Basic Algebra Math 30710kamenskm/teaching/algebra.pdf · 2014-10-27 · 8 1. GROUP THEORY Example...

Basic Algebra

Math 30710

Moshe Kamensky

Department of Math, University of Notre-DameE-mail address: mailto:[email protected]: http://mkamensky.notlong.com

mailto:[email protected]

http://mkamensky.notlong.com

Contents

Chapter 1. Group Theory 51. Symmetry 52. Sets and functions 63. Definition of a group 84. First properties of groups 105. Subgroups 116. Cyclic groups 157. Homomorphisms 158. The classification of cyclic groups 199. The symmetric groups 2110. Group actions 2811. Normal subgroups and quotients 31

Chapter 2. Galois theory 411. Statement of the problem, and its solution 412. First properties of fields 443. Polynomials in one variable 474. Linear algebra 505. Finite extensions 526. The Galois correspondence 547. Solvability of equations 60

Bibliography 65

Index 67

3

CHAPTER 1

Group Theory

This chapter presents some basic results on group theory. Good references withmore 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 1. circle Figure 2. square

Figure 3. rectangleFigure 4. penguin

Intuitively, the square is more symmetric than the rectangle, the circle is muchmore symmetric than them both, and the penguin has no symmetries at all. Howcan we formulate these observations precisely? We shall decide what do we meanby a ‘symmetry’, and then show that, e.g., the circle has more of them than thesquare.

So, what is a symmetry? Different definitions can be given, depending on thecontext. We will (loosely) define a symmetry of a shape to be a transformation ofthe shape into itself that preserves distances and (unoriented) angles (so that thedistance between any two points remains the same after the transformation.)

For example, rotating the square by 90◦ is such a symmetry. It brings thesquare as a whole into itself. So is the reflection along any of the diagonals. Everyshape has at least one symmetry: this is the transformation that doesn’t move theshape at all! It is called the identity transformation. This is the only symmetry ofthe penguin. But, as noted above, the square has some others. We note that we areonly interested in the “final outcome” of the transformation: the transformationthat consists of rotating the shape to the right, and then by the same amount tothe left is still the identity.

5

6 1. GROUP THEORY

What are the symmetries of the rectangle? In any symmetry, a short side hasto go to a short side, and a long side to a long side. Once a decision is madewhich side goes where, there is precisely one symmetry that does it (these facts areclear intuitively, and are easy to prove if the precise definitions are given.) Hencethere are precisely four symmetries: a horizontal reflection, a vertical reflection, acombination of them, and the identity.

What about the square? Since a square is a rectangle, any symmetry of therectangle is also a symmetry of the square. But there are more: the right rotationby 90 and 270 degrees are also symmetries, as well as reflections by the two diag-onals (what about the rotation by 180◦?) It can be checked that these are all thesymmetries of the square.

Finally, what can we say about the circle? The circle has infinitely manysymmetries: 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 anumber, the number of symmetries (which may be infinity), that tells us everythingabout the symmetry of the shape. One shape is more symmetric than another ifthe 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 foursymmetries, the four rotations (including the identity.) So now the new square andthe rectangle have the same number of symmetries, but they are clearly symmetricin different ways. How can one capture this difference?

Given two symmetries of some shape, we may transform the shape by the firstone, and then apply the second one to the result. The operation obtained in thisway is again a symmetry. For example, if we rotate the (original) square, and thenflip along the diagonal, this is again a symmetry. Since we have listed all of thesymmetries of the square, this should be one of the symmetries in the list (whichone?) In general, the process of applying one symmetry after another defines anoperation between the symmetries of the shape: for any two symmetries f, g, weget a new symmetry f ·g. The set of all symmetries of a given shape, together withthe composition operation described above is an example of a group.

Using the operation we can distinguish between the cases of the coloured squareand 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◦ twicewill not give the identity. A basic feature of this argument is that, after definingthe operation ·, it didn’t use the geometry at all! It is formulated in terms of thealgebraic properties of the operation. This is the kind of arguments that one applieswhen studying abstract group.

We will define a group abstractly as a set with an operation, satisfying certainproperties. Though we will not forget the geometric examples, it should be stressedthat the theory happens on the abstract level of the group operation. In fact, itis often used that the same group can be the group of “symmetries” of completelydifferent objects.

2. Sets and functions

Before defining what a group is, we should understand sets. We shall notdefine 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 itselements; Two sets A and B are equal if and only if any element of A is an elementof 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 elementof B. Thus A = B if and only if A ⊆ B and B ⊆ A. The set whose elements areprecisely 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 inA (for finite sets.) There is a unique set with no elements, the empty set , denotedby ∅.

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 isan ordered set of two elements: if a = b then (a, b) and (b, a) are different pairs (incontrast, {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 wayto 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 betweenthem.

2.1. Functions. A function f from a set A to another set B is a rule thatassigns to every element a of A, a unique element f(a) ∈ B. The set A is called thedomain of f and B is called the range. The subset of B consisting of all elementsof the form f(a) for some a ∈ A is called the image of f . We may view f as amachine, that takes elements of A and produces elements of B. The fact that f isa function from A to B is written as f : A −→ B. If A is any set, there is a functionIdA : 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 newfunction g ◦ f : A −→ C, defined by (g ◦ f)(a) = g(f(a)). Thus we apply f to a, andapply 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 isof the form f(a) for some a ∈ A. It is bijective if it is injective and surjective. Afunction g : B −→ A is a left inverse of f if g ◦ f = IdA and it is a right inverse iff ◦ 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 ithas a left inverse. If f : A −→ B has a right inverse, then it is surjective. A functionis bijective if and only if it has a right and left inverse. In this case, the right andleft 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|, thenf 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 canbe pictured as curves in the plane. For example, a continuous function is injectiveif and only if it is monotone. The inverse of a function (if it exists) is obtained byreflecting along the diagonal. □


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 kmis given by an n×m matrix, and that composition of functions is given by matrixmultiplication. If n = m, such a function is injective if and only if it is surjective ifand only if its determinant is non-zero. □

Example 6. The function f : N −→ N from the set of natural numbers to itselfgiven by f(n) = 2n is injective, but not surjective. The function g : N −→ N definedby “g(n) = n/2 if n is even, and g(n) = (n+ 1)/2 if n is odd” is a left inverse, butnot a right inverse of f . Accordingly, g is surjective, but not injective. Note that fhas other left inverses.

Any natural number n can be presented uniquely as pn11 pn2

2 . . . pnk

k , where piis the i-th prime, and nk = 0. The rule n 7→ (n1, . . . , nk) defines a function fromN to the set of all finite sequences of non-negative integers. This function is abijection! □

End lecture 1, Jan 19

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 byg∗h. For instance, G may be a set of symmetries, as in section 1, and the operationwill be that of applying one symmetry after another. The definition of a group willtry 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 = aG3 (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 clearfrom 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 multiplicativeterminology for the operation. However, it should be stressed that ingeneral, the elements of G are not numbers, and the operation has nothingto do with multiplication of numbers.

(3) The associativity axiom implies that it makes sense to write a ∗ b ∗ cwithout parentheses. It is easy to prove that, more generally, for anyelements a1, . . . , an of G, the expression a1 . . . an makes sense (i.e., we mayinterpret it by putting the parentheses wherever we like.) In particular, ifa is any element, and n is a natural number we write an for the productof a with itself n times (thus, a1 = a.)

(4) An element e as in axiom G2 is called a unit or an identity. Axiom G3refers to this element, and thus only makes sense if we know that such anelement is unique. This is indeed the case, as is proved in proposition 19.


3. DEFINITION OF A GROUP 9

Likewise, we will show in proposition 23 that for any element a, the ele-ment b promised by axiom G3 is unique. The inverse of a will be denoteda−1, and we abbreviate a−1n by a−n. Finally, we set a0 = e.

Example 9. The sets Z, Q, R and C of integer, rational, real and complexnumbers, with addition as an operation are groups (with identity element 0.) Theset N of natural numbers (with addition) is not a group: there are no inverses (andpossibly, depending on the definition, no unit.)

The same sets with multiplication as the operation are not groups: the number1 serves as the unit, but 0 has no inverse. If we remove 0, then all sets except forthe integers become groups.

The sets of positive rational or real numbers is also a group under multiplica-tion. 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 ex-

ample, let G = {e, a, b, c}, and define the operation by the following table:

e a b ce e a b ca a b c eb b c e ac c e 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 fol-

lowing examples are all of this kind:

Example 13. If X is any set, the set Sym(X) of bijective functions from X toitself forms a group, with composition of functions as an operation, forms a group.This is because composition of functions is associative, and any bijective functionhas a unique inverse. Note that in this group there are elements f and g such thatfg and gf are different (in other words, Sym(X) is not Abelian.) This group iscalled 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 toitself can be viewed as a “symmetry”. In this sense Sym(X) is the set of symmetriesof X. □

Example 14. If V is any vector space, the set GL(V ) of invertible linear mapsfrom V to itself is a group under composition. It is called the general linear groupof V . If V = kn, this set can be identified with the set of invertible n× n matricesover k. The composition corresponds to multiplication of matrices, and the set ofn× 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 seeis the vector space structure. Recall that a matrix A maps the unit cube into aparallelepiped of volume ∥det(A)∥. Thus, if we are observant enough to measurevolumes (and orientations), we will only consider A to be a symmetry if det(A) = 1.


10 1. GROUP THEORY

The set of all such matrices indeed forms a group (under multiplication), called thespecial linear group, denoted SL(n,k). More generally, we may defined SL(V ) forany finite dimensional vector space V .

Likewise, we may consider the group O(n) of (real) invertible matrices thatpreserve lengths and angles. This is again a group under multiplication, and it isthe symmetry group if you can see the lengths and angles. In fact, O(2) is preciselythe 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 bijectivemaps from X to itself is a group. In contrast, the set of continuously differentiablemaps 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 Xto itself whose inverse is of the same kind is a group. It is the group of symmetriesof the “smooth structure” on X. □

Example 16. The set of symmetries of any subset of the plane, as we consideredin section 1 is a group under composition. Here we defined symmetry to mean atransformation of the set into itself that preserves lengths and angles.

We already identified O(2) as the group of symmetries of the circle, and thepenguin has the trivial group as a group of symmetries. The other two groups arefinite, and their multiplication can be described explicitly: the non-trivial symme-tries of the rectangle are the horizontal flip H, the vertical flip V , and the rotationin 180◦ R. The operation is given by H2 = V 2 = R2 = e, HV = V H = R,HR = RH = V and V R = RV = H. The group of symmetries of the square isdescribed in Milne [2, sec. 1.17]. □

Example 17. More generally, for any n > 2, there is a regular n-gon in theplane. 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 definetwo new operations. These operations are defined on the set Zn of residues modn, where n is a natural number bigger than 1. The set Zn consists of the numbers0, . . . , n− 1. For x, y ∈ Zn, we define x⊕ y to be the remainder (residue) of x+ ywhen divided by n. In other words, it is the unique element z of Zn such thatx+y−z is divisible by n. Likewise, we define x⊙y to be the remainder of xy whendivided by n.

Example 18. The set Zn with the operation ⊕ of addition mod n forms agroup (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, thesubset 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 abstractlyfrom the definition.

Proposition 19. A group has exactly one identity element.


