2.9. Suppose that θ : R −→ S is a homomorphism. Then: (i) θ(0) = 0; (ii) θ(−r) = −θ(r) for every r ∈ R; (iii) if r ∈ R is invertible then θ(r) ∈ S is invertible and, in that case, (θ(r))−1 = θ(r−1 ); (iv) if r ∈ R is nilpotent then θ(r) is nilpotent (and the index of nilpotence of θ(r) is less than or equal to that of r); (v) the image of θ is a subring of S. The converse to (iii) above fails: in Exercise 2.10 you are asked to give an example of an element which is not invertible being sent by a homomorphism to an element which is invertible. Exercise** 2.10. Give an example to show that if θ : R −→ S is a homomorphism and if r ∈ R is such that θ(r) is invertible in S, it need not be the case that r is invertible in R. Examples 2.11. (1) The map Z −→ Zn defined by taking an integer a to its congruence class modulo n, [a]n = {b ∈ Z : b ≡ a mod n}, is a homomorphism. (2) The map θ : Znk −→ Zn which takes [a]nk to [a]n is well-defined and is a homomorphism of rings. Note that proving well-definedness is necessary since the map θ was defined in terms of representatives of equivalence classes. Exercise* 2.12. Prove that there is no homomorphism from Zn to Zl if l is not an integer factor of n. [Hint, make use of the notion of characteristic] 10 An embedding, or monomorphism, is an injective (i.e. one-to-one) homomorphism. Example 2.13. (3) There is a string of natural embeddings Z −→ Q −→ R −→ C −→ C[X] −→ C[X, Y ] −→ C(X, Y ). Check that you see what the embeddings are and why they are embeddings. Lemma 2.14. If θ : R −→ S and β : S −→ T are homomorphisms (of rings) then so is the composition βθ : R −→ T . If θ : R −→ S and β : S −→ T are embeddings then so is the composition βθ : R −→ T . If θ : R −→ S and β : S −→ T are homomorphisms and if βθ : R −→ T is an embedding then θ is an embedding. Exercise 2.15. Show that if θ : R −→ S and β : S −→ T are homomorphisms such that the composition βθ : R −→ T is an embedding then it need not be the case that β is an embedding. If θ : R −→ S is a homomorphism of rings then the kernel of θ, ker(θ), is the set, {r ∈ R : θ(r) = 0}, of elements which θ sends to 0S . For instance, referring to the natural homomorphisms, Z −→ Zn and Z6 −→ Z2 , as in 2.11, ker(Z −→ Zn ) = {nk : k ∈ Z} and ker(Z6 −→ Z2 ) = {[0]6 , [2]6 , [4]6 }. Lemma 2.16. If θ : R −→ S is a homomorphism then θ is injective iff ker(θ) = {0}. Example 2.17. (4) If r ∈ R then the map from R[X] to R defined by sending p(X) to p(r) is termed evaluation at r and is√easily checked to be a homomorphism. For instance, the map evaluation at 2 is determined (given that it is a homomorphism) by sending each √ constant polynomial s ∈ R[X] to s ∈ R and sending the polynomial X √ to 2. The kernel of this map is the set of those polynomials p(X) such that p( 2) = 0, that is, those polynomials which have √ (X − 2) as a factor. Proposition 2.18. If R is any commutative ring and r ∈ R then the map θr : R[X] −→ R given by θr (p(X)) = p(r) is a homomorphism and ker(θr ) = {p ∈ R[X] : (X − r)|p}. (The symbol “|” means “divides”.) Example 2.19. (5) The map from Z[ω], where ω is a primitive cube root of 1, to Z[ω] given by a + bω + cω 2 7→ a + bω 2 + cω is a well-defined homomorphism (note that, given that it is a homomorphism, it is determined by the fact that it fixes each integer and sends ω to ω 2 ). It is an isomorphism, hence has zero kernel, but it is not the identity map. An automorphism of a ring is an isomorphism from the ring to itself. The identity map always is an automorphism and the example just above shows that there might be√others. In Exercise 2.20 you are asked to find all the automorphisms of Z[ 2]. √ Exercise** 2.20. Show that the map from Z[ 2] to itself given by sending a + √ √ b 2 to √ a−b 2 is an automorphism. Show that there are no more automorphisms of Z[ 2] apart from the identity map. Example 2.21. (6) Let R1 × R2 be the product of rings R1 and R2 . The projection map, π1 : R1 × R2 −→ R1 , onto the first coordinate is a homomorphism and its kernel is {0} × R2 . 11 The injection map ι1 : R1 −→ R1 × R2 which takes r1 to (r1 , 0) is not a homomorphism, even though it commutes with addition and multiplication, because it fails to send 1 to 1 (if we were dealing with rings which don’t necessarily have an identity element then it would count as a homomorphism). Example 2.22. (7) Consider the map from R[X] to R[X] (where R is any ring) which substitutes X 2 for X, that is, which sends p(X) to p(X 2 ). Then this is a monomorphism which is not surjective. More generally if q(X) is any polynomial then substitution of X by q(X), p(X) 7→ p(q(X)), is a homomorphism. Lemma 2.23. Suppose that θ : R −→ S is a homomorphism. Then ker(θ) is a subgroup of (R, +). Let r, r0 ∈ R. Then θ(r) = θ(r0 ) iff r − r0 ∈ ker(θ) iff r and r0 belong to the same coset of ker(θ) in R. That is, the fibres of a homomorphism θ are the cosets of ker(θ) in the additive group (R, +). (If f : A −→ B is any map and if b ∈ B then the fibre of f above b is f −1 b = {a ∈ A : f (a) = b}.) An ideal of a ring R is a subset I ⊆ R such that: • 0 ∈ I; • a, b ∈ I implies a + b ∈ I; • a ∈ I and r ∈ R implies ar ∈ I and ra ∈ I. That is, an ideal of R is a subset of R which contains 0, is closed under addition and is closed under (right and left) multiplication by every element of the ring (not just under multiplication by elements of I). We write I C R to mean that I is an ideal of R. Exercise** 2.24. Give an example of a ring R and a subset H of R which contains 0 and is closed under addition and multiplication but which is not an ideal of R. Example 2.25. If a ∈ R then {r1 as1 + · · · + rn asn : n ≥ 1, ri , si ∈ R} is, you should check, an ideal which contains a and is the smallest ideal of R containing a. It is called the principal ideal generated by a and is denoted hai. If R is commutative then its description simplifies: hai = {ar : r ∈ R}. A principal ideal is one which can be generated by a single element. Examples 2.26. (1) In every ring h0i = {0} is the smallest ideal and may be referred to as the trivial ideal. (2) In every ring h1i = R is the largest ideal and every other ideal is referred to as a proper ideal. (3) If n ∈ Z then the principal ideal, hni = {nk : k ∈ Z}, generated by n consists of all multiples of n. (4) If p ∈ R[X] then hpi is the set of all polynomials with p as a factor. Note that if I is an ideal of R then a ∈ I implies −a ∈ I, so every ideal also is closed under subtraction. The more general notion of right ideal is defined as for ideal but with the third condition replaced by the weaker condition: a ∈ I and r ∈ R implies ar ∈ I (and left ideals are defined similarly). Then, if a ∈ R the principal right ideal generated by a ∈ R is defined to be the set {ar : r ∈ R} and is denoted aR. The difference between right ideals, left ideals and (two-sided) ideals is wellillustrated in matrix rings - do the next exercise to see this. 12 Exercise** 2.27. Let R = M2 (Z) be the ring of 2 × 2 matrices with integer 1 0 entries. Let a = . Compute the right ideal generated by a, the left 0 0 ideal generated by a and the (two-sided) ideal generated by a. Proposition 2.28. A commutative ring R is a field iff the only ideals of R are {0} and R. Exercise 2.29. Show that if R is any ring then R is a division ring iff the only right ideals of R are {0} and R (and equivalently for left ideals). [Comment: it is the case (and showing this is a substantial exercise) that the first Weyl algebra, defined at 1.32, is a (non-commutative) ring which has no ideals except {0} and R yet which is not a division ring. So the analogue of 2.28 for non-commutative rings must refer to one-sided ideals: the condition on two-sided ideals is not enough to give a division ring.] If A ⊆ R is any subset of the ring R then the ideal generated by A is the set {r1 a1 s1 + · · · + rn an sn : n ≥ 1, ai ∈ A, ri , si ∈ R} and is denoted hAi. For instance, if A = {a} is a singleton set then hAi, which is also written as hai, consists of all elements of R which can be written as a sum of terms of the form ras as r, s range over R. If A = {a, b} then hAi, also written ha, bi, is the set of all elements of the form r1 as1 + . . . rn asn + t1 bu1 + . . . tm bum where the ri , si , tj , uj can be any elements of R (and n, m are arbitrary natural numbers). Exercise 2.30. Show that if A is a subset of a ring R then hAi is the smallest ideal of R which contains A. Example/Exercise 2.31. (5) Let R = Z[X] and let I = h2, Xi be the ideal of Z[X] generated by {2, X}. Prove that I is not a principal ideal. [Hint: suppose, for a contradiction, that it is, choose a generator, ...] Exercise** 2.32. The ideal h4, 6i of Z generated by 4 and 6 together is principal - find a generator for it. Generalise this. Proposition 2.33. If θ : R −→ S is a homomorphism of rings then ker(θ) is an ideal of R. Corollary 2.34. If θ : R −→ S is a homomorphism of rings and R, S both are fields then θ is a monomorphism. Later, 3.4, we will prove a kind of converse to 2.33, showing that every ideal does occur as the kernel of some homomorphism. We finish the section with some general properties of ideals. Proposition 2.35. Suppose that I and J are ideals of the ring R. Then I +J = {a + b : a ∈ I, b ∈ J} (their sum) and I ∩ J (their intersection) T are ideals. If {Iλ }λ is any collection of ideals of R then their intersection, λ Iλ , is an ideal. If R is commutative then (I : J) = {r ∈ R : rJ ⊆ I} is an ideal. A product of ideals also may be defined: if I, J C R then their product is defined to be the ideal generated by the set {ab : a ∈ I, b ∈ J} (this set will not in general be closed under addition, which is why we need to say “generated by”). Then powers of an ideal may be defined inductively by I n+1 = I n I. Exercises 2.36, 2.37, 2.38 are concerned with these operations on ideals. 