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?

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    $\begingroup$ Olshanskii constructed non-abelian simple groups generated by 2 elements in which every proper subgroup is cyclic (infinite or finite, as you like), known under the fuzzy name "Tarski monsters". Since these are finitely generated groups, they have maximal subgroups, which are cyclic. $\endgroup$ – YCor May 6 '14 at 14:06
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    $\begingroup$ Another exercise is to extend the finite case (by deduction from the finite case): if $G$ is a residually finite group with a maximal subgroup that is abelian, then $G$ is metabelian. $\endgroup$ – YCor May 6 '14 at 14:10
  • $\begingroup$ What if we assume the group is finitely presented? $\endgroup$ – i. m. soloveichik May 6 '14 at 14:11
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    $\begingroup$ For finitely presented groups it's probably an open question (not asked anywhere I guess). $\endgroup$ – YCor May 6 '14 at 14:20
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    $\begingroup$ Another example, far less pathological than Tarski monsters, is BH Neumann's subgroup $G$ of permutations of $\mathbf{Z}$ generated by $c(n)=n+1$ and the group $A_\infty$ of finitely supported even permutations. Then the cyclic subgroup $C$ generated by $c$ is maximal (because it normalizes no nontrivial proper subgroup of $A_\infty$, exercise) and $G=A_\infty\rtimes C$. It's infinitely presented but it might suggest that a finitely presented example is not out of reach. $\endgroup$ – YCor May 7 '14 at 12:10

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).

  • $\begingroup$ In case $G$ is not simple, how do you check $A\cap [G,G]$ normal in $G$? more precisely, if $c\in [G,G]$ how do you check $c(A\cap [G,G])c^{-1}\subset A$? $\endgroup$ – YCor May 6 '14 at 16:51
  • $\begingroup$ Now amended ( with much weaker conclusion, of course) $\endgroup$ – Geoff Robinson May 6 '14 at 17:02
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    $\begingroup$ Another simple remark is that if $A$ is not normal, then $Core_G(A)$ equals the center of $G$. $\endgroup$ – YCor May 7 '14 at 9:09
  • $\begingroup$ @Yves Cornulier : Yes, thanks, I had not noticed that. $\endgroup$ – Geoff Robinson May 7 '14 at 9:12

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