5. SUBGROUPS 11

Proof. Assume that e and f are two identity elements of a group. Since e isan 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 isany set with more than one element, let G be the set of constant functions from Xto itself (i.e., for any function f in G there is an element c ∈ X with f(x) = c forall x ∈ X), with composition as an operation. Then any element g of G satisfiesfg = f for all f ∈ G (but, of course, not gf = f !) □

End lecture 3, Jan 24Before proving that inverses are unique, it is convenient to derive the cancella-

tion law:

Proposition 21. If a, b, c are elements of a group such that ab = ac, thenb = c. Likewise, if ba = ca, then b = c.

Proof. Assume that ab = ac, and let d be an inverse of a, as promised byaxiom G3. Then b = eb = (da)b = d(ab) = d(ac) = (da)c = ec = c. The proof forthe 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 weget b = c. □

Example 24. Let G = Mat2(k) be the set of all 2 × 2 matrices, with theoperation 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 agroup (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 thenumber 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 ∗ andan 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 themap 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 showthat lx is injective. But this is precisely the left cancellation law. □

Example 26. The set N of natural numbers (including 0) with addition satisfiesthe 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 isgiven by the operation on G.


12 1. GROUP THEORY

Example 28. The set 2Z of even integers is a subgroup of the group (Z,+) ofintegers under addition. Indeed, it is a subset of Z, the sum of two even numbersis 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 thetrivial group consisting of only the identity. □

Example 30. The group (Q∗, ·) of rational numbers under multiplication isnot a subgroup of (Q,+). Although it is a subset and a group, the operations donot coincide. Likewise, (Un, ·) is not a subgroup of (Zn,+). □

Example 31. The group SL(V ) of linear maps of determinant 1 from a finite-dimensional vector space V to itself is a subgroup of the group GL(V ) of all invert-ible linear maps on V . □

If H is a subset of a group G, the condition that the operation restricts to Hmeans that whenever a, b ∈ H, ab ∈ H as well. If this is the case, the associativityaxiom 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 followingare 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 bydefinition, we need to show that if x ∈ H, then x−1 ∈ H. First, if e is the identityof H, then e2 = e. By corollary 22, e is the unit of G. Since we are assuming thatH 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 non-empty, it has some element a ∈ H. By assumption, e = aa−1 ∈ H. Applying theassumption 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 (nota group), the subset H consisting of matrices

[0 00 c

], where c = 0, is a “subgroup” (a

subset closed under the operation, which is a group), whose unit[0 00 1

]is different

from the unit in G. □

Example 34. The subset Bn of GLn consisting of upper triangular matrices isa subgroup. Indeed, it is non-empty, and closed under inverses and products. Thesame 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 asubgroup of the group Sym(X) of bijections from X to itself. Indeed, the inverseof a continuous function is continuous. □

Example 36. The subset of “colour preserving” symmetries of a regular n-gonis a subgroup of the group Dn of all symmetries: the inverse of a colour preservingsymmetry is again colour preserving. □

For finite subgroups, the situation is simpler:


5. SUBGROUPS 13

Proposition 37. If H is a finite non-empty subset of a group G that is closedunder 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 inG. Hence, by proposition 25, H1 is a subgroup. Now, if a ∈ H, then a ∈ H1 hencea−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 familyof subsets of a set A, the intersection of the Ai is the subset of A consisting of allelements that belong to all the Ai.

Theorem 38. If Hi is a family of subgroups of a group G, their intersectionH is again a subgroup

Proof. We use 32.2. Since all the Hi contain the identity, so does H, and inparticular, it is non-empty. If a, b ∈ H, then for any i, a, b ∈ Hi, hence, since Hi isa subgroup, ab−1 ∈ Hi. Hence ab−1 ∈ H. □

Example 39. The subsets Bn and SLn consisting of upper-triangular and vol-ume 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 Ci(X)be the set of bijections from X to itself such that both it and its inverse havecontinuous i-th derivative. Then each Ci(X) is a subgroup of Sym(X), and theirintersection, 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, ingeneral, a subgroup. For example, the product of a matrix of determinant 1 by anupper triangular matrix will in general not be upper triangular or of determinant1. □

The theorem allows us to make the following definition.

Definition 42. Let A be a subset of a group G. The intersection of all sub-groups 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 generatorsfor G.

Example 43. If A is a subgroup of G, then ⟨A⟩ = A. Indeed, it is one of thegroups containing A, and any other such group contains it. □

It helps to have a more concrete description of the subgroup generated by asubset A.

Proposition 44. The subgroup generated by a subset A of a group G is theset of all elements that can be presented as a finite product of elements of A andtheir 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 theelements of A will contain the finite products as well, so we have equality. □


14 1. GROUP THEORY

Example 45. Let v1, . . . , vn be a set spanning a vector space V over Q. Thenthe set of vectors of the form vi/m, where m is an integer, generates V . □

Example 46. The group Dn is generated by two elements: a rotation r by2π/n, and a reflection s. The order of s is 2, and the order of r is n. We can alsopresent 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, togetherwith one reflection. □End lecture 6, Jan 28

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 morethan 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 anam = an+m = aman.

(2) If G has an element a of order n, then ⟨a⟩ is a subgroup of G of ordern, which is thus equal to G. Conversely, if G is cyclic, the order of anygenerator 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 theminimal such n. In particular, if m is the order of a and 0 < n < m, thenan = 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.) Whatare the orders of its elements? An element of O is determined uniquely by an angleof rotation 2πx, where x is a real number in the interval [0, 1). Let gx be theelement of corresponding to x. If x = p/q is rational, then gqx = e. Conversely,if gqx = e, then x = p/q for some p, so x is rational. Hence the elements of finiteorder correspond to rational numbers, and the order is the denominator (in reducedform.) □

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 subsetof elements x of G that commute with a: xa = ax. Then centre of G, Z(G), isdefined 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. Thecentre Z(G) is an abelian subgroup.

Proof. If x, y commute with a, then xya = xay = axy, so xy also commuteswith a. Likewise, x−1a = x−1axx−1 = x−1xax−1 = ax−1. This proves that CG(a)is a subgroup. Z(G) is the intersection of all CG(a). □


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 inthe proof of Proposition 48 that if a is a generator of a cyclic group G that has finiteorder 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 theorder of a divides i − j. In particular, if a has infinite order, then all the ai aredistinct.

Proof. By inverting ai, it is enough to consider the case i = 0. Let n be theorder of a. If n is finite and j = nm + r, where 0 ≤ r < n, then aj = (an)mar =emar = 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,+)

Proof. Assume A is a subgroup, and let i, j ∈ p(A). Then ai, aj ∈ A, soai+j = aiaj ∈ 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 thelemma, p(H) is a subgroup of Z, so is of the form nZ for some n (exercise.) Thenb = an is a generator of H. □


Corollary 56. If G is a finite cyclic group of order n, then G has precisely onesubgroup 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, thenby theorem 55 it is cyclic. Hence k is the order of some element b, which thereforedivides n. If m is the smallest power of a such that am ∈ H, then m characterisesH, so H is unique. □

7. Homomorphisms

We now consider homomorphisms — maps between groups. These are mapsthat respect the structure of the groups, namely, the operation.

Definition 57. A homomorphism from a group G to another group H is amap 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 inG, but the operation between f(x) and f(y) is in H.

The rest of the group structure is automatically preserved:


16 1. GROUP THEORY

Proposition 58. If f : G −→ H is a group homomorphism, then f(e) = e, andfor 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 ho-momorphism from G to itself. Also, the map given by f(x) = e for all x is ahomomorphism from G to any group H. If H is a subgroup of G, the inclusion ofH in G is a homomorphism. □

Example 60. For any finite dimensional vector space V over k, the determinantmap det : GL(V ) −→ k∗ is a homomorphism. □

Example 61. If V and U are vector spaces, then any linear map from V to Uis a homomorphism. □

Example 62. Let V be the set of vertices of the regular n-gon. For anyg ∈ Dn, let fg : V −→ V be the restriction of g to the vertices. Then g 7→ fg is ahomomorphism from Dn to Sym(V ). □

Example 63. If G is any group, and X is any set, let H be the group offunctions from X to G. Then the map from G to H sending g to the constantfunction g is a homomorphism. □

As with functions of sets, we are interested to know when a homomorphism isinjective or surjective or invertible.

Definition 64. A homomorphism f : G −→ H is injective (respectively surjec-tive, bijective) if it has the same property as a function of sets. It is invertible ifthere is a homomorphism g : H −→ G such that fg is the identity on H, and gf isthe 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, xis 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 asubgroup of G, and the image Im(f) of f is a subgroup of H. f is injective if andonly 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 functionof sets. If u, v ∈ H, there are, by assumption, elements x, y ∈ G such that f(x) = uand f(y) = v. Then g(uv) = g(f(x)f(y)) = g(f(xy)) = xy = g(u)g(v). The otherclaims are easy. □

Example 67. The kernel of the determinant map on GL(V ) is the subgroupof 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. □


7. HOMOMORPHISMS 17

Example 69. Let V be a vector space, and v ∈ V a non-zero vector. The subsetGv 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 mape : Gv −→ k∗ assigning to each linear map in Gv the eigenvalue corresponding to vis 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 determi-nant 1 in GL2. If v is the vector (1, 0), then Gv is the group of all upper triangularmatrices in GL2, and so G < Gv. The kernel of the corresponding map from theabove example is the set of matrices of the form

[1 x0 1

]. □

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 isenough 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 witha 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 anisomorphism.) □

Example 73. The group of matrices of the form[1 x0 1

]from example 70 is

isomorphic to the additive group (k,+): the function that takes the above elementto x is an isomorphism. □

The point of an isomorphism is that it demonstrates that two groups lookexactly the same as groups. In other words, any property of the groups that isdefined just in terms of the group structure is preserved under isomorphisms. Hereis a (non-exhaustive) list of such properties:

Proposition 74. Let f : G −→ H be a group isomorphism

(1) G and H have the same order(2) G is abelian if and only if H is abelian(3) G is cyclic if and only if H is cyclic(4) A ⊆ G is a subgroup if and only if f(A) is a subgroup(5) A ⊆ G generates G if and only if f(A) generates H(6) For any a ∈ G, f(C(a)) = C(f(a))(7) a ∈ G is in the centre of G if and only if f(a) is in the centre of H.(8) 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 followsfrom applying the first to the inverse of f . Also, 3 follows from 5, 2 follows from7, 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, twosets have the same size if there is a bijection between them.) It remains to prove 4and 6, which are easy. □

This proposition allows us to prove that certain group are not isomorphic: Oneshows that one group has a property that the other doesn’t.


18 1. GROUP THEORY

Example 75. The group D4 is not isomorphic to S4. Indeed, D4 has lesselements. This can be shown as follows: We have seen that if we enumerate thevertices of the regular n-gon by the numbers 1, . . . , n, we get a homomorphismh from Dn to Sn, sending a symmetry to its restriction to the vertices. Thishomomorphism is injective: if a symmetry fixes all the vertices, then it is theidentity. Since both groups are finite, to show that Sn has more elements, it isenough to show that h is not surjective. If n > 3, any permutation that exchangesonly two vertices does not come from a symmetry. Note, that for infinite groups Hand G, the existence of an injective, non-surjective, homomorphism from H to Gdoes not guarantee that H and G are not isomorphic!