13 Exercise** 2.36. Let R = Z and let I = h3i. What is I 2 ? What is I n ? Let J = h12i. What is IJ? Exercise* 2.37. Let I, J be ideals of a ring R. Show that IJ ⊆ I ∩ J. Give an example to show that this inclusion may be proper. Exercise** 2.38. Let R = Z. Show that h2i + h5i = Z. Compute h2i ∩ h5i and the product h2ih5i, writing each of these as a principal ideal, that is, in the form hni for some integer n. Now replace 2 and 5 by arbitrary integers a and b. What are hai + hbi, hai ∩ hbi and haihbi? [Hint: unless you’ve guessed the answers this might be difficult to answer, let alone prove, so try with some other pairs of integers in place of 2 and 5. In other words, explore until you see what’s happening.] Compute (h2i : h5i) and (h6i : h8i). What about the general case (hai : hbi)? 2.1 More exercises Exercise* 2.39. Find all ring homomorphisms from Z × Z to itself. √ a 2b : a, b ∈ Z . Define the Exercise* 2.40. Let R = Z[ 2] and let S = b a √ a 2b map θ : R −→ S by θ(a + b 2) = . Prove that θ is an isomorphism. b a Exercise** 2.41. Prove that Z6 ' Z2 × Z3 . Is Z10 ' Z2 × Z5 ? Is Z8 ' Z2 × Z4 ? Justify your answers. Exercise** 2.42. Prove that if θ : R −→ S is a surjective homomorphism of rings and R is commutative then S must be commutative. Give an example to show that the conclusion may be false if θ is not surjective. Exercise 2.43. Let R be the ring of all infinitely differentiable functions from the unit interval [0, 1] to R (with pointwise addition and multiplication as the operations). Let D : R −→ R be the map which takes a function f to its derivative f 0 . Is D a ring homomorphism? Exercise* 2.44. Suppose that R is a commutative ring of characteristic 3. Prove that the map θ : R −→ R defined by θ(r) = r3 is a homomorphism. Suppose also that R has no non-zero nilpotent elements: prove that θ is injective. Exercise** 2.45. Show that if I is a right ideal of a ring R and if there is an element a ∈ I with a right inverse then I = R. Exercise* 2.46. Let R be a commutative ring and set N (R) = {r ∈ R : ∃n, rn = 0} to be the set of all nilpotent elements of R. Prove that N (R) is an ideal of R. Compute N (R) for: (i) R = Z12 ; (ii) R = Z24; a b (iii) R = : a, b, c ∈ Q . 0 c Exercise 2.47. Prove that every non-zero ideal of the ring Z[i] contains a nonzero integer. Exercise* 2.48. Determine whether or not each of the following statements about ideals in the polynomial ring Q[X, Y ] is true: (i) hXY + Y 2 , X 2 + XY + Y 2 , X 2 + Y 2 i = hX 2 , XY, Y 2 i; (ii) hX 2 + XY, XY − Y 2 , X 2 + Y 2 i = hX 2 , XY, Y 2 i. 14 Exercise 2.49. Suppose that I, J are ideals of the commutative ring R such that I + J = R. Prove that IJ = I ∩ J. Exercise 2.50. Let R = RhX, Y : Y X = XY + 1i be the first Weyl algebra (as defined in the notes). Use the defining relation Y X = XY + 1 to write each of the following elements of R as a sum of monomials of the form X i Y j (i, j ≥ 0): Y X 2; Y 2X 2; Y X 3; Y X 2 − X 2Y ; Y X 3 − X 3Y . Prove that, for every n ∈ N, the ideal generated by X n is equal to R. Exercise* 2.51. Let R = R[X, Y ] and let S ⊆ R2 be any subset of the real plane. Set I(S) = {p ∈ R : ∀(r, s) ∈ S, p(r, s) = 0}. Prove that I(S) is an ideal of R. [Note that this is the kernel of the two-dimensional version of the evaluation map for polynomials which is defined in the printed notes.] Exercise** 2.52. Prove that if θ : R −→ S is a surjective homomorphism of rings and I is a right ideal of R then θ(I) is a right ideal of S. Give an example to show that the conclusion may be false if θ is not surjective. Exercise 2.53. Let a be an element of a ring R. Show that the right annihilator of a, rann(a) = {r ∈ R : ar = 0} is a right ideal of R. If also b ∈ R show that rann(a + b) ⊇ rann(a) ∩ rann(b). Show that can be proper. the inclusion 2 3 Compute rann(a) where R = M2 (Z6 ) and a = . 0 0 Show that rann(a) need not be a left ideal of R. Exercise 2.54. Let p be a prime integer and set Z(p) = m n ∈ Q : p - n}. Prove that Z(p) is a subring of Z. Show that the only proper non-trivial ideals of Z(p) are those generated by powers of p. 3 Factor Rings Given a ring R and an ideal I of R we will define the factor ring (also called the quotient ring) R/I and the canonical surjective homomorphism π : R −→ R/I with ker(π) = I. Informally, R/I is the “largest ring obtained from R by collapsing I to 0.” An example of this process which you know is to start with the ring, Z, of integers, take the ideal, h5i say, generated by 5: the resulting quotient ring is the ring, Z5 , of integers modulo 5 and the canonical surjection π is the map which takes an integer to its congruence class modulo 5. We give the precise general definition now. Let R be a ring and let I be a proper ideal. Let R/I denote the set of cosets of I in the additive group (R, +): R/I = {r + I : r ∈ R}. Define operations + and × on R/I by: (r + I) + (s + I) = (r + s) + I and (r + I) × (s + I) = (r × s) + I (so we’re adding and multiplying cosets, these being the elements of R/I). As always, when defining something in terms of representatives, the issue of welldefinedness has to be addressed. For a number of the results in this section we give the, slighly tedious but necessary, details here in the notes, so as to allow time for explanation and illustration in lectures. Lemma 3.1. The operations + and × on R/I are well-defined. That is, if r + I = r0 + I and s + I = s0 + I then (r + s) + I = (r0 + s0 ) + I and (r × s) + I = (r0 × s0 ) + I. 15 Proof. Recall that if H is a subgroup of an abelian group G and if a, b ∈ G then a + H = b + H iff a − b ∈ H. We apply this with G being the group (R, +) and H = I. Since r + I = r0 + I we have r − r0 ∈ I. Since s + I = s0 + I we have s − s0 ∈ I. Since I is closed under + we deduce that (r − r0 ) + (s − s0 ) ∈ I, that is (r + s) − (r0 + s0 ) ∈ I, hence (r + s) + I = (r0 + s0 ) + I. Also, since I is an ideal (r − r0 ) × (s − s0 ) ∈ I, that is, rs − r0 s − rs0 + r0 s0 ∈ I. Now, rs − r0 s − rs0 + r0 s0 = (rs − r0 s0 ) + 2r0 s0 − r0 s − rs0 = (rs − r0 s0 ) − r0 (s − s0 ) + (r0 − r)s0 . Since I is an ideal this is in I so, since both s − s0 and r0 − r belong to I, so does rs − r0 s0 . Hence rs + I = r0 s0 + I, as required. Lemma 3.2. Let R be a ring and let I be a proper ideal. With the operations defined above, the set R/I is a ring. The zero element of this ring is the coset 0 + I(= I) and the identity element is 1 + I. Furthermore −(r + I) = (−r) + I and, if r has an inverse r−1 in R then (r + I) is invertible in R/I with inverse r−1 + I. Proof. For instance, to check that the addition we have defined in R/I is commutative, take r + I, s + I - two typical elements of R/I. By definition (r + I) + (s + I) = (r + s) + I. Since addition is commutative in R this equals (s + r) + I which, by definition of + in R/I, equals (s + I) + (r + I), as required. Similarly for other properties. For instance to check distributivity, take elements r + I, s + I, t + I ∈ R/I. Then (r + I) (s + I) + (t + I) = (r + I) (s + t) + I) (definition of +) = r(s + t) + I (definition of ×) = (rs + rt) + I (since + is distributive over × in R) = (rs + I) + (rt + I) (definition of +) = (r + I)(s + I) + (r + I)(t + I) (definition of ×). To check that 0 + I is the zero element: given r + I ∈ R/I we have (0 + I) + (r + I) = (0 + r) + I = r + I. To check that 1 + I is the identity element: given r + I ∈ R/I we have (1 + I) × (r + I) = (1 × r) + I = r + I. Also given r + I ∈ R/I we have (r + I) + (−r + I) = (r + −r) + I = 0 + I so −r + I is the negative of r + I. Etc. etc. This ring is the factor ring (or quotient ring) of R by I. Example 3.3. As stated above, Z/h5i = Z5 . Theorem 3.4. Let I be a proper ideal of the ring R. The map π : R −→ R/I defined by π(r) = r + I is a surjective ring homomorphism with kernel I (π is called the canonical surjection, or canonical projection). It is the “smallest” homomorphism with domain R and kernel I. Indeed, if θ : R −→ S is a homomorphism with ker(θ) ⊇ I then there is a unique map θ0 : R/I −→ S with θ0 π = θ. This map θ0 is a homomorphism. The map θ0 is injective iff ker(θ) = I. If θ is surjective and ker(θ) = I then 0 θ is an isomorphism. Proof. Since every element of R/I has the form r +I, = π(r)+I for some r ∈ R, clearly the map π is surjective. We check that it is a