# Proving that group is not simple using Sylow's theorems

I'm having a problem with this following question:

Let $$G$$ be finite group of order $$p^rm$$ when $$p$$ is a prime number, $$r\ge1$$ and $$\gcd(m,p)=1$$. Prove that if $$p^r\nmid(m-1)!$$ then $$G$$ is not simple.

What I've tried so far: using Sylow's third theorem and come to conclusion that $$G$$ has a unique sylow-$$p$$ subgroup, and therefore $$G$$ is not simple.

Any hint would be appreciated.

Hints. Let $$G$$ act on the set $$E$$ of $$p$$-Sylow subgrous by conjugation, and consider the corresponding morphism $$f:G\to \mathfrak{S}(E)$$.
• If the kernel is the whole $$G$$ , you are done (why???)
• If $$f$$ is injective $$\vert G\vert\mid N_p!\mid m!,$$ where $$N_p$$ is the number of $$p$$-Sylow subgroups. Deduce a contradiction.
• If the kernel is neither trivial or $$G$$, you are also done (why???)