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

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Well I just remembered that we don't need character theory here. Still interested to see other approaches though. – user641 Aug 4 '10 at 17:01
Is this site functioning correctly? Where did Jack Schmidt's answer go? – user641 Aug 4 '10 at 22:06
Jack decided to delete it. – Zev Chonoles Jul 10 '11 at 8:42
John Thompson sharpened this to proving that if a finite group $G$ has a maximal subgroup which is nilpotent of odd order, then $G$ is solvable. There are finite simple groups whose Sylow $2$-subgroups are maximal. – Geoff Robinson Jul 10 '11 at 9:15

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.

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in the first line of your proof,you take a normal p complement N for M in G?but to do that we must know M is a sylow p-subgp of G.How do we know that? – user13159 Jul 10 '11 at 8:37
@7115763: You're a new user, so the confusion is understandable, but for future reference, when you want to make a comment about a post, you should use the comment feature - click the "add comment" button on the lower left of the comment area (it is in light grey text). The way you had originally posted was as an answer. – Zev Chonoles Jul 10 '11 at 8:41
@7115763: You have to apply Burnside's normal $p$-complement theorem to each $p$-Sylow of $M$ (which are also $p$-Sylows of $G$ as $M$ is a Hall subgroup). – j.p. Jul 15 '11 at 13:36