homomorphism. First, π(r + s) = (r + s) + I = (r + I) + (s + I) = π(r) + π(s). Similarly π(rs) = (rs) + I = (r + I)(s + I) = π(r)π(s). And π(1) = 1 + I which, by the result above, is the identity of R/I. 16 Next, the kernel of π: ker(π) = {r ∈ R : π(r) = 0R/I }. By the result above the zero, 0R/I of R/I is the coset I = 0 + I so ker(π) = {r ∈ R : π(r) = 0 + I} = {r ∈ R : r + I = 0 + I} = {r ∈ R : r ∈ I} = I. Now suppose that θ : R → S is a homomorphism with ker(θ) ⊇ I. Define the map θ0 : R/I → S by θ0 (r + I) = θ(r). First we have to show that this map is well-defined, because the definition used a choice of coset representative. So suppose r + I = r0 + I; then r − r0 ∈ I so, by assumption, θ(r − r0 ) = 0. Since θ is a homomorphism this gives θ(r) = θ(r0 ) - so we get the same result for θ0 (r + I) whatever representative of this coset we use. 0 Then We have θ0 (r + I) + we0 have to show that θ is a homomorphism. (s + I) = θ (r + s) + I = θ(r + s) = θ(r) + θ(s) = θ0 (r + I) + θ0 (s + I). And similarly for multiplication. The homomorphism θ0 is injective iff θ0 (r + I) = 0 implies r + I = 0 + I that is, iff θ(r) = 0 implies r ∈ I, that is, iff ker(θ) is contained in, hence equals, I, as required. Finally, if θ is surjective and ker(θ) = I then it is, by the previous paragraph, also injective, hence an isomorphism. This is a place where there are significant conceptual hurdles to get over and I will spend time on discussion and illustration. Example 3.5. Take R = Z and I = h6i = 6Z so we have the canonical surjection π : Z −→ Z6 . Let θ be the canonical projection from Z to Z/h2i = Z2 (which takes a ∈ Z to [a]2 ). Then ker(θ) = h2i ⊇ h6i so, by the theorem, there is a unique factorisation of θ through π. This is the map Z6 −→ Z2 already seen in 2.11. Some notation: if I ⊇ J are ideals then usually I write I ≥ J instead of just using the subset notation (but it means the same). Example 3.6. Let R = R[X] and take I = hX 2 + 1i to be the principal ideal generated by X 2 +1. Let π : R[X] −→ R[X]/hX 2 +1i be the canonical surjection. Now let θ : R[X] −→ C be the homomorphism which is evaluation at i ∈ C, that is, θ(p(X)) = p(i) (note that, even though i is not in R this does make sense and it is a homomorphism). A polynomial p is in the kernel of θ iff it has X 2 + 1 for a factor, that is, ker(θ) = hX 2 + 1i. By the last part of 3.4 the unique map, θ0 , from R[X]/hX 2 + 1i to C through which θ factorises, that is, such that θ = θ0 π, is injective. It is easily checked that θ0 also is surjective and hence is an isomorphism. Thus R[X]/hX 2 + 1i ' C. Theorem 3.7. Let I be an ideal of the ring R. Then there is a natural, inclusionpreserving, bijection between the set of ideals of R which contain I and the set of ideals of the factor ring R/I: • to an ideal J ≥ I there corresponds πJ = {r + I : r ∈ J} = {π(r) : r ∈ J}; • to an ideal K C R/I there corresponds π −1 K = {r ∈ R : π(r) ∈ K}. The notation J/I is also used instead of πJ for the image of J in R/I. Proof. First suppose that J ≥ I is an ideal of R: we check that πJ is an ideal of R/I. • Since 0 ∈ J, π(0) ∈ πJ, but π(0) = 0 (more accurately, π(0R ) = 0R/I ), so 0 ∈ πJ. 17 • Suppose that b, b0 ∈ πJ, say b = π(a), b0 = π(a0 ) for some a, a0 ∈ J. Since J is an ideal a + a0 ∈ J, so π(a + a0 ) ∈ πJ. Since π is a homomorphism (Theorem 3.4), π(a) + π(a0 ) = π(a + a0 ) ∈ πJ, as required. • Suppose that b ∈ πJ and s ∈ R/I. Say b = π(a) with a ∈ J. Also, π is surjective so there is r ∈ R with π(r) = s. Since J is an ideal both ar, ra ∈ J. Hence bs = π(a)π(r) = π(ar) ∈ πJ and similarly sb ∈ πJ, as required. Thus we have checked the three properties for πJ to be an ideal of R/I. (You might notice that we did not need to assume that J ≥ I: that assumption only comes in when we prove the bijection.) Next, let K be an ideal of R/I and consider its inverse image π −1 K = {a ∈ R : π(a) ∈ K}. We show that π −1 K is an ideal of R and contains I. • Since 0R/I ∈ K and π(0R ) = 0R/I we have 0R ∈ π −1 K. • Let a, a0 ∈ π −1 K. So π(a), π(a0 ) ∈ K. Since K is an ideal π(a + a0 ) = π(a) + π(a0 ) ∈ K. So a + a0 ∈ π −1 K, as required. • Let a ∈ π −1 K and let r ∈ R. Then π(ar) = π(a)π(r) ∈ K and π(ra) = π(r)π(a) ∈ K since π(a) ∈ K and K is an ideal. Therefore ar, ra ∈ π −1 K, as required. Thus we have shown that π −1 K is an ideal of R. • Let a ∈ I, then π(a) = 0 ∈ K, so a ∈ π −1 K. Thus π −1 K contains I. It remains to show that (1) if we start with an ideal J ≥ I, map it across to R/I to get πJ and then pull that back to R, to get π −1 πJ then we end up where we started, with J. And, similarly, that, (2) starting with an ideal K of R/I, pulling back to π −1 K and then mapping across again to get ππ −1 K, that we end up with K again. The uncommented steps in the arguments which follow are all direct from the definitions (of πJ and π −1 K etc.). (1) If a ∈ J then π(a) ∈ πJ so a ∈ π −1 πJ. Thus J ⊆ π −1 πJ. For the converse, take a ∈ π −1 πJ: then π(a) ∈ πJ, say π(a) = π(a0 ) for some a0 ∈ J. Therefore π(a−a0 ) = 0 and hence a−a0 ∈ ker(π) = I (by Theorem 3.4). At last we use the assumption that J ≥ I, to deduce a−a0 ∈ J. Then we have a = a0 +(a−a0 ) ∈ J (since both a0 and a − a0 are in J). Thus π −1 πJ ⊆ J and hence the two sets are equal. (2) If b ∈ K then, since π is surjective, there is r ∈ R with π(r) = b. So r ∈ π −1 K. Then b = π(r) ∈ ππ −1 K. Thus K ⊆ ππ −1 K. For the converse, take b ∈ ππ −1 K. So there is r ∈ π −1 K such that b = π(r). But the fact that r ∈ π −1 K says that π(r) ∈ K. Thus b ∈ K. Therefore ππ −1 K ⊆ K and so the two sets are equal. Example 3.8. The ideals of Z6 = Z/h6i are in bijection with the ideals of Z which contain h6i and these are Z, h2i, h3i, h6i, giving four ideals of the quotient ring Z6 (you should write them down explicitly). There are some exercises (Exercises 3.9, 3.10) around this. Exercise** 3.9. Determine the ideals of Z24 and match these up with the ideals of Z which contain h24i. Exercise 3.10. In the situation of 3.7 there is a similar correspondence for right (respectively left) ideals J containing I and right (resp. left) ideals of R/I (but I itself should still be a two-sided ideal in order that R/I be a ring). Check this. 18 An ideal I of a ring R is maximal if it is proper (i.e. I 6= R) and if there is no ideal between it and R, more formally, if any ideal J with I ≤ J ≤ R is equal to I or to R. Corollary 3.11. If R is a commutative ring then an ideal I C R is maximal iff the quotient ring R/I is a field. Theorem 3.12. If I ≤ J are ideals of R, so J/I is an ideal of R/I, then (R/I)/(J/I) ' R/J. Exercise 3.14 illustrates this. Example 3.13. We have h6i ≤ h2i and the theorem says that Z/h6i / h2i/h6i is isomorphic to Z/h2i (this is already essentially in 2.11, 3.5). Exercise* 3.14. Describe explicitly the isomorphism between Z/h10i / h5i/h10i and Z5 . Exercise 3.15. An ideal I of a commutative ring R is prime if whenever r, s ∈ R and rs ∈ I then either r ∈ I or s ∈ I. What are the prime ideals of Z? Prove that an ideal I of the commutative ring R is prime iff the factor ring R/I is a domain. 3.1 More exercises Exercise* 3.16. Show that the factor ring Z[i]/h1 + 3ii is isomorphic to Z10 . Exercise 3.17. Prove that if K is a field then K[X, Y ]/hXY i is isomorphic to the subring of the product K[X]×K[Y ] consisting of all pairs (p(X), q(Y )) with the same constant term (i.e. such that p(0) = q(0)). Exercise 3.18. Let R = K[X] where K is a field. Let I = hX(X − 1)i. Define a map θ from R to S = K × K as follows. Given p ∈ R, write p = q · X(X − 1) + r where q, r ∈ R and r is linear, so r has the form λ + µX for some λ, µ ∈ K. Then define θ(p) = (λ, λ + µ). Prove that θ is a homomorphism. Prove that θ induces an isomorphism from R/I to S. 4 4.1 Polynomial Rings and Factorisation Division of polynomials Throughout this subsection K is a field and polynomials will be in a single variable X, thus members ofP the polynomial ring K[X]. Recall that the canonical n form of f ∈ K[X] is f = i=0 ai X i with ai ∈ K. If an 6= 0 then the degree of f is n, deg(f ) = n. Theorem 4.1. (Division Theorem for Polynomials) Let K be a field and take f, g ∈ K[X] with g 6= 0. Then there are (unique) q, r ∈ K[X] with f = qg + r and deg(r) < deg(g). (q is the quotient and r the remainder when f is divided by g) (This should remind you of the Division Theorem for integers. So should the proof.) 