Generalized Cayley's Theorem

Cayley's theorem states that every group is isomorphic to a subgroup S(X) of the symmetric group. To prove the theorem we have taken X as G itself and noted that S(G) is much bigger than G. We try now to find an X which is smaller than G, so that S(X) gets smaller too.

7.18.1 Generalized Cayley's Theorem. Let G a group and H a subgroup of G. Let X = {xH}_{x ∈ G}, then there exists a homomorphism Ψ: G ⟶ S(X) whose kernel is the largest normal subgroup of G contained in H.

Proof. Define a permutation T_g on the left cosets of H by

a bijection for X onto itself and thus an element of S(X). We definite this other mapping

For any xH ∈ X, it results

that is T_gg' = T_g ∘ T_g' thus Ψ is a homomorphism between G and S(X). The kernel is

Let K = Ker Ψ, we shall show that K is the largest normal subgroup of G contained in H.

1. K ⊲̲ G, being the kernel a homomorpshim.

2. K ⊆ H: since

   k ∈ K   ⇒   kxH = xH  ∀ x ∈ G

   In particular for x = e we obtain

   kH = H   ⇒   k ∈ H

3. If N ⊲̲ G, N ⊆ H, then N ⊆ K. Indeed, being N ⊲̲ G and N ⊆ H, we have ∀ n ∈ N and ∀ x ∈ G (and thus ∀ x⁻¹ ∈ G) the following chain of implications

   xnx⁻¹ ∈ N ⊆ H   ⇒   xnx⁻¹H = H
   ⇒   nx⁻¹ = x⁻¹H
   ⇒   N ⊆ K.  □

We the particular choice of X made in the theorem we managed to reduce the order of S(X), but losing (in general) the injectivity of the mapping Ψ. Nevertheless the characterization of Ker Ψ as the largest normal subgroup of G contained in H gives us the following information on the G group.

7.18.2 Consequences of the theorem.

1. Let G a simple group, that is not possessing non-trivial normal group. Then Ψ is injective.

2. Let G a finite group and H a subgroup of G. The index (G : H) is such that |S(X)| = (G : H)! since X = {xH} with x in G, is the set of left cosets of H in G, whose size is the index. If G and H were such that |G| ∤ (G : H), then certainly Ψ would not be injective: if it were, then

   |G| = |Im Ψ|,   divisor of (G : H)!

   Hence there exists in G a normal subgroup. □

7.18.3 Example. Let G a group of order 35 and let H a subgroup with order 7 (necessarily existing from Cauchy's theorem). Then it results

There exists in G a normal non-trivial subgroup contained in H. Being the order of H a prime number, this normal subgroup must coincide with H, that is H is normal. □

Exercises

1. Let G a group of order 42. Prove that G is not simple.

2. Prove that a group with order 77 possesses a non-trivial normal subgroup.

3. By applying the simplicity test (corollary 7.17.10) prove that for n ≤ 100, and n non prime for which it's not excluded (using the same criterion) that there exists a simple group of order n are the following: 12, 24, 30, 48, 56, 60, 72, 80, 90, 96. Then using Cayley's generalized theorem verify that the only possible simple groups are those with order 56 and 100.

Solutions

1. Let H a group of order 5 (which exists by Cauchy's theorem). It results |G| ∤ (G : H)!. Thus by the generalized Cayley theorem, X the set of left cosets of H in G, the mapping ψ ⟶ S(X) given by ψ(g) = T_g, with T_g(xH) = gxH is a homomorphism and not an isomorphism , hence it possesses a non-trivial kernel, which is the largest normal subgroup of G contained in H. It follows that H itself is normal in G. ■

2. As in the exercise above considering a group of order 11 (which exists by Cauchy's theorem).

3. Applying Corollary 7.17.10 that is n = p^a and (m,p) = 1 and 1 is the only divisor of n, then the group has a normal subgroup and hence it's not simple. For example for a group of order 12 = 2² ⋅ 3, (2,3) = 1 and by the 3rd Sylow theorem the number of Sylow 2-subgroubs is such that it divides 3 and is congruent to 1 modulo 2. Since 3 ≡ 1 modulo 2, n₂ = 3 so, 2e cannot exclude it is simple.

   Let G be a group with order 56 then there exists a subgroup of order 7 by Cauchy's theorem. Applying the same reasoning of the previous exercises, it results

   (G : H) = 8,   8! = 40320, |G| = 56 | 40320 = 720

   Thus G is not simple. The same reasoning for the group with order 60. ■