Using Sylow Subgroups to show that a group is not simple

Question

Assume we have $G$ finite and $\vert G \vert = p^e m$, $p$ prime, $e, m \in \mathbb{N}$, $m > 1$ and $p \nmid m$.

Then it holds:

If $p^e \nmid (m - 1)! \Rightarrow G$ is not simple.

As the context are Sylow (sub-) groups there may be a way to show it using them. Any help would be appreciated.

Answer 1

Since $|G|=p^em$ with $(p,m)=1$ you have a Sylow p-subgroup of order $p^e$ and index $m$. So you have a group homomorphism $f:G\to S_m$. If $G$ is simple then $kerf\lhd G$ has to be $1$, so you can see $G$ as a subgroup of $S_m$, hence $p^em|m! \Rightarrow p^e|(m-1)!$