Alternatively, we can use (8): We have seen that the order of any element ofDn is either 2 or divides n. On the other hand, in Sn we have the permutationthat maps any i < n− 1 to i+ 1, n− 2 to 1, and (necessarily) leaves n fixed. Thispermutation has order n − 1, which (if n > 2) does not divided n, and which (ifn > 3) is not equal to 2. □

Example 76. The groups U35 and S4 are not isomorphic, even though theyhave 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 circlegroup has elements of finite order, but R doesn’t. □

Example 78. The group SLn(C) is not isomorphic to GLn(C): SLn(C) has afinite centre, GLn an infinite one. □

Example 79. (Z,+) is not isomorphic to (Q,+): the group Q is divisible: forany element x ∈ Q, and any natural number n, there is some y ∈ Q such thatny = 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 homomorphismgiven by 2 7→ 1 (from U5 to Z4) is an isomorphism. □

Example 82. The map from D3 to S3 is an isomorphism. We already sawthat this is an injective homomorphism. Since any permutation of the vertices of aregular triangle gives a symmetry, it is also surjective. □

Example 83. The subgroup of (Q∗, ·) consisting of powers of 2 is isomorphicto Z. □

Example 84. The circle group is isomorphic to the group SO(2) of real 2× 2orthogonal matrices with determinant 1: The function mapping the rotation by α

to the matrix[ cos(α) sin(α)− sin(α) cos(α)

]is an isomorphism. □

Example 85. The multiplicative group Q∗ is isomorphic to the centre ofGLn(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 linearisomorphism, then for any S ∈ GL(V ), the map TST−1 is a linear bijection of U toitself, i.e., an element of GL(U). It can be checked directly that S 7→ TST−1 is anisomorphism from GL(V ) to GL(U). In particular, this applies to any isomorphism


8. THE CLASSIFICATION OF CYCLIC GROUPS 19

T : kn −→ U (recall that such an isomorphism exists precisely when the dimensionof 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 setsX and Y gives an isomorphism betweenSym(X) and Sym(Y ), and in particular, if X is a finite set of size n, there is anisomorphism between Sym(X) and Sn. □

Example 87. The function x 7→ ex is an isomorphism from R onto the multi-plicative group R+ of positive real numbers. This does not hold form Q, since theexponent of a rational number is not rational. It also does not hold for C, since onC the exponent is not injective. □

As can be seen from these examples, it is sometimes hard to determine whentwo groups are isomorphic! End lecture 9, Feb 7

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 state-ment that we want to verify for cyclic groups, it is enough to verify for groups inthis list.

Our strategy in proving this theorem will be as follows: we will study theset Hom(C,G) of group homomorphisms from one of the groups C above, to anarbitrary group G. After obtaining an explicit description of this set, we will findwhich elements of this set are isomorphisms. Finally, we will show that when Gis cyclic, the corresponding set of isomorphisms is not empty. We start with thefollowing general statement.

Proposition 89. Let G be a group generated by a subset A, and let H be anarbitrary 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 easyto 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 issimilar 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 homomor-phisms from Zn to G is in one-to-one correspondence with elements in G whoseorder 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. Accordingto the proposition, each homomorphism F is determined by its value F (1) ∈ G on1. Conversely, if g ∈ G is an element, the association F (i) = gi is a well definedhomomorphism 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 Gof order a divisor of n. Which of them are injective?


20 1. GROUP THEORY

Corollary 92. A homomorphism from Zn to G is injective if the correspond-ing element has order n. A homomorphism from Z to G is injective if the order ofthe corresponding element is infinite.

Proof. An element in the homomorphism defined by g is in the kernel preciselyif gi = 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 elementsof order n in G. Since G is cyclic, it has such an element. Since both groups are ofsize n, any injective homomorphism is bijective.

Now assume that G is an infinite cyclic group, and let a be a generator. Againby 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. Thismeans that we have presented a list of cyclic groups, all non-isomorphic, such thatevery other cyclic group is isomorphic to one of them. Furthermore, we have astraightforward way, given a cyclic group, to determine to which group in our listthis group is isomorphic: this is determined by the number of elements. A morecomplete answer would be to compute the automorphism group of each group. Weshall do that in the next section.

We may use corollaries 91 and 92 to study the set of Hom(Zp,Zq) of homo-morphisms between Zp and Zq. Indeed, corollary 91 says that this set naturallycorresponds with the set of elements in Zq whose order divides p. Since the orderof each element in Zq divides q, we get that Hom(Zp,Zq) corresponds to the setof elements in Zq whose order divides gcd(p, q). This subset forms a subgroup (itis the set of elements x such that gcd(p, q)x = 0 in Zq), which is generated by

qgcd(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 ahomomorphism.

Furthermore, corollary 92 tells us that there are injective homomorphisms ifand only if p divides q. On the other hand, a homomorphism is surjective if andonly 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 ofany element in Z4 divides 4. On the other hand, 4 is a generator of Z5, so h istrivial. □

Example 95. Let p = 4 and q = 6. There is no homomorphism h from Z4 toZ6 with h(1) = 2. Indeed, such a homomorphism would map 3 to 0, but 3 = −1 inZ4, so this would imply h(1) = 0 as well. According to our observation, h(1) canonly by 0 or 3 = 6

gcd(4,6) = 62 . In the latter case, h(2) = 0, so there is no injective

homomorphism. □End lecture 10, Feb 9


9. THE SYMMETRIC GROUPS 21

8.1. Automorphisms of cyclic groups. We have started the study of groupsby looking at symmetries. It is now natural to ask what are the symmetries of agroup. Recall that we view a symmetry of an object to be a bijection of that objectto itself, that preserves its properties.

Definition 96. An automorphism of a group G is an isomorphism from G toitself. The set of all automorphisms of G is denoted Aut(G).

If f : G −→ H and h : H −→ K are two homomorphisms, their composi-tion is again a homomorphism: (h ◦ f)(g1g2) = h(f(g1g2)) = 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, associativityholds 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 mean-time, we compute the automorphism groups of the cyclic groups.

Example 98. According to the previous section, Hom(Zp,Zp) is in bijectionwith the set of elements of Zp. The injective homomorphisms correspond to ele-ments 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 ofZ are 1 and −1, so Aut(Z) consists of precisely two elements, the identity and theinverse. □

End lecture 11,Feb 11The 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, wecan do it directly for any cyclic group:

Proposition 99. If G is a finite cyclic group of order n, then Aut(G) isisomorphic to Un, by the isomorphism given by fk(g) = gk (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 isan automorphism. Hence k 7→ fk indeed defines a map of sets from Un to Aut(G).

Furthermore, this map is injective, and is a homomorphism since gkl = (gl)k.

Finally, it is an isomorphism since we have already computed that Aut(G) is inbijection with Un as sets. □

End lecture 12,Feb 14

9. The symmetric groups

In this section we will study an important class of groups, the symmetric groupof a set X. An element of such a group, a bijection from X to itself is called apermutation 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 groupof {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?


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 thats(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) thatfix 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 bysetting s′(y) = y for all y ∈ Y . The map s 7→ s′ is clearly a homomorphism ofgroups, 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 elementof Y is of the form s′ where s is the restriction to X − Y . Hence the image ofthe homomorphism is the set of all elements that fix all elements of Y (and inparticular, this set is a subgroup.) □

In particular, we may view Sn−1 as the subgroup in Sn of all elements that fixn. 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 ofsize 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 ofall permutations that take n to i (Ti is not a subgroup if i = n.) The Ti forma 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 sit fixesn, so is in Tn. Conversely, if r ∈ Tn, then sir takes n to i, so is in Ti. Thus siis 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, bythe induction hypothesis, we get |Sn| = n · (n− 1)! = n!. □

Example 102. Recall that S3 can be identified with D3, the group symmetriesof the regular triangle. If we identify the top vertex of the triangle with 3, the setTi in the proof corresponds to the set of symmetries that take this vertex to vertexi, namely, the rotation from 3 to i, and the same rotation followed by a reflectionaround vertex i. □

An arbitrary elements s of Sn can be written as[

1 2 ... ns(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 writeit as

[1 2 3 43 4 1 2

]. The top line seems redundant, but it helps when computing the

composition.End lecture 13,Feb 16 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 fors if s(Y ) = Y . Thus, every element s of Sn has the whole set {1, . . . , n} and theempty set as invariant subsets. Furthermore, any subset of the set of fixed elementsof s is invariant.



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 thata proper non-empty invariant subset that does not contain fixed elements, has tobe of size 2. □

Example 106. In S4, the element[1 2 3 43 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} isinvariant as well. On the other hand, the element

[1 2 3 43 2 4 1

]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 havesk(x) = x (why?), and any element not of the form si(x) is fixed (otherwise theset of si(x) is a proper non-empty invariant subset.) Therefore, s is completelydetermined 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, shiftedcyclically 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 to4, which is mapped to 1. Thus this cycle is represented by (1, 3, 4). If we wouldhave 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 4is written, and look what is the next number. If 4 is in the end (like in the firstrepresentation), the “next” is the first one, so in this case, s(4) = 1. If the numberdoes 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 (orthe s-orbit of k) is the set of elements of the form k, s(k), s2(k), . . .. Thus, k isfixed by s precisely if its s-orbit contains only k, and a cycle is an element thathas exactly one orbit of size bigger than one (and the elements in this orbit areprecisely the elements that appear in its representation in cycle notation.) We notethat the s-orbit of k can alternatively be defined as the smallest s-invariant subsetcontaining k. With this definition, we see that any two orbits are either equal ordisjoint.

Two permutations are called disjoint if their sets of non-fixed elements aredisjoint. Equivalently, any element is fixed by either one or the other. We willusually 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.


24 1. GROUP THEORY

(1) Let s and t be disjoint elements, and let x be an element fixed by t. Wewill show that s(t(x)) = t(s(x)). By symmetry it will follows for elementsfixed 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 notfixed 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 ofsize 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. Sincethe Xi are disjoint, so are the si. Finally, it is clear that their product isequal 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 decompositionof s into cycles corresponds to orbits of s, and the orbits are uniquelydetermined. □

Example 109. Let us present the permutation s =[1 2 3 4 5 6 7 8 91 5 2 8 7 9 3 4 6

]as a prod-

uct of disjoint cycles. We do this by simultaneously finding the orbits, and writingthe 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, wepick the next element not yet accounted for, namely, 4. We find s(4) = 8 ands(8) = 4, corresponding to the cycle (48). Similarly, we get the cycle (69). Thuss = (2573)(48)(69). □

End lecture 14,Feb 18 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 inter-section of ⟨g⟩ and ⟨h⟩ is trivial, then |gh| = n.

Proof. Since both orders divide n, it is clear that (gh)n = gnhn = e, so theorder of gh divides n.

Conversely, if (gh)m = e then gm = h−m. Since the left side belongs to ⟨g⟩ andthe 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 thenon-trivial orbit. Thus we get:

Corollary 111. The order of an element of Sn is the least common multipleof the sizes of its orbits

Proof. If s and t are disjoint permutations, then the subgroups they generateintersect trivially: indeed, any element in 1, . . . , n is fixed by either all elementsof ⟨s⟩ or of ⟨t⟩. The statement now follows from the lemma by induction on thenumber of cycles. □

Example 112. Compute all possible orders of elements of S4, and the numberof elements of each order. □



The uniqueness result in theorem 108 holds only for decompositions into disjointcycles. In fact, if we drop the disjointness requirement, we get a stronger result.

Corollary 113. Each group Sn is generated by transpositions (a transpositionis a cycle of order 2.)

We note that there is no uniqueness here, and that the transpositions are notdisjoint. In particular, they don’t commute.

