How to prove this?

Question

Let G be a finite group and H a proper subgroup such that |G| doesn't divide |G/H|!. Prove that H contains a nontrivial normal subgroup of G.

Answer 1

Let |G/H|=n, so |G/H|=n!. Let |G|=nm

Answer 2

Here's a more specific hint: Let G act on G/H = the set of left cosets of H: if x is in G and gH is a left coset, let x*gH = (xg)H. This gives a homomorphism G -> Aut (G/H). If g is an element of G which acts trivially, then show that g belongs to H. Thus, the set of elements which act trivally (the kernel) is a subgroup of H. So, if K = kernel and I = image,

1 -> K -> G -> I -> is an exact sequence. (Or if you are unfamiliar with this term, I'm just saying that by the first isomorphism theorem G is isomorphic to I/K.)

Now, you are asked to show that K is a nontrivial subgroup of H. Suppose not. Then K = 1 and G is isomorphic to I. But I is a subgroup of Aut(G/H) which is isomorphic to the symmetric group on [G:H] letters. (Think about what this says about the order of G and whether or not it has the stated divisibility property.)

Answer 3

What is a "nontrivial normal subgroup? Answer that and then I can deal with your question? I have taught math and do not recognize that expression so the definition of that expression may be your problem.