A solvable group is characteristically simple if and only if is abelian elementary

Question

Let $G$ a finite group. I know that

$G$ is solvable $\iff$ every principal factor is abelian elementary

I already showed that if $G$ is a abelian elementary group then $G$ is characteristically simple (and $G$ has structure of $\mathbb{F}_p$ vector-space).

Suppose then $G$ is solvable and characteristically simple. I tried to use the result quoted above for to show that $G$ is abelian elementary, but I only got it in the case where $G$ was simple. How to do in the characteristically simple case?

Any help is so appreciate.

Answer 1

Here is an outline argument.

1. $G$ finite and solvable implies $G$ has a nontrivial normal $p$-subgroup for some prime $p$, and the largest such subgroup $O_p(G)$ is characteristic in $G$, so $G=O_p(G)$ is a $p$-group.

2. Finite $p$-groups have nontrivial centre $Z(G)$ which is characteristic, so $Z(G)=G$ and $G$ is abelian.

3. Now $ \{ g : g \in G, g^p=1 \}$ is a characteristic subgroup and hence equal to $G$, so $G$ is elementary abelian.