Download Undergraduate algebra

Basic Algebra
Math 30710
Moshe Kamensky
Department of Math, University of Notre-Dame
E-mail address: mailto:[email protected]
URL: http://mkamensky.notlong.com
Contents
Chapter 1. Group Theory
1. Symmetry
2. Sets and functions
3. Definition of a group
4. First properties of groups
5. Subgroups
6. Cyclic groups
7. Homomorphisms
8. The classification of cyclic groups
9. The symmetric groups
10. Group actions
11. Normal subgroups and quotients
5
5
6
8
10
11
15
15
19
21
28
31
Chapter 2. Galois theory
1. Statement of the problem, and its solution
2. First properties of fields
3. Polynomials in one variable
4. Linear algebra
5. Finite extensions
6. The Galois correspondence
7. Solvability of equations
41
41
44
47
50
52
54
60
Bibliography
65
Index
67
3
CHAPTER 1
Group Theory
This chapter presents some basic results on group theory. Good references with
more details include Milne [2] and Rotman [4].
1. Symmetry
As a motivation for the notion of a group, we examine the idea of symmetry.
Consider the figures 1–4.
Figure 2. square
Figure 1. circle
Figure 3. rectangle
Figure 4. penguin
Intuitively, the square is more symmetric than the rectangle, the circle is much
more symmetric than them both, and the penguin has no symmetries at all. How
can we formulate these observations precisely? We shall decide what do we mean
by a ‘symmetry’, and then show that, e.g., the circle has more of them than the
square.
So, what is a symmetry? Different definitions can be given, depending on the
context. We will (loosely) define a symmetry of a shape to be a transformation of
the shape into itself that preserves distances and (unoriented) angles (so that the
distance between any two points remains the same after the transformation.)
For example, rotating the square by 90◦ is such a symmetry. It brings the
square as a whole into itself. So is the reflection along any of the diagonals. Every
shape has at least one symmetry: this is the transformation that doesn’t move the
shape at all! It is called the identity transformation. This is the only symmetry of
the penguin. But, as noted above, the square has some others. We note that we are
only interested in the “final outcome” of the transformation: the transformation
that consists of rotating the shape to the right, and then by the same amount to
the left is still the identity.
5
6
1. GROUP THEORY
What are the symmetries of the rectangle? In any symmetry, a short side has
to go to a short side, and a long side to a long side. Once a decision is made
which side goes where, there is precisely one symmetry that does it (these facts are
clear intuitively, and are easy to prove if the precise definitions are given.) Hence
there are precisely four symmetries: a horizontal reflection, a vertical reflection, a
combination of them, and the identity.
What about the square? Since a square is a rectangle, any symmetry of the
rectangle is also a symmetry of the square. But there are more: the right rotation
by 90 and 270 degrees are also symmetries, as well as reflections by the two diagonals (what about the rotation by 180◦ ?) It can be checked that these are all the
symmetries of the square.
Finally, what can we say about the circle? The circle has infinitely many
symmetries: any rotation, and any reflection along any diameter is a symmetry.
It seems that we have solved our problem: to any shape, we have attached a
number, the number of symmetries (which may be infinity), that tells us everything
about the symmetry of the shape. One shape is more symmetric than another if
the number of symmetries is of the first is bigger.
The following example shows that the situation is, in fact, more complicated.
Assume now that interior of the square in figure 2 is coloured on one side in green.
In other words, a reflection is no longer a symmetry. We are thus left with four
symmetries, the four rotations (including the identity.) So now the new square and
the rectangle have the same number of symmetries, but they are clearly symmetric
in different ways. How can one capture this difference?
Given two symmetries of some shape, we may transform the shape by the first
one, and then apply the second one to the result. The operation obtained in this
way is again a symmetry. For example, if we rotate the (original) square, and then
flip along the diagonal, this is again a symmetry. Since we have listed all of the
symmetries of the square, this should be one of the symmetries in the list (which
one?) In general, the process of applying one symmetry after another defines an
operation between the symmetries of the shape: for any two symmetries f, g, we
get a new symmetry f · g. The set of all symmetries of a given shape, together with
the composition operation described above is an example of a group.
Using the operation we can distinguish between the cases of the coloured square
and the rectangle: if f is any symmetry of the rectangle, then f · f is the identity.
This is not the case with the coloured square: applying a rotation by 90◦ twice
will not give the identity. A basic feature of this argument is that, after defining
the operation ·, it didn’t use the geometry at all! It is formulated in terms of the
algebraic properties of the operation. This is the kind of arguments that one applies
when studying abstract group.
We will define a group abstractly as a set with an operation, satisfying certain
properties. Though we will not forget the geometric examples, it should be stressed
that the theory happens on the abstract level of the group operation. In fact, it
is often used that the same group can be the group of “symmetries” of completely
different objects.
2. Sets and functions
Before defining what a group is, we should understand sets. We shall not
define what a set is, but rather assume it to be known. Intuitively, a set is simply
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
2. SETS AND FUNCTIONS
7
a collection of elements. The main property of a set is that it is determined by its
elements; Two sets A and B are equal if and only if any element of A is an element
of B, and vice versa. The statement that a is an element of A is written as a ∈ A.
A set A is a subset of B (written A ⊆ B) if every element of A is also an element
of B. Thus A = B if and only if A ⊆ B and B ⊆ A. The set whose elements are
precisely a, b, c, . . . is denoted by {a, b, c, . . . }.
A set can be finite or infinite. We will denote by |A| the number of elements in
A (for finite sets.) There is a unique set with no elements, the empty set, denoted
by ∅.
If A and B are two sets, the Cartesian product of A and B, denoted A × B,
is the set of all pairs of elements (a, b), where a ∈ A and b ∈ B. Here, a pair is
an ordered set of two elements: if a ̸= b then (a, b) and (b, a) are different pairs (in
contrast, {a, b} = {b, a}.)
Example 1. Let A and B be finite sets. Show that |A × B| = |A| · |B|
□
Note that, though A × B and B × A are different sets, there is an obvious way
to identify them. Likewise, if A, B and C are three sets, the sets (A × B) × C,
A × (B × C) and A × B × C (the last is the set of all triples) are formally distinct,
but can be identified in an obvious way. We will therefore not distinguish between
them.
2.1. Functions. A function f from a set A to another set B is a rule that
assigns to every element a of A, a unique element f (a) ∈ B. The set A is called the
domain of f and B is called the range. The subset of B consisting of all elements
of the form f (a) for some a ∈ A is called the image of f . We may view f as a
machine, that takes elements of A and produces elements of B. The fact that f is
a function from A to B is written as f : A →
− B. If A is any set, there is a function
IdA : A −
→ A defined by IdA (a) = a for all a ∈ A. It is called the identity function.
If f : A −
→ B and g : B −
→ C are functions, we may compose them to get a new
function g ◦ f : A −
→ C, defined by (g ◦ f )(a) = g(f (a)). Thus we apply f to a, and
apply g to the result. Note that f ◦ IdA = IdB ◦ f = f .
A function f : A −
→ B is injective (or one to one) if for any distinct a1 , a2 ∈ A,
f (a1 ) and f (a2 ) are also distinct. It is surjective (or onto) if any element b ∈ B is
of the form f (a) for some a ∈ A. It is bijective if it is injective and surjective. A
function g : B −
→ A is a left inverse of f if g ◦ f = IdA and it is a right inverse if
f ◦ g = IdB . The function f is invertible if it has both a left and a right inverse.
Example 2. If A is non-empty, a function from A is injective if and only if it
has a left inverse. If f : A −
→ B has a right inverse, then it is surjective. A function
is bijective if and only if it has a right and left inverse. In this case, the right and
left inverse coincide, and is unique, and is called simply the inverse of f .
□
Example 3. If f : A →
− B is a function between finite sets, and |A| = |B|, then
□
f is injective if and only if it is surjective. In particular, this holds if A = B.
Example 4. In analysis, one studies (particular kinds of) functions between
“nice” subsets of the set R of real numbers. These are generally functions that can
be pictured as curves in the plane. For example, a continuous function is injective
if and only if it is monotone. The inverse of a function (if it exists) is obtained by
reflecting along the diagonal.
□
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
8
1. GROUP THEORY
Example 5. In linear algebra, one studies functions between vector spaces,
which are linear. A basic theorem says that such a linear function from kn to km
is given by an n × m matrix, and that composition of functions is given by matrix
multiplication. If n = m, such a function is injective if and only if it is surjective if
and only if its determinant is non-zero.
□
End lecture 1, Jan 19
Example 6. The function f : N −
→ N from the set of natural numbers to itself
given by f (n) = 2n is injective, but not surjective. The function g : N −
→ N defined
by “g(n) = n/2 if n is even, and g(n) = (n + 1)/2 if n is odd” is a left inverse, but
not a right inverse of f . Accordingly, g is surjective, but not injective. Note that f
has other left inverses.
Any natural number n can be presented uniquely as pn1 1 pn2 2 . . . pnk k , where pi
is the i-th prime, and nk ̸= 0. The rule n 7→ (n1 , . . . , nk ) defines a function from
N to the set of all finite sequences of non-negative integers. This function is a
bijection!
□
3. Definition of a group
A binary operation ∗ on a set G is simply a function ∗ : G × G −
→ G. Thus,
it assigns to any pair (g, h) of elements of G, another element, which we denote by
g ∗ h. For instance, G may be a set of symmetries, as in section 1, and the operation
will be that of applying one symmetry after another. The definition of a group will
try to reflect the formal properties of that situation.
Definition 7. A Group (G, ∗) is a set G, together with a binary operation ∗
on it, satisfying the following conditions:
G1 (associativity): For any three elements a, b, c ∈ G, (a∗b)∗c = a∗(b∗c)
G2 (unit): There is an element e ∈ G such that for any a ∈ G, e ∗ a =
a∗e=a
G3 (inverses): For any element a ∈ G, there is an element b ∈ G such that
a∗b=b∗a=e
Remark 8.
(1) Thus, the data of a group consists of a set G and an operation ∗ on it —
knowing the set is not enough. Nevertheless, when the operation is clear
from the context, we will usually use just G to refer to the group.
(2) We will usually write simply ab instead of a ∗ b, and use multiplicative
terminology for the operation. However, it should be stressed that in
general, the elements of G are not numbers, and the operation has nothing
to do with multiplication of numbers.
(3) The associativity axiom implies that it makes sense to write a ∗ b ∗ c
without parentheses. It is easy to prove that, more generally, for any
elements a1 , . . . , an of G, the expression a1 . . . an makes sense (i.e., we may
interpret it by putting the parentheses wherever we like.) In particular, if
a is any element, and n is a natural number we write an for the product
of a with itself n times (thus, a1 = a.)
(4) An element e as in axiom G2 is called a unit or an identity. Axiom G3
refers to this element, and thus only makes sense if we know that such an
element is unique. This is indeed the case, as is proved in proposition 19.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
3. DEFINITION OF A GROUP
9
Likewise, we will show in proposition 23 that for any element a, the element b promised by axiom G3 is unique. The inverse of a will be denoted
n
a−1 , and we abbreviate a−1 by a−n . Finally, we set a0 = e.
Example 9. The sets Z, Q, R and C of integer, rational, real and complex
numbers, with addition as an operation are groups (with identity element 0.) The
set N of natural numbers (with addition) is not a group: there are no inverses (and
possibly, depending on the definition, no unit.)
The same sets with multiplication as the operation are not groups: the number
1 serves as the unit, but 0 has no inverse. If we remove 0, then all sets except for
the integers become groups.
The sets of positive rational or real numbers is also a group under multiplication. Likewise, the group of complex numbers of length 1.
Any vector space, with the addition operation, is also a group.
□
Example 10. The set of integer under subtraction is not a group — subtraction
is not associative.
□
Example 11. A finite group can be given by a multiplication table. For example, let G = {e, a, b, c}, and define the operation by the following table:
e
a
b
c
e a
e a
a b
b c
c e
b c
b c
c e
e a
a b
Note that it is not obvious at all that this table defines the operation of a group! □
Example 12. There is exactly one group of size 1. It is called the trivial
group.
□
The set of symmetries of “anything” is a group under composition. The following examples are all of this kind:
Example 13. If X is any set, the set Sym(X) of bijective functions from X to
itself forms a group, with composition of functions as an operation, forms a group.
This is because composition of functions is associative, and any bijective function
has a unique inverse. Note that in this group there are elements f and g such that
f g and gf are different (in other words, Sym(X) is not Abelian.) This group is
called the symmetric group on X. When X = {1, . . . , n} it is also denoted by Sn .
Since X is a set with no further structure, any bijective function from X to
itself can be viewed as a “symmetry”. In this sense Sym(X) is the set of symmetries
of X.
□
Example 14. If V is any vector space, the set GL(V ) of invertible linear maps
from V to itself is a group under composition. It is called the general linear group
of V . If V = kn , this set can be identified with the set of invertible n × n matrices
over k. The composition corresponds to multiplication of matrices, and the set of
n × n invertible matrices with multiplication thus forms a group GL(n, k). GL(V )
can be viewed as the group of symmetries of the vector space V , if all we can see
is the vector space structure. Recall that a matrix A maps the unit cube into a
parallelepiped of volume ∥det(A)∥. Thus, if we are observant enough to measure
volumes (and orientations), we will only consider A to be a symmetry if det(A) = 1.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
10
End lecture 2, Jan 21
1. GROUP THEORY
The set of all such matrices indeed forms a group (under multiplication), called the
special linear group, denoted SL(n, k). More generally, we may defined SL(V ) for
any finite dimensional vector space V .
Likewise, we may consider the group O(n) of (real) invertible matrices that
preserve lengths and angles. This is again a group under multiplication, and it is
the symmetry group if you can see the lengths and angles. In fact, O(2) is precisely
the group of symmetries of the circle that we considered in section 1!
□
Example 15. If X is an interval of real numbers, the set of continuous bijective
maps from X to itself is a group. In contrast, the set of continuously differentiable
maps from X to itself is not: for example if X = (−1, 1), the function f (x) = x3
is a continuously differentiable bijection, but its inverse is not differentiable at 0!
Nevertheless, the set of continuously differentiable (or smooth) bijections from X
to itself whose inverse is of the same kind is a group. It is the group of symmetries
of the “smooth structure” on X.
□
Example 16. The set of symmetries of any subset of the plane, as we considered
in section 1 is a group under composition. Here we defined symmetry to mean a
transformation of the set into itself that preserves lengths and angles.
We already identified O(2) as the group of symmetries of the circle, and the
penguin has the trivial group as a group of symmetries. The other two groups are
finite, and their multiplication can be described explicitly: the non-trivial symmetries of the rectangle are the horizontal flip H, the vertical flip V , and the rotation
in 180◦ R. The operation is given by H 2 = V 2 = R2 = e, HV = V H = R,
HR = RH = V and V R = RV = H. The group of symmetries of the square is
described in Milne [2, sec. 1.17].
□
Example 17. More generally, for any n > 2, there is a regular n-gon in the
plane. The group of all symmetries of this n-gon is called the (n-th) dihedral group,
and is denoted by Dn . See Milne [2, sec. 1.17] for further discussion.
□
3.1. Modular arithmetic. To give further examples of groups, we define
two new operations. These operations are defined on the set Zn of residues mod
n, where n is a natural number bigger than 1. The set Zn consists of the numbers
0, . . . , n − 1. For x, y ∈ Zn , we define x ⊕ y to be the remainder (residue) of x + y
when divided by n. In other words, it is the unique element z of Zn such that
x + y − z is divisible by n. Likewise, we define x ⊙ y to be the remainder of xy when
divided by n.
Example 18. The set Zn with the operation ⊕ of addition mod n forms a
group (also denoted by Zn ). The same set with ⊙ does not: 0 is not invertible.
After 0 is removed, we get a group if and only if n is prime. More generally, the
□
subset Un of elements of Zn prime to n is a group under ⊙.
4. First properties of groups
The actual theory of groups takes place in the abstract setting of definition 7,
rather than in any specific example. We will now see what can be deduced abstractly
from the definition.
Proposition 19. A group has exactly one identity element.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
5. SUBGROUPS
11
Proof. Assume that e and f are two identity elements of a group. Since e is
an identity, ef = f . Since f is an identity, ef = e. Hence f = e.
□
Note that in this proof, we only used axiom G2 in the definition of a group.
Example 20. In the proof, we used that a unit is two sided: if e is a unit,
both ex = x and xe = x for any element x. This is essential: for example, if X is
any set with more than one element, let G be the set of constant functions from X
to itself (i.e., for any function f in G there is an element c ∈ X with f (x) = c for
all x ∈ X), with composition as an operation. Then any element g of G satisfies
f g = f for all f ∈ G (but, of course, not gf = f !)
□
End lecture 3, Jan 24
Before proving that inverses are unique, it is convenient to derive the cancellation law:
Proposition 21. If a, b, c are elements of a group such that ab = ac, then
b = c. Likewise, if ba = ca, then b = c.
Proof. Assume that ab = ac, and let d be an inverse of a, as promised by
axiom G3. Then b = eb = (da)b = d(ab) = d(ac) = (da)c = ec = c. The proof for
the other case is similar.
□
Corollary 22. The unit is the only element a in a group satisfying a2 = a.
Proof. If a2 = a = ae, then cancelling a we get a = e.
□
Corollary 23. Any element in a group has a unique inverse.
Proof. If b and c are both inverses of a, then ab = e = ac. Cancelling a we
get b = c.
□
Example 24. Let G = M at2 (k) be the set of all 2 × 2 matrices, with the
operation of product of matrices. Then G is associative and has an identity element.
However, there are many elements A ∈ G with A2 = A. In particular, G is not a
group (of course, this can be easily seen directly!)
□
As we saw, a group can be finite or infinite. If G is finite, the order of G is the
number of elements in G. In the finite case, proposition 21 has a converse:
Proposition 25. If (G, ∗) is a finite set with an associative operation ∗ and
an identity, such that the cancellation laws hold, then G is a group.
Proof. We need to show that any element x ∈ G has an inverse. Consider the
map lx : G −
→ G given by lx (y) = xy. We need to show that lx (y) = 1 for some y,
so it is enough to show that lx is surjective. Since G is finite, it is enough to show
□
that lx is injective. But this is precisely the left cancellation law.
Example 26. The set N of natural numbers (including 0) with addition satisfies
the conditions of proposition 25, except for the finiteness, but is not a group. □
5. Subgroups
Definition 27. Let (G, ∗) be a group. A subgroup of G is a subset H of G,
such that ∗ restricts to an operation (also denoted ∗) on H, making (H, ∗) a group.
Note that the data of a subgroup is just the subset H of G, the operation is
given by the operation on G.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture 4, Jan 24
12
1. GROUP THEORY
Example 28. The set 2Z of even integers is a subgroup of the group (Z, +) of
integers under addition. Indeed, it is a subset of Z, the sum of two even numbers
is even, and it is a group under addition.
On the other hand, the set of odd integers is not a subgroup.
□
Example 29. Any group has at least two subgroups: the group itself, and the
trivial group consisting of only the identity.
□
Example 30. The group (Q∗ , ·) of rational numbers under multiplication is
not a subgroup of (Q, +). Although it is a subset and a group, the operations do
not coincide. Likewise, (Un , ·) is not a subgroup of (Zn , +).
□
Example 31. The group SL(V ) of linear maps of determinant 1 from a finitedimensional vector space V to itself is a subgroup of the group GL(V ) of all invertible linear maps on V .
□
If H is a subset of a group G, the condition that the operation restricts to H
means that whenever a, b ∈ H, ab ∈ H as well. If this is the case, the associativity
axiom of a group will hold automatically. Furthermore, we have the following result.
Theorem 32. Let H be a non-empty subset of a group G. Then the following
are equivalent:
(1) H is a subgroup
(2) For every a, b ∈ H, ab−1 ∈ H
(3) H is closed under the operation and the inverse
In particular, if H is a subgroup, the unit and inverses in H and in G coincide.
