# Finite simple groups.

Hi: Problem: Let U be a subgroup of G and 1 < |G:U| <5. Then |G| < 4 or G is not simple.

If |G:U| = 2 I can use a problem (problem A) that says: Let p be the smallest prime divisor of |G|. Then every subgroup of index p is normal in G. That is because in this case, 2 is a divisor of |G| and it is the smallest prime. For case |G:U| = 3 if |G| is not even then, by problem A I am done. If |G| is even, however, problem A is of no use. The same for |G:U| = 4. Any hint?

### Re: Finite simple groups.

Thank you very much, csguy. I realized I could use that theorem when I had already posted. Thanks again.

### Re: Finite simple groups.

Use the generalization of Cayley's representation theorem: If H is a subgroup of finite index n then there is a homomorphism
\rho: G \rightarrow S_n such that ker(\rho) \subseteq H. (Rotman, An Introduction to the Theory of Groups, 4th ed, Theorem 3.14, page 53 and its Corollary 3.15)

The case |G:U| = 3 follows immediately. For |G:U| = 4 you will narrow it down G to groups of order 8, 12 and there's only a few.