Group Actions

Equivalently, a permutation representation of G in X is a homomorphism G --> Sym(X). By taking X to be G itself, considered as a set, we can glean useful information about G. 

1. Let X = G and x^g = xg. This clearly [CHECK!] is an action of G on G, called the right multiplication action.Since the orbit of each x in G is the whole of G, the representation is faithful, so we have Cayley's Theorem: Every group is isomorphic to a subgroup of a symmetric group. In particuler, if |G| = n, then G is isomorphic to a subgroup of Sym_n.

 2. Let X = G and x^g = g^-1xg. Once again we have an action of G on G, called conjugation. For any x in G, [x], the orbit of x, is the conjugacy class of x in G. The fixed group of X, G_x = {y in G: y^-1xy} = {y in G: yx = xy} called the centralizer of x in G, denoted C_G(x). Lemma 4.3 now tells us that if G is finite, then for all x in G, |[x]| = |[G : C_G(x)]|. Hence the size of every conjugacy class, and the size of every centralizer, divides the order of G. 

 The kernel of conjugation is

 {g in G: for all x in G, g^-1xg = x} = 
 {g in G: for all x in G, xg = gx}

 which is the centre of G, denoted ZG. Thus the centre of G is a normal subgroup of G. 

3. Let H be a subgroup of G, and let X = [G : H], the set of right cosets of H in G. Let (Hx)^g = Hxg, another right coset, so this is an action of G on X. Hence we have a permutation representation of G in [G : H]. The kernel is

  {g in G: for all cosets Hx, Hxg = Hx} = 
  {g in G: for all x in G, xgx^-1 in H},

 which is just the intersection of all the conjugates of H in G. Note that K is a normal subgroup of G contained in H, in fact it is the maximum normal subgroup of G contained in H. By the fundamental homomorphism theorem, if |X| = n, then |G/K| is less or = n!, even if G is infinite.

 4. Once again let H be a subgroup of G, and let X be the set of conjugates of H in G. Define 

(x^-1Hx)^g = (xg)^-1 Hxg. 

Check that this defines an action of G on X. The orbit of H is the set of conjugates of H, all of which are subgroups of G. The fixed group of H is 

 {g in G: g^-1Hg = H},

called the normalizer of H in G, denoted N_G(H). It is a subgroup of G containing H as a normal subgroup, in fact it is the maximum subgroup of G in which H is normal. As usual, the index of N_G(H) in G is the number of conjugates of H in G. Hence if |G| = n, then the number of conjugates of any subgroup of G divides n.

Cauchy's Theorem

 Let G be a finite group, and let p be be a prime dividing the order of G. Then G has elements of order p. 

Cauchy guessed this result by examining the orders of many known groups and presented a long inductive proof. The following is a slick proof using group actions. 

 Let X = (x₁, x₂, ... , x_p) with x_i in G such that 
x₁.x₂ ... x_p = 1. 

Notice that X is not empty, since at least it contains (1,1,...1). The cyclic group Z_p acts on X by
(x₁, x₂, ... , x_p)⁰ = (x₁, x₂, ... , x_p), 
(x₁, x₂, ... , x_p)¹ = (x₂, x₃, ... , x_p, x₁) etc 

that is by cycling the subscripts. Note that this really is an action on X, because for example 
x₂.x₃. ... , x_p. x₁ = x₁^-1.x₁.x₂... x_p.x₁ = x₁^-1.1.x₁ = 1.

 Now we partition X up into orbits. Since size of each orbit divides the order of Z_p, they must have size 1 or p. What does an orbit of size 1 look like? Well, for each j = 0,1,...p-1,
(x_1+j, x_2+j,...,x_p+j) 
(subscripts calculated mod p) = 
(x₁, x₂,...x_p),
so the sequence is constant, say (a,a,...a) for some a in G. 

Let s be the number of orbits of size 1 and r the number of orbits of size p. Then s + rp = |G|. But this is a number divisible by p, so p divides s too. We have already seen that s is at least 1 (since (1,1,...1) forms one orbit of size 1). So s is at least p. Hence G has at least p elements a such that a^p = 1.

 The main idea behind this proof can be formalised as follows:

Lemma 10.1 Let G be a group acting on a finite set X. Denote by X_G the subset of X fixed by G, and let S be the set of orbits of length > 1.

 Then |X| = |X_G| + S _{x in S} |orb(x) |.

 Proof X_G is just the union of all the orbits of length 1, so the equation just expresses the fact that the orbits form a partition of X.

 Corollary 10.2 If G is a p-group acting on a finite set X, then p divides |X| - |X_G|.

 Proof The difference is just the total length of the orbits of length > 1. But the length of any such orbit divides |G| so is a positive p-power. 

To continue

To return to Page 1 .

 Last update Nov 24 1999

Author: Phill Schultz, schultz@maths.uwa.edu.au