G a finite group, M a maximal subgroup; M abelian implies G solvable?

Here is a classic theorem of Herstein: $G$ is a finite group, $M$ a maximal subgroup, which is abelian. Then $G$ is solvable.

The proof is pretty easy, but it uses character theory (specifically, Frobenius' theorem on Frobenius groups). Is there a character-theory-free proof?

To get things going, note that we can reduce to the case where:

i) $M$ is core-free and a Hall subgroup of $G$;

ii) $Z(G)=1$.

Steve

You can apply Burnside's normal $p$-complement theorem to get a normal complement $N$ of $M$.

Then take an element $m$ of $M$ with prime order.

Case 1: The centralizer $C_N(m)$ of $m$ in $N$ is nontrivial.

As $M$ centralizes $m$, it acts on the fixed points of $m$ (in the action by conjugation on $N$), i.e., $C_N(m)$ is an $M$-invariant subgroup of $N$. If $C_N(m) = N$, then $m \in Z(G)$ contradicts ii). Otherwise $M < C_N(m)M < G$ contradicting the maximality of $M$.

Case 2: $N$ has the fixed point free automorphism $m$ of prime order.

By Thompson's thesis $N$ is nilpotent, hence $G$ solvable.