19 Proof. Let A = {f − gh : h ∈ K[X]} be the set of all polynomials which can be obtained by subtracting polynomial multiples of g from f . This set is non-empty since it contains, for example, f = f − g · 0. Choose a polynomial, say r, in A of least degree, so r = f − qg for some q ∈ K[X]. Rearranging gives f = qg + r, so it has to be shown than deg(r) < deg(g). If it were not, then (think about it) some multiple of g could be used to reduce the degree of r: say deg(r − hg) < deg(g) for some h ∈ K[X]. But then r − hg = (f − qg) − hg = f − (q + h)g would be in A and would have smaller degree than r, contradicting the choice of r. It remains to show uniqueness. If we had qg − r = f = q 0 g − r0 with both deg(r), deg(r0 ) < deg(g) then (q − q 0 )g = r − r0 . Because deg(r0 − r) < deg(g) this can happen only if both sides are 0, that is, only if q = q 0 and r = r0 , as required. An element a ∈ K is a root (or zero) of f ∈ K[X] if f (a) = 0. Corollary 4.2. Let K be a field, let f ∈ K[X] and let a ∈ K. Then a is a root of f iff X − a is a factor of f . Proof. If X −a is a factor of f then certainly f (a) = 0. For the converse, suppose that f (a) = 0. Use the Division Theorem with g = X −a to deduce that there is an equation f = q · (X − a) + r for some q, r ∈ K[X] with deg(r) < deg(X − a). Since X − a has degree 1 it must be that r is a constant. So, substituting a for X we obtain f (a) = 0 + r. In particular, if f (a) = 0 then r = 0 and hence f = q · (X − a) is indeed a multiple of X − a. The greatest common divisor (or highest common factor) of polynomials f, g is a polynomial d such that d|f , d|g and, if h is any polynomial dividing both f and g then h divides d. Write d = gcd(f, g). This polynomial is defined only up to non-zero scalar multiple so, if we want a unique gcd then we can insist that d be monic (have leading coefficient 1). Division Algorithm This is entirely analogous to the Division Algorithm (Euclid’s Algorithm) for integers. The key observation is that if f = qg + r then gcd(f, g) = gcd(g, r) (because, by that equation and its rearrangement f − qg = r, the common divisors of f and g are the common divisors of g and r). So, given f, g ∈ K[X] with g 6= 0, apply the Division Theorem to obtain f = q0 g + r1 with deg(r1 ) < deg(g). Then apply it to divide r1 into g, obtaining: g = q1 r1 + r2 with deg(r2 ) < deg(r1 ) et cetera r1 = q2 r2 + r3 with deg(r3 ) < deg(r2 ) ... rn−2 = qn−1 rn−1 + rn with deg(rn ) < deg(rn−1 ) and, because the degrees of the remainders are strictly decreasing, eventually a remainder of 0 is reached: rn−1 = qn rn . Then gcd(f, g) = gcd(g, r1 ) = gcd(r1 , r2 ) = · · · = gcd(rn−1 , rn ) = rn (the last since rn divides rn−1 ). So the gcd of f and g is the last non-zero remainder in this algorithm. 20 Also, tracking back through the equations allows the expression of the gcd, rn , as a linear combination of f and g; “linear” in the polynomial sense, that is, there are polynomials k, l ∈ K[X] such that gcd(f, g) = kf + lg. And from this follows the next corollary. Corollary 4.3. Let K be a field and take f, g ∈ K[X]. Then the ideal generated by f together with g equals the ideal generated by their greatest common divisor: hf, gi = hgcd(f, g)i. Corollary 4.4. Let K be a field. Then every ideal of the polynomial ring K[X] is principal (i.e. can be generated by a single polynomial). Proof. It follows from the corollary above, and induction, that every ideal generated by finitely many polynomials can be generated by a single polynomial (their greatest common divisor) and then one could quote the Hilbert Basis Theorem which says that every ideal in K[X] is finitely generated. But that’s a bit heavy-handed and the idea in the proof of the Division Algorithm can be used: namely, given an ideal, choose a polynomial in it of least degree (ignoring the zero polynomial), and show that this generates the ideal (details left as an exercise). 4.2 Factorisation Throughout this section the ring R is assumed to be commutative. Most of the examples that we use will be polynomial rings but we set these in a more general context. Although we state various definitions and results about this general context we will, in practice, treat the general case rather lightly. In particular, when it comes to revising this subsection for the examination, concentrate on what was actually covered in lectures. An element r ∈ R is irreducible if r is not invertible and if, whenever r = st either s or t is invertible. For example, the irreducible elements of Z are the prime (positive and negative) integers. Exercise* 4.5. What are the irreducible elements of the polynomial ring R[X]? Elements r, s ∈ R are associated if s = ur for some invertible element u ∈ R. For instance two integers r, s are associated iff r = ±s. Exercise 4.6. Show that the relation “associated” is an equivalence relation on any commutative ring R. Exercise 4.7. What are the association equivalence classes in the polynomial ring R[X]? A commutative domain R is said to be a unique factorisation domain if every non-zero, non-invertible element of R has an essentially unique factorisation as a product of irreducible elements. More formally: the commutative domain R is a unique factorisation domain, or UFD for short, if for every nonzero r ∈ R which is not invertible there are irreducible elements r1 , . . . , rk ∈ R such that r = r1 × · · · × rk (existence of irreducible factorisation) and, if also r = s1 × · · · × sl where the sj are irreducible elements of R, then k = l and there is a permutation σ of {1, . . . , k} such that, for each i, sσ(i) is associated to ri (uniqueness of irreducible factorisation). 21 Example 4.8. Z is, as you know, a unique factorisation domain. For instance take r = −24. Then −24 = (−2)×2×(−2)×(−3) is one irreducible factorisation. There are others, such as −24 = 3 × (−2) × 2 × 2 but you can surely see how to permute the factors so that they match up as associated elements. √ Example 4.9. The ring Z[ −5], though a commutative domain, is not a UFD. √ √ For instance 21 = 3 × 7 = (1 + 2 −5)(1 − 2 −5) are, one may check, two factorisations into irreducibles but, again one may check, the irreducible factors do not match up into associated pairs. Example 4.10. By default, every field is a UFD. A principal ideal domain, or PID for short, is a commutative domain in which every ideal is principal (that is, is generated by some single element). Example 4.11. The ring Z is a PID since every ideal has the form hni for some integer n. If K is a field then K[X] is a PID (every ideal consists of all the multiples of a particular polynomial). In each case the proof makes use of the division theorem. It is a general result that every PID is a UFD so it follows that if K is a field then the polynomial ring K[X] is a UFD. A much harder result to prove is that K[X1 , . . . , Xn ] is a UFD: we won’t prove it in the course but it is, in fact, true that if R is a UFD then so is R[X] hence, by induction, so is R[X1 , . . . , Xn ] (for a proof see, e.g., Fraleigh, Section 45). Example 4.12. For an example of a ring which is a UFD but not a PID take either Z[X] or K[X, Y ] where K is a field. It is not difficult to show that these are not PIDs (cf. Example 2.31). The fact that they are UFDs follows from the, hard, general result mentioned earlier, that if R is a UFD then so is R[X]. Remark 4.13. Almost all the material that we present in this course is from the foundations of the subject and dates back around 100 years (or more). But this is a good place to mention a somewhat more recent result (from the late 1950s) of M. Auslander and D. Buchsbaum which says that regular local rings are UFDs: regular local rings are commutative rings which are of central importance in algebraic geometry. Remark 4.14. Another (this time, rather old) result which we should mention is the Hilbert Basis Theorem (named after D. Hilbert) which states that if K is a field then every ideal in a polynomial ring K[X1 , . . . , Xn ] is finitely generated (i.e. has a finite set of generators). Exercise* 4.15. Let K be a field and consider the polynomial ring R = K[X1 , . . . , Xn , . . . (n ∈ P)] in infinitely many generators. Find an ideal I of R which is not finitely generated. Can you prove that your choice for I isn’t finitely generated? There is a variety of tests for irreducibility in polynomial rings. Some examples will be given in lectures. 22 4.3 More exercises Exercise** 4.16. Find a generator, h, for the ideal hf, gi of Q[X], where f = X 3 − X 2 − X + 1, g = X 5 + X 2 − X − 1. Write h in the form af + bg for suitable a, b ∈ Q[X]. Exercise* 4.17. Find a generator, h, for the ideal hf, gi of C[X], where f = X 3 − iX 2 + 2X − 2i, g = X 2 + 1. Write h in the form af + bg for suitable a, b ∈ Q[X]. Exercise 4.18. Find a generator, k, for the ideal hf, g, hi of Q[X], where f = 25 2 X 3 + 4X 2 + 2X − 3, g = X 4 + 3X 3 − X 2 − 3X, h = X 4 − X 3 − 15 2 X + 2 X − 3. Write k in the form af + bg + ch for suitable a, b, c ∈ Q[X]. Exercise 4.19. For which integers n does the polynomial X 2 + X + 1 divide X 4 + 3X 3 + X 2 + 6X + 10 in the polynomial ring Zn [X]. Exercise** 4.20. Consider the field Z2 with two elements. Write down all the monic polynomials of degree 2 in R = Z2 [X]. For each of these, determine whether it is reducible or irreducible. Now do the same for monic polynomials of degree 3. Exercise** 4.21. Prove that the polynomial 7X 3 − 6X 2 + 2X − 1 ∈ Z[X] is irreducible. Then prove this using a different method. Exercise** 4.22. Prove that the polynomial X 5 + X 2 + X − 1 ∈ Z[X] is irreducible. Exercise* 4.23. Show that if K is a finite field then there is a polynomial in K[X] which has no root in K. Exercise 4.24. Show that the only invertible elements of the ring Z[i] are ±1, ±i. 5 Constructing Roots for Polynomials You know that some (non-constant) polynomials with rational coefficients, for example X 2 − 2, don’t have a rational root. That particular polynomial does have a solution in the field R which extends Q. For some other polynomials, for example, X 2 + 2, there is still no root in R but we could extend further, to C. In fact, we don’t have to go to such large field extensions. In this section we will see how, given a field K and a polynomial with coefficients in K, we can produce a root of that polynomial in a “minimal” field extension of K. That is, given a field K and p ∈ K[X], we will produce a field, L, extending K (i.e. into which K embeds) and containing a root for p. Our construction will be such that L is generated (as a ring) by K and a root of p; so the construction is an economical one. Example 5.1. Take p = X 2 − 2 ∈ Q[X]. For√an extension field containing a root we could take R or, less extravagantly, Q[ 2]. Example 5.2. Take p = X 2 + 1 ∈ R[X]: then for L we may take C (and that is the smallest possible choice since it is generated as a ring, so certainly as a field, by R and a root of p). 