Show that if $G$ is a simple group with a subgroup $H$ of index $n>1$, then $|G| \leq n!$.
Hence show that a group of order $2^k \times 3$ can never be simple for $k>1$.
So I have let $X$ be the set of all left cosets of $H$ in $G$, which has order $n$. I know I should look for a homomorphism from $G$ to $S_n$ but this at this point I am stuck.
For the second part Sylow III will tell me that there must be 1 or 3 Sylow 2-groups, so I'm guessing that I need to rule out there being 3?