Proof. We assume 1 and prove 3. Since H is closed under the operation by
definition, we need to show that if x ∈ H, then x−1 ∈ H. First, if e is the identity
of H, then e2 = e. By corollary 22, e is the unit of G. Since we are assuming that
H is a group, x has an inverse y in H, and by corollary 23, y = x−1 .
That 2 follows from 3 is trivial. We assume 2 and prove 1. Since H is nonempty, it has some element a ∈ H. By assumption, e = aa−1 ∈ H. Applying the
assumption again for e, a ∈ H, we get that also a−1 ∈ H.
□
Example 33. If G is the set of 2 × 2 matrices,
[ ] with matrix multiplication (not
a group), the subset H consisting of matrices 00 0c , where c ̸= 0, is[a “subgroup”
(a
]
subset closed under the operation, which is a group), whose unit 00 01 is different
from the unit in G.
□
Example 34. The subset Bn of GLn consisting of upper triangular matrices is
a subgroup. Indeed, it is non-empty, and closed under inverses and products. The
same holds for the subset of Bn consisting of matrices with 1 on the diagonal. □
Example 35. If X is an interval, the set of continuous bijective functions is a
subgroup of the group Sym(X) of bijections from X to itself. Indeed, the inverse
of a continuous function is continuous.
□
Example 36. The subset of “colour preserving” symmetries of a regular n-gon
is a subgroup of the group Dn of all symmetries: the inverse of a colour preserving
symmetry is again colour preserving.
□
For finite subgroups, the situation is simpler:
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
5. SUBGROUPS
13
Proposition 37. If H is a finite non-empty subset of a group G that is closed
under the operations, then H is a group.
Proof. Consider first H1 = H ∪ {e}. It is again closed under the operation,
and contains the identity. Furthermore, cancellation holds in H1 since it holds in
G. Hence, by proposition 25, H1 is a subgroup. Now, if a ∈ H, then a ∈ H1 hence
a−1 ∈ H1 so a−1 ∈ H. Hence e = aa−1 ∈ H, so H = H1 is a group.
□
5.1. Intersection of subgroups. Generators. Recall that if Ai is a family
of subsets of a set A, the intersection of the Ai is the subset of A consisting of all
elements that belong to all the Ai .
Theorem 38. If Hi is a family of subgroups of a group G, their intersection
H is again a subgroup
Proof. We use 32.2. Since all the Hi contain the identity, so does H, and in
particular, it is non-empty. If a, b ∈ H, then for any i, a, b ∈ Hi , hence, since Hi is
a subgroup, ab−1 ∈ Hi . Hence ab−1 ∈ H.
□
Example 39. The subsets Bn and SLn consisting of upper-triangular and volume preserving matrices, respectively, are subgroups (examples 34 and 31.) Hence,
so is their intersection, the set of upper triangular matrices of determinant 1. □
Example 40. Let X be an interval, and for any natural number i, let C i (X)
be the set of bijections from X to itself such that both it and its inverse have
continuous i-th derivative. Then each C i (X) is a subgroup of Sym(X), and their
intersection, the subset of smooth bijections with smooth inverse from X to itself,
is a subgroup.
□
Example 41. If H and K are subgroups of a group G, their union is not, in
general, a subgroup. For example, the product of a matrix of determinant 1 by an
upper triangular matrix will in general not be upper triangular or of determinant
1.
□
The theorem allows us to make the following definition.
Definition 42. Let A be a subset of a group G. The intersection of all subgroups of G containing A is called the subgroup generated by A. It is denoted by
⟨A⟩. If ⟨A⟩ = G, we say that G is generated by A, and that A is a set of generators
for G.
Example 43. If A is a subgroup of G, then ⟨A⟩ = A. Indeed, it is one of the
groups containing A, and any other such group contains it.
□
It helps to have a more concrete description of the subgroup generated by a
subset A.
Proposition 44. The subgroup generated by a subset A of a group G is the
set of all elements that can be presented as a finite product of elements of A and
their inverses (note: a product of 0 elements is e.)
Proof. Let H be the set of all such finite products. This is a subgroup of G,
and it contains A. Hence ⟨A⟩ < H. Conversely, every subgroup containing all the
□
elements of A will contain the finite products as well, so we have equality.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture 5, Jan 26
14
1. GROUP THEORY
Example 45. Let v1 , . . . , vn be a set spanning a vector space V over Q. Then
the set of vectors of the form vi /m, where m is an integer, generates V .
□
End lecture 6, Jan 28
Example 46. The group Dn is generated by two elements: a rotation r by
2π/n, and a reflection s. The order of s is 2, and the order of r is n. We can also
present the group via s and t = rs. Then s2 = t2 = e, and (st)n = e. The group D0
of symmetries of the circle is generated by the subgroup O of rotations, together
with one reflection.
□
A group that is generated by one element a is called cyclic. By proposition 44,
any element of such a group can be presented as an for some n ∈ Z (maybe in more
than one way.) Such groups will be discussed in more details in Section 6.
Example 47. The groups (Z, +) and (Zn , +) are cyclic.
□
If G is a group, and a ∈ G, then a generates a subgroup ⟨a⟩ of G, which is by
definition cyclic. The order of a is by definition the order (size) of ⟨a⟩.
Proposition 48.
(1) Any cyclic group is abelian
(2) If G is finite of order n, then it is cyclic if and only if it has an element
of order n.
(3) The order of an element a is the smallest positive n such that an = e.
Proof.
(1) If x, y ∈ G = ⟨a⟩, then for some n, m, x = an and y = am .
Then an am = an+m = am an .
(2) If G has an element a of order n, then ⟨a⟩ is a subgroup of G of order
n, which is thus equal to G. Conversely, if G is cyclic, the order of any
generator of G is n.
(3) Let n be a positive integer such that an = e. Then by proposition 37,
the subset {1, . . . , an−1 } is a subgroup. Hence, the order is at most the
minimal such n. In particular, if m is the order of a and 0 < n < m, then
an ̸= e. It follows that the elements ai , aj for 0 ≤ i ̸= j ≤ n are distinct,
and so am = e.
□
Example 49. The circle group O is the group of rotations of the plane around
a fixed point (equivalently, the group of coloured symmetries of the circle.) What
are the orders of its elements? An element of O is determined uniquely by an angle
of rotation 2πx, where x is a real number in the interval [0, 1). Let gx be the
element of corresponding to x. If x = p/q is rational, then gxq = e. Conversely,
if gxq = e, then x = p/q for some p, so x is rational. Hence the elements of finite
order correspond to rational numbers, and the order is the denominator (in reduced
form.)
□
We now consider another example of a subgroup obtained as an intersection.
If a is an element of a group G, the centraliser of a, denoted CG (a), is the subset
of elements x of G that commute with a: xa = ax. Then centre of G, Z(G), is
defined to be the set of elements that commute with all elements of G.
Proposition 50. For any element a, the subset CG (a) is a subgroup. The
centre Z(G) is an abelian subgroup.
Proof. If x, y commute with a, then xya = xay = axy, so xy also commutes
with a. Likewise, x−1 a = x−1 axx−1 = x−1 xax−1 = ax−1 . This proves that CG (a)
is a subgroup. Z(G) is the intersection of all CG (a).
□
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
7. HOMOMORPHISMS
15
Example 51. In any group G, a ∈ Z(G) if and only if CG (a) = G.
□
Example 52. The centre of GLn is the group of scalar matrices cI.
□
6. Cyclic groups
We now analyse in more details the structure of cyclic groups. We have seen in
the proof of Proposition 48 that if a is a generator of a cyclic group G that has finite
order n, then any element of G can be written uniquely as ai for some 0 ≤ i < n.
More generally we have:
Lemma 53. Let a be an element of a group G. Then ai = aj if and only if the
order of a divides i − j. In particular, if a has infinite order, then all the ai are
distinct.
Proof. By inverting ai , it is enough to consider the case i = 0. Let n be the
order of a. If n is finite and j = nm + r, where 0 ≤ r < n, then aj = (an )m ar =
em ar = ar . Thus we reduce to the previous case. Also, if aj = e for some j > 0,
then the order of a is finite, and this proves the infinite case.
□
Let a be a generator of a cyclic group G. If A is a subset of G, let p(A) = {i ∈
Z : ai ∈ A} (this depends on a.)
Lemma 54. The subset A is a subgroup if and only if p(A) is a subgroup of
(Z, +)
a
i+j
Proof. Assume A is a subgroup, and let i, j ∈ p(A). Then ai , aj ∈ A, so
= ai aj ∈ A and a−i ∈ A, hence i + j, −i ∈ A. The converse is similar.
□
We can now prove:
Theorem 55. Any subgroup of a cyclic group G is cyclic
Proof. Let H be a subgroup of G, and let a be a generator of G. By the
lemma, p(H) is a subgroup of Z, so is of the form nZ for some n (exercise.) Then
b = an is a generator of H.
□
Corollary 56. If G is a finite cyclic group of order n, then G has precisely one
subgroup of order k for any divisor k of n, which is cyclic, and no other subgroup.
Proof. Let a be a generator of G. If H is a subgroup of G of order k, then
by theorem 55 it is cyclic. Hence k is the order of some element b, which therefore
divides n. If m is the smallest power of a such that am ∈ H, then m characterises
H, so H is unique.
□
7. Homomorphisms
We now consider homomorphisms — maps between groups. These are maps
that respect the structure of the groups, namely, the operation.
Definition 57. A homomorphism from a group G to another group H is a
map f : G −
→ H such that for all x, y ∈ G, f (xy) = f (x)f (y).
Note that in this definition, the operation between x and y is the operation in
G, but the operation between f (x) and f (y) is in H.
The rest of the group structure is automatically preserved:
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture 7, Jan 31
16
1. GROUP THEORY
Proposition 58. If f : G −
→ H is a group homomorphism, then f (e) = e, and
for any x ∈ G, f (x−1 ) = f (x)−1 .
Proof. f (e) = f (e2 ) = f (e)2 , hence f (e) = e. For any x, f (x)f (x−1 ) =
f (xx−1 ) = f (e) = e, hence f (x−1 ) = f (x)−1 .
□
Example 59. If G is any group, then the identity map f (x) = x is a homomorphism from G to itself. Also, the map given by f (x) = e for all x is a
homomorphism from G to any group H. If H is a subgroup of G, the inclusion of
H in G is a homomorphism.
□
Example 60. For any finite dimensional vector space V over k, the determinant
map det : GL(V ) −
→ k∗ is a homomorphism.
□
Example 61. If V and U are vector spaces, then any linear map from V to U
is a homomorphism.
□
Example 62. Let V be the set of vertices of the regular n-gon. For any
g ∈ Dn , let fg : V →
− V be the restriction of g to the vertices. Then g 7→ fg is a
homomorphism from Dn to Sym(V ).
□
Example 63. If G is any group, and X is any set, let H be the group of
functions from X to G. Then the map from G to H sending g to the constant
function g is a homomorphism.
□
As with functions of sets, we are interested to know when a homomorphism is
injective or surjective or invertible.
Definition 64. A homomorphism f : G −
→ H is injective (respectively surjective, bijective) if it has the same property as a function of sets. It is invertible if
there is a homomorphism g : H →
− G such that f g is the identity on H, and gf is
the identity on G. The image of f is its image as a function of sets.
If f is injective, then, in particular, for any x ̸= e, f (x) ̸= e. In other words, x
is not in the set of elements of G that go to e under f .
Definition 65. If f : G −
→ H is a homomorphism, the set Ker(f ) = {x ∈ G :
f (x) = e} is called the kernel of f .
Thus, if f is injective, then Ker(f ) consists only of e. The converse also holds:
Proposition 66. Let f : G −
→ H be a homomorphism. Then Ker(f ) is a
subgroup of G, and the image Im(f ) of f is a subgroup of H. f is injective if and
only if Ker(f ) is trivial. It is invertible if and only if it is injective and surjective.
Proof. Assume that the kernel Ker(f ) is trivial, and suppose that f (x) =
f (y). Then f (xy −1 ) = f (x)f (y)−1 = e, hence by assumption xy −1 = e, so x = y.
Assume that f is injective and surjective, and let g be its inverse as a function
of sets. If u, v ∈ H, there are, by assumption, elements x, y ∈ G such that f (x) = u
and f (y) = v. Then g(uv) = g(f (x)f (y)) = g(f (xy)) = xy = g(u)g(v). The other
claims are easy.
□
Example 67. The kernel of the determinant map on GL(V ) is the subgroup
of matrices of determinant 1, SL(V ).
□
Example 68. The function z 7→ ∥z∥ is a group homomorphism from C∗ to R∗ .
The kernel can be identified with the circle group.
□
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
7. HOMOMORPHISMS
17
Example 69. Let V be a vector space, and v ∈ V a non-zero vector. The subset
Gv of GL(V ) of maps for which v is an eigenvector is a subgroup (geometrically,
it is the group of all linear maps that fix the line determined by v.) The map
e : Gv −
→ k∗ assigning to each linear map in Gv the eigenvalue corresponding to v
is a homomorphism. The kernel is the set of all linear maps that fix v.
□
Example 70. Let G be the group of all upper triangular matrices of determinant 1 in GL2 . If v is the vector (1, 0), then Gv is the group of all upper triangular
matrices in GL2 , and so G < Gv . The kernel of [the ]corresponding map from the
above example is the set of matrices of the form 10 x1 .
□
End lecture 8, Feb 4
7.1. Isomorphisms.
Definition 71. An invertible homomorphism is called an isomorphism.
By proposition 66, to check that a homomorphism is an isomorphism, it is
enough to show that is is onto, and the kernel is trivial.
Example 72. If G is a finite group, and f : G −
→ G is a homomorphism with
a trivial kernel, then it is an isomorphism. This is false for infinite groups (e.g.,
n 7→ 2n is a homomorphism with trivial kernel from Z to itself, but is not an
isomorphism.)
□
[1 x]
Example 73. The group of matrices of the form 0 1 from example 70 is
isomorphic to the additive group (k, +): the function that takes the above element
to x is an isomorphism.
□
The point of an isomorphism is that it demonstrates that two groups look
exactly the same as groups. In other words, any property of the groups that is
defined just in terms of the group structure is preserved under isomorphisms. Here
is a (non-exhaustive) list of such properties:
Proposition 74. Let f : G →
− H be a group isomorphism
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
G and H have the same order
G is abelian if and only if H is abelian
G is cyclic if and only if H is cyclic
A ⊆ G is a subgroup if and only if f (A) is a subgroup
A ⊆ G generates G if and only if f (A) generates H
For any a ∈ G, f (C(a)) = C(f (a))
a ∈ G is in the centre of G if and only if f (a) is in the centre of H.
The order of a is equal to the order of f (a) for all a ∈ G
Proof. In each case it is enough to prove one direction, since the other follows
from applying the first to the inverse of f . Also, 3 follows from 5, 2 follows from
7, and since both 5 and 7 are intersections of subgroups, they both follow from 4,
together with 6. Finally, 8 follows from 5 and 1, and 1 is trivial (by definition, two
sets have the same size if there is a bijection between them.) It remains to prove 4
and 6, which are easy.
□
This proposition allows us to prove that certain group are not isomorphic: One
shows that one group has a property that the other doesn’t.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
18
1. GROUP THEORY
Example 75. The group D4 is not isomorphic to S4 . Indeed, D4 has less
elements. This can be shown as follows: We have seen that if we enumerate the
vertices of the regular n-gon by the numbers 1, . . . , n, we get a homomorphism
h from Dn to Sn , sending a symmetry to its restriction to the vertices. This
homomorphism is injective: if a symmetry fixes all the vertices, then it is the
identity. Since both groups are finite, to show that Sn has more elements, it is
enough to show that h is not surjective. If n > 3, any permutation that exchanges
only two vertices does not come from a symmetry. Note, that for infinite groups H
and G, the existence of an injective, non-surjective, homomorphism from H to G
does not guarantee that H and G are not isomorphic!
Alternatively, we can use (8): We have seen that the order of any element of
Dn is either 2 or divides n. On the other hand, in Sn we have the permutation
that maps any i < n − 1 to i + 1, n − 2 to 1, and (necessarily) leaves n fixed. This
permutation has order n − 1, which (if n > 2) does not divided n, and which (if
n > 3) is not equal to 2.
□
Example 76. The groups U35 and S4 are not isomorphic, even though they
□
have the same number of elements (24): U35 is abelian, but S4 is not.
Example 77. The group (R, +) is not isomorphic to the circle group: the circle
group has elements of finite order, but R doesn’t.
□
Example 78. The group SLn (C) is not isomorphic to GLn (C): SLn (C) has a
finite centre, GLn an infinite one.
□
Example 79. (Z, +) is not isomorphic to (Q, +): the group Q is divisible: for
any element x ∈ Q, and any natural number n, there is some y ∈ Q such that
ny = x. This is false for Z.
□
Example 80. The groups U8 and Z4 are not isomorphic: both are abelian,
and have 4 elements, but U8 is not cyclic.
□
Here are some examples of groups which are isomorphic:
Example 81. The group U5 is isomorphic to Z4 : The only homomorphism
given by 2 7→ 1 (from U5 to Z4 ) is an isomorphism.
□
Example 82. The map from D3 to S3 is an isomorphism. We already saw
that this is an injective homomorphism. Since any permutation of the vertices of a
regular triangle gives a symmetry, it is also surjective.
□
Example 83. The subgroup of (Q∗ , ·) consisting of powers of 2 is isomorphic
to Z.
□
Example 84. The circle group is isomorphic to the group SO(2) of real 2 × 2
orthogonal matrices with determinant 1: The function mapping the rotation by α
[
sin(α) ]
to the matrix −cos(α)
is an isomorphism.
□
sin(α) cos(α)
Example 85. The multiplicative group Q∗ is isomorphic to the centre of
GLn (Q) via x 7→ xI. The same holds when Q is replaced by R, C, . . ..
□
Example 86. Let V and U be vector spaces. If T : V −
→ U is a linear
isomorphism, then for any S ∈ GL(V ), the map T ST −1 is a linear bijection of U to
itself, i.e., an element of GL(U ). It can be checked directly that S 7→ T ST −1 is an
isomorphism from GL(V ) to GL(U ). In particular, this applies to any isomorphism
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
8. THE CLASSIFICATION OF CYCLIC GROUPS
19
T : kn −
→ U (recall that such an isomorphism exists precisely when the dimension
of U is n, and in this case, such isomorphisms correspond to bases of U .) Thus,
GL(U ) is isomorphic to GLn , where n = dim(U ).
Likewise, any bijection between the sets X and Y gives an isomorphism between
Sym(X) and Sym(Y ), and in particular, if X is a finite set of size n, there is an
isomorphism between Sym(X) and Sn .
□
Example 87. The function x 7→ ex is an isomorphism from R onto the multiplicative group R+ of positive real numbers. This does not hold form Q, since the
exponent of a rational number is not rational. It also does not hold for C, since on
C the exponent is not injective.
□
As can be seen from these examples, it is sometimes hard to determine when
two groups are isomorphic!
8. The classification of cyclic groups
In this section we will prove the following theorem:
Theorem 88. Any cyclic group is isomorphic either to (Z, +) or to (Zn , +)
for some n.
It follows, using the principle of proposition 74 that any group theoretic statement that we want to verify for cyclic groups, it is enough to verify for groups in
this list.
Our strategy in proving this theorem will be as follows: we will study the
set Hom(C, G) of group homomorphisms from one of the groups C above, to an
arbitrary group G. After obtaining an explicit description of this set, we will find
which elements of this set are isomorphisms. Finally, we will show that when G
is cyclic, the corresponding set of isomorphisms is not empty. We start with the
following general statement.
Proposition 89. Let G be a group generated by a subset A, and let H be an
arbitrary group. Let f and g be two homomorphisms from G to H that agree on A
(f (a) = g(a) for all a ∈ A.) Then f = g.
Proof. Let K be the set of elements k of G such that f (k) = g(k). It is easy
to see that K is a subgroup. By assumption K contains A. Hence K = G.
□
Remark 90. We have already noticed that a set of generators for a group is
similar to a spanning set in a vector space. This is another instance of this fact:
Two linear maps that agree on a spanning set are equal.
