Divisivility and Sylow p-subgroups Hello I have been working in this problem for two days but i can't get the answer, I would appreciate any help or hint.
Let $p$ be prime, $m\ge 1$,     $r \ge 2$   and $(p,r) = 1$.
If there is a simple group $G$ such that $ |G| = p^ {m}r  $, then $ p^ {m}|(r-1)! $

Pick a Sylow $p$-subgroup $P$ and let $G$ act on $G/P$ by left multiplication: $p\cdot gP = pgP$. This gives a homomorphism $G\to S(G/P)=S_r$, which is clearly not identically equal to $1$. Since $G$ is simple, the kernel is thus trivial and hence $S_r$ contains an isomorphic copy of $G$ as a subgroup. So $p^m r| r!$.
