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???)


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