Corollary 91. If G is any group, the set Hom(Zn , G) of group homomorphisms from Zn to G is in one-to-one correspondence with elements in G whose
order divides n, and Hom(Z, G) corresponds to the set of all elements of G.
Proof. Let C be one of the groups Zn or Z. Each is generated by 1. According
to the proposition, each homomorphism F is determined by its value F (1) ∈ G on
1. Conversely, if g ∈ G is an element, the association F (i) = g i is a well defined
□
homomorphism from C to G precisely when the order of g divides n.
In other words, homomorphisms from Zn allow us to detect the elements of G
of order a divisor of n. Which of them are injective?
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture 9, Feb 7
20
1. GROUP THEORY
Corollary 92. A homomorphism from Zn to G is injective if the corresponding element has order n. A homomorphism from Z to G is injective if the order of
the corresponding element is infinite.
Proof. An element in the homomorphism defined by g is in the kernel precisely
if g i = e for i < n (in the case of Z, for some i.)
□
We can now prove the theorem:
proof of theorem 88. Let G be a cyclic group of order n. By the corollary,
the set of injective homomorphisms from Zn to G corresponds to the set of elements
of order n in G. Since G is cyclic, it has such an element. Since both groups are of
size n, any injective homomorphism is bijective.
Now assume that G is an infinite cyclic group, and let a be a generator. Again
by the corollary, the map n 7→ an is injective. Since a generates the whole group,
it is also surjective.
□
Remark 93. We have thus classified all cyclic group, up to isomorphism. This
means that we have presented a list of cyclic groups, all non-isomorphic, such that
every other cyclic group is isomorphic to one of them. Furthermore, we have a
straightforward way, given a cyclic group, to determine to which group in our list
this group is isomorphic: this is determined by the number of elements. A more
complete answer would be to compute the automorphism group of each group. We
shall do that in the next section.
We may use corollaries 91 and 92 to study the set of Hom(Zp , Zq ) of homomorphisms between Zp and Zq . Indeed, corollary 91 says that this set naturally
corresponds with the set of elements in Zq whose order divides p. Since the order
of each element in Zq divides q, we get that Hom(Zp , Zq ) corresponds to the set
of elements in Zq whose order divides gcd(p, q). This subset forms a subgroup (it
is the set of elements x such that gcd(p, q)x = 0 in Zq ), which is generated by
q
gcd(p,q) . In particular, if p and q are coprime, the only homomorphism is the trivial
one. On the other hand, if q divides p, then any element of Zq corresponds to a
homomorphism.
Furthermore, corollary 92 tells us that there are injective homomorphisms if
and only if p divides q. On the other hand, a homomorphism is surjective if and
only if 1 is in the image, if and only if gcd(p, q) = q, if and only if q divides p.
Example 94. Let p = 5 and q = 4. If h : Z5 −
→ Z4 is any homomorphism,
then h(4) = h(1 + 1 + 1 + 1) = h(1) + h(1) + h(1) + h(1) = 0, since the order of
any element in Z4 divides 4. On the other hand, 4 is a generator of Z5 , so h is
trivial.
□
Example 95. Let p = 4 and q = 6. There is no homomorphism h from Z4 to
Z6 with h(1) = 2. Indeed, such a homomorphism would map 3 to 0, but 3 = −1 in
Z4 , so this would imply h(1) = 0 as well. According to our observation, h(1) can
6
= 62 . In the latter case, h(2) = 0, so there is no injective
only by 0 or 3 = gcd(4,6)
homomorphism.
□
End lecture 10, Feb 9
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
9. THE SYMMETRIC GROUPS
21
8.1. Automorphisms of cyclic groups. We have started the study of groups
by looking at symmetries. It is now natural to ask what are the symmetries of a
group. Recall that we view a symmetry of an object to be a bijection of that object
to itself, that preserves its properties.
Definition 96. An automorphism of a group G is an isomorphism from G to
itself. The set of all automorphisms of G is denoted Aut(G).
If f : G −
→ H and h : H →
− K are two homomorphisms, their composition is again a homomorphism: (h ◦ f )(g1 g2 ) = h(f (g1 g2 )) = h(f (g1 )f (g2 )) =
h(f (g1 ))h(f (g2 )). In particular, if f and h are both automorphisms of some group,
their composition is again an automorphism of the same group.
Proposition 97. Aut(G) forms a group under composition of automorphisms.
It is also denoted Aut(G).
Proof. Since composition of any kind of functions is associative, associativity
holds in Aut(G). The identity automorphism is the identity for the operation.
Finally, for any automorphism, its inverse is the group inverse.
□
We will study automorphisms in more detail later in the course. In the meantime, we compute the automorphism groups of the cyclic groups.
Example 98. According to the previous section, Hom(Zp , Zp ) is in bijection
with the set of elements of Zp . The injective homomorphisms correspond to elements of order p, i.e., to elements prime to p. These are precisely the isomorphisms.
For Z, an automorphism must take 1 to a generator. The only generators of
Z are 1 and −1, so Aut(Z) consists of precisely two elements, the identity and the
inverse.
□
The last example presents Aut(Zp ) as a set. It does not describe the operation
of the group in terms of this set. We now describe the group structure. In fact, we
can do it directly for any cyclic group:
End lecture
Feb 11
11,
End lecture
Feb 14
12,
Proposition 99. If G is a finite cyclic group of order n, then Aut(G) is
isomorphic to Un , by the isomorphism given by fk (g) = g k (here fk ∈ Aut(G)
corresponds to k ∈ Un ).
Proof. Since k is prime to the order of G, the map fk is injective, hence fk is
an automorphism. Hence k 7→ fk indeed defines a map of sets from Un to Aut(G).
k
Furthermore, this map is injective, and is a homomorphism since g kl = (g l ) .
Finally, it is an isomorphism since we have already computed that Aut(G) is in
bijection with Un as sets.
□
9. The symmetric groups
In this section we will study an important class of groups, the symmetric group
of a set X. An element of such a group, a bijection from X to itself is called a
permutation of X. We will concentrate on the case that X is finite. We have seen
(example 86) that in this case the group is isomorphic to Sn , the symmetry group
of {1, . . . , n}, where n is the number of elements of X. Thus, up to isomorphism,
it is enough to study the Sn . We will be interested in the following questions:
(1) What is the order of Sn ?
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
22
1. GROUP THEORY
(2) How can we present the elements of Sn conveniently?
(3) What are “nice” sets of generators for Sn ?
(4) What interesting subgroups does Sn have?
We begin with some generalities. If s ∈ Sym(X) is a symmetry of X such that
s(x) = x for some x ∈ X, we say that s fixes x. We have the following result.
Proposition 100. Let Y ⊆ X be sets. The subset of elements of Sym(X) that
fix each element of Y is a subgroup, which is isomorphic to Sym(X − Y ).
Proof. If s ∈ Sym(X − Y ), we extend s to a bijection s′ from X to itself by
setting s′ (y) = y for all y ∈ Y . The map s 7→ s′ is clearly a homomorphism of
groups, which is injective since if s′ is the identity map, then so if s. Furthermore, s′
fixes all elements of Y . Conversely, every permutation of X that fixes any element
of Y is of the form s′ where s is the restriction to X − Y . Hence the image of
the homomorphism is the set of all elements that fix all elements of Y (and in
particular, this set is a subgroup.)
□
In particular, we may view Sn−1 as the subgroup in Sn of all elements that fix
n. We can now compute the order of Sn .
Proposition 101. The order of Sn is n! = 1 · 2 . . . n (by convention, 0! = 1.)
As explained above, it follows that the order of Sym(X) is n! for any set X of
size n.
Proof. By induction on n. The case n = 0 (symmetries of the empty set)
is left as an exercise. Fix n > 0. For 1 ≤ i ≤ n, let Ti be the subset in Sn of
all permutations that take n to i (Ti is not a subgroup if i ̸= n.) The Ti form
a partition of Sn : they are disjoin, and their union is the whole of Sn . Hence
|Sn | = |T1 | + |T2 | + · · · + |Tn |.
For any 1 ≤ i ≤ n, let si be the permutation such that si (i) = n, si (n) = i,
and that fixes all other elements (so sn is the identity.) If t ∈ Ti , then si t fixes
n, so is in Tn . Conversely, if r ∈ Tn , then si r takes n to i, so is in Ti . Thus si
is a bijection between Tn and Ti , and so they have the same number of elements.
Thus |Sn | = n|Tn |. Finally, we note that Tn was identified with Sn−1 . Thus, by
the induction hypothesis, we get |Sn | = n · (n − 1)! = n!.
□
End lecture
Feb 16
13,
Example 102. Recall that S3 can be identified with D3 , the group symmetries
of the regular triangle. If we identify the top vertex of the triangle with 3, the set
Ti in the proof corresponds to the set of symmetries that take this vertex to vertex
i, namely, the rotation from 3 to i, and the same rotation followed by a reflection
around vertex i.
□
[ 1 2 ... n ]
An arbitrary elements s of Sn can be written as s(1) s(2) ... s(n) . For example,
if s ∈[ S4 is ]defined by s(1) = 3, s(2) = 4, s(3) = 1 and s(4) = 2, we write
it as 13 24 31 42 . The top line seems redundant, but it helps when computing the
composition.
The element s just considered has the following property: if we restrict it to
the subset Y = {1, 3}, we get a permutation of Y . In other words, it maps Y into
(and, necessarily, onto) itself. In general, we say that Y is an invariant subset for
s if s(Y ) = Y . Thus, every element s of Sn has the whole set {1, . . . , n} and the
empty set as invariant subsets. Furthermore, any subset of the set of fixed elements
of s is invariant.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
9. THE SYMMETRIC GROUPS
23
Definition 103. Let s ∈ Sn be a non-identity element, and let Y ⊆ {1, . . . , n}
be the set of elements not fixed by s. Then s is called a cycle if Y has no proper,
non-empty s-invariant subsets.
Remark 104. Sometimes the identity is also considered to be a cycle.
Example 105. In S3 , any non-identity element is a cycle. The reason is that
a proper non-empty invariant subset that does not contain fixed elements, has to
be of size 2.
□
[1 2 3 4]
Example 106. In S4 , the element 3 4 1 2 considered above is not a cycle: it
has no fixed elements, but {1, 3} is an invariant set.
[ We note
] that the set {2, 4} is
invariant as well. On the other hand, the element 13 22 34 41 is a cycle.
□
Let s be a cycle, and let x be an element not fixed by s. Applying s iteratively,
we get a sequence of elements x, s(x), s(s(x)), . . .. For some k > 1, we must have
sk (x) = x (why?), and any element not of the form si (x) is fixed (otherwise the
set of si (x) is a proper non-empty invariant subset.) Therefore, s is completely
determined by the sequence, and we will represent s by the tuple (x, s(x), . . . , si (x)).
If we would start with s(x) rather than x, we would get the same sequence, shifted
cyclically by 1. This is the origin of the name “cycle”.
Example 107. To present the second permutation in example 106 in this form,
we start with the element 1, which is mapped to 3, which is in turn mapped to
4, which is mapped to 1. Thus this cycle is represented by (1, 3, 4). If we would
have started with 3, we would get (3, 4, 1) instead, which is the same tuple shifted.
To compute the value of s(4), for instance, using this notation, we find where 4
is written, and look what is the next number. If 4 is in the end (like in the first
representation), the “next” is the first one, so in this case, s(4) = 1. If the number
does not appear in the tuple, then it is fixed. In this example, 2 does not appear,
and indeed s(2) = 2.
□
In general, if s is an element of Sn and 1 ≤ k ≤ n, the orbit of k under s (or
the s-orbit of k) is the set of elements of the form k, s(k), s2 (k), . . .. Thus, k is
fixed by s precisely if its s-orbit contains only k, and a cycle is an element that
has exactly one orbit of size bigger than one (and the elements in this orbit are
precisely the elements that appear in its representation in cycle notation.) We note
that the s-orbit of k can alternatively be defined as the smallest s-invariant subset
containing k. With this definition, we see that any two orbits are either equal or
disjoint.
Two permutations are called disjoint if their sets of non-fixed elements are
disjoint. Equivalently, any element is fixed by either one or the other. We will
usually apply this terminology to cycles.
We can now describe a good generating set for Sn :
Theorem 108.
(1) Two elements in Sn commute if they are disjoint.
(2) Any element of Sn can be expressed as a product of disjoint cycles. The
set of cycles that appear in this expression is uniquely determined.
In other words, the set of cycles generates Sn .
Proof.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
24
End lecture
Feb 18
14,
1. GROUP THEORY
(1) Let s and t be disjoint elements, and let x be an element fixed by t. We
will show that s(t(x)) = t(s(x)). By symmetry it will follows for elements
fixed by s, and since, by assumption, any element is fixed by one of them,
we are done.
If x is also fixed by s, the claim is trivial. So assume that x is not
fixed by s. Then s(x) is also not fixed by s. Hence, it is fixed by t. Thus,
t(s(x)) = s(x) = s(t(x)).
(2) Let s ∈ Sn be a non-identity element. Let X1 , . . . , Xm be the s-orbits of
size bigger than 1. Let si be the permutation that agrees with s on Xi
and fixes all other elements. Since Xi is an invariant set, si is well defined.
Since Xi is the only orbit of si of size greater than 1, si is a cycle. Since
the Xi are disjoint, so are the si . Finally, it is clear that their product is
equal to s: an element x of Xi is not fixed only by si , and s(x) = si (x),
and an element not in any of the Xi is fixed by both sides.
The uniqueness statement follows from the fact that a decomposition
of s into cycles corresponds to orbits of s, and the orbits are uniquely
determined.
□
[1 2 3 4 5 6 7 8 9]
Example 109. Let us present the permutation s = 1 5 2 8 7 9 3 4 6 as a product of disjoint cycles. We do this by simultaneously finding the orbits, and writing
the corresponding cycles. We start with 1: since s(1) = 1, 1 is a fixed point,
and does not contribute any cycle. The next element is 2, and we find s(2) = 5,
s(5) = 7, s(7) = 3 and s(3) = 2. We thus get the cycle (2573). To continue, we
pick the next element not yet accounted for, namely, 4. We find s(4) = 8 and
s(8) = 4, corresponding to the cycle (48). Similarly, we get the cycle (69). Thus
s = (2573)(48)(69).
□
As an application, we can compute the order of an arbitrary permutation. First,
we prove the following general fact.
Lemma 110. Let g, h ∈ G be commuting elements. Then |gh| divides n =
lcm(|g|, |h|) (where the lcm is infinity if either of the arguments is.) If the intersection of ⟨g⟩ and ⟨h⟩ is trivial, then |gh| = n.
Proof. Since both orders divide n, it is clear that (gh)n = g n hn = e, so the
order of gh divides n.
Conversely, if (gh)m = e then g m = h−m . Since the left side belongs to ⟨g⟩ and
the right side to ⟨h⟩, it follows from the assumption that they are both equal to e.
Hence the orders of both g and h divide m, so n divides m.
□
We note that the order of a cycle is simply its length, i.e., the size of the
non-trivial orbit. Thus we get:
Corollary 111. The order of an element of Sn is the least common multiple
of the sizes of its orbits
Proof. If s and t are disjoint permutations, then the subgroups they generate
intersect trivially: indeed, any element in 1, . . . , n is fixed by either all elements
of ⟨s⟩ or of ⟨t⟩. The statement now follows from the lemma by induction on the
number of cycles.
□
Example 112. Compute all possible orders of elements of S4 , and the number
of elements of each order.
□
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
9. THE SYMMETRIC GROUPS
25
The uniqueness result in theorem 108 holds only for decompositions into disjoint
cycles. In fact, if we drop the disjointness requirement, we get a stronger result.
Corollary 113. Each group Sn is generated by transpositions (a transposition
is a cycle of order 2.)
We note that there is no uniqueness here, and that the transpositions are not
disjoint. In particular, they don’t commute.
Proof. Since the cycles generate Sn , it is enough to show that a cycle is a
product of transpositions. We do this by induction on the length k of the cycle.
When k = 2, there is nothing to prove. If s = (a1 . . . ak ) is a cycle with k > 2,
it can be checked directly that s = (a1 . . . ak−1 )(ak−1 ak ). By induction, the first
factor is a product of transpositions.
□
Example 114. Let us write present the permutation in example 109 as a product of transpositions. The cycle (2573) is equal to (257)(73), and (257) = (25)(57).
The other cycles are already transpositions, so the whole element is equal to
(25)(57)(73)(48)(69).
□
9.1. Cayley’s theorem. We now consider what possible subgroups a permutation group can have. Obviously, a subgroup of Sn has to be finite. Conversely,
we have:
End lecture 15,
Feb 21 (Midterm 1)
Theorem 115 (Cayley’s theorem). Any group of order n is isomorphic to a
subgroup of Sn
Proof. Since Sn is isomorphic to Sym(X) for any set X of size n, it is enough
to show that there is an injective homomorphism from G to such a group. We take
X = G, and the homomorphism assigns to each element g ∈ G the permutation
lg of G defined by lg (x) = gx. This is a homomorphism since lgh (x) = (gh)x =
g(hx) = lg (hx) = lg (lh (x)) = lg ◦ lh (x). It is injective since if lg is the identity
bijection, then e = lg (e) = ge = g.
□
This theorem can be visualised by writing down the multiplication table of G.
The elements of the permuted set, namely, the elements of G, can be read from the
top line of the table. If g ∈ G is an element, the row that has g as the leftmost
element consists of the elements of G in some order. This is the permutation
induced by G.
]
[
116.[ for the
group in example 11, we see that la = ae ab cb ec , lb =
[ e a Example
]
]
b c
ea b c
□
b c e a and lc = c e a b .
We can use this description to extend a result we know for cyclic groups.
Corollary 117. The order of an element in a group of order n divides n.
Proof. Let g ∈ G be an element in a group of order n. Since the Cayley
homomorphism is injective, it is enough to prove that the order of lg divides n.
Consider the lg -orbits: we claim that the size of any orbit is the order k of g.
Indeed, if h ∈ G is any element, the lg orbit of h is the set of elements g i h. Since
the orbits form a partition of G, the sum of their sizes is n. Hence k divides n. □
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture
Feb 23
16,
26
1. GROUP THEORY
9.2. The sign homomorphism. We have stated that the presentation of an
element as a product of transpositions is not unique. Indeed, even the number
of transpositions in different presentations of the same element may be different.
For example, (13) = (23)(12)(23). However, the following theorem shows that the
parity of this number is preserved.
Theorem 118. For any n, there is a unique homomorphism sgn : Sn →
− {−1, 1}
with the property that sgn(x) = −1 for every transposition x. sgn(s) is called the
sign of s.
The uniqueness follows directly from the fact that the transpositions generate
Sn . The existence will be proved later.
Thus, to compute the sign of s, we write it as a product of transpositions. The
sign will be −1 if and only if the number of transpositions is odd. The contents
of the theorem is that this will not depend on the way we decomposed. In other
words, the number of transpositions might not be the same, but it is either always
odd or always even. The permutation itself is called odd or even, accordingly.
Example 119. The sign of the permutation from example 109 is −1, i.e., it is
odd.
□
Another way compute the sign is as follows: in the two rows way of writing the
permutation, draw a line from every number in the top row to the same number
in the bottom row. The sign will be the parity of the number of intersections
(assuming that only two lines go through any intersection; this can always be
achieved by rescaling the picture.)
Example 120. For the permutation from example 109, we get
1
1
6 TTTT q 7 fffff 8 jjj 9
2>
3 WWWWW 4 WWWjWjWj 5
WWjWjWjW WWWWW
qT ff
j
>>
j
W
W
fqfqfqffTTTTTTjjjjjj
>>
j
W
W
f
W
W
j
f
W
W
q
f
j
W
W
f
j
j
T
W
W
f
q
>> jjj
f
W
W
j
T
f
W
W
q
j
T
f
W
W
TTTT
fff qqqWWWWWjWjjWjWjWWWW
>j
TT
fffff
WW
q
j WW
jjjj
5j
2
8
7 qq
9j
3
4
6
So the number of intersections is 13, which is odd, as expected.