23 Notice that, to produce a root for a polynomial p, it is enough to produce a root of one of its irreducible factors. Therefore we may concentrate on the case of an irreducible polynomial. Theorem 5.3. (Kronecker’s Theorem) Let K be a field and let f ∈ K[X] be irreducible of degree n. Then the canonical homomorphism π : K[X] −→ K[X]/hf i induces an embedding ι : K −→ K[X]/hf i of K into L = K[X]/hf i and L is a field. Also α = π(X) ∈ L is a root of f . The dimension of L as a vector space over K is n, with {1, α, α2 , . . . , αn−1 } being a basis of L over K, so every element of L has a unique representation of the form an−1 αn−1 + · · · + a1 α + a0 with an−1 , . . . , a1 , a0 ∈ K. (Note that in the latter part of the statement we have identified K with its image in L.) The last part of the theorem says that L, which we also write as K[α], is minimal, not just as a field but even as a ring, given that we have added a root of f . Examples 5.4. (1) K = R, f = X 2 + 1. The degree of f is 2, so every element of R[X]/hX 2 + 1i has the form a1 i + a0 with a1 , a0 ∈ R, where i, instead of α, has (for obvious reasons) been used as notation for π(X). Clearly L = R[X]/hX 2 + 1i ' C. (2)√K = Q, f = X 2 − 2. Then it’s easy to check that L = Q[X]/hX 2 − 2i ' Q[ 2]. Corollary 5.5. Let f ∈ K[X] be irreducible, where K is a field, and let π, ι and α be as in the statement of Kronecker’s Theorem. Let θ : K −→ L0 be an extension field of K and suppose that L0 contains a root, β, of f . Then there is a homomorphism ρ : L = K[X]/hf i −→ L0 extending θ (in the sense that ρι = θ) and taking α to β. (So there is a copy of L sitting between θ(K) and L0 .) 2 0 Example 5.6. √ Take K = Q, f = X − 2 and L = R. The three fields involved are: Q ≤ Q[ 2] ≤ R. Example 5.7. (3) K = Q, f = X 4 +1. First we will check that f is irreducible. That done, since deg(f ) = 4 the canonical form for elements of L = Q[X]/hX 4 + 1i is a3 α3 +a2 α2 +a1 α+a0 with the ai rational and where α is as in the theorem. Of course we can identify α with √ a primitive fourth root of −1 in C and see that Q[X]/hX 4 + 1i ' Q[(1 + i)/ 2]. This is continued in Exercise 5.8. √ √ Exercise 5.8. Show that the ring Q[(1+i)/ 2] (see Example 5.7) equals√ Q[i, 2], by which is meant the smallest subfield which contains Q, i and 2. Also √ of C √ show that this is equal to {a + bi + c 2 + di 2; a, b, c, d ∈ Q}). Exercise* 5.9. (quite long but worth doing) Show that the polynomial f = X 4√ +1 factorises completely (i.e. into linear factors) in the extension field Q[(1+ i)/ 2] constructed above: thus, adding a single root of f gives all the roots. By way of contrast, consider g = X 3 − 2 ∈ Q[X]. Check that the field L = Q[α] constructed in the theorem may be regarded as Q[21/3 ] but that the other two roots of g are not in this field. Find a field L0 , of degree (i.e. dimension as a vector space) 6 over Q, over which g does split into linear factors. Now check that the field L could have been taken to be Q[η] where η is one of the properly complex roots of g and find the factorisation of g as (X − η)g 0 24 where g 0 is a quadratic polynomial with coefficients in Q[η]. This emphasises that the three roots of g are, abstractly, equivalent (precisely, Q[21/3 ] ' Q[η]). If you want to understand better what’s happening here, pursue the topic called Galois Theory. In Kronecker’s Theorem the polynomial f is assumed to be irreducible: what happens if we drop that assumption? Example 5.10. Let K be any field and let f = X(X −1). Then K[X]/hX(X − 1)i is not a field. To see this, let α = π(X) where, as usual, π : K[X] −→ K[X]/hX(X − 1)i is the canonical projection. Then neither α nor α − 1 is zero (since neither is a multiple of f ) but their product α(α − 1) is 0 = π(X(X − 1)). So the ring K[X]/hX(X − 1)i is not even a domain. Examples 5.11. We will consider some finite field extensions of F2 , F3 , F5 , including the degree 2 extension of F2 . In particular we’ll use X 3 +X+1 ∈ Z2 [X], X 2 + X + 2 ∈ Z3 [X] and look at a quadratic extension of Z5 . Example 5.12. Let p be an odd prime. Then −1 is a quadratic residue modulo p (that is, a square modulo p) iff p has the form 4k + 1 for some integer k, so Fp [X 2 + 1] is a proper extension of Fp iff p ∼ = 3 mod 4. In any case, for any prime p, there always will be an irreducible polynomial in Fp [X] of the form X 2 + aX + b (just by counting) and hence some proper extension of Fp of degree 2. The remainder of this section is purely for your edification: likely we will not have time to cover it and it will not be examined. So: purely for interest! A field K is algebraically closed if every non-constant polynomial with coefficients from K has a solution from K. Exercise 5.13. Prove that the following are equivalent for a field K: (i) K is algebraically closed; (ii) every irreducible polynomial over K is linear; (iii) every non-constant polynomial with coefficients from K has a factorisation as a product of linear factors. Examples 5.14. Neither Q nor R is algebraically closed but C is (this is the so-called Fundamental Theorem of Algebra). If K ≤ L is an extension of fields then an element λ ∈ L is said to be algebraic over K if it is a root of some non-zero polynomial with coefficients from K. For example every element in C is algebraic over R, indeed is a root of a quadratic polynomial with coefficients in R (exercise: describe such a polynomial explicitly, given a complex number λ = a + bi). On the other hand C is not algebraic over Q. There is a fairly short “counting” argument to prove this. From that argument it follows that many (indeed “most”) elements of C are transcendental (not algebraic) over Q. It is, however, surprisingly difficult to prove that a particular real or complex number is transcendental over Q. For instance neither π nor e is a root of any (non-zero) polynomial with rational coefficients but neither of these facts is easy to prove (Hermite in 1873 showed it for e and Lindemann in 1882 for π). And, although one might guess that π + e and eπ are transcendental over Q, this has never been proved. (It is not difficult to show that at least one of them must be, probably both are, but no-one knows. See, e.g., the article on mathworld about transcendental numbers.) 25 e containing K which is Theorem 5.15. Let K be a field. There is a field K e is algebraic over K. This algebraically closed and such that every element of K field is unique up to isomorphism (over K) and is called the algebraic closure of K. Example 5.16. The algebraic closure of Q is not the field of complex numbers. In fact it is much smaller: if you trace through the proof of the above result with e is countable, hence is much K being Q then you can see that the end result, Q, smaller than C. This is a fundamental mathematical object which, nevertheless, you are unlikely to have encountered before. 5.1 More exercises Exercise** 5.17. Let F4 be the field with 4 elements (so F4 ' Z2 [X]/hX 2 + X + 1i), say F4 = {0, 1, α, α + 1} where α2 + α + 1 = 0. (i) Show that the polynomial p = X 2 + αX + (α + 1) ∈ F4 [X] is reducible. (ii) Show that q = X 2 + αX + 1 ∈ F4 [X] is irreducible. (iii) Let F16 = F4 [X]/hX 2 + αX + 1i be the corresponding extension field of F4 , and let β be a root of X 2 + αX + 1 in F16 , so 1, β is a basis of F16 over F4 . Compute β 3 and β −1 in terms of this basis. Write down a basis of F16 over F2 . Exercise 5.18. Show that the map from R[X, Y ]/hX 2 +1, Y 2 +1i to C×C which takes a polynomial p(X, Y ) to (p(i, i), p(i, −i)) is an isomorphism. √ √ Exercise 5.19. Show that each of 7 + 21/3 and 3 + −5 is algebraic over Q. That is, for each, find a non-zero polynomial with rational coefficients of which it is a root. Exercise 5.20. Prove that Z2 [X]/hX 3 + X + 1i is a field and that Z3 [X]/hX 3 + X +1i is not. [Hint: this is not a long computation, it’s a case of seeing what the point is, quoting a result from the notes, and doing a very small computation.] Part II Groups 6 Groups, Isomorphisms and Group Actions Recall that a group is given by the following data: • a set G; • a binary operation, “multiplication”, denote it · say, on G; • a unary operation, “inverse”, denote it (−)−1 say, on G; • a specific element, the identity element, denote it e, in G; such that: ∗ the operation · is associative, a · (b · c) = (a · b) · c for all a, b, c ∈ G; ∗ a · a−1 = e = a−1 · a for all a ∈ G; ∗ a · e = a = e · a for all a ∈ G. In practice one usually gives just the set G together with an associative operation · on G and then checks existence of an identity and existence of inverses. Also, in practice, we write ab for a · b and just G for (G, ·). 