# Simple group with order $\geq n!$ cannot have subgroup of index $n$.

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?

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.
• Oh, I am sorry! :-) As a compensation I propose (if you want) to prove that $G$ does not contain a subgroup of index $n+1$. – Boris Novikov Oct 20 '13 at 16:32
• Uh please don't feel sorry about that. By the way I'm a bit confused about the $n+1$ problem you proposed. What about alternating group $A_{n+1}\ (n \geq 4)$? It is simple, and has an index $n+1$ subgroup $A_n$ (fix $n+1$). Its order is $(n+1)!/2 > n!$. Anything wrong here? – 4ae1e1 Oct 20 '13 at 18:06
• Yes, you are right and I mistaked. What one can prove instead of my question is: if $G$ contains a subgroup of index $n+1$ then $G$ is embedded into $A_{n+1}$. Sorry onсe more. – Boris Novikov Oct 20 '13 at 19:49