If G is a finite group which contains a maximal subgroup M which is abelian, then it is an exercise to show that G is solvable and that the third term in the derived series equals 1.
What happens for infinite groups?
If G is a finite group which contains a maximal subgroup M which is abelian, then it is an exercise to show that G is solvable and that the third term in the derived series equals 1.
What happens for infinite groups?
Here are a few (presumably well-known) things one can say about an arbitrary group $G$ with an Abelian maximal subgroup $A.$ We discount the trivial situation when $A \lhd G,$ which is easy to analyse. We might as well suppose that $A$ contains no non-identity normal subgroup $N$ of $G,$ for otherwise we can just factor out the intersection of the $G$-conjugates of $A$, usually denoted by ${\rm core}_{G}(A).$ In that case, we note that if $G$ is not simple, then $G^{\prime} = [G,G]$ is the unique minimal proper non-identity normal subgroup of $G.$ For if $1 \neq N \lhd G \neq N,$ then $G = AN \neq N$ as $A$ is maximal. Hence $[G,G] \leq N.$ Furthermore, in that case, $G = A[G,G].$ If the product is not semidirect and $A \cap [G,G] \neq 1,$ then we have $D \cap D^{y} = 1$ for all $y \in [G,G] \backslash D$, where $D = A \cap [G,G].$ For we have $A = N_{G}(D)$ when $D \neq 1,$ so that $N_{[G,G]}(D) = D.$ Also, for any $y \in [G,G] \backslash D$, we have $D \cap D^{y} \lhd \langle D,D^{y} \rangle$, while $A = N_{G}(D \cap D^{y}) $ when $D \cap D^{y} \neq 1.$ Thus $D^{y} \leq A \cap [G,G] = D.$ Likewise, $D^{y^{-1}} \leq D,$ so that $D^{y} = D,$ a contradiction. This also covers the case that $G = [G,G]$ is simple, in which case $D = A.$ In summary, here are the possibilities including the case that $A \lhd G :$
a) $A \lhd G$ and $G/A$ is cyclic of prime order. b) $H = G/{\rm core}_{G}(A)$ has an Abelian maximal subgroup $B = A/{\rm core}_{G}(A)$ and either $H$ is a semi-direct product $[H,H]B,$ and $B$ normalizes no proper non-identity subgroup of $[H,H],$ or else, setting $C = B \cap [H,H],$ we have $C \cap C^{x} = 1$ for all $x \in [H,H]\backslash C,$ and every subgroup of $[H,H]$ normalized by $B$ is contained in $C.$ (Thanks to Yves Cornulier for pointing out an error in an earlier version of this post).