26 Recall how to draw up the group table for a (small) group G: arrange the elements of G along the top and left-hand side and, at the entry in the a-labelled row and b-labelled column, place the product ab. Recall that the order of an element a ∈ G, we write o(a), is the least k ≥ 1 such that ak = e and is ∞ if there is no such k. For instance the identity element of G is the only one of its elements with order 1. Elements of order 2 sometimes are called involutions. The order of the group G is quite a different notion and means simply the number of elements in G: we denote this by |G|. Example 6.1. (1) Sn - the symmetric group on n letters, that is, the group of all permutations of n objects. Recall that every permutation may be written, in an essentially unique way, as a product of disjoint cycles. The order of an element is obtained by inspection from its cycle decomposition (and that is easy to compute). Exercise** 6.2. Draw up the group table for S3 . List, giving each as a product of disjoint cycles, all the permutations in S4 . Determine the order of each element of S4 . Recall that if (G, ·) is a group, with identity element e, then a subset H of G forms a subgroup if: • a, b ∈ H implies a · b ∈ H; • a ∈ H implies a−1 ∈ H; • e ∈ H. If these conditions are satisfied then H, with the restriction of “·” to H forms a group in its own right. The trivial group, {e} has just one subgroup (namely itself). Every other group G has at least two subgroups: the whole group G (all other subgroups are proper) and the trivial subgroup {e}. We write H ≤ G when H is a subgroup of G. If a ∈ G then the cyclic subgroup generated by a is the set of all, positive and negative (and 0), powers hai = {ak : k ∈ Z} and this is a subgroup of G (if the order of a is n < ∞ then only positive powers are needed since a−1 = an−1 ). A group is said to be cyclic if it can be generated by a (suitable) single element. Recall that if H is a subgroup of G and a ∈ G then the corresponding left coset is aH = {ah : h ∈ H} and the right coset of a with respect to H is Ha = {ha : h ∈ H}. The sets H, aH and Ha all have the same number of elements and the, say left, cosets partition G into disjoint pieces. The number of distinct left cosets of H in G is the index of H in G (and equals the number of distinct right cosets of H in G). This is written [G : H]. Lagrange’s theorem says that |G| = |H|[G : H]. Exercise** 6.3. Give an example of a group G, a subgroup H and an element a ∈ G such that the left and right cosets aH and Ha are not equal. Lemma 6.4. Let H be a subgroup of the group G and let a, b ∈ G. (i) aH = H iff a ∈ H iff Ha = H (ii) aH = bH iff a−1 b ∈ H iff b−1 a ∈ H. A group G is abelian (or commutative) if ab = ba for all a, b ∈ G. 27 Example 6.5. (3) If (R; +, ×, 0, 1) is a ring then (R, +) is an abelian group. The most basic example is perhaps (Z, +). Note that every non-zero element of this group has infinite order. Exercise* 6.6. Write down the group tables, where the operation is addition, for Z4 and also for Z5 . Do the same for the field F4 with 4 elements (see Section 5). Example 6.7. (4) If (R; +, ×, 0, 1) is a ring then the set, U (R) of invertible (with respect to multiplication) elements of R, with the operation × forms a group (exercise: prove this), called the group of units of R. Exercise** 6.8. Find U (R) and draw up the group table for each of the following rings R: Z; Z5 ; Z8 ; F4 . Exercise 6.9. Let R be the ring M2 (Z) of 2 × 2 matrices with integer entries. Find some elements in the group of units of R and compute their orders. a b Now replace Z by R. What is a criterion for ∈ U (M2 (R)? Given c d n can you find an element of order n in U (M2 (R))? Example 6.10. (5) Groups of symmetries of geometric figures (composition being the operation) is another good source of groups. Exercise** 6.11. Denote by Dn the group of symmetries of a regular n-gon. Show that the dihedral group Dn has 2n elements, of which n are rotations and n are reflections. Draw up the group table for D3 . We say that two groups G and H are isomorphic if they have the same shape as abstract structures. Precisely, an isomorphism from G to H is a bijection θ : G −→ H such that θ(ab) = θ(a)θ(b) for all a, b ∈ G. Note the similarity with the notion of isomorphism for rings (this is, if anything, simpler). Similar comments apply: for example in the equation above, the operation on the left hand side is the multiplication in G whereas that on the right is the multiplication in H. The next lemma is analogous to 2.3. Lemma 6.12. Suppose that θ : G −→ H is an isomorphism between groups. Then: (1) θ(eG ) = eH ; (2) θ(a−1 ) = (θ(a))−1 ; (3) o(θ(a)) = o(a) for every a ∈ G; (4) |G| = |H|. We say that G and H are isomorphic, and write G ' H, if there is an isomorphism from G to H. As with rings, this is an equivalence relation (exercise). To say that G and H are isomorphic is to say that they are the same group “up to re-naming elements”. In terms of the group tables, this means that, reordering elements if necessary, the group tables are the same up to re-naming elements (θ is the “renaming function”). Exercise** 6.13. Show that the symmetric group S3 and the dihedral group D3 are isomorphic by bringing their group tables to the same form. Define an isomorphism from S3 to D3 . Now define another isomorphism from S3 to D3 . 28 Suppose that G and H are groups. The direct product, G × H, of G and H is the group with underlying set G × H = {(g, h); g ∈ G, h ∈ H} and with operation defined by (g, h) · (g 0 , h0 ) = (gg 0 , hh0 ) (i.e. (g ·G g 0 , h ·H h0 )). Exercise** 6.14. Check that G × H is a group. Prove that G × H ' H × G (note that, to do this, you should produce an isomorphism θ). Exercise** 6.15. Draw up the multiplication table for the group of symmetries of a rectangle which is not a square. Show that this group is isomorphic to Z2 × Z2 (where addition is the operation). Exercise 6.16. Show that the group of rotations of a cube is isomorphic to S4 . How many symmetries of a cube are there? Show, by considering the relation between an octahedron and a cube, that an octahedron has the same group of symmetries. 6.1 The sign of a permutation Recall that Sn denotes the symmetric group on n objects: the set of all (n!) permutations of, say, {1, 2, . . . , n} with the operation being composition of permutations. We use the convention that if σ, τ ∈ Sn then στ means “do τ then do σ” (so the usual order for composition of functions). Thus (1 2)(2 3) = (1 2 3). To every permutation σ ∈ Sn we associate a matrixPσ (a permutation ma- 1 2 ... n trix) as follows. Write σ using “two-row notation”, as . σ(1) σ(2) . . . σ(n) Take the n × n identity matrix In and rearrange its columns according to the second row: so the σ(1)-th column of In becomes the first column of Pσ ; the σ(2)-th column of In becomes the second column of Pσ , et cetera. We define the sign, sgn(σ), of σ to be the determinant of Pσ . Therefore sgn(σ) = ±1 since, recall, switching two columns of a matrix multiplies its determinant by −1 and since Pσ can be obtained from In (which has determinant 1) by a sequence of such switches. This shows the following. Lemma 6.17. If a permutation σ is written as a product of k transpositions then sgn(σ) = (−1)k . For instance take (1 2), (2 3) in S3. Then the corresponding matrices are: 0 1 0 1 0 0 P(1 2) = 1 0 0 and P(2 3) = 0 0 1 . Note that P(1 2) P(2 3) = 0 0 1 0 1 0 0 0 1 1 0 0 = P(2 3 1) = P(1 2)(2 3) . This is a special case of the next result. 