GTnotes2.pdf

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2 Group Actions

Definition. Let G be a group, and a set. A (left) action ofG on is a map : G, such that1. (e,x) = x for all x , and

2. (g,(h,x)) = (gh,x) for all g,h G and x .

We shall generally write gx for (g,x), except where this leads to ambiguities, or where other notation

is more convenient. By the second axiom, we may unambiguously write ghx without bracketing.

[A right action ofG on is defined similarly, but with the group elements written on the right instead of

the left; this is not purely a notational difference, as the second axiom in this case becomes ((x,g),h) =

(x,gh). It can easily be checked that ifx gx is a left action, then x xg1 defines a right action, and

vice versa.]

Examples.

1. The dihedral group D2n acts on a regular n-gon. (Take to be the set of vertices, for instance.)

2. The symmetric group Sn acts on the set {1, . . . ,n}.

3. The general linear group GLn(R) acts on Rn, considered as column vectors. (It also has a right action

on row vectors.)

Definition. Let G be a group acting on a set . Let x .

1. The orbit ofx, written OrbG(x), is the set {gx | g G}.

2. The stabilizer ofx, written StabG(x), is the set {g G | gx = x}.

Note that OrbG(x) is a subset of, while StabG(x) is a subset ofG.

Proposition 9. LetG be a group acting on a set, and letx . Then StabG(x) G.

Proof. Suppose that g,h StabG(x). Then ghx = g(hx) = gx = x, and so gh StabG(x). Also g1x =

g1(gx) = ex = x, and so g1 StabG(x). So StabG(x) is a subgroup ofG.

Theorem 10. Orbit-Stabilizer Theorem

LetG be a finite group acting on a set, and letx . Then

|G| = |OrbG(x)||StabG(x)|.

Proof. Let S= StabG(x). Let g,h G. We observe that

gx = hx h1gx = x h1g S gS= hS.

It follows that the number of distinct elements in OrbG(x) is equal to the number of cosets gSfor g G. But

this is |G|/|S|, and the theorem follows.

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Proposition 11. LetG be a group acting on a set. Then the relation R on defined byR(x,y) if and only

ify OrbG(x) is an equivalence relation, whose equivalence classes are the orbits of G on . (Briefly: the

orbits ofG partition.)

Proof. We need to show that R is reflexive, symmetric, and transitive.

Reflexive. We have ex = x for all x , and so R(x,x).

Symmetric. Suppose R(x,y). Then y = gx for some g G. Now g1y = g1gx = x, and so R(y,x).

Transitive. Suppose R(x,y) and R(y,z). Then there are g,h G such that y = gx and z = hy. But now

(hg)x = h(gx) = h = z, and so R(x,z).

Recall that for a set , the group Sym() consists of all permutations on (i.e. invertible functions

) under composition.

Proposition 12. LetG be a group acting on a set. Then for allg G the mapg : x gx is a permutation

of. Moreover, the map : g g is a homomorphismG Sym().

Certainly g is a function , and it is a permutation since it has the inverse g1 . We note that

gh(x) = (gh)x = g(hx) = g h(x), and so is a homomorphism as required.

Definition. Let G be a group acting on a set .

1. We say that the action ofG is transitive if OrbG(x) = for any x .

2. The kernel of the action is the kernel of the homomorphism from Proposition 12; that is to say, the

set {g G | gx = x for all x }.

3. We say that the action is faithful if its kernel is {e}.

A. Four important group actions

In this section, let G be any group. We introduce four important actions of G. Every one of these actions

plays an important part in the general theory of groups.

Action 1: action ofG on itself by (left) translation.

This is an action ofG on itself, i.e. we have = G; it is sometimes called the left regular action. It is given

by (g,x) gx, where gx here denotes multiplication in G. So for this action, our usual notation for a group

action is in happy agreement with our standard notation for group multiplication. It is easy to show that this

is indeed an action, since ex =x for all x G, and since (gh)x = g(hx) by the associativity axiom for groups.

There is a similar right action given by (x,g) = xg; this is the right regular action.

Let x,y G. Ifg = yx1, then y = gx. Hence OrbG(x) = G, and so this action is transitive. The action is

faithful, since clearly its kernel is {e}; in fact StabG(x) = {e} for every x G. This gives an easy proof of

the following theorem.

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Theorem 13. Cayleys Theorem

LetG be a finite group. Then G is isomorphic to a subgroup of Sn for somen.

Proof. By Proposition 12 there is a homomorphism : G Sym(G) whose kernel is {e}, the kernel of the

action defined above. Now G = G/{e} = Im by the First Isomorphism Theorem. But Im is a subgroup

of Sym(G), and Sym(G) = Sn, where n = |G|.