□
This works for, roughly, the following reason: looking at the bottom row, the
fact that s is not the identity means that there is a pair of adjacent elements whose
order is wrong. Switching these two elements removes precisely one intersection
point, and is achieved by composing with a transposition. This gives a way to
write s as a product of transpositions, where the number of transpositions is the
number of intersections.
The set of even permutations is the kernel of the sign homomorphism, and is
therefore a subgroup of Sn .
Definition 121. The subgroup of Sn of even elements is called the alternating
group of order n. It is denoted by An .
For example, A4 can be viewed as the symmetry group of the regular tetrahedron.
End lecture
Feb 25
17,
Example 122. Compute the possible orders of elements of A5 , and the number
of elements of each order.
□
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
9. THE SYMMETRIC GROUPS
27
9.3. Symmetric groups and linear groups. We now outline a connection
between the symmetric group Sn and linear algebra. If (x1 , . . . , xn ) is a tuple of (say,
rational) numbers and s ∈ Sn , we may form a new tuple Ts (x̄) by permuting the
xi according to s: Ts (x̄) = (xs−1 (1) , . . . , xs−1 (n) ). It is clear that Ts is a linear map
from the space of n-tuples to itself. Note that our definition, which might appear
confusing, simply means that if, say, s(1) = 3, then Ts moves the first coordinate
to the third one. In particular, if ei = (0, . . . , 1, . . . , 0) is the i-th standard basis
element, then Ts (ei ) = es(i) .
Proposition 123. The map T : s 7→ Ts is an injective homomorphism from
Sn to GLn .
Proof. We first note that Ts−1 is the inverse of Ts , so each Ts is indeed in
GLn . To show that T is a homomorphism, we need to show that Trs = Tr ◦ Ts .
Since these are linear maps, it is enough to show this on a basis. By the remark
above,
Tr ◦ Ts (ei ) = Tr (es(i) ) = er(s(i)) = Trs (ei )
(1)
To show that T is injective, we show that the kernel is trivial. Indeed, if Ts is
the identity, then es(i) = Ts (ei ) = ei for all i, so s is the identity.
□
In general, a homomorphism from a group G to a linear group GL(V ) is called
a linear representation of G. Combined with Cayley’s theorem, we get for any finite
group a representation, called the regular representation:
Corollary 124. Any group of order n is isomorphic to a subgroup of GLn .
What is the image of T ? To find the matrix representation of Ts we apply Ts
to the elements of the standard basis. We get that the Ts is represented by the
matrix whose columns are es(1) , . . . , es(n) . In particular, all entries of this matrix
are either 0 or 1, and every column or row contains exactly one entry whose values
is 1. Conversely, it is easy to see that any matrix of this form represents Ts for some
permutation s. These matrices are called permutation matrices (or rook matrices).
Example 125. Compute the matrix Ts , where s = (145)(23) (in S5 ).
□
Since every matrix Ts is orthogonal, the inverse is computed by taking the
transpose: Ts−1 = Tst . The decomposition into disjoint cycles corresponds to decomposing the space into invariant subspaces (subspaces U such that Ts (U ) = U .)
Finally, we can prove the existence of the sign homomorphism:
Proof of theorem 118. The map s 7→ det(Ts ) satisfies all the properties:
it is a homomorphism, its values on matrices of the form Ts is either 1 or −1, and,
since each transposition is obtained by switching two columns from the identity
matrix, its value on the transpositions is −1. The uniqueness was already noted to
□
follow from the fact that the transpositions generate Sn .
Thus, An is the subgroup of elements whose matrix is in SLn .
Example 126. Note that Tr(Ts ) is the number of fixed elements of s. Since for
any two matrices A and B, Tr(AB) = Tr(BA), we get that for any two permutations
s and r, sr and rs have the same number of fixed points.
□
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture
Mar 4
18,
28
1. GROUP THEORY
10. Group actions
The definition of a group resulted from looking at symmetry. The first examples
where the group of symmetries of some object X (a set, a polygon, a vector space,
etc.) Although the group was obtained in this way, the object X plays no role in
the definition of a group. We now explain how to bring X back into the game.
If g ∈ G = Sym(X) is a symmetry of X, then g is a certain kind of invertible
function from X to itself. Thus, to any element x ∈ X, it assigns another element,
gx. If h is another symmetry, we may apply h to the new element gx, and obtain
h(gx). Since the operation of the group is the composition of these functions, we
get that this element is also equal to (hg)x. Also, since e ∈ G is the identity
symmetry, we have ex = x for any x ∈ X. The definition of an action axiomatises
these observations.
Definition 127. An action of a group G on a set X is a function m : G × X −
→
X that satisfies m(e, x) = x and m(g, m(h, x)) = m(gh, x) for all g, h ∈ G and
x ∈ X. A G-set is a set together with an action of G on it.
We will usually write gx instead of m(g, x), as for the group operation. We
stress that a G-set (or an action) is an additional information, it is not part of the
datum of a group.
Example 128. The group Sn acts, by its very definition, on the set [n].
□
Example 129. The group Dn acts, again by definition, on the set of points of
a regular n-gon. Since the symmetries preserve the geometric structure, the same
group acts on the set of vertices, and on the set of edges.
□
Example 130. The operation of a group can be viewed as an action of G on
itself.
□
Example 131. If V is a vector space over Q, then Q∗ acts on V by multiplication. The same holds when Q is replaced by R, C, etc. More generally, if k is an
integer, we may define another action of Q∗ on V by m(x, v) = xk v.
□
Example 132. Let P1 (R) be the set R ∪ {∞} (where ∞ is simply a new
element.) Then SL2 (R) acts on P1 (R) via the following formula:
[ ]
ax + b
m( ac db , x) =
cx + d
Where the right hand side is interpreted as follows: first, if x is a number and
cx + d ̸= 0, then it is evaluated in the usual way. If cx + d = 0 then ax + b ̸= 0
(since the matrix is invertible), and we declare the value to be ∞. Finally, if x = ∞
then the value is ac (which is equal to ∞ if c = 0.)
□
Example 133. An invertible function f : X −
→ X from a set to itself gives rise
to an action of Z on X: m(k, x) = f k (x), where f k (x) = f (f (. . . (x)) . . . ) applied
□
k times (if k < 0, we apply −k times the inverse of f .)
Example 134. If X is any set, and G is any group, the trivial action of G on
X is the one where m(g, x) = x for all g and x.
□
Example 135. If X is a G-set, and H is a subgroup of G, then the restriction
→ G is a
of the action to H gives an action of H on X. More generally, if f : H −
group homomorphism, then the function m(h, x) = f (h)x gives an action of H on
X.
□
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
10. GROUP ACTIONS
29
If X is a G-set, every element g ∈ G gives an invertible function fg : X →
− X.
Hence we get a function g 7→ fg , from G to Sym(X). The axioms of an action imply
that this map is a group homomorphism. Conversely, if we are given a function
g 7→ fg from G to Sym(X), then we may define a map m : G × X −
→ X by
m(g, x) = fg (x), and if the function g 7→ fg is a group homomorphism, then this
map is an action. Thus we have the following corollary:
Corollary 136. An action of a group G on a set X is “the same” as a group
→ Sym(X).
homomorphism G −
For example, the homomorphism corresponding to the action in example 128
is the identity homomorphism from Sn to itself. For other examples, example 129
(with the vertices) corresponds to the usual embedding of Dn in Sn , example 130
corresponds to the Cayley homomorphism, example 133 to the unique homomorphism from Z to Sym(X) that takes 1 to f , and example 134 to the trivial homomorphism.
Groups are often studied via their actions. For example, all our treatment of
the symmetric groups was done by considering its action. The very definition of the
notions of cycle, disjoint cycles and so on was done through this action. Though it
is possible to define these notions in purely group theoretic terms, it is substantially
more complicated. The disadvantage of our approach is that we did not prove, for
example, that the notion of a “cycle” depends only on the structure of the group.
Again, this can be done, but not very easily.
We will generalise the notions we associated to the action of Sn on [n] to
arbitrary group actions. As our goal, we will use the following generalisation of
corollary 117, due to Lagrange.
Theorem 137 (Lagrange). If H is a subgroup of a finite group G, then the
order of H divides the order of G.
Corollary 117 asserts this for cyclic subgroups H.
We begin with the notion of an orbit. Recall that if s ∈ Sn and x ∈ [n], the
s-orbit of x was defined as the set of elements of the form si (x). In other words,
it is the set of elements gx, where g is an element of the subgroup generated by s.
We generalise this definition as follows:
Definition 138. Let x be an element of a G-set X. The orbit of x, denoted
Gx, is the set of elements of the form gx, where g ∈ G. If H is a subgroup of G,
the H-orbit of x is the set of elements hx, where h ∈ H. The action of G on X is
called transitive if it has only one orbit.
Example 139. In example 131 above, if v is a non-zero vector, then the orbit
of v is the one-dimensional subspace of V generated by v, with 0 removed. Geometrically, such an orbit is a line through the origin (again, with the origin removed.)
In addition, 0 is an orbit by itself (a fixed point.) In most of the other examples
above, the action is transitive.
□
Remark 140. Most other notions and statements we proved about elements s
of Sn apply to more general group actions, by considering the group in place of the
subgroup generated by s. For example, given a G-set X:
(1) G fixes x ∈ X if gx = x for all g ∈ G
(2) For any subset Y ⊆ X, the set of elements of G that fix any element of Y
is a subgroup, which acts on X − Y .
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture
Mar 7
19,
30
1. GROUP THEORY
(3) A subset Y ⊆ X is G-invariant if gy ∈ Y for all g ∈ G and y ∈ Y . Y is G
invariant if and only if the action on X restricts to an action on Y .
(4) The orbit of x is the smallest invariant subset of X containing x. Any
minimal non-empty invariant subset is the orbit of any of its elements.
(5) The union, intersection and complement of invariant subsets is again invariant
(6) Any two orbits are either equal or disjoint.
The proofs are all direct generalisations of the corresponding statements for Sn ,
and are left as an exercise.
For the purpose of Lagrange’s theorem, we are interested in the action of a
group G on itself (example 130) and its restriction to the subgroup H. Explicitly,
the action is given by m(h, g) = hg for h ∈ H and g ∈ G. An orbit Hg of this
action is called a (right) coset of H. We note that H itself is a coset: it is the orbit
of e.
End lecture
Mar 9
20,
Example 141. If G = R2 , and H is a line (one-dimensional subspace), then
the cosets of H are the lines parallel to H.
□
Example 142. If G = R, and H = Z, then the cosets can be visualised
as follows: consider the set of real numbers as a helix over the (complex) plane,
projecting onto the unit circle via the map x 7→ e2πix . Then Z is the fibre of this
map over 1, and any other coset is the fibre over some other point in the circle. □
Example 143. If H is the trivial group, then the cosets of H are simply the
elements of G.
□
Whereas for a general G-set, distinct orbits can be very different (consider the
orbit 0 and another orbit in example 139), for cosets we have the following result.
Proposition 144. If H is a subgroup of G, there is a bijection between any
two cosets of H.
Proof. It is enough to prove that there is a bijection between H and any
other coset Hg. We claim that the function h 7→ hg is such a bijection. Indeed, by
the definition of the orbit it is a well defined function from H onto Hg, and it is
injective since multiplication by g is injective on the whole group g.
□
The proof of Lagrange’s theorem follows directly:
Proof of theorem 137. Since the orbits (cosets) are all disjoint, and every
element of G lies in some coset, we get that G is the disjoint union of the cosets
of H. In particular, the order of G is the sum of the sizes of the cosets. Since,
by proposition 144, the size of any coset is equal to the size of H, we get that
|G| = k|H|, where k is the number of cosets.
□
The number of cosets of H in G is called the index of H in G, and is denoted
G : H. Thus, the proof of Lagrange’s theorem shows that (when |G| is finite)
|G| = (G : H)|H|. Since |H| = (H : 1) (where 1 is the trivial group), we can also
write it as (G : 1) = (G : H)(H : 1).
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
11. NORMAL SUBGROUPS AND QUOTIENTS
31
10.1. Applications of Lagrange’s theorem.
Example 145. If p is a prime bigger than n, then Sn has no subgroup of order
p (actually, this follows already from Corollary 117).
□
Example 146. Any group G of order 2p, where p is an odd prime, is isomorphic
to either Dp or Z2p . Indeed, if G has no element of order 2p, then any element is
of order either 2 or p. If all elements are of order 2 then the group is commutative
(exercise), hence the subset consisting of e, a, b, ab, where a and b are distinct elements of order 2, is a subgroup of order 4, which contradicts Lagrange’s theorem.
Hence G has a subgroup R of order p.
Furthermore, R is the only such subgroup: if a is not in R, then a2 R = R,
hence a2 ∈ R. If a2 ̸= e, then the order of a2 is p, so the order of a is also p, so
⟨a⟩ = ⟨a2 ⟩ = R, contradicting the assumption. Hence all elements outside of R
have order 2.
We now know that G is the disjoint union of R and aR, and all elements of the
latter have order 2. This determines completely the multiplication of G, and so, G
is isomorphic to Dp .
□
11. Normal subgroups and quotients
11.1. Direct products. Recall that if G and H are arbitrary groups, we
defined a group structure on G × H by performing the multiplication pointwise:
(g1 , h1 )(g2 , h2 ) = (g1 g2 , h1 h2 ). We have seen that much of the information about
G × H can be deduced from corresponding information on G and on H. For
example, the order of an element (g, h) is the lcm of the corresponding orders. It
is thus natural to ask when a given group can be presented as the product of two
non-trivial groups.
We have seen that G × H has subgroups G′ and H ′ isomorphic to G and
H (namely i1 (G) and i2 (H), where i1 (g) = (g, 1) and i2 (h) = (1, h)). These
subgroups have a trivial intersection, the elements of G′ commute with those of H ′ ,
and G′ H ′ = G × H. All of these properties do not change under isomorphism, so a
group isomorphic to a direct product has two subgroup with the above properties.
Conversely, we have the following statement.
Theorem 147. Let P be a group with two subgroups G and H such that:
(1) G ∩ H = {e}
(2) Any element of G commutes with any element of H.
(3) GH = P
The P is isomorphic to G × H.
Proof. Define f : G × H →
− P by f (g, h) = gh. Then f is a group homomorphism because of 2, it is injective because of 1 and it is surjective because
of 3.
□
The theorem requires us to find two subgroups of P . In fact, we can get one
− G
of them for free, as follows: If P = G × H, we have a projection map π : P →
(given by π(g, h) = g.) The kernel of this map is the subgroup H ′ isomorphic to
H. As before this is preserved under isomorphisms. Thus, if P is isomorphic to
G × H, it has a surjective homomorphism π to G, with kernel isomorphic to H.
What additional conditions on π are required for the converse to be true? A section
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture
Mar 11
21,
32
1. GROUP THEORY
→ P such that π ◦ s is the identity on G. For
of π is a group homomorphism s : G −
example, if P = G × H, and π is the projection onto G, then i1 : G −
→ G × H is a
section. We note that a section is automatically injective.
Theorem 148. Let π : P −
→ G be a surjective homomorphism between two
groups, with kernel H. Assume that there is a section s : G −
→ P of π, such that
for all g ∈ G and h ∈ H, s(g) commutes with h. Then P is isomorphic to G × H.
Proof. Let G′ = s(G). Then by assumption, all elements of G′ commute
with the elements of H. If g ′ ∈ G′ , then g ′ = s(g) for some g ∈ G, and so
π(g ′ ) = g. Hence G′ ∩ H is trivial. Finally, if p ∈ P , let h = s(π(p))−1 p. Then
π(h) = π(s(π(p)))−1 π(p) = π(p)−1 π(p) = e, so h ∈ H. Since p = s(π(p))h, we
have shown that an arbitrary element p can be written as a product of an element
of G′ and an element of H. Using theorem 147, we see that P is isomorphic to
G′ × H. But s is an isomorphism of G with G′ (with inverse π), so P is isomorphic
to G × H.
□
Remark 149. Not every surjective homomorphism has a section. For example,
→ Z2 given by taking the residue mod 2. If s
consider the homomorphism π : Z4 −
is an injective homomorphism from Z2 , then s(1) must be an element of order 2.
The only such element in Z4 is 2, but 2 goes to 0 under π.
Remark 150. The theorem can be strengthened as follows: instead of requiring that s(g)h = hs(g) for all g ∈ G and h ∈ H, it is enough to require that
s(g)h = hs(g ′ ) for some g ′ ∈ G (exercise.) In the terminology introduced below,
this requirement says that s(G) is a normal subgroup of P .
Of course, even if a section exists, the commutativity condition of theorem 148
need not hold. However, it holds trivially if P is abelian:
End lecture
Mar 21
22,
Corollary 151. If a surjective homomorphism π : P −
→ G from an abelian
group P has a section, then P is isomorphic to G × Ker(π)
We can use this description to prove an important theorem about the structure
of finite abelian groups. Given a finite abelian group G and a prime number p, we
define the p-free part of G, G(p) , to be the set of elements whose order is prime to
p, and the p-part of G, Gp , to be the set of elements whose order is a power of p.
It is easy to see that G(p) and Gp are subgroups of G.
Corollary 152. If G is a finite abelian group, then it is isomorphic to G(p) ×
Gp .
Proof. Let q = pk be the highest power of p dividing the order of G. Then
x 7→ xq is a homomorphism from G to itself, whose kernel is Gp and whose image
is G(p) (both by Lagrange’s theorem.) Furthermore, this homomorphism is an
automorphism of G(p) , and so it has an inverse s, which is by definition a section.
□
Thus the corollary follows from corollary 151.
Corollary 153. Any finite abelian group is the direct product of groups Gp ,
where Gp has only elements whose order is a power of p.
Remark 154. We will see below that each group Gp is of order pk , the highest
power of p that divides the order of G.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
11. NORMAL SUBGROUPS AND QUOTIENTS
33
Example 155. The group U19 has 18 elements. Hence, the theorem says that
it is isomorphic to a product of two groups, one of which is of order 2 and the
other of order 9. In fact, it is isomorphic to Z2 × S, where S is the subgroup of
U19 consisting of elements that are squares of another element. The isomorphism
is given by x 7→ (i, x2 ), where i is 0 is x is a square, and is 1 otherwise.
□
11.2. Normal subgroups. We have seen that being isomorphic to a product
is a strong condition on a group. Unfortunately, a group is rarely a product of two
groups. In view of theorem 148, we may try replacing this condition by the weaker
condition that a group P has a surjective map onto a non-trivial group G. In more
detail, a plan to study all groups might look as follows:
• Find conditions when a group has a non-trivial surjective homomorphism
onto another group
• Study the groups that have no such homomorphisms (such groups are
called simple.)
• Determine in what ways the simple groups can be combined to give other
groups, and what can be deduced about a group from such a composition
For example, we have seen in 152 that a finite Abelian group always has a
surjective homomorphism onto another such group, which is non-trivial if the group
is not a p-group. Furthermore, the latter group is then a direct factor of the original
group, so combining groups is easy in this case. The full statement (which we shall
not prove) is that any finite Abelian group can be presented in a unique way as a
direct sum of cyclic groups. On the other hand, it is easy to see that the simple
Abelian groups are the finite cyclic groups of prime order.
It is in fact possible to classify all finite simple groups: there is a list of nonisomorphic finite simple groups, such that each finite simple is isomorphic to one
in the list. The proof is somewhat harder: it occupies around 15,000 pages!
If π : P −
→ G is a surjective homomorphism of abelian groups, it follows from
the previous section that either G is a direct factor of P , or π has no section. When
P (and possibly G) are not abelian, there is an intermediate case: there might be
a section that does not commute with the kernel. It turns out that this situation
is almost as good as having a product, and there is a rather rich theory (which we
shall not pursue) of both the existence of such sections and the consequences.
Currently we shall concentrate on the existence of surjective homomorphisms.
We have seen that a surjective homomorphism π : P −
→ G holds information about
another group: the kernel H of π. We now want to invert this observation and ask:
can we recover π (and G) from H? And: for which subgroups H of P is there a
homomorphism π : P −
→ G with kernel H?