Proof. Since the cycles generate Sn, it is enough to show that a cycle is aproduct 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−1ak). By induction, the firstfactor is a product of transpositions. □

Example 114. Let us write present the permutation in example 109 as a prod-uct 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). □

End lecture 15,Feb 21 (Midterm 1)

9.1. Cayley’s theorem. We now consider what possible subgroups a permu-tation group can have. Obviously, a subgroup of Sn has to be finite. Conversely,we have:

Theorem 115 (Cayley’s theorem). Any group of order n is isomorphic to asubgroup of Sn

Proof. Since Sn is isomorphic to Sym(X) for any set X of size n, it is enoughto show that there is an injective homomorphism from G to such a group. We takeX = G, and the homomorphism assigns to each element g ∈ G the permutationlg 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 identitybijection, 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 thetop line of the table. If g ∈ G is an element, the row that has g as the leftmostelement consists of the elements of G in some order. This is the permutationinduced by G.

Example 116. for the group in example 11, we see that la =[e a b ca b c e

], lb =[

e a b cb c e a

]and lc =

[e a b cc 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 Cayleyhomomorphism 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 gih. Sincethe orbits form a partition of G, the sum of their sizes is n. Hence k divides n. □



26 1. GROUP THEORY

9.2. The sign homomorphism. We have stated that the presentation of anelement as a product of transpositions is not unique. Indeed, even the numberof transpositions in different presentations of the same element may be different.For example, (13) = (23)(12)(23). However, the following theorem shows that theparity 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 thesign of s.

The uniqueness follows directly from the fact that the transpositions generateSn. The existence will be proved later.

Thus, to compute the sign of s, we write it as a product of transpositions. Thesign will be −1 if and only if the number of transpositions is odd. The contentsof the theorem is that this will not depend on the way we decomposed. In otherwords, the number of transpositions might not be the same, but it is either alwaysodd 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 isodd. □

Another way compute the sign is as follows: in the two rows way of writing thepermutation, draw a line from every number in the top row to the same numberin 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 beachieved by rescaling the picture.)

Example 120. For the permutation from example 109, we get

1 2

>>>>

>>>>

3

WWWWWWWWWW

WWWWWWWWWW

WWWWWWW 4

WWWWWWWWWW

WWWWWWWWWW

WWWWWWW 5

jjjjjjjj

jjjjjjjj

jjjj6

TTTTTTTT

TTTTTTTT

TTTT 7

qqqqqq

qqqqqq

qq8

ffffffffff

ffffffffff

ffffffff 9

jjjjjjjj

jjjjjjjj

jjjj

1 5 2 8 7 9 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, thefact that s is not the identity means that there is a pair of adjacent elements whoseorder is wrong. Switching these two elements removes precisely one intersectionpoint, and is achieved by composing with a transposition. This gives a way towrite s as a product of transpositions, where the number of transpositions is thenumber of intersections.

The set of even permutations is the kernel of the sign homomorphism, and istherefore a subgroup of Sn.

Definition 121. The subgroup of Sn of even elements is called the alternatinggroup of order n. It is denoted by An.

For example, A4 can be viewed as the symmetry group of the regular tetrahe-dron.

Example 122. Compute the possible orders of elements of A5, and the numberof elements of each order. □




9.3. Symmetric groups and linear groups. We now outline a connectionbetween 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(x) by permuting thexi according to s: Ts(x) = (xs−1(1), . . . , xs−1(n)). It is clear that Ts is a linear mapfrom the space of n-tuples to itself. Note that our definition, which might appearconfusing, simply means that if, say, s(1) = 3, then Ts moves the first coordinateto the third one. In particular, if ei = (0, . . . , 1, . . . , 0) is the i-th standard basiselement, then Ts(ei) = es(i).

Proposition 123. The map T : s 7→ Ts is an injective homomorphism fromSn to GLn.

Proof. We first note that Ts−1 is the inverse of Ts, so each Ts is indeed inGLn. 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 remarkabove,

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 isthe 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 calleda linear representation of G. Combined with Cayley’s theorem, we get for any finitegroup 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 thematrix whose columns are es(1), . . . , es(n). In particular, all entries of this matrixare either 0 or 1, and every column or row contains exactly one entry whose valuesis 1. Conversely, it is easy to see that any matrix of this form represents Ts for somepermutation 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 thetranspose: Ts−1 = T t

s . The decomposition into disjoint cycles corresponds to de-composing 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 identitymatrix, its value on the transpositions is −1. The uniqueness was already noted tofollow 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 forany two matrices A and B, Tr(AB) = Tr(BA), we get that for any two permutationss and r, sr and rs have the same number of fixed points. □

End lecture 18,Mar 4


28 1. GROUP THEORY

10. Group actions

The definition of a group resulted from looking at symmetry. The first exampleswhere 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 inthe 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 invertiblefunction 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 obtainh(gx). Since the operation of the group is the composition of these functions, weget that this element is also equal to (hg)x. Also, since e ∈ G is the identitysymmetry, we have ex = x for any x ∈ X. The definition of an action axiomatisesthese 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 andx ∈ 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. Westress that a G-set (or an action) is an additional information, it is not part of thedatum 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 samegroup 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 onitself. □

Example 131. If V is a vector space over Q, then Q∗ acts on V by multipli-cation. The same holds when Q is replaced by R, C, etc. More generally, if k is aninteger, we may define another action of Q∗ on V by m(x, v) = xkv. □

Example 132. Let P1(R) be the set R ∪ {∞} (where ∞ is simply a newelement.) Then SL2(R) acts on P1(R) via the following formula:

m([a bc d

], x) =

ax+ b

cx+ d

Where the right hand side is interpreted as follows: first, if x is a number andcx + 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 a

c (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) = fk(x), where fk(x) = f(f(. . . (x)) . . . ) appliedk 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 onX 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 restrictionof the action to H gives an action of H on X. More generally, if f : H −→ G is agroup homomorphism, then the function m(h, x) = f(h)x gives an action of H onX. □


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 implythat this map is a group homomorphism. Conversely, if we are given a functiong 7→ fg from G to Sym(X), then we may define a map m : G × X −→ X bym(g, x) = fg(x), and if the function g 7→ fg is a group homomorphism, then thismap 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 grouphomomorphism G −→ Sym(X).

For example, the homomorphism corresponding to the action in example 128is 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 130corresponds to the Cayley homomorphism, example 133 to the unique homomor-phism from Z to Sym(X) that takes 1 to f , and example 134 to the trivial homo-morphism. End lecture 19,

Mar 7Groups are often studied via their actions. For example, all our treatment ofthe symmetric groups was done by considering its action. The very definition of thenotions of cycle, disjoint cycles and so on was done through this action. Though itis possible to define these notions in purely group theoretic terms, it is substantiallymore complicated. The disadvantage of our approach is that we did not prove, forexample, 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] toarbitrary group actions. As our goal, we will use the following generalisation ofcorollary 117, due to Lagrange.

Theorem 137 (Lagrange). If H is a subgroup of a finite group G, then theorder 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, denotedGx, 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 iscalled transitive if it has only one orbit.

Example 139. In example 131 above, if v is a non-zero vector, then the orbitof v is the one-dimensional subspace of V generated by v, with 0 removed. Geomet-rically, 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 examplesabove, the action is transitive. □

Remark 140. Most other notions and statements we proved about elements sof Sn apply to more general group actions, by considering the group in place of thesubgroup 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 .


30 1. GROUP THEORY

(3) A subset Y ⊆ X is G-invariant if gy ∈ Y for all g ∈ G and y ∈ Y . Y is Ginvariant 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. Anyminimal non-empty invariant subset is the orbit of any of its elements.

(5) The union, intersection and complement of invariant subsets is again in-variant

(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 agroup 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 thisaction is called a (right) coset of H. We note that H itself is a coset: it is the orbitof e.

Example 141. If G = R2, and H is a line (one-dimensional subspace), thenthe cosets of H are the lines parallel to H. □


Example 142. If G = R, and H = Z, then the cosets can be visualisedas 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 thismap 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 theelements of G. □

Whereas for a general G-set, distinct orbits can be very different (consider theorbit 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 anytwo cosets of H.

Proof. It is enough to prove that there is a bijection between H and anyother coset Hg. We claim that the function h 7→ hg is such a bijection. Indeed, bythe definition of the orbit it is a well defined function from H onto Hg, and it isinjective 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 everyelement of G lies in some coset, we get that G is the disjoint union of the cosetsof 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 denotedG : 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 alsowrite it as (G : 1) = (G : H)(H : 1).


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 orderp (actually, this follows already from Corollary 117). □

Example 146. Any group G of order 2p, where p is an odd prime, is isomorphicto either Dp or Z2p. Indeed, if G has no element of order 2p, then any element isof 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 ele-ments 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 a2R = 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 Rhave order 2.

We now know that G is the disjoint union of R and aR, and all elements of thelatter have order 2. This determines completely the multiplication of G, and so, Gis isomorphic to Dp. □


11. Normal subgroups and quotients

11.1. Direct products. Recall that if G and H are arbitrary groups, wedefined a group structure on G × H by performing the multiplication pointwise:(g1, h1)(g2, h2) = (g1g2, h1h2). We have seen that much of the information aboutG × H can be deduced from corresponding information on G and on H. Forexample, the order of an element (g, h) is the lcm of the corresponding orders. Itis thus natural to ask when a given group can be presented as the product of twonon-trivial groups.

We have seen that G × H has subgroups G′ and H ′ isomorphic to G andH (namely i1(G) and i2(H), where i1(g) = (g, 1) and i2(h) = (1, h)). Thesesubgroups 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 agroup 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 homo-morphism because of 2, it is injective because of 1 and it is surjective becauseof 3. □

The theorem requires us to find two subgroups of P . In fact, we can get oneof them for free, as follows: If P = G ×H, we have a projection map π : P −→ G(given by π(g, h) = g.) The kernel of this map is the subgroup H ′ isomorphic toH. As before this is preserved under isomorphisms. Thus, if P is isomorphic toG × 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


32 1. GROUP THEORY

of π is a group homomorphism s : G −→ P such that π ◦ s is the identity on G. Forexample, if P = G×H, and π is the projection onto G, then i1 : G −→ G×H is asection. We note that a section is automatically injective.

Theorem 148. Let π : P −→ G be a surjective homomorphism between twogroups, with kernel H. Assume that there is a section s : G −→ P of π, such thatfor 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′ commutewith the elements of H. If g′ ∈ G′, then g′ = s(g) for some g ∈ G, and soπ(g′) = g. Hence G′ ∩ H is trivial. Finally, if p ∈ P , let h = s(π(p))−1p. Thenπ(h) = π(s(π(p)))−1π(p) = π(p)−1π(p) = e, so h ∈ H. Since p = s(π(p))h, wehave shown that an arbitrary element p can be written as a product of an elementof G′ and an element of H. Using theorem 147, we see that P is isomorphic toG′ ×H. But s is an isomorphism of G with G′ (with inverse π), so P is isomorphicto G×H. □

Remark 149. Not every surjective homomorphism has a section. For example,consider the homomorphism π : Z4 −→ Z2 given by taking the residue mod 2. If sis 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 requir-ing that s(g)h = hs(g) for all g ∈ G and h ∈ H, it is enough to require thats(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 148need not hold. However, it holds trivially if P is abelian:

Corollary 151. If a surjective homomorphism π : P −→ G from an abeliangroup P has a section, then P is isomorphic to G×Ker(π)

End lecture 22,Mar 21 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, wedefine the p-free part of G, G(p), to be the set of elements whose order is prime top, 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. Thenx 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 anautomorphism 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 highestpower of p that divides the order of G.