Example. Cayleys Theorem predicts that the action ofS3 on itself by translation should give an injective

homomorphism ofS3 into S6 (since |S3| = 6). To avoid a notational clash, we let S3 act on {a,b,c} instead

of the usual {1,2,3}. We shall assign each element ofS3 a number, 1, . . . ,6, as follows:

e = 1; (a b) = 2; (a c) = 3; (b c) = 4; (a b c) = 5; (a c b) = 6.

The map x ex clearly leaves each x fixed, and so the permutation effected on {1, . . . ,6} is the identity.

Next consider the map x (a b)x; we calculate

(a b)1 = 2; (a b)2 = 1; (a b)3 = 6; (a b)4 = 5; (a b)5 = 4; (a b)6 = 3.

So translation by (a b) effects the permutation (1 2)(3 6)(4 5) on the (numbered) elements of S3. The re-

maining elements ofS3 can be treated similarly; we obtain the homomorphism

eS3 eS6, (a b) (1 2)(3 6)(4 5), (a c) (1 3)(2 5)(4 6),

(b c) (1 4)(2 5)(3 6), (a b c) (1 5 6)(2 3 4), (a c b) (1 6 5)(2 4 3).

(It should be pointed out that this is far being from the only injective homomorphism ofS3 into S6.)

Action 2: action ofG on the (left) cosets of a subgroup.

Let H be a subgroup ofG. For this action we take= {xH |x G}; the action is given by (g,xH) gxH.

It is easy to show that this is indeed an action. (There is a similar right action ofG on the rightcosets, given

by (Hx,g) = Hxg.)

Let xH,yH. Then putting g = yx1, we have gxH = yH. Hence yH OrbG(xH) for all x,y, and so

the action is transitive.

For any xH, we have

gxH = xH gx xH g xHx1.

So we have StabG(xH) = xHx1.

Action 3: action ofG on itself by conjugation.

This is another action for which = G, but it is very different from the left regular action. It is given by

(g,x) gxg1. This is an action for which we cannot use our standard notation gx for the action of g on

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x, since this would conflict with group multiplication. Instead we write gx for gxg1, which we call the

conjugate ofx by g.

This defines an action, since ex = exe1 = x, and

ghx = (gh)x(gh)1 = ghxh1g1 = g(hxh1 = g(hx).

There is a similar right action given by (x,g) g1xg. We may write xg for g1xg; note that xg = g1

x.

The orbits of the conjugacy action are known as conjugacy classes. We write Gx, or else ConG(x), for

the conjugacy class of x. The stabilizer of x is the subgroup {g G | gx = xg}. This is known as the

centralizer of x in G, and written CentG(x). By the Orbit-Stabilizer Theorem, if G is finite, then |G| =

|ConG(x)||CentG(x)|.

This action is never transitive if |G| > 1, since the identity e lies in a conjugacy class on its own. Thekernel of the action is the subgroup {g G | gx = xg for all x G}. This is known as the centre of G, and

written Z(G).

Action 4: action ofG on its subgroups by conjugation.

Let be the set of all subgroups of G. Then G acts on by (g,H) gHg1. We have seen that gHg1 is

the stabilizer ofgH under Action 2; so it is a subgroup of G, called. We write gH for gHg1, and call it the

conjugate ofH by g. It is easy to check that the map is an action ofG.

A subgroup H of G is normal if and only if OrbG(H) = {H}. The stabilizer of a subgroup H is known

as the normalizer of H in G, and written NG(H). It is the largest subgroup of G of which H is a normalsubgroup.

3 Sylows Theorems

Let G be a group with finite order n. Lagranges Theorem tells us that the order of any subgroup of G

is a divisor of n. It is not in general the case that G has a subgroup of order d for every divisor d of n. For

instance, the group A4, which has order 12, has no subgroup of order 6. In this section, however, we show

that in the special case that d is a power of a prime number, G does have a subgroup of order d.

Definition. Let p be a prime number.

1. A finite group whose order is a power ofp is said to be a p-group.

2. If G is a group, and H a subgroup ofG which is a p-group, then H is a p-subgroup ofG.

3. An element ofG whose order is a power of p is a p-elementofG.

We deal first with subgroups of prime order.

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Theorem 14. Cauchys Theorem

Let p be a prime, andG a finite group whose order is divisible by p. Then G has an element of order p.

Note that ifg has order p, then g is a subgroup of order p. So an equivalent statement of this theorem

is that a group whose order is divisible by p has a subgroup of order p.

Proof. Let be the set of p-tuples of elements ofG whose product is the identity; i.e.