Let H be a subgroup of P . Since we want H to be the kernel of a homomorphism, we may try to construct G by simply removing H from P , and replacing
→ G: π(x) = e
it by the identity. We will then have the following function π : P −
if x ∈ H, and π(x) = x otherwise. Unfortunately, G is not a group: if h ∈ H is
non-trivial, and g ∈ P − H, then the product of g −1 and gh is no longer defined in
G.
Example 156. Let P = S3 , and let H = {e, (12)}. Is there a group homomorphism π : P →
− G, to some group G, with kernel H? In the process above, we
would try to construct G be removing (12) from P , and defining pi((12)) = e and
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
34
1. GROUP THEORY
π(x) = x otherwise. However, G is not a group: the product of (23) and (132) is
no longer defined.
□
To fix the problem above, we could define the multiplication on G to be such
that (the image of) g −1 multiplied by gh is the identity. Since this should hold for
all g ∈ P and all h ∈ H, we need to identify any two elements of the form gh1 and
gh2 . In other words, each left coset should become one element. Thus we redefine
G to be the set of (left) H cosets, and the function π to be the function that send
g ∈ P to gH.
Example 157. Continuing the previous example, we now wish to define G to
be the set of left cosets of H. These are: e = H, a = (23)H = {(23), (132)} and b =
(13)H = {(13), (123)} (we know by Lagrange’s theorem that there are three cosets,
but you may wish to check directly that there are no others). Thus, G = {e, a, b}
is a set of three elements. We have π(e) = π((12)) = e, π((23)) = π((132)) = a,
π((13)) = π((13)) = π((123)) = b.
□
What is the group structure on G? If we want π to be a group homomorphism,
we have no choice but to define g1 H · g2 H = π(g1 )π(g2 ) = π(g1 g2 ) = g1 g2 H.
However, we must make sure that this is well defined, and gives a group structure.
We note that if f : P →
− G is a group homomorphism with kernel H, then for
any g ∈ P and h ∈ H, f (g −1 hg) = f (g −1 )f (h)f (g) = f (g)−1 f (g) = e. Thus,
g −1 hg ∈ H.
Definition 158. A subgroup H < P is called normal if for any g ∈ P ,
g −1 Hg ⊆ H. The notation H ⊴ P means that H is a normal subgroup of P .
Remark 159. It follows that in fact, g −1 Hg = H, by applying the condition
with g −1 instead of G. However, it is not true, in general, that g −1 Hg < H implies
that g −1 Hg = H.
End lecture
Mar 23
23,
Example 160. In the example above, if we want to define a group structure
on G making π a homomorphism, we get:
a ∗ b = f ((23)) ∗ f ((13)) = f ((23)(13)) = f ((123)) = b
but also,
a ∗ b = f ((132)) ∗ f ((13)) = f ((132)(13)) = f ((12)) = e
so we have a problem. The problem can be rephrased as follows: we have (132) =
(23)(12), which accounts for (23) and (132) going to the same element a. So for
any g ∈ S3 , we should have some h ∈ H with (23)(12)g = (23)gh. Cancelling (23),
we get (12)g = gh or g −1 (12)g = h, i.e., g −1 (12)g should be in H. However, for
g = (13), we get (13)(12)(13) = (23), which is not in H. This shows that H is not
normal in S3 .
□
The discussion just before the definition shows:
Corollary 161. The kernel of any homomorphism is a normal subgroup
In particular, since (as we saw above) there are subgroups which are not normal,
this shows that there exist subgroups that are not the kernel of any homomorphism.
However, it turns out that this is the only obstacle:
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
11. NORMAL SUBGROUPS AND QUOTIENTS
35
Theorem 162. Let H ⊴ P be a normal subgroup. Then the set P/H of left
H cosets has a group structure, such that the function π : P −
→ P/H given by
π(g) = gH is a surjective group homomorphism with kernel H. Furthermore, if
f :P −
→ G is another group homomorphism whose kernel contains H, then there is
a unique homomorphism f¯ : P/H −
→ G such that f = f¯ ◦ π.
The group is P/H is called the quotient group (also the factor group) of P
by H. The second part of the theorem implies that this is (essentially) the only
solution to the problem: find a surjective homomorphism from P with kernel H.
− P/H is surjective, and
Proof. We already know that the function π : P →
by definition, the fibre over the coset H (which will serve as the identity of P/H)
is H. Furthermore, we noticed that the requirement that π is a homomorphism
forces us to define the group law by g1 H · g2 H = g1 g2 H. It remains to show that
this formula determines a well defined group structure. However, the normality
condition implies that in P , the set g1 Hg2 H = {g1 h1 g2 h2 |h1 , h2 ∈ H} is equal to
g1 (Hg2 )H = g1 (g2 H)H = g1 g2 H. In other words, if X, Y ∈ P/H then the product
XY , in P , of the sets X and Y , is again a coset. This shows that the product is
well defined, and it follows in the same way that it defines a group.
Let f : P −
→ G be a homomorphism whose kernel contains H. Define f¯ :
P/H −
→ G by f¯(gH) = f (g). This is well defined, since if gH = g1 H, then
−1
g g1 ∈ H, hence f (g −1 g1 ) = e, so f (g) = f (g1 ). It is clearly a homomorphism
that satisfies the requirements, and since its definition is forced on us by these
requirements, it is unique.
□
Corollary 163. Let f : P →
− G be a surjective homomorphism with kernel
H. Then G is isomorphic to P/H. Furthermore there is a unique isomorphism
f¯ : P/H −
→ G such that f = f¯ ◦ π.
Proof. By the theorem, there is a unique group homomorphism f¯ with the
required property. It remains to show that it is an isomorphism. If g ∈ G, let p ∈ P
be such that f (p) = g. Then f¯(pH) = f¯(π(p)) = f (p) = g. This shows that f¯ is
surjective. If f¯(pH) = e then f (p) = f¯(π(p)) = e, hence p ∈ H, so pH = H.
□
Because of this corollary, any surjective map P −
→ G, as well as the group G
itself are referred to as quotients of P .
11.3. Examples and applications.
Example 164. Any subgroup of an Abelian group is normal, The quotient is
Abelian as well. The quotient of a cyclic group is cyclic.
□
Example 165. We have seen in example 160 that the subgroup {e, (12)} of S3
is not normal. On the other hand, the subgroup generated by (123) is normal. □
Example 166. The subgroup SLn of GLn is normal:
it is the kernel of the
[ ]
determinant. The subgroup of matrices of the form 10 xy is not normal[in GL
] 2 (or
0 1
in SL2 ). This can be shown directly by conjugating with the element −1 0 , but
also follows from example 169, since this subgroup is precisely the stabiliser of the
□
vector (1, 0) in the action of GL2 on the plane.
Example 167. The group Z/nZ is isomorphic to Zn : the function from Z to
Zn assigning to each number its residue modulo n is a surjective homomorphism
whose kernel is nZ.
□
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
36
End lecture
Mar 25
24,
1. GROUP THEORY
Example 168. If H < G is a subgroup of index 2, then it is normal. Indeed,
if g ̸∈ H, then gH is the complement of H. But then any element ghg −1 , where
h ∈ H, must be either in H or in gH. It cannot be in gH, so it is in H.
□
Example 169. Let G be a group acting on a set X. The subset H of G
consisting of elements that act trivially (gx = x for all x ∈ X) is a normal subgroup,
since it is the kernel of the map G −
→ Sym(X). It follows that we have an induced
map G/H −
→ Sym(X), i.e., an induced action of G/H on X. This action is faithful :
the only element that acts trivially is the identity.
Let G × X −
→ X be a transitive and faithful action, and let x ∈ X be an
element. If Gx is the stabiliser of x, and g ∈ G, then gGx g −1 = Gg(x) . Since the
action is transitive, any point of X has the form g(x) for some g ∈ G. Hence all
the stabilisers are conjugate, and all conjugates of a stabiliser are stabilisers. In
particular, if Gx is normal, then all the stabilisers are equal. But then any element
in Gx fixes all points, and thus (since the action is faithful) is equal to the identity.
Hence the only way a stabiliser can be normal is if it is trivial.
□
Example 170. Recall that A4 has order 12. We prove that it has no subgroup
of order 6. Indeed, by example 168, such a group would be normal. It follows that
the square of any element of A4 is in H. However, A4 has 8 elements of order 3,
whose squares are therefore distinct, and so cannot all be in H.
On the other hand, the subgroup of A4 consisting of elements of order (at most)
2 is a subgroup, which is normal, since conjugation in Sn preserves the form of the
disjoint cycle decomposition (see proposition 179 below.) Hence A4 has a normal
(and even characteristic) proper subgroup. In contrast, we will see in theorem 180
that for n ≥ 5, An is simple.
□
Example 171. If n > 1 is a number, then the rotation R by a half circle is
an element of D2n . The subgroup {e, R} is normal, since R commutes with all
rotations, and is the only rotation of order 2. The quotient can be identified with
Dn by identifying any two opposite points of the regular 2n-gon, turning it into an
n-gon.
□
The following application proves that the group of inner automorphisms (homework) is not cyclic (unless it is trivial.)
Proposition 172. If G/Z(G) is cyclic then G is Abelian.
End lecture
Mar 28
25,
Proof. We prove something stronger: if x, y ∈ G commute, and x′ , y ′ represent the same cosets in G/Z(G) as x and y, respectively, then they commute as well.
In other words, the commutator depends only on the class in G/Z(G). Indeed, if
x′ = xc and y ′ = yd, where c and d are in the centre, then x′ y ′ = xcyd = xycd =
yxdc = ydxc = y ′ x′ .
This proves the proposition, since the assumption implies that there is an element g ∈ G such that any class is represented by some g i , and all such elements
commute.
□
The following theorem can be viewed as a partial converse to Lagrange’s theorem (we have seen in example 170 that the full converse is false.)
Theorem 173 (Cauchy’s theorem). If a prime p divides the order of a finite
group G, then G contains an element of order p.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
11. NORMAL SUBGROUPS AND QUOTIENTS
37
Proof. We prove the theorem in the case that G is abelian; the general case
will be proved after corollary 178. Let G be a counterexample of minimal order,
and let a ∈ G be a non-trivial element. By assumption, the order of a is prime to
p (otherwise a power of a has order p.) Hence the order of the quotient G/⟨a⟩ is
again divisible by p. By minimality, it contains an element b of order p, but any
pre-image of b will then have order divisible by p.
□
Remark 174. In fact, a stronger result is true: if pk divides the order of G,
then G has a subgroup of order pk . This is known as the first Sylow theorem, and
a subgroup of maximal order pk is called a p-Sylow subgroup. The other Sylow
theorems say that every p-subgroup of G is contained in a p-Sylow subgroup, and
that all p-Sylow subgroups are conjugate.
Remark 175. Combined with corollary 153, it follows from Cauchy’s theorem
that any finite abelian group is isomorphic (canonically) to a product Gp1 ×. . .×Gpk ,
where the pi are distinct primes, and the order of Gpi is a power of pi .
End lecture
Mar 30
26,
End lecture
More generally if the group action is not necessarily transitive, then X is a Apr 1
disjoint union of the G orbits, and the action on each orbit is transitive. Therefore,
if X is finite, we get the following formula for the size of X:
∑
|X| =
(G : Gx )
(2) {eq:orbits}
27,
11.4. Conjugacy classes. To prove the general case of Cauchy’s theorem,
we recall the following fact from the homework: If G × X −
→ X is a transitive group
action, and x ∈ X, then the map G →
− X given by g 7→ gx induces a function
G/Gx −
→ X, which is a bijection. In particular, If G is finite (and hence so is X),
|G|
then |X| = |G
.
x|
Example 176. The action of D4 on the set of vertices of the square is transitive.
The stabiliser of a vertex consists of the identity and the reflection around the axis
that passes through that axis. Since D4 has 8 elements, we get that a square has
four vertices.
□
Example 177. The action of Sn on [n] is transitive. Hence Sn /H has size n,
where H is the stabiliser of n. We saw that H is isomorphic to Sn−1 , so this gives
another proof that the order of Sn is n!.
□
X/G
where the sum is over the set of orbits, and x is any element of the corresponding
orbit.
We apply this formula in the following example: recall that any group G acts
−
on itself by conjugation (inner automorphisms): ch (g) = hgh 1 (exercise.) An orbit
for this action is called a conjugacy class. Thus, the conjugacy class of an element
g is the set of all elements of the form hgh−1 , where h ∈ G. The stabiliser of g is
the set of all h such that hgh−1 = g. In other words, it is the centraliser Cg of g.
Substituting this in equation (2), we get:
Corollary 178 (Class equation). If G is a finite group, then
∑
∑
|G| =
(G : Cg ) = |Z(G)| +
(G : Cg )
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
(3)
{eq:class}
38
1. GROUP THEORY
where the first sum is over all conjugacy classes, and the second is over conjugacy
classes of size bigger than 1 (and in each case g is an element in the corresponding
class.)
The second formula follows from the first one since the elements of Z(G) are
precisely the conjugacy classes of size 1. We can now prove the general case of
Cauchy’s theorem (Theorem 173.)
End lecture
Apr 4
28,
Proof of Cauchy’s theorem, general case. Let G be a counterexample
of minimal order. By minimality, G cannot contain a proper subgroup whose order
is divisible by p. In particular, if x ∈ G is not in the centre, then the order of Cx
cannot be divisible by p. Hence any term in the sum in the second form of the
class formula is divisible by p. Since the order of G is also divisible by p, it follows
that so is the order of Z(G). Hence G = Z(G), and we are back in the previous
case.
□
We next compute the conjugacy classes in Sn :
Proposition 179.
(1) If s is an element of Sn and c = (a1 a2 . . . ak ) is
a cycle, then scs−1 = (s(a1 )s(a2 ) . . . s(ak ))
(2) Two elements of Sn are conjugate if and only if their disjoint cycle decomposition has the same form.
End lecture 29,
Apr 6 (Midterm 2)
Proof.
(1) Since s is a bijection, any element of [n] can be written
(uniquely) as s(x) for some x. Applying scs−1 we get s(x) if x is not
one of the ai , and s(c(ai )) if x = ai . This is precisely the function given
by the cycle in the statement.
(2) If g is any element, conjugating by s amounts, by the previous part, to
applying s to each element in the disjoint cycle decomposition. After
applying s, we get a product of disjoint cycles, which is therefore the
disjoint cycle decomposition of the conjugate. Conversely, if the cycle
decomposition is has the same shape, we may find a permutation s that
maps the elements of corresponding cycles to each other.
□
Recall that a group is simple if it has no non-trivial quotients. Given that
quotients correspond to normal subgroups, a group is simple if and only if it has no
proper non-trivial normal subgroups. For example, an Abelian group is simple if
and only if it is cyclic of prime order. The following theorem was prove by Galois,
and combined with Galois theory (see section 1) it shows polynomial equations of
degree greater than 4 cannot be solved.
Theorem 180. For n > 4, An is simple
Sketch of proof. Let G be a normal subgroup of An . If G contains a 3-cycle
(abc), we will show that it contains any other 3-cycle. This will show that G = An ,
since the 3-cycles generate An .
Since G is normal, it contains any conjugate of (abc). By proposition 179, such
conjugates have the form (s(a)s(b)s(c)), where s ∈ An . Thus we need to find, for
any cycle (xyz) an element s ∈ An that maps (abc) to (xyz). We would like to take
s = (ax)(by)(cz). This element works, but might be odd. However, since n > 4,
there are at least two elements d, e distinct from a,b and c, and we may use s · (de)
instead.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
11. NORMAL SUBGROUPS AND QUOTIENTS
39
It remains to show that G must contain at least one 3-cycle. This can be proven
directly by induction on the maximal number of elements fixed by an element of
G. See Milne [2, Lemma 4.36] for details.
□
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture
Apr 8
30,
CHAPTER 2
Galois theory
In this chapter, we will study an application of group theory to a classical problem — solving polynomial equations by radicals. Along the way we will introduce
a new kind of algebraic structure, namely fields. The problem was studied by Abel
and Galois (among others), and in the course of the solution they came up with
both the notions of groups and fields, and the connection between them.
The route we take is rather direct, omitting many important notions, and
leaving a lot to exercises. See Milne [1] or Rotman [3] (for example) for more
detailed accounts of Galois theory, and field theory in general.
1. Statement of the problem, and its solution
In the current section we introduce the problem, and the main algebraic structure associated to it, field.
To describe the problem, we recall that the solutions
of a quadratic equation
√
2
x2 +ax+b = 0 can always be written in the form −a± 2a −4b . In general, a function
of the form p(x) = xn + an−1 xn−1 + · · · + a0 is called a (monic) polynomial , and
an equation of the form p(x) = 0 is called a polynomial equation. n is called the
degree of the polynomial (or the equation). There are explicit formulas, similar to
the above, for the solutions of equations of degrees 3 and 4, which were known since
the 16th century. For example, a solution to the equation
x3 + ax2 + bx + c = 0
is given by
√
x=
3
−r +
√
√
√
3 −r −
r2 + 4q 3 /27
r2 + 4q 3 /27
+
2
2
(4) {eq:cubic}
where
q =b−
a2
3
2a3
ab
−
27
3
and the other two solutions involve similar formulas.
However, there was no formula for equations of degree 5 and above. Abel and
Galois proved that no such formula exists.
To formulate the problem more precisely, we first note that, if we start with
a polynomial equation (even quadratic) with coefficients in Q, the solutions above
are not rational. They lie in some bigger “set of numbers”. For the formula above
to make sense, in this bigger set of numbers it should be possible to use addition,
r =c+
41
42
2. GALOIS THEORY
multiplication and division, just as in the rationals. Such a set of numbers is called
a field. More precisely:
Definition 181. A field is a set L together with two operations + and ·, and
two elements 0 and 1, such that
(1) (L, +) is an Abelian group, with identity 0.
(2) · is a commutative operation such that (L∗ , ·) is an Abelian group, with
identity 1 (L∗ = L − {0})
(3) For any x ∈ L∗ , the map given by multiplication by x is a homomorphism
of the additive group
If L is a field, and K is a subset of L which is closed under the operations, and
which is itself a field, then K is called subfield of L, and L is a field extension of K.
Example 182. The sets Q, R and C are all fields with the usual operations,
and Q is a subfield of R, which is a subfield of C. The set Z of integers is not a
field, since multiplication does not give a group.
□
Exercise 183. Prove that (Zn , +, ·), where + and · are multiplication mod n
is a field if and only if n is prime. In this case, the field is denoted by Fn .
Exercise 184. Let s be a non-zero rational number, and let K be the set of
pairs (a, b) of rational numbers, written as a + br, where r is a fixed symbol. Define
operations on K by
(a + br) + (c + dr) = (a + c) + (b + d)r
(a + br)(c + dr) = (ac + sbd) + (ad + bc)r
Show that K with these operations is a field if and only if s does not have a rational
square root, and that in K, r2 = s (where
[ a b ] r is 0 + 1r). Hint: one way to do it is to
identify K with the set of matrices sb
a .
Note that if we replace rational numbers by real numbers, and take s = −1, we
have constructed the complex numbers.
The construction in the last exercise can be described as follows: we start
with a polynomial equation x2 − s = 0 that does not have a solution in the field,
and produce a new field K that contains all possible solutions of the equation.
Furthermore, it is the smallest extension of Q with this property. This kind of
construction is central to the theory, and in the terminology below, K is the splitting
field of x2 − s. To discuss this more precisely, we first note that the notions of
polynomials and rational functions makes sense for any field.
Definition 185. Let K be a field. A polynomial (in one variable t) over K is
an expression of the form p(t) = an tn + · · · + a0 , where each ai is an element of K.
We identify an expression as above with the expression 0tn+m +· · · +0tn+1 + an tn +
· · · + a0 for any m (i.e., they define the same polynomial), so that the coefficient ai
above is defined for all i (and is equal to 0 for i > n). Likewise, we sometimes omit
terms whose coefficient is 0.
We identify an element x of K with the polynomial for which a0 = x and ai = 0
for i > 0. The degree of a non-zero polynomial is the largest i for which ai ̸= 0. A
non-zero polynomial is monic if ad = 1, where d is the degree.