0 1 0 Lemma 6.18. If σ, τ ∈ Sn then Pσ Pτ = Pστ . Proof. (Sketch) Consider the matrices Pσ , Pτ which are to be multiplied. The 1 in the first row of Pτ is in the τ −1 (1)-th column (that’s where the first column of In was moved to). The 1 in the first column of Pσ is in the σ(1)-th row (the σ(1)-th column is the one that was moved into the 1-st position). When we multiply to obtain Pσ Pτ the 1 in the first column of the product is going to be obtained from this product, of the σ(1)-th row of Pσ with the τ −1 (1)-th column of Pτ , so we end up with a 1 in the (σ(1), τ −1 (1)) position. Similarly for all 29 i = 1, . . . , n. So the non-zero entries (all 1) of Pσ Pτ are in the (σ(i), τ −1 (i)) positions. Hence the permutation, ρ say, represented by Pσ Pτ puts the σ(i)-th column in the τ −1 (i)-th place. That is (think about it) ρ takes τ −1 (i) to σ(i). Therefore (again, think about it) ρτ −1 = σ. Therefore (multiply by τ on the right) ρ = στ . So Pστ does represent the permutation στ . Corollary 6.19. If σ, τ ∈ Sn then sgn(στ ) = sgn(σ)sgn(τ ). Proof. sgn(στ ) = det(Pστ ) = det(Pσ Pτ ) (by the lemma above) = det(Pσ )det(Pτ ) = sgn(σ)sgn(τ ). Corollary 6.20. For any permutation σ, sgn(σ) = sgn(σ −1 ). Lemma 6.21. Any cycle of length t can be written as the product of t − 1 transpositions. (A proof will be given but you might like to try it for yourself first.) Corollary 6.22. If σ is a cycle of length t then sgn(σ) = (−1)t−1 . So, if a permutation is written as a product of disjoint cycles then its sign can be calculated immediately. A permutation is said to be even if its sign is +1 and is odd if its sign is −1. So, for instance, the identity permutation is even, every transposition is odd, any product of two transpositions is even, every 3-cycle is even, in general a cycle of even length is odd and vice versa. Let An denote the set of even permutations in Sn : An = {σ ∈ Sn : sgn(σ) = 1}. This is the alternating group (on n symbols). Proposition 6.23. An is a subgroup of Sn and is of index 2 in Sn . Proof. The identity permutation is even, the inverse of any even permutation is even (by 6.20) and a product of even permutations is even (by 6.19) so An is a subgroup. If σ, τ are in Sn but not in An then both are odd, so the product σ −1 τ is even, that is σ −1 τ ∈ An , hence the cosets σAn and τ An are equal. Thus all elements of Sn \ An form a single coset and so there are just two cosets of An in Sn (namely the set, An of even permutations and the set, Sn \ An , of odd permutations). Example 6.24. (2) An denotes the subgroup of Sn consisting of the even permutations (recall the division of the elements of Sn into the even and odd permutations and, as an exercise, prove that An is a subgroup). The sign of a permutation, even and odd permutations will be discussed on a separate handout. Exercise* 6.25. Draw up some of the group table for A4 . Exercise 6.26. Show that the group of symmetries of a tetrahedron is S4 . Show that the subgroup of those symmetries which are realisable as rotations (i.e. without going into a mirror 3-space) is isomorphic to A4 . (In each case, use reasoning rather than try to draw up the (rather large) group table.) 30 6.2 Group actions Let G be a group. A G-set is a set, X, together with an action of G on X, that is, a map α : G × X −→ X which satisfies α(g, α(h, x)) = α(gh, x) and α(e, x) = x where g, h ∈ G and x ∈ X. Normally we write g · x or just gx for α(g, x) (generalising the notation for the operation in a group), so then the conditions become: g(hx) = (gh)x and ex = x (perhaps more readily comprehensible). Thus, for every element g ∈ G we have a function x 7→ gx from X to X (which can be though of as the action of g on X) such that composition of functions corresponds to multiplication in G and the identity element of G corresponds to the identity function on X. Example 6.27. G acting on itself: the group multiplication in G is a function G × G −→ G satisfying the condition for being a G-action. Exercise* 6.28. Show that, in the action of G on itself described in Example 6.27, the orbit of each element is the whole group and the stabiliser of each element is the trivial subgroup. Example 6.29. If G is a group of “symmetries” of a set or object X (e.g. the group of symmetries of a geometric object or the group of permutations of a set) then the natural action of G on this set or object is a G-action (in the case of symmetries of a polygon or polyhedron, think of the action on, say, the set of vertices to get an action on an actual set X). Suppose that G is a group and X is a G-set (so that means a set with a specified G-action). Let x ∈ X. The orbit of x is {gx : g ∈ G}. This is written Gx. If Gx = {x} then x is a fixed point of the G-action. The stabiliser of x, stab(x), is {g ∈ G : gx = x}. So x is a fixed point iff Stab(x) = G. Exercise** 6.30. Compute some orbits and stabilisers in geometric examples and also for the action of the symmetric group Sn on {1, . . . , n}. For example, compute the stabiliser, in the group of symmetries of a cube, of a vertex of the cube. Also consider the action of this group of symmetries on the set of edges of the cube and compute the stabiliser of an edge. Compute the orbit, under the group of symmetries, of each vertex of a square-based regular (as far as it can be) pyramid. Compute the stabiliser of 4 under the natural action of S4 on {1, 2, 3, 4}. Compute the stabiliser of the set {1, 2} under the action of S5 on two-element subsets of {1, . . . , 5} which is induced by the action of S5 on {1, . . . , 5}. Theorem 6.31. (Orbit-Stabiliser Theorem) Suppose that G is a finite group and that X is a G-set. Let x ∈ X. Then its stabiliser, Stab(x), is a subgroup of G and the index of Stab(x) in G is equal to the number of elements in the orbit of x. Hence |G| = |Gx| |Stab(x)|. Exercise** 6.32. Check the statement of the Orbit-Stabiliser Theorem in the examples you looked at in Exercise 6.30. If G is a group and g ∈ G then conjugation by g is the map, (−)g , from G to itself defined by a 7→ g −1 ag, the notation ag is used as a slight abbreviation for g −1 ag. This map will be the identity map from G to G iff g belongs to the centre, ZG = {b ∈ G : ba = ab for all b ∈ G}, of G (exercise). The centraliser, of a ∈ G is defined to be C(a) = {g ∈ G : ag = ga} = {g ∈ G : ag = a} 31 Lemma 6.33. If G is a group and g ∈ G then conjugation by g is an automorphism of G (that is, an isomorphism from G to G). The action of G on G given by α(g, a) = gag −1 is a G-action. The stabiliser of a ∈ G under this action is its centraliser C(a). The orbit of a under this action, i.e. the set of elements of the form g −1 ag as g varies over G, is the set of conjugates of a. If H, H 0 are subgroups of a group G then we say that H and H 0 are conjugate if there is an element g ∈ G such that g −1 Hg = H 0 , where g −1 Hg = {g −1 hg : h ∈ H}. Note that, if this is the case, then the map h 7→ g −1 hg is an isomorphism from H to H 0 . Exercise 6.34. Show that being conjugate is an equivalence relation on the set of subgroups of G. Proposition 6.35. Suppose that X is a G-set. Let x, y ∈ X lie in the same orbit of G, say gx = y. Then g −1 Stab(y)g = Stab(x), so the map, θg , from Stab(y) to Stab(x) which takes h ∈ Stab(y) to g −1 hg is an isomorphism Stab(y) ' Stab(x). In particular Stab(x) and Stab(y) are conjugate subgroups of G. (This applies, for example if G is the group of symmetries of a geometric figure X and if x and y are two vertices of X which lie in the same orbit of G: their stabilisers will be conjugate subgroups of G.) Example 6.36. (6) If (−) is any mathematical object then an automorphism of (−) is a structure-preserving bijection from (−) to itself. The set, Aut(−) of automorphisms of (−) forms a group under composition (the composition of two automorphisms is an automorphism, the inverse of an automorphism is an automorphism, the identity map is an automorphism). Below are some specific examples. Example 6.37. Let K be a field and let p ∈ K[X] be an irreducible polynomial. Let L = K[α] be the minimal field containing a root of p, as constructed in 5.3. A K-automorphism of L is an automorphism, θ, of the field L which fixes each element of K. It is easy (exercise) to show that θ(α) must be a root of p and that θ is completely determined by the value θ(α). It is somewhat less easy (a more extended exercise) to show that, if β ∈ L is any root of p then mapping α to β and fixing K pointwise determines a K-automorphism of L. Thus the K-automorphisms of L are in bijection with the roots of p which lie in L. Earlier examples, 2.20 and 5.9, show that there may be just one or more than one such, depending on the polynomial chosen. Example 6.38. Let K be a field and consider the polynomial ring R = K[X1 , . . . , Xn ]. There is a natural action of the symmetric group Sn on R: if σ ∈ Sn and p(X1 , . . . , Xn ) ∈ R then σp, more usually written pσ , is the polynomial p(Xσ(1) , . . . , Xσ(n) ) obtained by permuting the variables according to σ. The set of fixed points is usually denoted K[X1 , . . . , Xn ]Sn and is the set of all symmetric polynomials (those polynomials unchanged by every permutation of the variables) Exercise** 6.39. Consider the natural action of S3 on K[X1 , X2 , X3 ]. Compute the orbit and stabiliser of each of the following polynomials: X1 , X1 + X2 , X1 + X2 + X3 , X1 X32 + X3 X12 , X1 X22 + X2 X32 + X3 X12 , X1 + X22 + X33 . 