TITLE: Given a finite solvable group $G$, prove that a minimal normal subgroup $H$ is a $p$-group
QUESTION [5 upvotes]: Given a finite solvable group $G$, and a minimal normal subgroup $H$, prove that $H$ is a $p$-subgroup for some prime $p$.
My Attempt:
I am trying to write this proof without using the term "characteristic subgroup". I'm aware to the fact that by proving that $p$-subgroup of $M$ is a characteristic subgroup will finish the proof.
Let $p$ be a prime number such that $p$ is a divisor of the order of $M$. By Sylow theorem, there exists a $p$-subgroup of $M$, let it be called $S$.
I'd like to prove that every element of $M$ is of order $p^n$, for some $n \in \mathbb{N}$.
let $s\in S, \  g\in G: g^{-1}sg \in M$, as $S\le M$ and $M$ is normal in $G$. but why and how I can prove that $=g^{-1}sg$ is of order $p^m$, for some $m\in \mathbb{N}: m\le n$.

REPLY [2 votes]: Since $G$ is solvable, $H$ is solvable, so its derived series must eventually end at $1$ and since $H$ is non-trivial we have $H' \lt H$. But $H' \text { char } H \unlhd G$, so $H' \unlhd G$. But $H$ is minimal normal, whence $H'=1$, that is $H$ is abelian. Since $H$ is finite we can find an element in $H$ of order $p$, a prime (Cauchy!). Since $H$ is abelian the set $S=\{h \in H : h^p=1\}$ is in fact a (non-trivial) characteristic subgroup of $H$ and hence it is normal in $G$. So we must have $H=S$ and thus $H$ is an elementary abelian $p$-group.