Doubt in proof of Herstein Theorem (Solvable Groups) I'm trying to prove this result due to Herstein (1958):

If $G$ is a finite group and $A \le G$ is a maximal subgroup. If $A$ is abelian, then G is solvable.

Since $A$ is maximal, we have only 2 possible cases:

*

*$N_G(A) = G$

*$N_G(A) = A$
In the first case, we prove that $\dfrac{G}{A}$ is cyclic and so solvable, then $G$ is solvable, since $A$ is abelian.
In the second case, we consider two new cases:
a) There exists $x \notin A$ such that $A^x \cap A \neq 1$
b) $A^x \cap A = 1$, for all $x \notin A$
For the letter (b) we use the Frobenius Theorem, since $A$ is a Frobenius complement of $G$ and obtain the result.
In the letter (a), letting $w \in A^x \cap A$ and using that $A$ is abelian and maximal, we obtain that $w \in Z(G)$ and so $\langle w \rangle$ is a abelian normal subgroup of $G$ contained in $A$. So far so good. But at this point the author says it follows that $\dfrac{G}{\langle w \rangle}$ is solvable, arguing by induction in group order and that's what I didn't understand.
 A: Note that in general, "maximal X group" can means either "maximal subgroup which has property X", or "subgroup which is maximal among all subgroups with property X". Most of the time it means the latter. (Thus, for instance, Sylow $p$-subgroups are maximal $p$-subgroups).
Here, it appears that you actually mean "maximal subgroup which is abelian", not "subgroup which is maximal among abelian subgroups". To avoid ambiguity, you should say that $A$ is "an abelian maximal subgroup", because that will be interpreted as $\text{abelian}(\text{maximal subgroup})$, i.e., a maximal subgroup which is abelian; whereas "maximal abelian subgroup" is interpreted as $\text{maximal}(\text{abelian subgroup})$, i.e., a subgroup which is maximal among abelian subgroups.
(This ambiguity would also happen in Spanish, and I suspect the same would happen in Portuguese... to describe an abelian maximal subgroup in Spanish I would say "subgrupo maximal abeliano", whereas a maximal abelian subgroup would be a "subgroup abeliano maximal", which could lead to the mistranslation because of the reverse word order...)

On to the substance of your question:
This is a pretty standard argument, though it is not explicitly phrased as an induction.
We proceed by induction on $|G|$. Assume that $|G|=n$, and that for all subgroups $\mathfrak{G}$ with $|\mathfrak{G}|\lt n$, if $\mathfrak{G}$ has an abelian maximal subgroup, then $\mathfrak{G}$ is solvable.
In your final situation, you have a nontrivial subgroup $\langle w\rangle\leq Z(G)$; let $\mathfrak{G}=G/\langle w\rangle$. Then $|\mathfrak{G}|\lt n$; moreover, $A/\langle w\rangle$ is a maximal subgroup of $\mathfrak{G}$, and since $A/\langle w\rangle$ is a quotient of an abelian group, it is abelian. Thus, $\mathfrak{G}$ has an abelian maximal subgroup. By the induction hypothesis, $\mathfrak{G}$ is solvable.
But then $\langle w\rangle$ is abelian, and $G/\langle w\rangle$ is solvable, therefore $G$ is solvable as well. QED