Example 155. The group U19 has 18 elements. Hence, the theorem says thatit is isomorphic to a product of two groups, one of which is of order 2 and theother of order 9. In fact, it is isomorphic to Z2 × S, where S is the subgroup ofU19 consisting of elements that are squares of another element. The isomorphismis 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 productis a strong condition on a group. Unfortunately, a group is rarely a product of twogroups. In view of theorem 148, we may try replacing this condition by the weakercondition that a group P has a surjective map onto a non-trivial group G. In moredetail, a plan to study all groups might look as follows:

• Find conditions when a group has a non-trivial surjective homomorphismonto another group

• Study the groups that have no such homomorphisms (such groups arecalled simple.)

• Determine in what ways the simple groups can be combined to give othergroups, 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 asurjective homomorphism onto another such group, which is non-trivial if the groupis not a p-group. Furthermore, the latter group is then a direct factor of the originalgroup, so combining groups is easy in this case. The full statement (which we shallnot prove) is that any finite Abelian group can be presented in a unique way as adirect sum of cyclic groups. On the other hand, it is easy to see that the simpleAbelian 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 non-isomorphic finite simple groups, such that each finite simple is isomorphic to onein 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 fromthe previous section that either G is a direct factor of P , or π has no section. WhenP (and possibly G) are not abelian, there is an intermediate case: there might bea section that does not commute with the kernel. It turns out that this situationis almost as good as having a product, and there is a rather rich theory (which weshall 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 aboutanother 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 ahomomorphism π : P −→ G with kernel H?

Let H be a subgroup of P . Since we want H to be the kernel of a homomor-phism, we may try to construct G by simply removing H from P , and replacingit by the identity. We will then have the following function π : P −→ G: π(x) = eif x ∈ H, and π(x) = x otherwise. Unfortunately, G is not a group: if h ∈ H isnon-trivial, and g ∈ P −H, then the product of g−1 and gh is no longer defined inG.

Example 156. Let P = S3, and let H = {e, (12)}. Is there a group homo-morphism π : P −→ G, to some group G, with kernel H? In the process above, wewould try to construct G be removing (12) from P , and defining pi((12)) = e and


34 1. GROUP THEORY

π(x) = x otherwise. However, G is not a group: the product of (23) and (132) isno longer defined. □

To fix the problem above, we could define the multiplication on G to be suchthat (the image of) g−1 multiplied by gh is the identity. Since this should hold forall g ∈ P and all h ∈ H, we need to identify any two elements of the form gh1 andgh2. In other words, each left coset should become one element. Thus we redefineG to be the set of (left) H cosets, and the function π to be the function that sendg ∈ P to gH.

Example 157. Continuing the previous example, we now wish to define G tobe 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 g1H · g2H = π(g1)π(g2) = π(g1g2) = g1g2H.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 forany g ∈ P and h ∈ H, f(g−1hg) = f(g−1)f(h)f(g) = f(g)−1f(g) = e. Thus,g−1hg ∈ H.

Definition 158. A subgroup H < P is called normal if for any g ∈ P ,g−1Hg ⊆ H. The notation H ⊴ P means that H is a normal subgroup of P .

Remark 159. It follows that in fact, g−1Hg = H, by applying the conditionwith g−1 instead of G. However, it is not true, in general, that g−1Hg < H impliesthat g−1Hg = H.


Example 160. In the example above, if we want to define a group structureon 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 forany 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, forg = (13), we get (13)(12)(13) = (23), which is not in H. This shows that H is notnormal 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:



Theorem 162. Let H ⊴ P be a normal subgroup. Then the set P/H of leftH cosets has a group structure, such that the function π : P −→ P/H given byπ(g) = gH is a surjective group homomorphism with kernel H. Furthermore, iff : P −→ G is another group homomorphism whose kernel contains H, then there isa 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 Pby H. The second part of the theorem implies that this is (essentially) the onlysolution to the problem: find a surjective homomorphism from P with kernel H.

Proof. We already know that the function π : P −→ P/H is surjective, andby 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 homomorphismforces us to define the group law by g1H · g2H = g1g2H. It remains to show thatthis formula determines a well defined group structure. However, the normalitycondition implies that in P , the set g1Hg2H = {g1h1g2h2|h1, h2 ∈ H} is equal tog1(Hg2)H = g1(g2H)H = g1g2H. In other words, if X,Y ∈ P/H then the productXY , in P , of the sets X and Y , is again a coset. This shows that the product iswell 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 = g1H, theng−1g1 ∈ H, hence f(g−1g1) = e, so f(g) = f(g1). It is clearly a homomorphismthat satisfies the requirements, and since its definition is forced on us by theserequirements, it is unique. □

Corollary 163. Let f : P −→ G be a surjective homomorphism with kernelH. Then G is isomorphic to P/H. Furthermore there is a unique isomorphismf : P/H −→ G such that f = f ◦ π.

Proof. By the theorem, there is a unique group homomorphism f with therequired property. It remains to show that it is an isomorphism. If g ∈ G, let p ∈ Pbe such that f(p) = g. Then f(pH) = f(π(p)) = f(p) = g. This shows that f issurjective. 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 Gitself are referred to as quotients of P .

11.3. Examples and applications.

Example 164. Any subgroup of an Abelian group is normal, The quotient isAbelian 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[1 x0 y

]is not normal in GL2 (or

in SL2). This can be shown directly by conjugating with the element[

0 1−1 0

], but

also follows from example 169, since this subgroup is precisely the stabiliser of thevector (1, 0) in the action of GL2 on the plane. □

Example 167. The group Z/nZ is isomorphic to Zn: the function from Z toZn assigning to each number its residue modulo n is a surjective homomorphismwhose kernel is nZ. □


36 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, whereh ∈ H, must be either in H or in gH. It cannot be in gH, so it is in H. □

End lecture 24,Mar 25 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 inducedmap 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 anelement. If Gx is the stabiliser of x, and g ∈ G, then gGxg

−1 = Gg(x). Since theaction is transitive, any point of X has the form g(x) for some g ∈ G. Hence allthe stabilisers are conjugate, and all conjugates of a stabiliser are stabilisers. Inparticular, if Gx is normal, then all the stabilisers are equal. But then any elementin 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 subgroupof order 6. Indeed, by example 168, such a group would be normal. It follows thatthe 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 thedisjoint cycle decomposition (see proposition 179 below.) Hence A4 has a normal(and even characteristic) proper subgroup. In contrast, we will see in theorem 180that for n ≥ 5, An is simple. □

Example 171. If n > 1 is a number, then the rotation R by a half circle isan element of D2n. The subgroup {e,R} is normal, since R commutes with allrotations, and is the only rotation of order 2. The quotient can be identified withDn by identifying any two opposite points of the regular 2n-gon, turning it into ann-gon. □

The following application proves that the group of inner automorphisms (home-work) is not cyclic (unless it is trivial.)

Proposition 172. If G/Z(G) is cyclic then G is Abelian.

Proof. We prove something stronger: if x, y ∈ G commute, and x′, y′ repre-sent 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, ifx′ = 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 el-ement g ∈ G such that any class is represented by some gi, and all such elementscommute. □

End lecture 25,Mar 28 The following theorem can be viewed as a partial converse to Lagrange’s theo-

rem (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 finitegroup G, then G contains an element of order p.



Proof. We prove the theorem in the case that G is abelian; the general casewill 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 top (otherwise a power of a has order p.) Hence the order of the quotient G/⟨a⟩ isagain divisible by p. By minimality, it contains an element b of order p, but anypre-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, anda subgroup of maximal order pk is called a p-Sylow subgroup. The other Sylowtheorems say that every p-subgroup of G is contained in a p-Sylow subgroup, andthat all p-Sylow subgroups are conjugate.

Remark 175. Combined with corollary 153, it follows from Cauchy’s theoremthat any finite abelian group is isomorphic (canonically) to a productGp1×. . .×Gpk

,where the pi are distinct primes, and the order of Gpi is a power of pi.


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 groupaction, and x ∈ X, then the map G −→ X given by g 7→ gx induces a functionG/Gx −→ X, which is a bijection. In particular, If G is finite (and hence so is X),

then |X| = |G||Gx| .

Example 176. The action ofD4 on the set of vertices of the square is transitive.The stabiliser of a vertex consists of the identity and the reflection around the axisthat passes through that axis. Since D4 has 8 elements, we get that a square hasfour 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 givesanother proof that the order of Sn is n!. □

End lecture 27,Apr 1More generally if the group action is not necessarily transitive, then X is a

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| =∑X/G

(G : Gx) (2) {eq:orbits}

where the sum is over the set of orbits, and x is any element of the correspondingorbit.

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 elementg is the set of all elements of the form hgh−1, where h ∈ G. The stabiliser of g isthe 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) (3) {eq:class}


38 1. GROUP THEORY

where the first sum is over all conjugacy classes, and the second is over conjugacyclasses of size bigger than 1 (and in each case g is an element in the correspondingclass.)

The second formula follows from the first one since the elements of Z(G) areprecisely the conjugacy classes of size 1. We can now prove the general case ofCauchy’s theorem (Theorem 173.)

Proof of Cauchy’s theorem, general case. Let G be a counterexampleof minimal order. By minimality, G cannot contain a proper subgroup whose orderis 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 theclass formula is divisible by p. Since the order of G is also divisible by p, it followsthat so is the order of Z(G). Hence G = Z(G), and we are back in the previouscase. □

End lecture 28,Apr 4 We next compute the conjugacy classes in Sn:

Proposition 179. (1) If s is an element of Sn and c = (a1a2 . . . ak) isa cycle, then scs−1 = (s(a1)s(a2) . . . s(ak))

(2) Two elements of Sn are conjugate if and only if their disjoint cycle de-composition has the same form.

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 notone of the ai, and s(c(ai)) if x = ai. This is precisely the function givenby the cycle in the statement.

(2) If g is any element, conjugating by s amounts, by the previous part, toapplying s to each element in the disjoint cycle decomposition. Afterapplying s, we get a product of disjoint cycles, which is therefore thedisjoint cycle decomposition of the conjugate. Conversely, if the cycledecomposition is has the same shape, we may find a permutation s thatmaps the elements of corresponding cycles to each other. □

End lecture 29,Apr 6 (Midterm 2) 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 noproper non-trivial normal subgroups. For example, an Abelian group is simple ifand 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 ofdegree 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, suchconjugates have the form (s(a)s(b)s(c)), where s ∈ An. Thus we need to find, forany cycle (xyz) an element s ∈ An that maps (abc) to (xyz). We would like to takes = (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.



It remains to show that G must contain at least one 3-cycle. This can be provendirectly by induction on the maximal number of elements fixed by an element ofG. See Milne [2, Lemma 4.36] for details. □

End lecture 30,Apr 8


CHAPTER 2

Galois theory

In this chapter, we will study an application of group theory to a classical prob-lem — solving polynomial equations by radicals. Along the way we will introducea new kind of algebraic structure, namely fields. The problem was studied by Abeland Galois (among others), and in the course of the solution they came up withboth the notions of groups and fields, and the connection between them.

The route we take is rather direct, omitting many important notions, andleaving a lot to exercises. See Milne [1] or Rotman [3] (for example) for moredetailed 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 struc-ture associated to it, field.