= {(g1, . . . ,gp) Gp | g1 . . .gp = e}.

Notice that the tuple (g1, . . . ,gp) belongs to if and only ifgp = (g1 . . .gp1)1. This observation has two

important consequences. Firstly, it allows us to calculate the size of as |G|p1, since for any choice of

g1, . . . ,gp1 there is a unique gp such that (g1, . . . ,gp)

. Secondly, it shows that

(g1, . . . ,gp) (gp,g1, . . . ,gp1) ,

since inverses in groups are two-sided. Let be a generator of the cyclic group Cp. Then there is an action

ofCp on given by (g1, . . . ,gp) = (gp,g1, . . . ,gp1). By the Orbit-Stabilizer Theorem, every orbit of this

action has size divisible by p. If there are M orbits of size 1, and N orbits of size p, then we clearly have

|| = M+ pN. But || = |G|p1 is divisible by p (since |G| is), and it follows that p divides M. Clearly the

orbit of(e, . . . ,e) has size 1, and so M> 0. Therefore there exists at least one other orbit of size 1. But the

unique element of this orbit is fixed by , and so it has the form (g, . . . ,g) for some g G such that gp = e.

Now g is an element of order p in G, as required.

The principal theorems of this chapter, named collectively after the Norwegian mathematician Ludwig

Sylow, deal with the case of subgroups whose order is the largest power of a prime p which divides |G|.

These are some of the most important theorems in group theory. As with the Isomorphism Theorems in

Section 1, there is no standard way of numbering these results; some authors choose to collect them together

into a single theorem.

Theorem 15. First Sylow Theorem

Let G be a finite group of order n, let p be prime, and let pa be the largest power of p dividing n. Then G

has a subgroup of order pa.

Definition. Let G be a finite group. If p is prime, and pa the largest power of p dividing |G|, then a subgroup

of G whose order is pa is known as a Sylow p-subgroup of G. We write Sylp(G) for the set of Sylow p-

subgroups ofG.

Theorem 16. Second Sylow Theorem

LetG be a finite group, and p a prime. Then |Sylp(G)| 1 mod p.

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Remark. It is clear that Theorem 16 implies Theorem 15, since if |Sylp(G)| is congruent to 1 mod p then

it cannot be 0. However the logical structure of our argument will be to establish Theorem 15 first, and to

derive Theorem 16 and the other Sylow Theorems as consequences.

Theorem 17. Third Sylow Theorem

LetG be a finite group, and p a prime. Every p-subgroup ofG is contained in at least one Sylow p-subgroup.

Theorem 18. Fourth Sylow Theorem

Let G be a finite group, and p a prime. The Sylow p-subgroups of G form a single conjugacy class of

subgroups, i.e. ifP,Q Sylp(G) then there exists g G such thatQ =gP.

Before embarking on the proof of the First Sylow Theorem, we recall that Z(G), the centre of G, is

defined to be {g G | gh = hg for all h g}. It is clear that an element g of G lies in a conjugacy class

of size 1 if and only if g Z(G), since for any h G we have hg = g gh = hg. For this reason, the

conjugacy classes of size 1 are known as the central conjugacy classes.

If G is a finite group, then it has finitely many conjugacy classes, and so finitely many non-central

classes. Suppose that C1, . . .Ck are the non-central conjugacy classes ofG; so |Ci|> 1 for all i. Then since G

is partitioned by its conjugacy classes, and since the central conjugacy classes are those which lie in Z(G),

we have the Class Equation for G:

|G| = |Z(G)| +k

i=1

|Ci|.

We are now in a position to begin the proof.

Proof of First Sylow Theorem. We proceed by induction on n, the order of G. Our inductive hypothesis is

that every group whose order is less than n possesses a Sylow p-subgroup. We show that a group of order n

has a Sylow p-subgroup.

If p does not divide n, then {e} is a Sylow p-subgroup (of order p0) for G. So we may suppose that the

highest power of p dividing n is pa where a > 0. We consider two cases:

Case i. p divides |Z(G)|. Then by Cauchys Theorem, Z(G) has a subgroup K of order p. Now observe that

if k K then gk= kg for all g G (since k is central), and so gK = Kg for all g. So K G. The quotient

G/K has order n/p, which is less than n, and so by the inductive hypothesis, G/K has a Sylow p-subgroup

Q of order pa1. By Propositions 2 and 3 we see that there is a subgroup of P of G which contains K such

that P/K= Q. Now |P| = |K||Q| = pa, and so P is a Sylow p-subgroup ofG.

