$G$ can't be a simple group if $|G| > n!$ and $|G:H| <n$.

Question

Prove that if the order of the group $G$ is bigger than $n!$ and $H < G$ is a subgroup with $|G:H| <n$, then $G$ cannot be a simple group.

We got the hint that we should represent $G$ on the right cosets of $H$ with right multiplication, but I can't really begin.

Any help appreciated.

Answer 1

Hint: find a normal subgroup of $G$ by considering the kernel of a homomorphism from $G$ to somewhere. What is the "somewhere" that might be useful, bearing in mind that $n! < |G|$?

Second hint: you already know you need to make $G$ act on the set of cosets of $H$. This action induces a homomorphism into a symmetric group.