The set of all polynomials is denoted by K[t]. We define operations of addition
and multiplication on the polynomials in the usual way: if p(t) = am tm + · · · + a0
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
1. STATEMENT OF THE PROBLEM, AND ITS SOLUTION
43
and q(t) = bn tn + · · · + b0 , with m ≥ n, then
p + q = q + p = (am + bm )tm + · · · + (a0 + b0 )
(5)
p · q = am bn tm+n + (am bn−1 + am−1 bn tm+n−1 + · · · + a0 b0
(6)
Polynomials in several variables are defined analogously, and the operations between
them are defined in a similar way.
If p(t) is a polynomial over K, and x is an element of K (or, more generally,
a field L extending it), we may substitute x for t and get a new element of the
same field. The polynomial thus determines a function from L to L, and viewed
in this way, the addition and multiplication operations correspond to addition and
multiplication of functions.
This construction provides another class of examples of fields.
Example 186. Given a field K, the set of polynomials K[t] is not a field (with
the operations above), since there are no multiplicative inverses. The set of rational
functions over K (in one variable t) is defined as the set of expressions
p
K(t) = { ∥p, q ∈ K[t], q is monic}
(7)
q
With addition and multiplication extended to K(t) in the usual way, K(t) is a
field, extending K. We may apply this construction when K = L(s) is itself a field
of rational functions, to get the field L(s, t) of rational functions in two variables,
and so on.
□
We now go back to the question solutions to polynomial equations. Though
the original question was formulated for polynomials with rational coefficients, it
is essential to discuss it more generally. Thus, we are interested in solutions to the
equation p(x) = 0, where p is a polynomial over a field K. As we saw, one cannot
expect to find the solutions in the field K itself, so we would like to find a nice field
extension where these solutions exist. One of our main goals will be to prove the
following theorem.
Theorem 187. Let K be a field, and let p be a polynomial over K. Then there
is a field L with the following properties:
(1) If M is a field extending L that contains a solution a to p(x) = 0, then
a ∈ L.
(2) If K ⊆ M ⊂ L is a proper subfield of L, then there is a solution of p(x) = 0
in L, which is not in M .
Furthermore, the field L is unique up to isomorphism over K 1
Definition 188. Given a field K and a polynomial p over it, the field L given
by Theorem 187 is called the splitting field of p (over K).
Thus, the splitting field of p is the minimal field that contains all possible
solutions of p(x) = 0.
Example 189. The field constructed in Exercise 184 is the splitting field of
x2 − s (this will become apparent later).
□
1isomorphisms of fields are defined in Definition 204
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
44
2. GALOIS THEORY
Now that we have a field that contains all solutions, what does it mean to have
a formula for them? The formula for the quadratic equation involves elements of
the base field, as well as roots of such elements. Hence, they live in a splitting field
as considered in Example 189. For the cubic equation, we have (third) roots of
elements in a splitting field of that form. This motivates the following definition.
Definition 190. The equation p(x) = 0 (over a field K) is solvable by radicals
if there is a sequence of field extensions K0 = K ⊆ · · · ⊆ Kn , such that for i > 0,
Ki is the splitting field of a polynomial of the form xk − a with a ∈ Ki−1 , and Kn
contains a splitting field for p.
Example 191. Any quadratic equation x2 + ax + b = 0 (over Q) is solvable by
radicals: if we take K1 to be the splitting field of x2 −(a2 −4b) (as in Example 189),
both solutions of the original equation belong to K1 (in fact, K1 is the splitting
field of the original equation).
□
Example 192. A cubic equation x3 + ax2 + bx + c = 0 is also solvable by
radicals. Considering equation (4), we first note that q and r are in the base field.
The terms within the cube root are the two solutions u and v of the quadratic
q3
, so both lie in its splitting field K1 , which is an extension by
equation y 2 + ry − 27
a root. If we let K2 be the splitting field of x3 − u over K1 , and K3 the splitting
field of x3 − v over K2 , then K3 contains the solution given by Equation (4) (the
other two solutions are obtained similarly).
□
We may now finally state precisely the result mentioned informally before.
Theorem 193 (Abel–Ruffini). There is a polynomial of degree 5 over Q that
is not solvable by radicals.
End lecture
Apr 13
31,
To prove the result, one needs to analyse the relation between different sequences of field extensions. This is where group theory enters the picture. The full
theorem, due to Galois, describes which equations are solvable in terms of some
associated groups.
2. First properties of fields
In this section we mention some basic properties and definitions related to
fields. Much of the theory is parallel to (and uses) the theory of groups. Therefore,
we leave most of it as an exercise.
Exercise 194. Let x, y be elements of a field K. Prove the following facts.
(1) 0 · x = x · 0 = 0
(2) −x = −1 · x (−x is the inverse of x with respect to addition).
(3) If xy = 0 then x = 0 or y = 0 (or both).
Exercise 195. Let K be a field, Ki ⊆ K a collection of subfields. Prove that
∩i Ki is a subfield (you may use Theorem 38).
2.1. Generators of a field. As with subgroups, interesting subfields of a field
often arise as intersections.
Definition 196. Let K be a field, A ⊆ K a subset. The subfield generated
by A is the intersection of all subfields of K containing A (this is a subfield by
Exercise 195).
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
2. FIRST PROPERTIES OF FIELDS
45
If F ⊆ K is a subfield, the subfield generated by A over F is the subfield
generated by F ∪ A.
√
Example 197. The field C of complex numbers is generated over R by i = −1.
If K is any field, the field K(t) is generated over K by t.
□
As with subgroups, it is possible describe the subfield generated by A more
explicitly (compare Proposition 44).
Proposition 198. If F ⊆ K is a field extension, and A ⊆ K is a subset, then
the subfield generated by A over F consists of all elements of the form f (a1 , . . . , an )
where f is a rational function over F , and a1 , . . . , an are elements of A on which
f is defined.
Proof. For any such function f and elements a1 , . . . , an , f (a1 , . . . , an ) is a ratio p(a1 , . . . , an )/q(a1 , . . . , an ), where p and q are polynomials, and q(a1 , . . . , an ) ̸=
0. If the ai all belong to a subfield L containing F , then so do p(ā) and q(ā, since
these are sums of products of elements of L. Hence so does the ratio. This proves
that the subfield generated by A over F contains all elements as in the claim.
On the other hand, the set of all such elements forms a subfield, since the set
of rational functions over F is a field.
□
2.2. The characteristic of a field. If x is an element of a field K, and n
is an integer, we write nx and xn for the “product” of x with itself n times in the
corresponding group structure (so x2 = x · x, 2x = x + x, x0 = 1, x−1 is the inverse
of x with respect to ·, etc.). We abbreviate n · 1 as n (note that now nx a priori
has two different meanings, which in fact coincide). We note that we may have a
natural number n > 0 such that n = 0 in K. If this happens, there is a smallest n
with this property.
Definition 199. Let K be a field. If there is an integer n > 0 such that n = 0
in K, the smallest such integer is called the characteristic of K. Otherwise, we say
that K has characteristic 0. We denote by ch(K) the characteristic of K.
Example 200. The fields Q, R and C all have characteristic 0. The field F2
with two elements has characteristic 2.
□
Exercise 201. Prove that the characteristic of field is either 0 or a prime
number
Exercise 202. Prove that if K has characteristic p, then px = 0 for all x ∈ K.
Exercise 203. Prove that if K ⊂ L is a field extension, then K and L have
the same characteristic.
2.3. Field homomorphisms. We define the relevant maps between fields.
Definition 204. Let K and L be fields. A function f : K −
→ L is a field
homomorphism if it is a non-zero homomorphism of the additive groups, and
f (ab) = f (a)f (b) for all a, b ∈ K. A field isomorphism is an invertible homomorphism.
If F is a subfield of both K and L, then a homomorphism as above is over F
if f (x) = x for any x ∈ F .
Exercise 205. Prove the following facts about field homomorphisms.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
46
2. GALOIS THEORY
(1) Any field homomorphism is injective
(2) A field homomorphism is an isomorphism if and only if it is surjective
(you may use Proposition 66).
(3) The image of a homomorphism is a subfield.
(4) If there is a homomorphism from K to L, then K and L have the same
characteristic.
It follows from the last exercise that if there is a homomorphism from K to L,
then K is isomorphic to a subfield of L (namely, the image).
Exercise 206. Prove that any homomorphism from a finite field to itself is an
isomorphism
2.4. Prime fields. Another application of Exercise 195 is to find the smallest
subfield of a field.
Definition 207. Let K be a field. The prime subfield of K is the intersection
of all subfields of K.
Exercise 208. Show that the prime subfield of Q and of C is Q, and the prime
subfield of F2 is F2 .
The prime subfields are completely determined by the characteristic:
Proposition 209. Let K be a field. The following are equivalent (where, for
convenience, we write F0 for Q):
(1) The characteristic of K is p.
(2) There is a homomorphism from Fp to K.
(3) The prime subfield of K is isomorphic to Fp
Proof. We prove only that (1) implies (2), since the rest follows from previous
exercises. Assume first that p > 0. The assumption says the order of 1 in the
additive group of K is p. Hence, there is a unique injective group homomorphism
from the additive group Zp of Fp to K. Since (n1) · (m1) = (nm)1 in K, this is a
field homomorphism.
Now assume that p = 0. Then for any non-zero integer n, n1 ̸= 0 in K, and
1
n
. The map that send a rational number m
to the
therefore it has an inverse n1
1
element n1 · m1 of K is then a field homomorphism.
□
Exercise 210. Prove that the only isomorphism from Fp to itself is the identity.
Conclude that the homomorphism in the Proposition is unique (include the case
F0 = Q).
To summarise, any prime field is either Q or Fp for some prime p. Exactly
one of these field is contained in any field, depending on the characteristic, and
in a unique way. If K and L are fields of different characteristics, there are no
maps between them, they have no common subfields, and no common extensions.
Hence, the theories of fields of different characteristics are, from our point of view,
unrelated, and we may work with a fixed characteristic. We will mostly concentrate
on the characteristic 0 case, since it is simpler and contains the classical application.
End lecture
Apr 15
32,
Exercise 211. Prove that Fp is the only field of size p
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
3. POLYNOMIALS IN ONE VARIABLE
47
3. Polynomials in one variable
We now endeavour to prove Theorem 187. Hence we are interested in constructing a field extension L of a field K, containing a solution to an equation
p(x) = 0, with p a polynomial over K. By analogy with the case of usual roots, we
call any solution to such an equation a root of p. Our general strategy will be to
add a solution formally, as in Exercise 184. To make it work, we first need to study
polynomials in one variable in general. We fix a base field K. All polynomials will
be in one variable and over K, unless mentioned otherwise.
As explained in the definition, the polynomials are endowed with operations
of addition and multiplication. It can be easily checked that K[x] forms a group
under addition, but the non-zero polynomials do not, in general, have inverses under multiplication. In other words, K[x] is a (commutative) ring. This is another
interesting class of algebraic structures, but we will not study them in general, since
we will only need this example. On the other hand, as will be seen immediately, the
algebraic properties of K[x] are very similar to the properties of the integers, with
usual addition and multiplication. It was already shown above that the rational
functions are obtained from the polynomials in the same way as the rational numbers are obtained from Z. Other examples come from notions related to divisibility,
as introduced below.
Exercise 212. Let p and q be two polynomials.
(1) Show that deg(pq) = deg(p) + deg(q).
(2) Conclude that an element of K[x] has an inverse under multiplication if
and only if it is a non-zero element of K.
(3) Conclude that pq = 0 if and only if p = 0 or q = 0.
(4) Show that if r is a non-zero polynomial, and rp = rq, then p = q.
Definition 213. Let p and q be two polynomials. We say that q divides p if
there is a polynomial r with p = qr. If this is the case, we write q|p.
Exercise 214. Let p, q and r be polynomials. Show the following:
(1) If p|q and q|r, then p|r
(2) If p|q and q|p, then there is an element a of K ∗ such that p = aq. In
particular, if p and q are both monic, then p = q.
Because of the last part in the exercise, it is often convenient to restrict attention to monic polynomials. As for integer numbers, we get division with remainder.
Proposition 215. Let p and q be polynomials, q ̸= 0. Then there are unique
polynomials s and r, such that p = sq + r, and deg(r) < deg(q).
The proof is essentially long division, as with integers.
Proof. We prove existence by induction on deg(p). When deg(p) ≤ 0, this
is an exercise. Let deg(p) = n, deg(q) = k. If k > n, take s = 0 and r = p.
Otherwise, we may assume p and q are monic (and multiply s be a constant later).
Then p1 = p − xn−k q is a polynomial of smaller degree than p. By induction,
p1 = s1 q + r, so p = xn−k q + s1 q + r = (xn−k + s1 )q + r, and taking s = xn−k + s1
we are done.
For uniqueness, assume that s1 q + r1 = p = s2 q + r2 . Rearranging, we get
(s1 − s2 )q = r2 − r1 . However, the degree of q is greater than the degree of r2 − r1 ,
so the only way this can happen is if s1 − s2 = 0 = r2 − r1 .
□
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
48
2. GALOIS THEORY
The following corollary will be relevant for us.
Corollary 216. If p(a) = 0 for some field element a, then p is divisible by
(x − a).
Proof. By long division, p(x) = (x − a)s(x) + r, where deg(r) < 1, hence r is
a field element. Plugging a in both sides yields r = 0.
□
Corollary 217. Any non-zero polynomial p has at most deg(p) roots (in any
field).
Next, we have common divisors. Again, this is similar to the integers.
Proposition 218. Let p and q be non-zero polynomials. Then there is a unique
monic polynomial d dividing p and q, such that if t is some other polynomial dividing
p and q, then t|d.
Definition 219. The polynomial d given in Proposition 218 is called the greatest common divisor of p and q (abbreviated gcd ). The polynomials p and q are
coprime if gcd(p, q) = 1.
The proof uses Euclid’s algorithm, which also provides a way of computing the
greatest common divisor.
Proof of Proposition 218. We may assume deg(p) ≥ deg(q), and both
monic. By long division, p = sq + r, with deg(r) < deg(q). If r = 0, we let d = q.
Otherwise, by induction, there is an element d satisfying the requirements for q and
r. Hence, q = q1 d and r = r1 d, so p = sq + r = sq1 d + r1 d = (sq1 + r1 )d, hence d
divides p. Assume t divides both p and q. Then it divides r, so by the choice of d,
it divides d.
For uniqueness, if e is another polynomial satisfying the properties, then e|d,
d|e and both are monic, so e = d by Exercise 214.
□
As with the integers, Euclid’s algorithm also provides a way to write the gcd
of two polynomials as a “linear combination” of them.
Corollary 220. For any non-zero polynomials p and q, there are polynomials
r and s with gcd(p, q) = rp + sq.
Proof. Exercise
□
We next come to the analogue of prime numbers.
Definition 221. A non-zero polynomial p is irreducible (over K) if there are
no polynomials q and r of positive degree, such that p = qr.
Exercise 222. Show that and irreducible polynomial over K has no root in K
Recall that every polynomial over K can also be viewed as a polynomial over
L, for any extension L of K. Whereas the previous notions did not depend on the
base field (for instance, p|q as polynomials over K if and only if the same holds over
L), irreducibility depends on our base field. Hence when discussing it, we should
be careful about the base field. As before, unless we mention otherwise, we work
over the fixed base K.
As with integers, we have prime decomposition.
Proposition 223. Let p be an irreducible monic polynomial
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
3. POLYNOMIALS IN ONE VARIABLE
49
(1) If q is a non-zero polynomial, then either p|q or gcd(p, q) = 1.
(2) If p|p1 . . . pk , where each pi is monic irreducible, then p = pi for some i.
Proof.
(1) Exercise
(2) By induction on k. Let q = p1 . . . pk−1 . If p|q, then by induction p = pi
for some i. Otherwise by the first part, gcd(p, q) = 1. Hence by Euclid’s
algorithm, there are s and t with ps + qt = 1. Hence pk ps + pk qt = pk .
Since p|pk q, we get that p divides the left hand side, so p|pk . Since pk is
irreducible, we get p = pk .
□
Corollary 224. Any monic polynomial p is a product p1 . . . pk of monic irreducible polynomials of positive degree. The polynomials p1 , . . . , pk are unique, up
to reordering.
Proof. If p = qr is reducible, each of q and r are of smaller degree, so by
induction, each is a product of irreducibles. Otherwise, p itself is the decomposition.
For uniqueness, assume that p1 . . . pk = q1 . . . ql are two decompositions. Then
p1 divides q1 . . . ql , so by the proposition, p1 = qi . Cancelling, we get the result by
induction.
□
We may now construct the splitting field.
Proposition 225 (Kronecker’s Theorem). Let p(x) be a non-constant polynomial over a field K. Then K has a field extension L = K(α) generated by one
element α which is a root of p.
Proof. We note first that p may be assumed irreducible, since a root of p is
also a root of one of the irreducible components of p. Likewise, we may assume p
to be monic.
Let n be the degree of p. We define L as follows. As a set, L consists of
polynomials of degree smaller than n in a variable α. The additive group structure
is the same as for usual polynomials (note that the degree of a sum of polynomials
is bounded by the degrees of the summands).
If s(α) and t(α) are in L, we define s · t to be the product mod p, i.e., the
residue of the usual product st when divides by p(α). In other words, s(α)t(α) =
q(α)p(α) + (s · t)(α). By Proposition 215, s · t is well defined, and is an element of
L. It is easy to verify that this product is associative. Hence, to prove that L is a
field, we only need to show that each non-zero q(α) is invertible.
Since p is irreducible, and the degree of q is smaller than n, we have gcd(p, q) =
1. Hence, by Corollary 220, there are polynomials s(x) and t(x) with s(x)p(x) +
t(x)q(x) = 1, hence t(x)q(x) = 1 − s(x)p(x). Hence by definition, t(α) is the inverse
of q(α).
This shows that L is a field. To show that p(α) = 0, we note that p(x) − p(0)
has a root at x = 0, so p(x) − p(0) = xq(x). Since q has lower degree, we get
□
α · q(α) = −p(0). Hence p(α) = α · q(α) + p(0) = −p(0) + p(0) = 0.
Exercise 226. Verify that if s ∈ K has no square root in K, then the construction above recovers the field constructed in Exercise 184.
Corollary 227. Given a monic polynomial p over K, there is a field L generated over K by roots of p, such that in L, p is a product (x − b1 ) . . . (x − bk ) of
linear terms. The field L is a splitting field for p.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture
Apr 18
33,
50
2. GALOIS THEORY
Proof. If p = 1 the statement is obvious. Otherwise, by Proposition 225, there
is a field L = K(b1 ) such that p(b1 ) = 0. By Corollary 216, p(x) = (x − b1 )p1 (x)
for some polynomial p1 (x) over L. By induction, p1 is a product of linear terms in
a field generated over L by roots of p1 . This field is generated over K by roots of
p, and p is a product of linear factors there.
Any root of p is one of the bi , so all possible roots are contained in L. If M is
a subfield of L that contains all roots, then it is equal to L, since L is generated by
□
the roots. Hence L is a splitting field.
The name “splitting field” comes from the fact that p splits into linear factors
in L.
4. Linear algebra
To show the uniqueness of the splitting field, we recall some basic linear algebra.
Definition 228. A linear space (or a vector space) over a field K is an Abelian
group V (written additively), together with a map m : K × V −
→ V such that:
∗
(1) m restricts to an action of K on V by group automorphisms
(2) For any x, y ∈ K and v ∈ V , m(x + y, v) = m(x, v) + m(y, v).
As usual, we will write the action m as a product: m(x, v) = x · v = xv.
A subgroup of V is a linear subspace if it is closed under m.
Example 229. The field itself is a vector space over itself, with the action
given by the field multiplication. More generally, the set of tuples K n is a vector
space in the usual way.
□
Example 230. The set of continuous, differentiable, smooth, rational,... functions on the reals (or on an interval, or a domain,...) is a vector space over the real
numbers, with usual multiplication by numbers is a vector space. We note each of
these sets also has a product that is defined among the elements, which we ignore.
