If $G$ is non-abelian and simple then $|G|$ divides $n_p!/2$ 
If $G$ is non-abelian and simple then $|G|$ divides $n_p!/2$ where $n_p$ is the number of Sylow p-subgroups.

I am struggling to prove this. I am considering the conjugation action so I know $G$ must be a subgroup of $\mathrm {Sym}(\mathrm{Syl}_p(G))$ so $|G|$ divides $n_p!$ but I cannot see why it must be a subgroup of $A_{n_p}$. 
 A: Some would consider the trivial group as (unique) Sylow $p$-subgroup when $p$ does not divide $|G|$. However, in the current context, this won't work out (in fact, that would make $n_p!/2$ a non-integer). Thus we state the obvious: $p\mid |G|$.
The action of $G$ on the set of Sylow $p$-subgroups gives us a homomoprhism $\phi\colon G\to S_{n_p}$. Compose with the projection to get a homomorphism $\psi\colon G\to S_{n_p}/A_{n_p}\cong C_2$. As $G$ is simple, $\phi$ and $\psi$ must either be trivial or injective.
As $G$ is not abelian, we have $|G|>5$ and thus conclude that $\psi$ is not injective. 
If $\phi$ is trivial then $n_p=1$ by transitivity of the action. Then the unique Sylow $p$-subgroup $P$ is normal in $G$. It follows that $G=P$ is a $p$-group (the alternative that $P=1$ is excluded).Then $G$ has non-trivial centre, and as $G$ is not abelian, this contradicts simplicity of $G$.
We conclude that $\phi$ is injective and $\psi$ is trivial. That makes $G$ (isomorphic to) a subgroup of $A_{n_p}$, hence $|G|$ divides $|A_{n_p}|=n!/2$.