To describe the problem, we recall that the solutions of a quadratic equation

x2+ax+b = 0 can always be written in the form −a±√a2−4b2 . In general, a function

of the form p(x) = xn + an−1xn−1 + · · · + a0 is called a (monic) polynomial , and

an equation of the form p(x) = 0 is called a polynomial equation. n is called thedegree of the polynomial (or the equation). There are explicit formulas, similar tothe above, for the solutions of equations of degrees 3 and 4, which were known sincethe 16th century. For example, a solution to the equation

x3 + ax2 + bx+ c = 0

is given by

x =3

√−r +

√r2 + 4q3/27

2+

3

√−r −

√r2 + 4q3/27

2(4) {eq:cubic}

where

q = b− a2

3

r = c+2a3

27− ab

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 aboveare not rational. They lie in some bigger “set of numbers”. For the formula aboveto make sense, in this bigger set of numbers it should be possible to use addition,

41

42 2. GALOIS THEORY

multiplication and division, just as in the rationals. Such a set of numbers is calleda field. More precisely:

Definition 181. A field is a set L together with two operations + and ·, andtwo 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, andwhich 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 afield, since multiplication does not give a group. □

Exercise 183. Prove that (Zn,+, ·), where + and · are multiplication mod nis 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 ofpairs (a, b) of rational numbers, written as a+ br, where r is a fixed symbol. Defineoperations 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 rationalsquare root, and that in K, r2 = s (where r is 0 + 1r). Hint: one way to do it is toidentify K with the set of matrices

[a bsb a

].

Note that if we replace rational numbers by real numbers, and take s = −1, wehave constructed the complex numbers.

The construction in the last exercise can be described as follows: we startwith 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 ofconstruction is central to the theory, and in the terminology below, K is the splittingfield of x2 − s. To discuss this more precisely, we first note that the notions ofpolynomials and rational functions makes sense for any field.

Definition 185. Let K be a field. A polynomial (in one variable t) over K isan expression of the form p(t) = ant

n + · · ·+ a0, where each ai is an element of K.We identify an expression as above with the expression 0tn+m+ · · ·+0tn+1+ant

n+· · ·+ a0 for any m (i.e., they define the same polynomial), so that the coefficient aiabove is defined for all i (and is equal to 0 for i > n). Likewise, we sometimes omitterms whose coefficient is 0.

We identify an element x of K with the polynomial for which a0 = x and ai = 0for i > 0. The degree of a non-zero polynomial is the largest i for which ai = 0. Anon-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 additionand multiplication on the polynomials in the usual way: if p(t) = amtm + · · ·+ a0


1. STATEMENT OF THE PROBLEM, AND ITS SOLUTION 43

and q(t) = bntn + · · ·+ b0, with m ≥ n, then

p+ q = q + p = (am + bm)tm + · · ·+ (a0 + b0) (5)

p · q = ambntm+n + (ambn−1 + am−1bnt

m+n−1 + · · ·+ a0b0 (6)

Polynomials in several variables are defined analogously, and the operations betweenthem 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 thesame field. The polynomial thus determines a function from L to L, and viewedin this way, the addition and multiplication operations correspond to addition andmultiplication 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 (withthe operations above), since there are no multiplicative inverses. The set of rationalfunctions over K (in one variable t) is defined as the set of expressions

K(t) = {pq∥p, q ∈ K[t], q is monic} (7)

With addition and multiplication extended to K(t) in the usual way, K(t) is afield, extending K. We may apply this construction when K = L(s) is itself a fieldof 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. Thoughthe original question was formulated for polynomials with rational coefficients, itis essential to discuss it more generally. Thus, we are interested in solutions to theequation p(x) = 0, where p is a polynomial over a field K. As we saw, one cannotexpect to find the solutions in the field K itself, so we would like to find a nice fieldextension where these solutions exist. One of our main goals will be to prove thefollowing theorem.

Theorem 187. Let K be a field, and let p be a polynomial over K. Then thereis a field L with the following properties:

(1) If M is a field extending L that contains a solution a to p(x) = 0, thena ∈ L.

(2) If K ⊆ M ⊂ L is a proper subfield of L, then there is a solution of p(x) = 0in L, which is not in M .

Furthermore, the field L is unique up to isomorphism over K1

Definition 188. Given a field K and a polynomial p over it, the field L givenby Theorem 187 is called the splitting field of p (over K).

Thus, the splitting field of p is the minimal field that contains all possiblesolutions of p(x) = 0.

Example 189. The field constructed in Exercise 184 is the splitting field ofx2 − s (this will become apparent later). □

1isomorphisms of fields are defined in Definition 204


44 2. GALOIS THEORY

Now that we have a field that contains all solutions, what does it mean to havea formula for them? The formula for the quadratic equation involves elements ofthe base field, as well as roots of such elements. Hence, they live in a splitting fieldas considered in Example 189. For the cubic equation, we have (third) roots ofelements 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 radicalsif 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 byradicals: 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 splittingfield of the original equation). □

Example 192. A cubic equation x3 + ax2 + bx + c = 0 is also solvable byradicals. 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

equation y2 + ry− q3

27 , so both lie in its splitting field K1, which is an extension by

a root. If we let K2 be the splitting field of x3 − u over K1, and K3 the splittingfield of x3 − v over K2, then K3 contains the solution given by Equation (4) (theother 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 thatis not solvable by radicals.

To prove the result, one needs to analyse the relation between different se-quences of field extensions. This is where group theory enters the picture. The fulltheorem, due to Galois, describes which equations are solvable in terms of someassociated groups.End lecture 31,

Apr 132. First properties of fields

In this section we mention some basic properties and definitions related tofields. 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∩iKi is a subfield (you may use Theorem 38).

2.1. Generators of a field. As with subgroups, interesting subfields of a fieldoften arise as intersections.

Definition 196. Let K be a field, A ⊆ K a subset. The subfield generatedby A is the intersection of all subfields of K containing A (this is a subfield byExercise 195).


2. FIRST PROPERTIES OF FIELDS 45

If F ⊆ K is a subfield, the subfield generated by A over F is the subfieldgenerated 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, thenthe 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 whichf is defined.

Proof. For any such function f and elements a1, . . . , an, f(a1, . . . , an) is a ra-tio 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(a) and q(a, sincethese are sums of products of elements of L. Hence so does the ratio. This provesthat 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 setof rational functions over F is a field. □

2.2. The characteristic of a field. If x is an element of a field K, and nis an integer, we write nx and xn for the “product” of x with itself n times in thecorresponding group structure (so x2 = x ·x, 2x = x+x, x0 = 1, x−1 is the inverseof x with respect to ·, etc.). We abbreviate n · 1 as n (note that now nx a priorihas two different meanings, which in fact coincide). We note that we may have anatural number n > 0 such that n = 0 in K. If this happens, there is a smallest nwith this property.

Definition 199. Let K be a field. If there is an integer n > 0 such that n = 0in K, the smallest such integer is called the characteristic of K. Otherwise, we saythat 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 havethe 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 fieldhomomorphism if it is a non-zero homomorphism of the additive groups, andf(ab) = f(a)f(b) for all a, b ∈ K. A field isomorphism is an invertible homo-morphism.

If F is a subfield of both K and L, then a homomorphism as above is over Fif f(x) = x for any x ∈ F .

Exercise 205. Prove the following facts about field homomorphisms.


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 anisomorphism

2.4. Prime fields. Another application of Exercise 195 is to find the smallestsubfield of a field.

Definition 207. Let K be a field. The prime subfield of K is the intersectionof all subfields of K.

Exercise 208. Show that the prime subfield of Q and of C is Q, and the primesubfield 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, forconvenience, 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 previousexercises. Assume first that p > 0. The assumption says the order of 1 in theadditive group of K is p. Hence, there is a unique injective group homomorphismfrom the additive group Zp of Fp to K. Since (n1) · (m1) = (nm)1 in K, this is afield homomorphism.

Now assume that p = 0. Then for any non-zero integer n, n1 = 0 in K, andtherefore it has an inverse 1

n1 . The map that send a rational number nm to the

element n1 · 1m1 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 caseF0 = Q).

To summarise, any prime field is either Q or Fp for some prime p. Exactlyone of these field is contained in any field, depending on the characteristic, andin a unique way. If K and L are fields of different characteristics, there are nomaps 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 concentrateon the characteristic 0 case, since it is simpler and contains the classical application.

Exercise 211. Prove that Fp is the only field of size pEnd lecture 32,Apr 15


3. POLYNOMIALS IN ONE VARIABLE 47

3. Polynomials in one variable

We now endeavour to prove Theorem 187. Hence we are interested in con-structing a field extension L of a field K, containing a solution to an equationp(x) = 0, with p a polynomial over K. By analogy with the case of usual roots, wecall any solution to such an equation a root of p. Our general strategy will be toadd a solution formally, as in Exercise 184. To make it work, we first need to studypolynomials in one variable in general. We fix a base field K. All polynomials willbe in one variable and over K, unless mentioned otherwise.

As explained in the definition, the polynomials are endowed with operationsof addition and multiplication. It can be easily checked that K[x] forms a groupunder addition, but the non-zero polynomials do not, in general, have inverses un-der multiplication. In other words, K[x] is a (commutative) ring . This is anotherinteresting class of algebraic structures, but we will not study them in general, sincewe will only need this example. On the other hand, as will be seen immediately, thealgebraic properties of K[x] are very similar to the properties of the integers, withusual addition and multiplication. It was already shown above that the rationalfunctions are obtained from the polynomials in the same way as the rational num-bers 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 ifthere 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 atten-tion 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 uniquepolynomials 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, thisis 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−kq is a polynomial of smaller degree than p. By induction,p1 = s1q+ r, so p = xn−kq+ s1q+ r = (xn−k + s1)q+ r, and taking s = xn−k + s1we are done.

For uniqueness, assume that s1q + r1 = p = s2q + 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. □


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 isa field element. Plugging a in both sides yields r = 0. □

Corollary 217. Any non-zero polynomial p has at most deg(p) roots (in anyfield).

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 uniquemonic polynomial d dividing p and q, such that if t is some other polynomial dividingp and q, then t|d.

Definition 219. The polynomial d given in Proposition 218 is called the great-est common divisor of p and q (abbreviated gcd). The polynomials p and q arecoprime if gcd(p, q) = 1.

The proof uses Euclid’s algorithm, which also provides a way of computing thegreatest common divisor.

Proof of Proposition 218. We may assume deg(p) ≥ deg(q), and bothmonic. 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 andr. Hence, q = q1d and r = r1d, so p = sq + r = sq1d + r1d = (sq1 + r1)d, hence ddivides 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 gcdof two polynomials as a “linear combination” of them.

Corollary 220. For any non-zero polynomials p and q, there are polynomialsr 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 areno 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 overL, for any extension L of K. Whereas the previous notions did not depend on thebase field (for instance, p|q as polynomials over K if and only if the same holds overL), irreducibility depends on our base field. Hence when discussing it, we shouldbe careful about the base field. As before, unless we mention otherwise, we workover the fixed base K.

As with integers, we have prime decomposition.

Proposition 223. Let p be an irreducible monic polynomial


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’salgorithm, there are s and t with ps + qt = 1. Hence pkps + pkqt = pk.Since p|pkq, we get that p divides the left hand side, so p|pk. Since pk isirreducible, we get p = pk.

□Corollary 224. Any monic polynomial p is a product p1 . . . pk of monic ir-

reducible polynomials of positive degree. The polynomials p1, . . . , pk are unique, upto reordering.

Proof. If p = qr is reducible, each of q and r are of smaller degree, so byinduction, each is a product of irreducibles. Otherwise, p itself is the decomposition.

For uniqueness, assume that p1 . . . pk = q1 . . . ql are two decompositions. Thenp1 divides q1 . . . ql, so by the proposition, p1 = qi. Cancelling, we get the result byinduction. □

End lecture 33,Apr 18We may now construct the splitting field.

Proposition 225 (Kronecker’s Theorem). Let p(x) be a non-constant polyno-mial over a field K. Then K has a field extension L = K(α) generated by oneelement α which is a root of p.

