Why is a finite group, with a maximal subgroup which is abelian, soluble? I've come across an exercise saying the following:
If $G$ is a finite group which contains a maximal subgroup $M$ which is abelian, show that $G$ is solvable and that $G^{(3)}$ (the third term in the derived series) equals 1.
I can see that it's sufficient to show that $G^{(2)}$ is a subgroup of $M$, but am not sure how to show this. (Or even if it's true!) Please could you help me out?
Thanks in advance for your help :)
P.S. I don't know if it helps, but this is exercise 3.4.7 in Dixon and Mortimer's 'Permutation Groups'.
 A: In the book by Dixon and Mortimer, you can use the previous exercise 3.4.6. in which states the following:

Let $G$ be a finite primitive permutation group with abelian point stabilizers. Show that $G$ has a regular normal elementary abelian $p$-subgroup for some prime $p$.

Now for the claim, it is enough to prove the following:

Let $G$ be a finite group with a maximal subgroup $M$ which is abelian. Then $G'' \leq M$.

Proof: By induction. If $M$ contains a normal subgroup $N \neq 1$ of $G$, then $M/N$ is a maximal subgroup of $G/N$ that is abelian, so $G''N/N \leq M/N$ by induction and thus $G'' \leq M$. Suppose then that $M$ does not contain any nontrivial normal subgroups of $G$. Then via the conjugation action on $M$, we may consider $G$ as a finite primitive permutation group with abelian point stabilizers. By exercise 3.4.6. above, $G$ has a normal abelian $p$-subgroup $N$. Now $N$ is not contained in $M$, so $G = MN$. Then $G/N$ is abelian, so $G' \leq N$ and $G'' = 1$.
A: This is a theorem of Herstein, and the proof can be found here, see Theorem $5.53$. One considers the normalizer $N_G(M)$. The easy case is that $N_G(M)>M$. Then by maximality of $M$ we have $N_G(M)=G$. Hence $M\triangleleft G$ and $G/M$ is of prime order. It follows that $M$ and $G/M$ are both solvable, hence $G$ is solvable, too.