32 6.3 More exercises Exercise 6.40. Let R = M2 (Z) be the ring of 2 × 2 matrices with entries from the ring Z of integers. Determine the condition on a matrix to be in the group of units of this ring and give some examples of units. Exercise* 6.41. Let R = M2 (Z4 ) be the ring of 2 × 2 matrices with entries from the ring Z4 of integers modulo 4. Find some elements of the group of units of this ring. Is this group U (R) of units of R abelian? What is a criterion for a matrix in M2 (Z8 ) to be invertible? Exercise 6.42. Let R = M2 (F4 ) be the ring of 2 × 2 matrices with entries from the field F4 with 4 elements. Find some elements of the group of units of this ring. How many elements are in the group U (R) of units of R? How many of these have determinant 1? Exercise* 6.43. Identify the groups of symmetries of the following geometric figures: (i) a triangle; (ii) a square; (iii) a rectangle which is not a square; (iv) a hexagon; (v) a tetrahedron; (vi) a cube. Why is the group of symmetries of a cube the same as that of a regular octahedron? (Or, for that matter, why is the group of symmetries of a dodecahedron the same as that of an icosahedron?) Exercise 6.44. Let X be a square-based pyramid with all sides of equal length. Find the group G of symmetries of X and compute the stabiliser of each vertex. 7 Homomorphisms of Groups, Normal Subgroups and Factor Groups The idea of a homomorphism is, for groups as for rings, that of a structurepreserving map. A homomorphism from the group G to the group H is a map θ : G −→ H such that θ(ab) = θ(a)θ(b) for all a, b ∈ G. We have a lemma completely analogous to 2.9. Lemma 7.1. Suppose that θ : G −→ H is a homomorphism between groups. Then: (1) θ(eG ) = eH ; (2) θ(a−1 ) = (θ(a))−1 for every a ∈ G; (3) o(θ(a)) ≤ o(a) for every a ∈ G; (4) the image of θ, im(θ) = {θ(a) : a ∈ G} is a subgroup of H. The kernel of a homomorphism θ : G −→ H of groups is the set of elements which map to the identity of H: ker(θ) = {a ∈ G : θ(a) = e}. A subgroup N of G is said to be normal if it is stable as a set under conjugation, that is, if a−1 N a = N for every a ∈ G (this is not to say that every element of N is fixed under conjugation - the elements of N might well be permuted around, but they won’t be sent outside N ). We write N C G to indicate that N is a normal subgroup of G. The following lemma slightly simplifies checking that a subgroup is normal. Lemma 7.2. Suppose that N is a subgroup of the group G. If a−1 N a ⊆ N for every a ∈ G then N is a normal subgroup of G. 33 Proof. Let a ∈ G; we have to show that N ⊆ a−1 N a. So let n ∈ N . By assumption (a−1 )−1 na−1 = n0 for some n0 ∈ N , that is, ana−1 = n0 , so an = n0 a, hence n = a−1 n0 a ∈ a−1 N a, as required. Note that if G is finite then the proof is even easier: because conjugation by a is a bijection, the number of elements in a−1 N a is the same as the number of elements in N so, if one is a subset of the other, then they must be equal. The “manipulating sets” argument is also short (and valid): if a−1 N a ⊆ N then multiply on the left by a and on the right by a−1 to get aa−1 N aa−1 ⊆ aN a−1 , that is, N ⊆ aN a−1 which, noting that it’s for all a and so we can replace a by a−1 , gives what we want. Lemma 7.3. The following conditions are equivalent for a subgroup N of the group G. (i) N is a normal subgroup of G; (ii) aN = N a for every a ∈ G; (ii) every right coset of N in G is also a left coset of N in G. Proposition 7.4. Suppose that θ : G −→ H is a homomorphism of groups. Then ker(θ) is a normal subgroup of G. In Exercises 7.5 and 7.6 you are asked to prove that the centre of G is a normal subgroup and also to prove that any subgroup of index 2 must be normal. Exercise** 7.5. Show that the centre of G is a normal subgroup. Exercise** 7.6. Prove that any subgroup of index 2 is normal. Thus normal subgroups of groups are analogous to ideals of rings. And the analogy does not stop there. Let N be a normal subgroup of the group G. Consider the set, written G/N, of cosets (right or left, it doesn’t matter by the lemma above) of N in G. We make this set into a group by defining the product of two cosets: aN · bN = (ab)N. As usual, because the definition uses particular representatives, we have to show that this is well-defined, but that is straightforward enough. We refer to G/N as the factor group of G by N (or as the quotient of G by N ). See Exercise 7.7 for an example of what goes wrong if you try to do this construction with a non-normal subgroup. Exercise* 7.7. If H is a subgroup of G which is not normal then trying to define a product as above won’t work. For instance, let G = S3 and let H be the (cyclic) subgroup generated by the transposition (1 2). Compute the left and right cosets of H in G to see that H is not normal in G. Show that the product of the cosets (1 3)H and (2 3)H is not well-defined (note that (1 3)H = (1 3 2)H). Theorem 7.8. Let N be a normal subgroup of the group G. Defining a product on the set, G/N, of cosets of N in G by (aN )(bN ) = (ab)N gives a well-defined structure of a group to G/N. There is a canonical projection π : G −→ G/N defined by π(a) = aN and this is a surjective homomorphism of groups, with kernel N. Any group homomorphism θ : G −→ H with ker(θ) ≥ N factors uniquely through π. 34 Examples 7.9. (1) Consider the map from Sn to the group {−1, +1} (with multiplication as the operation) given by assigning to a permutation its sign. This is a homomorphism with kernel the alternating group, An . (2) If R is any ring and I is an ideal of R then the canonical surjection onto the factor ring R → R/I is, forgetting the multiplication, a homomorphism of additive groups, with kernel I. For instance, consider Z6 → Z6 /([0]6 , [2]6 , [4]6 ) ' Z2 . As for rings there is a correspondence between (normal) subgroups of the quotient G/N and (normal) subgroups of G which contain N . Theorem 7.10. Suppose that N is a normal subgroup of the group G. Then there is a bijection between subgroups H of G which contain N and subgroups K of the quotient group G/N given by: H 7→ H/N = π(H); K 7→ π −1 K = {a ∈ G : π(a) ∈ K}. Furthermore H is normal in G iff H/N is normal in G/N, in which case the quotient (G/N )/(H/N ) is isomorphic to G/H. If H1 and H2 are subgroups of a group G then we define H1 H2 to be the subgroup of G generated by H1 and H2 : this is the smallest subgroup containing H1 and H2 and is more explicitly defined as the set of all products of the form h11 h21 h12 h22 . . . h1k h2k for some k, where hij ∈ Hi . That is (think about it), form all possible products of elements from the set H1 ∪ H2 . If one of these groups is normal then this becomes much simpler to describe. Exercises** 7.11. (1) Let G = S4 , H1 = h(1 2)i, H2 = h(2 3)i, H3 = h(3 4)i. Compute H1 H2 and H1 H3 . (2) Find the subgroup of D6 (the group of symmetries of a regular hexagon) generated by H1 = hρi where ρ is rotation about the centre by 2π/3 and H2 = hσi where σ is one of the reflections. (3) (additive notation) Find the subgroup of Z generated by 6 and 4 (i.e. by the cyclic subgroups h6i and h4i). Lemma 7.12. Suppose that H is a subgroup of the group G and that N is a normal subgroup. Then HN = {hn : h ∈ H, n ∈ N } = {nh : n ∈ N, h ∈ H}. Exercise* 7.13. In Exercise 7.11(2) above show that H1 (the subgroup generated by rotation through 2π/3 is normal in D6 and verify the statement of Lemma 7.12 in this case, that is, show that H1 H2 can be described as in the Lemma. Proposition 7.14. Suppose that H is a subgroup of the group G and that N is a normal subgroup. Then N is a normal subgroup of HN , H ∩ N is a normal subgroup of H and HN/N ' H/(H ∩ N ). 7.1 More exercises Exercise* 7.15. Find all the group homomorphisms from Z12 to itself. Which of these are automorphisms? Exercise 7.16. Show that if G is a group T and if {Ni }i∈I is any set of normal subgroups of G then their intersection, i Ni , is a normal subgroup of G. Deduce that if A is any subset of G then there is a smallest normal subgroup of G which contains A (this is referred to as the normal subgroup of G generated by A). 35 Show that the normal subgroup generated by A is the set of all elements which are products of elements of the form g −1 a g with g ∈ G, a ∈ A and = ±1 (allowing the identity element in as being the “empty product”). Exercise 7.17. Suppose that N is a subgroup of the group G and that N is abelian. Suppose also that N 0 is a subgroup of G which is conjugate to N . Prove that N 0 is abelian. Exercise 7.18. (i) Let α be an automorphism of the group G. Suppose that N is a normal subgroup of G and let N 0 = α(N ). Prove that N 0 is a normal subgroup of G. (ii) Suppose now only that N and N 0 are subgroups of G with N normal in G and N 0 isomorphic to N . Show, by giving an appropriate example, that N 0 need not be normal in G. Exercise** 7.19. Let H be a subgroup of the group G. Show that the set NG (H) = {g ∈ G : g −1 Hg = H} is a subgroup of G, the normaliser of H in G. Show that NG (H) contains H and that H is a normal subgroup of NG (H). Show that NG (H) is the largest subgroup of G in which H is normal (in the sense that if H1 is a subgroup of G which strictly contains NG (H) then H is not normal in H1 ). Exercise** 7.20. Let G be the group S(3) and let H be the subgroup generated by the transposition (1 2). Find the normaliser of H in G. Find the normal subgroup of G generated by H. Repeat the exercise using the subgroup, H 0 , generated by the 3-cycle (1 2 3). 36