Proof. We note first that p may be assumed irreducible, since a root of p isalso a root of one of the irreducible components of p. Likewise, we may assume pto be monic.

Let n be the degree of p. We define L as follows. As a set, L consists ofpolynomials of degree smaller than n in a variable α. The additive group structureis the same as for usual polynomials (note that the degree of a sum of polynomialsis 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., theresidue 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 ofL. It is easy to verify that this product is associative. Hence, to prove that L is afield, 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 inverseof 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 getα · 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 con-struction above recovers the field constructed in Exercise 184.

Corollary 227. Given a monic polynomial p over K, there is a field L gen-erated over K by roots of p, such that in L, p is a product (x − b1) . . . (x − bk) oflinear terms. The field L is a splitting field for p.


50 2. GALOIS THEORY

Proof. If p = 1 the statement is obvious. Otherwise, by Proposition 225, thereis 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 ina field generated over L by roots of p1. This field is generated over K by roots ofp, 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 isa subfield of L that contains all roots, then it is equal to L, since L is generated bythe roots. Hence L is a splitting field. □

The name “splitting field” comes from the fact that p splits into linear factorsin 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 Abeliangroup 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 actiongiven by the field multiplication. More generally, the set of tuples Kn is a vectorspace in the usual way. □

Example 230. The set of continuous, differentiable, smooth, rational,... func-tions on the reals (or on an interval, or a domain,...) is a vector space over the realnumbers, with usual multiplication by numbers is a vector space. We note each ofthese 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 characteristicp, 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 rightside 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 grouphomomorphism T : U −→ V is a linear map if T (xu) = xT (u) for any x ∈ K andu ∈ U .

Exercise 235. Show that the kernel and the image of a linear map are linearspaces. Show that if a linear map is invertible as a map of groups, then the inverseis also a linear map.

Exercise 236. Show that the intersection of any collection of subspaces of Vis again a subspace


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 spacestructure making the two projections linear maps.

Exercise 238. Let U be a subspace of a vector space V . Show that the quotientgroup V/U has a unique vector space structure making the map π : V −→ V/U linear.

End lecture 34,Apr 20Definition 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 vectorspace, can be extended uniquely to a linear map from U to V (i.e., there is a uniquelinear 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 anexpression of the form x1v1 + · · · + xnvn. The linear combination is non-trivial ifnot all the xi are 0.

A subset B of a vector space U is linearly independent if no non-trivial linearcombination 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 linearcombination 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 toa 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 Bis 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 skipit.

Definition 245. The cardinality of any basis of U is called the dimension ofU . U is finite dimensional if it has a finite basis.

Example 246. The space Kn has dimension n. A basis is given by (ei)i, whereei has i-th coordinate 1, and the others 0.

It follows that a linear map T from Kn to Km can be give by a matrix, namelythe matrix whose columns are T (ei), for ei the standard basis on Kn. □

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 Kd (a similar statement holdsfor infinite dimensional spaces).


52 2. GALOIS THEORY

Proof. Let B be a basis of U . Then B is linearly independent also in V , socan be extended to a basis B1 of V . It is easy to see that the subspace W generatedby 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 thestandard 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 isa scalar a ∈ K such that Tv = av. The element a is then called the eigenvalueassociated to v.

If p(x) is a polynomial over K, and T : V −→ V is a linear map, we may “eval-uate” p at T to get a new linear map p(T ) (where multiplication is interpreted ascomposition). 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 characteristicpolynomial of T ). The minimal polynomial of T is defined to be the polynomial pof least positive degree, such that p(T ) = 0 (hence it always exists, and divides thecharacteristic polynomial).

Proposition 248. Let T : V −→ V be a linear map on a finite dimensionalvector space V over K. Assume that the minimal polynomial p of T has a root ain 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 somepolynomial 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, contradictingthe minimality of p. Since T − a is not invertible, it has a non-zero kernel. Anynon-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 overK as a K-vector space. The dimension of L over K is called the degree of theextension, denoted [L : K].

Example 250. The complex numbers are spanned by 1 and i as a vector spaceover R. Hence [C : R] = 2. On the other hand, R is an infinite extension of Q, bycardinality.

For any field K, the field K(t) of rational functions over K is an infinite exten-sion, 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 knowthat the splitting field is unique, we start with the particular splitting field we haveconstructed.

Exercise 251. Show that the field L constructed in Proposition 225 is finiteover 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 finiteextension of E. Then F is a finite extension of K, and [F : K] = [F : E][E : K].


5. FINITE EXTENSIONS 53

Exercise 253. Prove Proposition 252 (Hint: if e1, . . . , en is a basis of E overK, and f1, . . . , fm is a basis of F over E, show that (eifj) is a basis of F over K.)

Exercise 254. Assume that p is a polynomial of degree n over K. Show thatthe splitting field of p constructed in Corollary 227 is finite over K, of degree atmost n! (Use previous exercises)

We now go back to the proof of Theorem 187, and prove the main step in theuniqueness result.

Proposition 255. Let p be an irreducible polynomial over a field K, and letK(a) and K(b) two field extensions generated by roots a and b of p. Then there isa 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 fieldconstructed in Kronecker’s Theorem. For any polynomial q(α) ∈ K(α), defineT (q(α)) = q(b). This is clearly an additive group homomorphism. If q1 and q2 aretwo polynomials in K(α), T (q1q2) = r(b), where q1(x)q2(x) = s(x)p(x) + r(x), sor(b) = r(b)+ s(b)p(b) = q1(b)q2(b) = T (q1)T (q2), since p(b) = 0. Hence, T is a fieldhomomorphism. 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 pover K with p(b) = 0, then L has finite degree over K (such an element b is said tobe 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. End lecture 35,Apr 27

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 thedegree, using Proposition 255 and Exercise 252. □

Exercise 257. Fill in the details in the last proof (Note that if L is a splittingfield of p over K, and a is one of the root, then L is also a splitting field of p overK(a))

Remark 258. We have been using that if p is a polynomial over K, and Lis an extension, then p may be viewed as a polynomial over L. More generally, ift : 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 appliesto embeddings as well: given two embeddings ti of K into fields L1 and L2, suchthat each Li is a splitting field for the corresponding polynomial ti(p), there is anisomorphism 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 mostn roots. It is convenient to have a criterion when there are precisely n of them.

Definition 259. Let p(x) = anxn + · · · + a0 be a polynomial over K. The

derivative of p is the polynomial p′(x) = nanxn−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 derivativewhen K = R).


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 andonly if p is constant (i.e., has degree at most 0). Show that this is false in positivecharacteristic.

If two polynomials p and q have a common root a in some extension L, thenover L, both are divisible by (x − a), so they are not coprime. In fact, this holdsover the original field as well.

Proposition 262. Assume that polynomials p and q over K have a commonroot 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, acontradiction. □

Proposition 263. Let p be a polynomial of degree n over a field K. Assumethat p and p′ are coprime. Then p has n different roots (in the splitting field). Inparticular, if K has characteristic 0, then any irreducible polynomial of degree nover 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)2q(x) (over the splitting field). Hence p′(x) = 2(x −a)q(x) + (x− a)2q′(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 is0, p′ is non-zero. Hence the degree of gcd(p, p′) is at most deg(p′) < deg(p) so is 1since p is irreducible. □

Exercise 264. Prove that in a field of characteristic 0, every non-zero elementhas n distinct n-th roots in some extension.

Exercise 265. Let K be a field of positive characteristic p. Show that thefunction f : K −→ K given by f(x) = xp is a homomorphism from K to itself. Thishomomorphism is called the Frobenius endomorphism. (Hint: use the binomialexpansion).

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 groupof symmetries.

Definition 266. The automorphism group of an extension L of K is the groupAut(L/K) of field automorphisms of L over K.

Proposition 267. Let L be a field extension of K, let a ∈ L, and assume thatp(a) = 0 for some non-zero polynomial p over K. Then τ(a) is also a root of p, forany 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.


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 Xgenerates L. □

Example 268. The equation x3 = 2 has a unique real solution a = 3√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, andtherefore a. Hence Aut(L/K) is trivial. □

It may seem from the last example that the group Aut(L/K) does not carrymuch information about the extension. We will see below that the situation isdifferent 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 isprime. The set of solutions in L of the equation forms a multiplicative subgroup ofL, 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 ofthe 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 anelement that does not have a p-th root. The splitting field L of xp − a over Kcontains the unique p-th root. □

The examples above suggest that the group of automorphisms is most meaning-ful for splitting fields. A splitting field is attached to a particular polynomial, butit will be more convenient to have a condition that is independent of a particularpolynomial. We thus make the following definition.

Definition 271. Let L be a finite extension of K. We say that L is a normalextension of K if any irreducible polynomial over K that has a root in L, splits inL. 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 Lover K.

By Exercise 264, if an irreducible polynomial of degree n over K has a root ina 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 char-acteristic (we have seen in Example 270 that normality is insufficient to have ameaningful Galois group). Since it is slightly more complicated, and irrelevant tothe 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 havecharacteristic 0. However, the statements below remain true in positive character-istic.

It might not be obvious that Galois extensions exist at all. In fact, any splittingfield is normal.

Proposition 273. A finite extension L of K is normal if and only if it is thesplitting field of some polynomial.

End lecture 36,Apr 29If 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.


56 2. GALOIS THEORY

Corollary 274. Let L be a Galois extension of K, and E and F be twointermediate extensions. If T : E −→ F is an isomorphism over K, then T extendsto an automorphism of L over K.

Proof. We know this when L is a splitting field, by Remark 258, hence forGalois extensions by Proposition 273. □

We will need to know that the notion of a Galois extension behaves well withrespect to intermediate extensions. The proofs of Propositions 273 and 275 aregiven below, after some more tools are developed.

Proposition 275. If L is a Galois extension of K, and K ⊆ E ⊆ L is anintermediate extension, then L is a Galois extension of E.

The fundamental theorem of Galois theory establishes a connection interme-diate extensions K ⊆ E ⊆ L where L/K is a Galois extension, and subgroups ofAut(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 isany 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)

Exercise 276. Show that LA is always a subfield of L containing K. Showalso that LA = LH , where H is the subgroup generated by A.

Thus we have a way of translating between intermediate extensions and sub-groups: 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 thatthis is a bijection.

Theorem 277 (The fundamental theorem of Galois theory). Let L be a finiteGalois extension of a field K, and let G = Aut(L/K) be the Galois group.

(1) The correspondence between subgroups of G and intermediate extensionsis 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 subgroupH 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-extensionK ⊆ E ⊆ L is a normal extension of K if and only if Aut(L/E) is anormal subgroup of G. In this case, Aut(E/K) = G/Aut(L/E).

This theorem provides a full translation between intermediate field extensionsand subgroups of the Galois group. It allows to transfer properties and tools fromone 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 extensionis cyclic, Abelian, etc. if G = Aut(L/K) has the corresponding property.

Exercise 279. Let L be the splitting field of x3−2 over Q. Compute the Galoisgroup G of L over Q, find all subfields of L, and the corresponding subgroups of G.

End lecture 37,May 2



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 theminimal 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 leta be a root of p in an extension L. Then p divides any polynomial q over K withq(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 uniquemonic polynomial p over K, such that for any polynomial q over K, p divides q ifand 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 polynomialsover K. At least one of the factors p has a as a root. Since p is irreducible, it hasthe required property by Proposition 280. □

Definition 283. The polynomial given by Corollary 282 is called the minimalpolynomial of a over K.

