Let $G$ be a simple group and let $n_p$ be the number of Sylow $p$-subgroups, $p$ prime. Show that $|G|$ divides $(n_p)!$ (factorial).
If i start off by assuming G is abelian then G is isomorphic to Z/pZ. So |G| = p, by Sylow's third theorem n_p is of the form n_p = 1+pk, k some integer. n_p must divide order of G witch implies k=0 and n_p = 1. But this would mean that the Sylow-p-subgroup is normal in G, contradicting the simplicity of G. Worse, |G|=p does not divide 1.
So i guess G is not abelian, not really sure how to continue from here.
The number of Sylow $p$-subgroups equals the index $n=n_p$ of the normalizer $H$ of a Sylow $p$-subgroup. Now $G$ acts by left multiplication on the left cosets of $H$. If $K$ is the kernel of this action, then $G/K$ embeds homomorphically in $S_n$. But $G$ is simple so $K=1$, and hence $|G|$ divides $n!$.