# 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

• 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. 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. 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.

• 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. 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