We may now return some debts.

Proof of Proposition 275. Let p be an irreducible polynomial over E thathas a root a in L. The minimal polynomial q of a over K is also a polynomial overE, and since q(a) = 0, p divides q. Hence every root of p is also a root of q (in anyextension). But q splits in L, hence so does p. □

Proof of Proposition 273. Assume first that L/K is normal, and let a ∈ Lbe an element not inK. Let E = K(a). By Proposition 275, L is a normal extensionof E. Since E is a proper extension of K, [L : E] < [L : K], so by induction, L isthe splitting field of some polynomial q over E. Let b be a root of q in L, and let pbe the minimal polynomial of b over K. By definition, p splits in L. Any root of qis also a root of p, so L is generated over E by the roots of p. Hence L is generatedover 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 pover K, let a ∈ L, and let q be the minimal polynomial of a over K. If q doesnot split in L, let E be the splitting field of q over L, and let b ∈ E be a rootof q not in L. Since q is irreducible over K, we have, by Proposition 255, anisomorphism 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 aregenerated 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 somemore linear algebra. In this subsection, L is a field of characteristic 0, H is a finitegroup 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, wemay view En as a subset of V , and in particular, V has a basis consisting of vectors


58 2. GALOIS THEORY

over E (i.e., with all entries in E). If U ⊂ V is an L-subspace of Ln, this no longerneeds to be the case.

Example 284. Let U be the subspace of C2 spanned by (1, i). Then the onlyvector with real entries in U is 0. □

The group H acts on V , by acting on each coordinate. This action respectsthe addition on V , but it is not L-linear: h(xv) = h(x)h(v) for h ∈ H, x ∈ L andv ∈ V . The elements with coordinates in E are precisely those fixed by H. So theexample 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 elementh of H to a general element u = x1v1 + · · · + xkvk of U , we get h(u) = h(x1v1 +· · · + xkvk) = 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 underthe action of H. It turns out that this is also sufficient: if U is invariant, then ithas 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 ofautomorphisms 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 En.

Proof. By assumption, there is a non-zero vector u ∈ U . We may assume thatthe 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 firstcoordinate of v is |H|. Since L has characteristic 0, this is non-zero, so v is non-zero. Finally, for any h ∈ H, h(v) is given by the same sum, with the summandspermuted. Hence h(v) = v, so v ∈ En. □

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 theidentity. 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 permutationcorresponding to h under the Cayley homomorphism). In other words, there is apermutation 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 a1e(x1)+· · ·+ ame(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(a1e(x1) + · · ·+ ame(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 aias above where all the ai are in E. But the first coordinate of e(x) is x, so we geta1x1 + · · · + amxm = 0 for a tuple ai in E, contradicting the linear independenceof the xi over E. □

We are now in position to prove one inequality in (2).



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 provingthe main theorem, so we will now assume to know that [L : E] = |H|.

Corollary 289 (Dedekind). The elements of H are linearly independent overL: if ai ∈ L are such that a1h1(x) + · · · + akhk(x) = 0 for all x ∈ L, then ai = 0for all i.

Proof. Let l1, . . . , lk be a basis of L over E, and consider the matrix A whoserows are e(li). We showed that the rows are linearly independent. Since A is asquare matrix, it follows that the columns are also linearly independent (over L).Hence there are no non-zero ai with a1h1(li) + · · ·+ akhk(li) = 0 for all i. □

Exercise 290. Assume that H is a cyclic group of order n, generated by τ .Show that Tn − 1 is the minimal polynomial of τ , viewed as a linear map from Lto itself over E.

We have proven Proposition 285 using a particular action of H on V . We maynow 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 vectorsspace over L, and assume that H acts on V by additive group homomorphisms, andsatisfying h(xv) = h(x)h(v) for h ∈ H, v ∈ V and x ∈ L. Then there is a non-zerovector 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 actionthere to U satisfies the assumption.

Proof. If u ∈ V is any non-zero vector, the vector v =∑

h∈H h(u) is clearlyinvariant. Hence it is enough to show that there is a vector of this form whichis 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 thate(v) = 0, and hence v = 0. □

6.3. Proof of the Fundamental Theorem. We now proceed to prove themain 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 par-ticular” 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] (9)


60 2. GALOIS THEORY

The last equality by induction on the degree. Hence it is enough to show that

[E(a) : E] ≥ |H||H1| = (H : H1).

Since H1 is by definition the stabiliser of a, by 11.4 of Chapter 1 (H : H1) isequal to the size of the orbit Ha of H acting on L. Hence we need to show thatthat |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 theother hand, by Proposition 255, Corollary 274 and Proposition 267, the orbit Haconsists of all the roots of p. By Proposition 263, the number of such roots is alsoequal 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 elementsof E, we have E ⊆ LH . Assume there is an element a fixed by H, that is not inE. By Proposition 275 and Exercise 264, the minimal polynomial of a over E hasanother root b in L. We thus have an isomorphism E(a) −→ E(b) over E, whichextends to an automorphism τ of L over E, by Corollary 274. This contradicts theassumption that a is fixed by H.

Now, let H be a subgroup of G, and let E = LH . We need to show thatH = Aut(L/E). Again, one inclusion is clear: H ≤ Aut(L/E). However, we havejust shown that the latter group has E as its fixed field. Hence, by part (2), theyhave the same size, so they are equal. □

Proof of (4). Assume E is normal. By Proposition 273, E is the splittingfield of some polynomial p. Any automorphism of L would have to take any root ofp 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 agroup homomorphism. According to Corollary 274, any automorphism of E over Kextends to an automorphism of L. Hence the map r is surjective. The kernel is theset of automorphisms in Aut(L/K) whose restriction to E is the identity. Hence itis precisely Aut(L/E). Thus, Aut(L/E) is normal, with quotient Aut(E/K) (andthe quotient map is the restriction).

Conversely, assume that Aut(L/E) is normal. Let a ∈ E have minimal polyno-mial p over K, and let b be another root of p (since L is normal, b ∈ L). We mustshow that b ∈ E. By Proposition 255 and Corollary 274, there is an automorphism τof L over K with τ(a) = b. If σ ∈ Aut(L/E), then τ−1στ ∈ Aut(L/E) as well, sinceit is normal. Since a ∈ E, we have a = τ−1στ(a) = τ−1σ(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. Wecontinue to assume that all fields have characteristic 0.

7.1. Cyclic extensions. Recall (Definition 190) that we are interested insplitting fields of equations of the form xn − a = 0. We will now show that theyare precisely the cyclic extensions.

We have seen in Example 268 that such an extension is not, in general, obtainedby adding one root of a. However, if b and c are two roots, then ( bc )

n = bn

cn = 1, so


7. SOLVABILITY OF EQUATIONS 61

c = rb, where r is an n-th root of unity (i.e., rn = 1). Hence we have the followingresult.

Lemma 294. Assume K contains all n-th roots of unity. Then for any n andany 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 Galoisgroup. 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 homomor-phism. 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 Lbe an extension of degree n. Then L is a cyclic Galois extension of K if and onlyif 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 ofdegree n, and let τ be a generator of the Galois group. According to Exercise 290,the minimal polynomial of τ acting on L is Tn−1. SinceK contains all n-th roots of1, 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(vl) = τ(v)l = αlvl,hence vl ∈ K precisely if l is divisible by n. □

We note also that the arguments above show that, if L is the splitting field ofxn − a, then L contains all n-th roots of unity.

7.2. Solvability criterion. We may now provide a group theoretic criterionfor solvability.

Theorem 297. Let p(x) be a polynomial over a field K of characteristic 0, letL be its splitting field, and let G = Aut(L/K) be the Galois group. Then p(x) issolvable 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 iscyclic.

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 andonly if its Galois group is solvable.

Exercise 298. Let p : G −→ H be a surjective group homomorphism. Showthat if G is solvable, then so is H. Show that any subgroup of a solvable group issolvable.

For the proof, we will need the following lemma.

Lemma 299. Let p be a polynomial over K with splitting field L and Galoisgroup G = Aut(L/K), let K ′ be an extension of K, and let L′ be a splitting fieldof 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 theroots of p are in L. If an automorphism goes to e in G, it fixes all the roots, sofixes all elements of L′ (which is generated by the roots). □


62 2. GALOIS THEORY

Proof of Theorem 297. Assume that p is solvable by radicals. Then thereis a sequence of field extensions K = L0 ⊂ L1 ⊂ · · · ⊂ Ln, such that each Li isthe splitting field of xki − ai, with ai ∈ Li−1, and such that L ⊆ Ln. Since eachextension by roots of unity is an extension by radicals, we may assume that thefirst 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), aswell as a quotient map from Hn to G. By Exercise 298, it is enough to prove thatHn is solvable. But since each Li is Galois, we have by the main Theorem thatHi+1 is normal in Hi, and Hi/Hi+1 = Aut(Li+1/Li). The last group is cyclic byProposition 296 (and Example 269).

In the other direction, assume that G is solvable. Let K ′ be an extension ofK 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. Hencewe may assume that K itself already had all roots of unity. Let Li = LGi . ThenAut(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 andonly if it is Abelian.

We now give an example of a particular equation that cannot be solved. Letp(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 (theother case is similar). Then, from the second to last equation we get that b mustbe even, then that a has to be even, and we get a contradiction from the secondequation, 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 suitablepower 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, calledthe Eisenstein criterion


7. SOLVABILITY OF EQUATIONS 63

Proof. The polynomial is irreducible by the discussion above. Hence, if Lis 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 groupG = Aut(L/Q). By Cauchy’s Theorem (Theorem 173), G contains an element oforder 5. Viewing G as a subgroup of S5 (by enumerating the roots), it follows thatG 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 3real roots. An explicit calculation show that the two extreme points have differentsigns. It follows that complex conjugation determines a transposition in G. HenceG 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 byradicals.

Proof. The Galois group of p is S5. If it were solvable, then so would itssubgroup A5. But A5 is simple by Theorem 180, hence is not solvable by Exer-cise 300. □

This concludes the proof of Theorem 193. End lecture 38,May 4


Bibliography

[1] James S. Milne. Fields and Galois theory. Course lecture notes. 2008. url:http://jmilne.org/math/ (cit. on p. 41).

[2] James S. Milne. Group Theory. Course lecture notes. 2010. url: http://jmilne.org/math/ (cit. on pp. 5, 10, 39).

[3] Joseph Rotman. Galois theory. Second. Universitext. New York: Springer-Verlag, 1998, pp. xiv+157. isbn: 0-387-98541-7 (cit. on p. 41).

[4] Joseph J. Rotman. An introduction to the theory of groups. Fourth. GraduateTexts in Mathematics 148. New York: Springer-Verlag, 1995, pp. xvi+513.isbn: 0-387-94285-8 (cit. on p. 5).

65

http://jmilne.org/math/



Index

action, 28

faithful, 36

transitive, 29

alternating group, 26

automorphism, 21

automorphism group, 54

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

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

domain, 7

image, 7

injective, 7

invertible, 7

left inverse, 7

one to one, 7

onto, 7

range, 7

right inverse, 7

surjective, 7

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

homomorphism, 15

kernel of, 16

section of, 31

index (of a subgroup), 30

irreducible, 48

isomorphism, 17

67

68 INDEX

kernel, 16

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

vector space, 50


Date post:	24-Jul-2020
Category:	Documents
Upload:	others
View:	1 times
Download:	0 times

Basic Algebra Math 30710kamenskm/teaching/algebra.pdf · 2014-10-27 · 8 1. GROUP THEORY Example...

Documents