Case ii. p does not divide |Z(G)|. Then consider the class equation: by assumption p divides |G|, but it

does not divide |Z(G)|. Therefore p does not dividei |Ci|, and so there is at least one non-central conjugacy

class Ci whose size is not divisible by p. Let x Ci. We have |Ci||CentG(x)| = |G| by the Orbit-Stabilizer

Theorem, and it follows that pa divides |CentG(x)|. But since |Ci|> 1, we see that |CentG(x)|< n, and so by

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the inductive hypothesis, CentG(x) has a Sylow p-subgroup P. But |P|= pa, and so P is a Sylow p-subgroup

ofG.

For the proofs of the other Sylow Theorems, the following proposition will be needed.

Proposition 19. LetG be a finite group, and let p be prime. LetP be a Sylow p-subgroup ofG, andQ any

p-subgroup. Then eitherQ P, or else there exists q Q such thatqPq1 = P.

An equivalent statement is: ifP Sylp(G), and ifQ is a p-subgroup ofNG(P), then Q P.

Proof. Suppose qPq1 = P for all q Q. Then qP = Pq for all q, and so QP = PQ. Hence PQ is a

subgroup of G of order |P||Q|/|PQ|, by Proposition 5. Clearly |P||Q| is a power of p; but since P is a

Sylow p-subgroup ofG, the order ofPQ cannot be greater than |P|. Hence |PQ| = |Q|, and so Q P as

required.

Proof of Second, Third and Fourth Sylow Theorems. Let G be a finite group, and p a prime. By the First

Sylow Theorem, G has a Sylow p-subgroup P. Let be the set of subgroups ofG which are conjugate to P

in G. Then every element of is a Sylow p-subgroup ofG.

Consider the action ofP on by conjugation. Clearly OrbP(P) has size 1. IfP is another element of

then P is not a subgroup of P, and so by Proposition 19 it does not normalize P. But since |OrbP(P)|

divides |P|, it is a power of p greater than 1, and so it is divisible by p. Therefore is the union ofP-orbits

whose sizes are divisible by p, together with one part of size 1, and it follows that || 1 mod p.

Let Q be a p-subgroup ofG, and consider the action ofQ on . Since the size of any Q-orbit is a powerof p, and since the union of the orbits is , whose size is 1 mod p, we see that there must be an orbit of size

1. So there exists some P such that Q NG(P). Now by Proposition 19 it follows that Q P. We

have shown that every p-subgroup lies in one of the Sylow p-subgroups in the set, and this establishes the

Third Sylow Theorem. Furthermore, if the subgroup Q is a Sylow p-subgroup, then clearly we must have

Q = P , and so = Sylp(G). It follows that Sylp(G) is a single conjugacy class of subgroups of G,

whose size is 1 mod p, and this establishes the Second and Fourth Sylow Theorems.

We note that the Fourth Sylow Theorem has the following corollary.

Corollary 20. LetG be a finite group and p a prime. Then |Sylp(G)| divides |G|.

Proof. The Sylow p-subgroups form a single orbit under the conjugacy action of G; the size of an orbit

divides |G| by the Orbit-Stabilizer Theorem.

It remains to deal with p-subgroups which are not Sylow subgroups. We shall need the following fact.

Proposition 21. Let p be a prime, and letG be a non-trivial p-group. Then the centerZ(G) is non-trivial.

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Proof. The size of any conjugacy class of G divides |G|, and is therefore a power of p. Consider the class

equation for G (given before the proof of the First Sylow Theorem). Since p divides |G|, and also |Ci| for

each i, we see that p divides |Z(G)|. This implies that |Z(G)|> 1.

Proposition 22. LetG be a finite group, and p a prime, and let pb be a power of p dividing |G|. Then G has

a subgroup of order pb.

Proof. Since any group has a Sylow p-subgroup, it will be sufficient to prove the theorem in the case that G

is a p-group. So let us suppose that |G| = pa. We proceed by induction on a; our inductive property P(a) is

that any group of order pa has a subgroup of order pb for all b < a. We observe that P(1) is certainly true;

this is the base case for the induction. Suppose as our inductive hypothesis that P(a) holds, and let G be a

group of order pa+1. Let b < a + 1. Ifb = 0 then {e} is a subgroup ofG of order pb, so we may assume that

b > 0.

By Proposition 21, the centre Z(G) is a non-trivial p-group. We now argue as in Case i. of the proof

of the First Sylow Theorem. By Cauchys Theorem, Z(G) has a subgroup K of order p, and K G. The

quotient G/K has order pa, and so by the inductive hypothesis, G/K has a subgroup Q of order pb1. Now

by Propositions 2 and 3, there is a subgroup of P ofG containing K, such that P/K= Q, and clearly P has

order pb as required.

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