The set of positive real functions is not a vector space in the same way.
□
Example 231. The trivial group is a vector space over any field.
□
Example 232. If L is a field extension of K, then L is a vector space over K,
using the field multiplication.
□
Exercise 233. Show that if V is a vector space over a field of characteristic
p, then pv = 0 for all v ∈ V . More generally, for any element v ∈ V , m(n, v) = nv
(the left side is the action of n, viewed as an element of the field, on v; the right
side is the sum of v with itself in the group V ).
Definition 234. Let U and V be two linear spaces over a field K. A group
homomorphism T : U −
→ V is a linear map if T (xu) = xT (u) for any x ∈ K and
u ∈ U.
Exercise 235. Show that the kernel and the image of a linear map are linear
spaces. Show that if a linear map is invertible as a map of groups, then the inverse
is also a linear map.
Exercise 236. Show that the intersection of any collection of subspaces of V
is again a subspace
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
4. LINEAR ALGEBRA
51
Exercise 237. Show that if U and V are linear spaces, then x(u, v) = (xu, xv)
defines a linear space structure on the group U × V . It is the unique vector space
structure making the two projections linear maps.
Exercise 238. Let U be a subspace of a vector space V . Show that the quotient
group V /U has a unique vector space structure making the map π : V −
→ V /U linear.
Definition 239. Let U be a vector space. A basis of U is a subset B of U −{0}
with the property that any function (of sets) t : B −
→ V , where V is another vector
space, can be extended uniquely to a linear map from U to V (i.e., there is a unique
linear map T : U →
− V with T (b) = t(b) for all b ∈ B).
Exercise 240. Show that U and V have bases that have the same cardinality,
then U and V are isomorphic.
Definition 241. Let v1 , . . . , vn ∈ V . A linear combination of the vi is an
expression of the form x1 v1 + · · · + xn vn . The linear combination is non-trivial if
not all the xi are 0.
A subset B of a vector space U is linearly independent if no non-trivial linear
combination of elements of B is 0.
The subset B spans U if it is not contained in any proper subspace of U .
Exercise 242. Show that B spans U if and only if any element of U is a linear
combination of elements of B.
Proposition 243. Let B be a subset of V − {0}. The following are equivalent.
(1) B is a basis of V
(2) B spans V and is linearly independent
(3) B is a maximal linearly independent subset
(4) B is a minimal spanning set
Proof. Exercise
□
Theorem 244. Any two bases of the same space have the same cardinality (i.e.,
there is a bijection between them). Any linearly independent set can be extended to
a basis. Any spanning set has a subset that is a basis.
Given any set B, there is a vector space over K containing B, and in which B
is a basis.
Since the empty set is linearly independent, the Theorem asserts, in particular,
that any linear space has a basis. The proof depends on some set theory, so we skip
it.
Definition 245. The cardinality of any basis of U is called the dimension of
U . U is finite dimensional if it has a finite basis.
Example 246. The space K n has dimension n. A basis is given by (ei )i , where
ei has i-th coordinate 1, and the others 0.
It follows that a linear map T from K n to K m can be give by a matrix, namely
the matrix whose columns are T (ei ), for ei the standard basis on K n .
□
Corollary 247. If U is a subspace of V , then V is isomorphic to U × V /U .
If V has finite dimension d, then V is isomorphic to K d (a similar statement holds
for infinite dimensional spaces).
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture
Apr 20
34,
52
2. GALOIS THEORY
Proof. Let B be a basis of U . Then B is linearly independent also in V , so
can be extended to a basis B1 of V . It is easy to see that the subspace W generated
by B1 − B is isomorphic to V /U .
If B is a basis of V is a basis of size d, there is a bijection between it and the
standard basis ei of K d . Any such bijection extends to an isomorphism.
□
Finally, we need to recall some facts about eigenvectors and eigenvalues. If
T : V →
− V is a linear map, a non-zero v ∈ V is an eigenvector of T if there is
a scalar a ∈ K such that T v = av. The element a is then called the eigenvalue
associated to v.
If p(x) is a polynomial over K, and T : V −
→ V is a linear map, we may “evaluate” p at T to get a new linear map p(T ) (where multiplication is interpreted as
composition). The Cayley–Hamilton theorem says that if V has finite dimension n,
there is a polynomial p of degree n such that p(T ) = 0 (p is called the characteristic
polynomial of T ). The minimal polynomial of T is defined to be the polynomial p
of least positive degree, such that p(T ) = 0 (hence it always exists, and divides the
characteristic polynomial).
Proposition 248. Let T : V −
→ V be a linear map on a finite dimensional
vector space V over K. Assume that the minimal polynomial p of T has a root a
in K. Then T have an eigenvector with eigenvalue a.
Proof. Since a ∈ K is a root of p, we may write p(x) = (x − a)q(x) for some
polynomial q over K. Hence, for all v ∈ V ,
0 = 0v = p(T )v = (T − a)q(T )v
Hence, if T −a is invertible, then q(T )v = 0 for all v ∈ V , so q(T ) = 0, contradicting
the minimality of p. Since T − a is not invertible, it has a non-zero kernel. Any
non-zero element of the kernel is an eigenvector with eigenvalue a.
□
5. Finite extensions
Recall that if L is an extension of K, then L is a vector space over K.
Definition 249. An extension L of K is finite if L has finite dimension over
K as a K-vector space. The dimension of L over K is called the degree of the
extension, denoted [L : K].
Example 250. The complex numbers are spanned by 1 and i as a vector space
over R. Hence [C : R] = 2. On the other hand, R is an infinite extension of Q, by
cardinality.
For any field K, the field K(t) of rational functions over K is an infinite extension, since, for example, the ti are linearly independent over K.
□
We would like to show that splitting fields are finite. Since we don’t yet know
that the splitting field is unique, we start with the particular splitting field we have
constructed.
Exercise 251. Show that the field L constructed in Proposition 225 is finite
over K, of degree deg(p) (when p is irreducible).
To make inductive arguments, the following statement is very useful.
Proposition 252. Let E be a finite extension of a field K, and let F be a finite
extension of E. Then F is a finite extension of K, and [F : K] = [F : E][E : K].
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
5. FINITE EXTENSIONS
53
Exercise 253. Prove Proposition 252 (Hint: if e1 , . . . , en is a basis of E over
K, and f1 , . . . , fm is a basis of F over E, show that (ei fj ) is a basis of F over K.)
Exercise 254. Assume that p is a polynomial of degree n over K. Show that
the splitting field of p constructed in Corollary 227 is finite over K, of degree at
most n! (Use previous exercises)
We now go back to the proof of Theorem 187, and prove the main step in the
uniqueness result.
Proposition 255. Let p be an irreducible polynomial over a field K, and let
K(a) and K(b) two field extensions generated by roots a and b of p. Then there is
a unique field isomorphism T : K(a) −
→ K(b) over K such that T (a) = b.
Proof. It is enough to prove the statement when K(a) = K(α) is the field
constructed in Kronecker’s Theorem. For any polynomial q(α) ∈ K(α), define
T (q(α)) = q(b). This is clearly an additive group homomorphism. If q1 and q2 are
two polynomials in K(α), T (q1 q2 ) = r(b), where q1 (x)q2 (x) = s(x)p(x) + r(x), so
r(b) = r(b) + s(b)p(b) = q1 (b)q2 (b) = T (q1 )T (q2 ), since p(b) = 0. Hence, T is a field
homomorphism. The image of T is a field containing b, so is equal to K(b).
□
Exercise 256. Show that if L = K(b), and there is a non-zero polynomial p
over K with p(b) = 0, then L has finite degree over K (such an element b is said to
be algebraic over K). You may use Proposition 255 and Exercise 251.
Conclude that the same holds for K(b1 , . . . , bn ), where each bi is algebraic.
We now deduce the uniqueness using induction on the degree.
Proof of Theorem 187. The existence is Corollary 227. For the uniqueness,
we note that any splitting field is generated by a finite number of algebraic elements,
so has finite degree over K by Exercise 256. Now proceed by induction on the
degree, using Proposition 255 and Exercise 252.
□
Exercise 257. Fill in the details in the last proof (Note that if L is a splitting
field of p over K, and a is one of the root, then L is also a splitting field of p over
K(a))
Remark 258. We have been using that if p is a polynomial over K, and L
is an extension, then p may be viewed as a polynomial over L. More generally, if
t:K−
→ L is an embedding of fields, then we may view p as a polynomial over L,
by applying t to the coefficients. In this sense, the uniqueness statement applies
to embeddings as well: given two embeddings ti of K into fields L1 and L2 , such
that each Li is a splitting field for the corresponding polynomial ti (p), there is an
isomorphism t : L1 −
→ L2 over K (i.e., t(t1 (x)) = t2 (x) for all x ∈ K.)
5.1. The derivative. We have seen that a polynomial of degree n has at most
n roots. It is convenient to have a criterion when there are precisely n of them.
Definition 259. Let p(x) = an xn + · · · + a0 be a polynomial over K. The
derivative of p is the polynomial p′ (x) = nan xn−1 + · · · + a1 .
We note that the derivative is define as a formal operation on polynomials,
there is no analytic content (though, of course, it agrees with the usual derivative
when K = R).
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture
Apr 27
35,
54
2. GALOIS THEORY
Exercise 260. Verify that the derivative satisfies the usual properties, namely,
(p + q)′ = p′ + q ′ and (pq)′ = p′ q + pq ′ (the Leibniz rule).
Exercise 261. Assume that K has characteristic 0. Show that p′ = 0 if and
only if p is constant (i.e., has degree at most 0). Show that this is false in positive
characteristic.
If two polynomials p and q have a common root a in some extension L, then
over L, both are divisible by (x − a), so they are not coprime. In fact, this holds
over the original field as well.
Proposition 262. Assume that polynomials p and q over K have a common
root a in some extension field L. Then p and q are not coprime over K.
Proof. Otherwise, there are polynomials s and t over K with s(x)p(x) +
t(x)q(x) = 1. This equation remains true in L, so plugging in a, we get 0 = 1, a
contradiction.
□
Proposition 263. Let p be a polynomial of degree n over a field K. Assume
that p and p′ are coprime. Then p has n different roots (in the splitting field). In
particular, if K has characteristic 0, then any irreducible polynomial of degree n
over K has n distinct roots.
Proof. Assume that p has less than n roots. Then for some root a, (x − a)2
divides p: p(x) = (x − a)2 q(x) (over the splitting field). Hence p′ (x) = 2(x −
a)q(x) + (x − a)2 q ′ (x). So a is a root of p′ (x) as well. By Proposition 262, p and p′
cannot be coprime.
For the second statement, we need to show that if p is irreducible, then p and
p′ are coprime. We may assume p is non-constant, and since the characteristic is
0, p′ is non-zero. Hence the degree of gcd(p, p′ ) is at most deg(p′ ) < deg(p) so is 1
since p is irreducible.
□
Exercise 264. Prove that in a field of characteristic 0, every non-zero element
has n distinct n-th roots in some extension.
Exercise 265. Let K be a field of positive characteristic p. Show that the
function f : K −
→ K given by f (x) = xp is a homomorphism from K to itself. This
homomorphism is called the Frobenius endomorphism. (Hint: use the binomial
expansion).
Conclude that in K, every element has at most one p-th root.
6. The Galois correspondence
We now introduce the main tool in the study of field extensions — the group
of symmetries.
Definition 266. The automorphism group of an extension L of K is the group
Aut(L/K) of field automorphisms of L over K.
Proposition 267. Let L be a field extension of K, let a ∈ L, and assume that
p(a) = 0 for some non-zero polynomial p over K. Then τ (a) is also a root of p, for
any τ ∈ Aut(L/K). In particular, if L is the splitting field of p, then Aut(L/K)
maybe identified with a subgroup of S(X), where X is the set of roots of p.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
6. THE GALOIS CORRESPONDENCE
55
Proof. If τ fixes K, then it fixes the coefficients of p. Hence,
0 = τ (0) = τ (p(a)) = p(τ (a))
For the second statement, τ 7→ τ |X is the homomorphism. It is injective since X
generates L.
□
√
3
Example 268. The equation x3 = 2 has a unique real solution a = 2 (this
can be seen with basic analysis). Let K = Q, L = Q(a). Since L is a subfield of R,
a is the unique solution of the equation, any automorphism of L must fix a, and
therefore a. Hence Aut(L/K) is trivial.
□
It may seem from the last example that the group Aut(L/K) does not carry
much information about the extension. We will see below that the situation is
different when L is the splitting field of a polynomial.
Example 269. Let L be the splitting of the equation xp = 1 over Q, where p is
prime. The set of solutions in L of the equation forms a multiplicative subgroup of
L, which has order p according to Exercise 264. It follows that the group is cyclic.
Hence Aut(L/K) is canonically a subgroup of Up . Since L is generated by any of
□
the non-trivial roots, the automorphism group is, in fact, the whole of Up .
Example 270. Let K be a field of characteristic p > 0, and let a ∈ K be an
element that does not have a p-th root. The splitting field L of xp − a over K
contains the unique p-th root.
□
The examples above suggest that the group of automorphisms is most meaningful for splitting fields. A splitting field is attached to a particular polynomial, but
it will be more convenient to have a condition that is independent of a particular
polynomial. We thus make the following definition.
Definition 271. Let L be a finite extension of K. We say that L is a normal
extension of K if any irreducible polynomial over K that has a root in L, splits in
L. If K has characteristic 0, we also call it a Galois extension.
If L is a Galois extension, the group Aut(L/K) is called the Galois group of L
over K.
By Exercise 264, if an irreducible polynomial of degree n over K has a root in
a Galois extension L of K, then it has n distinct roots there.
Remark 272. There is a notion of Galois extensions for fields of positive characteristic (we have seen in Example 270 that normality is insufficient to have a
meaningful Galois group). Since it is slightly more complicated, and irrelevant to
the problem of solvability of polynomial equations over Q, we will not discuss it,
and instead assume from now on (at least in the proofs) that all our fields have
characteristic 0. However, the statements below remain true in positive characteristic.
It might not be obvious that Galois extensions exist at all. In fact, any splitting
field is normal.
Proposition 273. A finite extension L of K is normal if and only if it is the
splitting field of some polynomial.
If p is a polynomial over K, the Galois group of p over K is the Galois group
of the splitting field of p over K.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture
Apr 29
36,
56
2. GALOIS THEORY
Corollary 274. Let L be a Galois extension of K, and E and F be two
intermediate extensions. If T : E →
− F is an isomorphism over K, then T extends
to an automorphism of L over K.
Proof. We know this when L is a splitting field, by Remark 258, hence for
Galois extensions by Proposition 273.
□
We will need to know that the notion of a Galois extension behaves well with
respect to intermediate extensions. The proofs of Propositions 273 and 275 are
given below, after some more tools are developed.
Proposition 275. If L is a Galois extension of K, and K ⊆ E ⊆ L is an
intermediate extension, then L is a Galois extension of E.
The fundamental theorem of Galois theory establishes a connection intermediate extensions K ⊆ E ⊆ L where L/K is a Galois extension, and subgroups of
Aut(L/K). We now describe this connection.
Let L be a finite Galois extension of K, and let G = Aut(L/K). If A ⊆ G is
any subset, we denote by LA the set of elements of L fixed by all elements of A:
LA = {x ∈ L|τ (x) = x
∀τ ∈ A}
(8)
A
Exercise 276. Show that L is always a subfield of L containing K. Show
also that LA = LH , where H is the subgroup generated by A.
Thus we have a way of translating between intermediate extensions and subgroups: to an intermediate extension M we assign the subgroup Aut(L/M ) of G,
while to a subgroup H we attach the fixed field LH . The main theorem states that
this is a bijection.
Theorem 277 (The fundamental theorem of Galois theory). Let L be a finite
Galois extension of a field K, and let G = Aut(L/K) be the Galois group.
(1) The correspondence between subgroups of G and intermediate extensions
is inclusion reversing: If H1 ≤ H2 ≤ G, then LH2 ⊆ LH1 , and if K ⊆
E ⊆ F ⊆ L, then Aut(L/F ) ≤ Aut(L/E).
(2) Degrees of extensions correspond to subgroups: For any H1 ≤ H2 ≤ G,
[LH1 : LH2 ] = (H2 : H1 ). In particular, [L : LH ] = |H| for any subgroup
H of G.
(3) The two operations are inverse to each other: for any subgroup H ≤ G,
Aut(L/LH ) = H, and for any subfield K ⊆ E ⊆ L, LAut(L/E) = E.
(4) Normal extensions correspond to normal subgroups: The sub-extension
K ⊆ E ⊆ L is a normal extension of K if and only if Aut(L/E) is a
normal subgroup of G. In this case, Aut(E/K) = G/Aut(L/E).
This theorem provides a full translation between intermediate field extensions
and subgroups of the Galois group. It allows to transfer properties and tools from
one side to the other. In particular, we have the follow terminology.
Definition 278. Let L be a Galois extension of K. We say that the extension
is cyclic, Abelian, etc. if G = Aut(L/K) has the corresponding property.
End lecture
May 2
37,
Exercise 279. Let L be the splitting field of x3 −2 over Q. Compute the Galois
group G of L over Q, find all subfields of L, and the corresponding subgroups of G.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
6. THE GALOIS CORRESPONDENCE
57
6.1. Minimal polynomials. We now aim to prove Propositions 275 and 273.
To that end, we show that if p is an irreducible polynomial over K, then it is the
minimal polynomial of any of its roots. This again follows from Euclid’s algorithm.
Proposition 280. Let p be an irreducible monic polynomial over K, and let
a be a root of p in an extension L. Then p divides any polynomial q over K with
q(a) = 0. In particular, any non-zero such polynomial has degree at least deg(p).
Proof. Since p is irreducible, we have either gcd(p, q) = 1 or gcd(p, q) = p.
The first option is excluded by Proposition 262.
□
Exercise 281. Show that in the situation of Proposition 280, if deg(q) =
deg(p) and q is monic, then p = q.
Corollary 282. If L is an extension of K, and a ∈ L, there is a unique
monic polynomial p over K, such that for any polynomial q over K, p divides q if
and only if q(a) = 0.
Proof. We may assume that there is a polynomial s over K with s(a) = 0
(otherwise take p = 0). s can be written as a product of irreducible polynomials
over K. At least one of the factors p has a as a root. Since p is irreducible, it has
the required property by Proposition 280.
□
Definition 283. The polynomial given by Corollary 282 is called the minimal
polynomial of a over K.
We may now return some debts.
Proof of Proposition 275. Let p be an irreducible polynomial over E that
has a root a in L. The minimal polynomial q of a over K is also a polynomial over
E, and since q(a) = 0, p divides q. Hence every root of p is also a root of q (in any
extension). But q splits in L, hence so does p.
□
Proof of Proposition 273. Assume first that L/K is normal, and let a ∈ L
be an element not in K. Let E = K(a). By Proposition 275, L is a normal extension
of E. Since E is a proper extension of K, [L : E] < [L : K], so by induction, L is
the splitting field of some polynomial q over E. Let b be a root of q in L, and let p
be the minimal polynomial of b over K. By definition, p splits in L. Any root of q
is also a root of p, so L is generated over E by the roots of p. Hence L is generated
over K by the roots of p and a. Thus, if r is the minimal polynomial of a over K,
then L is the splitting field of pr.
In the other direction, assume that L is the splitting field of a polynomial p
over K, let a ∈ L, and let q be the minimal polynomial of a over K. If q does
not split in L, let E be the splitting field of q over L, and let b ∈ E be a root
of q not in L. Since q is irreducible over K, we have, by Proposition 255, an
isomorphism T : K(a) −
→ K(b) over K. Let L1 be the splitting field of p over K(b).
By Remark 258, T extends to an embedding of L into L1 . But both L and L1 are
generated by the roots of p, so L = L1 . In particular, b ∈ L.
□
6.2. More linear algebra. To prove part (2) of Theorem 277, we need some
more linear algebra. In this subsection, L is a field of characteristic 0, H is a finite
group of automorphisms of L, and E = LH is the fixed field.
