[Math] Simple group with order $\geq n!$ cannot have subgroup of index $n$.

Question

My problem is as seen in the title:

For positive integer $n>1$, prove that a simple group with order $\geq n!$ cannot have subgroup of index $n$.

Could anyone give me some hints on how to approach this?

Answer 1

If $[G:H]=n$ then $G$ acts on $n$ cosets of $H$. Hence there is a homomorphism $G$ into the symmetric group $S_n$. Since $|S_n|=n!$ then either this homomorphism has the non-trivial kernel or $G=S_n$. But in the last case $G$ also is not simple.