Let $G$ be a finite group and $p$ a prime s.t. $p\big||G|$, and let $P$ be a normal p-subgroup of $G$, with $|P|=p^m$. I want to prove the following:
If $M$ is a maximal subgroup of $G$, then $P\leq M$ or $[G:M]=p^k$, with $k\leq m$.
What I got so far:
If $M$ happens to be a p-subgroup then it clearly follows that $P\leq M$, so lets suppose $M$ is not a p-subgroup. Now, $P$ is normal in $MP$ and $P\cap M$ normal in $P$ we can apply the second isomorphism theorem and get that $[MP:P]=[P:P\cap M]$. And got stucked here, can't see how relate this last result with $|P|$ and $[G:M]$ to get that $[G:M]=p^k$, with $k\leq m$. Any help would be appreciated.