We consider the vector space V = Ln over L. Since E is a subfield of L, we
may view E n as a subset of V , and in particular, V has a basis consisting of vectors
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
58
2. GALOIS THEORY
over E (i.e., with all entries in E). If U ⊂ V is an L-subspace of Ln , this no longer
needs to be the case.
Example 284. Let U be the subspace of C2 spanned by (1, i). Then the only
vector with real entries in U is 0.
□
The group H acts on V , by acting on each coordinate. This action respects
the addition on V , but it is not L-linear: h(xv) = h(x)h(v) for h ∈ H, x ∈ L and
v ∈ V . The elements with coordinates in E are precisely those fixed by H. So the
example shows that V may have a subspace U with no non-zero fixed elements.
If U does have a basis (vi ) consisting of fixed elements, then applying an element
h of H to a general element u = x1 v1 + · · · + xk vk of U , we get h(u) = h(x1 v1 +
· · · + xk vk ) = h(x1 )v1 + · · · + h(xk )vk . Hence h(u) is again in U . In other words,
U is invariant under the action of H.
So a necessary condition for U to have such a basis is that it is invariant under
the action of H. It turns out that this is also sufficient: if U is invariant, then it
has a basis with coordinates in E. We will need only a step in this direction.
Proposition 285. Let L be a field of characteristic 0, H a finite group of
automorphisms of L, E = LH . Let U ⊆ Ln be a non-zero linear subspace over L,
such that h(U ) = U for all h ∈ H. Then U contains a non-zero vector in E n .
Proof. By assumption, there is a non-zero vector u ∈ U . We may assume
that
∑
the first coordinate is non-zero, and, after dividing by it, 1. Let v = h∈H h(u).
Since U is invariant, v ∈ U . The first coordinate of each summand is 1, so the first
coordinate of v is |H|. Since L has characteristic 0, this is non-zero, so v is nonzero. Finally, for any h ∈ H, h(v) is given by the same sum, with the summands
permuted. Hence h(v) = v, so v ∈ E n .
□
If x is an element of L, we will denote by e(x) the tuple (h1 (x), . . . , hk (x)),
where H = {h1 , . . . , hk } is some fixed enumeration of H. We assume h1 is the
identity. Note that if h ∈ H is any element, then there is a permutation τ ∈ Sk ,
such that h(e(x)) = (hτ (1) (x), . . . , hτ (k) (x)) for all x (namely, τ is the permutation
corresponding to h under the Cayley homomorphism). In other words, there is a
permutation matrix Ah , such that h(e(x)) = Ah (e(x)).
Corollary 286. If x1 , . . . , xm are elements of L linearly independent over E,
then the vectors e(x1 ), . . . , e(xm ) ∈ Lk are linearly independent over L.
Proof. Consider the set U of all tuples (a1 , . . . , am ) ∈ Lm such that a1 e(x1 )+
· · · + am e(xm ) = 0. Clearly, U is a linear subspace. We need to show that U = 0.
Applying an element h to the equality, we get
0 = h(a1 e(x1 ) + · · · + am e(xm )) = h(a1 )h(e(x1 )) + · · · + h(am )h(e(xm )) =
= h(a1 )Ah (e(x1 )) + · · · + h(am )Ah (e(xm )) = Ah (h(a1 )e(x1 ) + · · · + h(am )e(xm ))
Since Ah is invertible, we get that (h(a1 ), . . . , h(am )) ∈ U as well. In other words,
U is invariant. If U is non-zero, by Proposition 285, there is a non-zero tuple ai
as above where all the ai are in E. But the first coordinate of e(x) is x, so we get
a1 x1 + · · · + am xm = 0 for a tuple ai in E, contradicting the linear independence
□
of the xi over E.
We are now in position to prove one inequality in (2).
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
6. THE GALOIS CORRESPONDENCE
59
Corollary 287. With L, H and E as above, [L : E] ≤ |H|.
Exercise 288. Deduce Corollary 287 from Corollary 286.
We draw some more conclusions. These conclusions will not be used in proving
the main theorem, so we will now assume to know that [L : E] = |H|.
Corollary 289 (Dedekind). The elements of H are linearly independent over
L: if ai ∈ L are such that a1 h1 (x) + · · · + ak hk (x) = 0 for all x ∈ L, then ai = 0
for all i.
Proof. Let l1 , . . . , lk be a basis of L over E, and consider the matrix A whose
rows are e(li ). We showed that the rows are linearly independent. Since A is a
square matrix, it follows that the columns are also linearly independent (over L).
Hence there are no non-zero ai with a1 h1 (li ) + · · · + ak hk (li ) = 0 for all i.
□
Exercise 290. Assume that H is a cyclic group of order n, generated by τ .
Show that T n − 1 is the minimal polynomial of τ , viewed as a linear map from L
to itself over E.
We have proven Proposition 285 using a particular action of H on V . We may
now extend the result to an arbitrary action of the same kind. This is a version of
“Hilbert’s Theorem 90”.
Corollary 291 (Hilbert 90). Let V be a finite dimensional non-zero vectors
space over L, and assume that H acts on V by additive group homomorphisms, and
satisfying h(xv) = h(x)h(v) for h ∈ H, v ∈ V and x ∈ L. Then there is a non-zero
vector v ∈ V , such that h(v) = v for all h ∈ H.
We note that Proposition 285 is a special case, since the restriction of the action
there to U satisfies the assumption.
∑
Proof. If u ∈ V is any non-zero vector, the vector v = h∈H h(u) is clearly
invariant. Hence it is enough to show that there is a vector of this form which
is non-zero.∑In particular, it is enough to show that there is x ∈ L with vx =
∑
h(xv) =
h(x)h(v) non-zero.
Assume that all such sums are zero. Applying the equation to elements x of a
basis of L over E, we get Ae(v) = 0, where A is the matrix whose rows are e(li ),
for li elements of the basis. We have seen that A is invertible, so this implies that
e(v) = 0, and hence v = 0.
□
6.3. Proof of the Fundamental Theorem. We now proceed to prove the
main theorem.
Exercise 292. Prove part (1) of Theorem 277.
For the second part, we have the following reduction.
Exercise 293. Show that the full statement of (2) follows from the “In particular” part (i.e., from [L : LH ] = |H|).
Proof of (2). Let E = LH . By Exercise 293, we need to show [L : E] = |H|.
One direction is proved in Corollary 287, so we need to prove that [L : E] ≥ |H|.
Let a ∈ L − E, let H1 = {h ∈ H|h(a) = a} be the stabiliser of a, and let F = LH1 .
Then E(a) ⊆ F , so
[L : E] = [L : F ][F : E] ≥ [L : F ][E(a) : E] = |H1 |[E(a) : E]
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
(9)
60
2. GALOIS THEORY
The last equality by induction on the degree. Hence it is enough to show that
|H|
[E(a) : E] ≥ |H
= (H : H1 ).
1|
Since H1 is by definition the stabiliser of a, by 11.4 of Chapter 1 (H : H1 ) is
equal to the size of the orbit Ha of H acting on L. Hence we need to show that
that |Ha| = [E(a) : E]. We have seen in Exercise 251 (and using Proposition 255)
that [E(a) : E] is equal to the degree of the minimal polynomial p of a. On the
other hand, by Proposition 255, Corollary 274 and Proposition 267, the orbit Ha
consists of all the roots of p. By Proposition 263, the number of such roots is also
equal to the degree of p.
□
Proof of (3). We first prove that if E is an intermediate extension, and H =
Aut(L/E), then LH = E. Since, by definition, any element of H fixes the elements
of E, we have E ⊆ LH . Assume there is an element a fixed by H, that is not in
E. By Proposition 275 and Exercise 264, the minimal polynomial of a over E has
another root b in L. We thus have an isomorphism E(a) −
→ E(b) over E, which
extends to an automorphism τ of L over E, by Corollary 274. This contradicts the
assumption that a is fixed by H.
Now, let H be a subgroup of G, and let E = LH . We need to show that
H = Aut(L/E). Again, one inclusion is clear: H ≤ Aut(L/E). However, we have
just shown that the latter group has E as its fixed field. Hence, by part (2), they
have the same size, so they are equal.
□
Proof of (4). Assume E is normal. By Proposition 273, E is the splitting
field of some polynomial p. Any automorphism of L would have to take any root of
p to another such root, which is also in E. It follows that for any τ ∈ Aut(L/K),
τ (E) = E.
Hence we have a restriction map r : Aut(L/K) →
− Aut(E/K), which is clearly a
group homomorphism. According to Corollary 274, any automorphism of E over K
extends to an automorphism of L. Hence the map r is surjective. The kernel is the
set of automorphisms in Aut(L/K) whose restriction to E is the identity. Hence it
is precisely Aut(L/E). Thus, Aut(L/E) is normal, with quotient Aut(E/K) (and
the quotient map is the restriction).
Conversely, assume that Aut(L/E) is normal. Let a ∈ E have minimal polynomial p over K, and let b be another root of p (since L is normal, b ∈ L). We must
show that b ∈ E. By Proposition 255 and Corollary 274, there is an automorphism τ
of L over K with τ (a) = b. If σ ∈ Aut(L/E), then τ −1 στ ∈ Aut(L/E) as well, since
it is normal. Since a ∈ E, we have a = τ −1 στ (a) = τ −1 σ(b), so σ(b) = τ (a) = b.
It follows that any σ ∈ Aut(L/E) fixes b, so Aut(L/E) = Aut(L/E(b)). By (3),
E = E(b), so b ∈ E.
□
7. Solvability of equations
We now have the tools to analyse the solvability of polynomial equations. We
continue to assume that all fields have characteristic 0.
7.1. Cyclic extensions. Recall (Definition 190) that we are interested in
splitting fields of equations of the form xn − a = 0. We will now show that they
are precisely the cyclic extensions.
We have seen in Example 268 that such an extension is not, in general, obtained
n
by adding one root of a. However, if b and c are two roots, then ( cb )n = cbn = 1, so
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
7. SOLVABILITY OF EQUATIONS
61
c = rb, where r is an n-th root of unity (i.e., rn = 1). Hence we have the following
result.
Lemma 294. Assume K contains all n-th roots of unity. Then for any n and
any a ∈ K, the splitting field of xn − a is of the form K(b), where bn = a.
Let L = K(b), where bn = a, be a splitting field as above. Let G be the Galois
group. If g ∈ G, then g(b) is another n-th root of a, so by the calculation above,
there is a root of unity t(g) such that g(b) = t(g)b. Thus we get a map g 7→ t(g)
from G to the group µn of n-th roots of unity.
Exercise 295. Show that the map t above is an injective group homomorphism. Show that it is surjective if and only if no smaller power of b is in K.
The group µn is a cyclic group of order n. It follows that G is a cyclic group.
This proves one side of the following proposition.
Proposition 296. Assume that K contains the n-th roots of unity, and let L
be an extension of degree n. Then L is a cyclic Galois extension of K if and only
if L = K(b) for some b such that bn ∈ K, and n is minimal with this property.
Proof. One direction was proved above. Let L be a cyclic Galois extension of
degree n, and let τ be a generator of the Galois group. According to Exercise 290,
the minimal polynomial of τ acting on L is T n −1. Since K contains all n-th roots of
1, this polynomial splits in K, so L contains an eigenvector of τ , with eigenvalue α,
a primitive root of unity: τ (v) = αv (Proposition 248). Then τ (v l ) = τ (v)l = αl v l ,
hence v l ∈ K precisely if l is divisible by n.
□
We note also that the arguments above show that, if L is the splitting field of
xn − a, then L contains all n-th roots of unity.
7.2. Solvability criterion. We may now provide a group theoretic criterion
for solvability.
Theorem 297. Let p(x) be a polynomial over a field K of characteristic 0, let
L be its splitting field, and let G = Aut(L/K) be the Galois group. Then p(x) is
solvable by radicals if and only if there is a sequence of groups {e} = Gn < . . . <
G1 < G0 = G such that for each i < n, Gi+1 is normal in Gi , and Gi /Gi+1 is
cyclic.
Finite groups that satisfy the condition in the Theorem are called solvable.
Thus, the theorem can be re-stated as saying: A polynomial p is solvable if and
only if its Galois group is solvable.
Exercise 298. Let p : G −
→ H be a surjective group homomorphism. Show
that if G is solvable, then so is H. Show that any subgroup of a solvable group is
solvable.
For the proof, we will need the following lemma.
Lemma 299. Let p be a polynomial over K with splitting field L and Galois
group G = Aut(L/K), let K ′ be an extension of K, and let L′ be a splitting field
of p over K ′ . Then the restriction map Aut(L′ /K ′ ) −
→ G is injective.
Proof. We note that restriction does give a map as above, because all the
roots of p are in L. If an automorphism goes to e in G, it fixes all the roots, so
fixes all elements of L′ (which is generated by the roots).
□
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
62
2. GALOIS THEORY
Proof of Theorem 297. Assume that p is solvable by radicals. Then there
is a sequence of field extensions K = L0 ⊂ L1 ⊂ · · · ⊂ Ln , such that each Li is
the splitting field of xki − ai , with ai ∈ Li−1 , and such that L ⊆ Ln . Since each
extension by roots of unity is an extension by radicals, we may assume that the
first extension is an extension by all roots of unity that we will need.
Translating to group theory, we get a sequence of group Hi = Aut(Ln /Li ), as
well as a quotient map from Hn to G. By Exercise 298, it is enough to prove that
Hn is solvable. But since each Li is Galois, we have by the main Theorem that
Hi+1 is normal in Hi , and Hi /Hi+1 = Aut(Li+1 /Li ). The last group is cyclic by
Proposition 296 (and Example 269).
In the other direction, assume that G is solvable. Let K ′ be an extension of
K by enough roots of unity, L′ the splitting field of p over K ′ . Then Aut(L′ /K ′ )
is a subgroup of G by Lemma 299, so by Exercise 298 is solvable as well. Hence
we may assume that K itself already had all roots of unity. Let Li = LGi . Then
Aut(Li+1 /Li ) = Gi /Gi+1 , hence Li+1 /Li is a cyclic extension. By Proposition 296,
it is an extension by a root.
□
Exercise 300. Let G be a finite simple group. Show that G is solvable if and
only if it is Abelian.
We now give an example of a particular equation that cannot be solved. Let
p(x) = x5 − 4x + 2. We first claim that p is irreducible. Otherwise,
x5 − 4x + 2 = (x3 + ax2 + bx + c)(x2 + dx + e) =
= (x5 + (d + a)x4 + (e + ad + b)x3 + (ae + bd + c)x2 + (be + dc)x + ce
Hence we get the equations
a+d=0
e + ad + b = 0
ae + bd + c = 0
be + cd = −4
ce = 2
The parameters a–e are a-priori rational, but it is easy to see they must be integers.
It follows from the last equation that exactly one of c, e is even. Assume it is c (the
other case is similar). Then, from the second to last equation we get that b must
be even, then that a has to be even, and we get a contradiction from the second
equation, since e is odd and a, b are even.2
We next need the following result.
Exercise 301. Show that if g is a 5-cycle in S5 and h is any transposition,
then S5 is generated by g and h (Hint: consider the conjugate of h by a suitable
power of g to produce a 3-cycle.)
We may now compute the Galois group of p.
Proposition 302. The Galois group of p(x) = x5 − 4x + 2 over Q is S5
2This method can be generalised to a general criterion for testing irreducibility over Q, called
the Eisenstein criterion
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
7. SOLVABILITY OF EQUATIONS
63
Proof. The polynomial is irreducible by the discussion above. Hence, if L
is a splitting field, and a ∈ L is a root of p, then Q(a) have degree 5 over Q.
Hence [L : Q] = [L : Q(5)][Q(5) : Q] is divisible by 5, and so is the Galois group
G = Aut(L/Q). By Cauchy’s Theorem (Theorem 173), G contains an element of
order 5. Viewing G as a subgroup of S5 (by enumerating the roots), it follows that
G contains a 5-cycle.
We next claim that p has exactly 3 real roots. This follows from analysis:
the derivative of p is 5x4 − 4, so has only two real roots, hence p has at most 3
real roots. An explicit calculation show that the two extreme points have different
signs. It follows that complex conjugation determines a transposition in G. Hence
G contains a 5-cycle and a transposition, so G = S5 be Exercise 301.
□
Corollary 303. The equation p(x) = x5 − 4x + 2 = 0 is not solvable by
radicals.
Proof. The Galois group of p is S5 . If it were solvable, then so would its
subgroup A5 . But A5 is simple by Theorem 180, hence is not solvable by Exercise 300.
□
This concludes the proof of Theorem 193.
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk
End lecture
May 4
38,
Bibliography
[1]
[2]
[3]
[4]
James S. Milne. Fields and Galois theory. Course lecture notes. 2008. url:
http://jmilne.org/math/ (cit. on p. 41).
James S. Milne. Group Theory. Course lecture notes. 2010. url: http://jmi
lne.org/math/ (cit. on pp. 5, 10, 39).
Joseph Rotman. Galois theory. Second. Universitext. New York: SpringerVerlag, 1998, pp. xiv+157. isbn: 0-387-98541-7 (cit. on p. 41).
Joseph J. Rotman. An introduction to the theory of groups. Fourth. Graduate
Texts in Mathematics 148. New York: Springer-Verlag, 1995, pp. xvi+513.
isbn: 0-387-94285-8 (cit. on p. 5).
65
Index
action, 28
faithful, 36
transitive, 29
alternating group, 26
automorphism, 21
automorphism group, 54
domain, 7
image, 7
injective, 7
invertible, 7
left inverse, 7
one to one, 7
onto, 7
range, 7
right inverse, 7
surjective, 7
basis, 51
binary operation, 8
Cartesian product, 7
Cayley–Hamilton, 52
centraliser, 14
centre, 14
characteristic, 45
characteristic polynomial, 52
circle group, 14, 16, 18
conjugacy class, 37
coprime, 48
coset, 30
cycle, 23
length of, 24
G-set, 28
Galois extension, 55
Galois group, 55
gcd, 48
general linear group, 9
greatest common divisor, 48
Group, 8
Abelian, 9
action, 28
alternating, 26
automorphism, 21
cyclic, 14
Dihedral, 10
divisible, 18
general linear, 9, 27
generators of, 13
homomorphism, 15
isomorphism, 17
order of, 11
quotient, 35
simple, 33, 33, 38
solvable, 61
special linear, 10
symmetric, 9, 21
trivial, 9
degree, 42, 52
derivative, 53
dihedral group, 10
dimension, 51
eigenvalue, 52
eigenvector, 52
Eisenstein criterion, 62
empty set, 7
factor group, see also quotient group
field, 42
homomorphism
over a subfield, 45
isomorphism, 45
field extension, 42
field homomorphism, 45
finite dimensional, 51
finite extension, 52
Frobenius endomorphism, 54
function, 7
bijective, 7
homomorphism, 15
kernel of, 16
section of, 31
index (of a subgroup), 30
irreducible, 48
isomorphism, 17
67
68
INDEX
kernel, 16
vector space, 50
Leibniz rule, 54
linear combination, 51
linear map, 50
linear representation, 27
linear space, 50
linear subspace, 50
linearly independent, 51
minimal polynomial, 52, 57
monic, 42
normal extension, 55
normal subgroup, 34
orbit, 29
orbit (of a number under a permutation),
23
order (of a group), 11
order (of an element), 14
p-free part, 32
p-part, 32
pair, 7
permutation, 21
disjoint (from another permutation), 23
even, 26
fixes (an element), 22
invariant subset, 22
odd, 26
sign of, 26
polynomial, 41, 42
degree of, 41
quotient group, 35
regular representation, 27
residues, 10
ring, 47
root, 47
root of unity, 61
section (of a homomorphism), 31
simple group, 33
solvable by radicals, 44
spans, 51
special linear group, 10
splitting field, 42, 43
subfield, 42
subfield generated, 44
subgroup, 11
generated by (a set), 13
index of, 30
normal, 34
subset, 7
symmetric group, 9, 21
transposition, 25
trivial group, 9
Id: notes.tex,v 1.14 2011/05/06 14:48:30 mkamensk Exp mkamensk