# solvable groups

Let $G$ be a finite group which has a cyclic maximal subgroup. Is $G$ solvable?

-
mathoverflow.net/howtoask – Yemon Choi Nov 12 '12 at 17:38
@soroosh: you should accept the answer if it satisfies you, so that the question is known to be answered by other users (and the answerer is rewarded). – Benoît Kloeckner Nov 13 '12 at 10:17
Why is this not getting closed? – Igor Rivin Jan 7 '13 at 0:04
Many thanks for your useful comments and statements. – sebastian Jan 13 '13 at 20:50

Yes. Let $M$ be the cyclic maximal subgroup (actually the proof works for an Abelian maximal subgroup). We may suppose by induction that $M$ contains no non-trivial normal subgroup of $G.$ Then for each non-identity subgroup $X$ of $M,$ we have $M = N_{G}(X),$ as $M$ is maximal and $X \lhd M.$ It follows easily that $M$ is a Hall subgroup of $G,$ and by Burnside's normal $p$-complement theorem, there is a normal complement $K$ to $M$ in $G.$ Furthermore, we have $M = C_{G}(m)$ for each non-identity $m \in M.$ Hence $G$ is a Frobenius group with Frobenius kernel $K$ and Frobenius complement $M.$ Since Frobenius kernels are nilpotent by Thompson's theorem, we see that $G$ is solvable as $K$ and $G/K \cong M$ both are. In fact, I think this argument (for the case of $M$ Abelian) is due to Thompson.
Can you see also Exercise 3.4.7 of Permutation Groups, Dixon and Mortimer. To prove the result, just consider the action of $G$ on the right cosets of its (abelian) maximal subgroup and use its previous exercise. – majid arezoomand Jan 6 '13 at 7:03
Thanks for the comment. I haven't read that book, so I don't know what the previous exercise says. However, I think you do need a transfer argument and/or Frobenius's theorem to prove that a group with an Abelian maximal subgroup is solvable. I can see now how use of Thompson's theorem on the nilpotence of Frobenius kernels can be avoided, because the Frobenius complement normalizes a Sylow subgroup of the Frobenius kernel, so the maximality of the complement implies that the Frobenius kernel is a $q$-group for some prime $q.$ – Geoff Robinson Jan 6 '13 at 7:29
Every primitive permutation group with an abelian point stabilizer is regular of prime degree or a Frobenius group. Also by the prevous exercise (mentioned above) every finite primitive permutation group has a regular elementary abelian normal $p$-subgroup which is the same Frobenius kernel. – majid arezoomand Jan 6 '13 at 7:51
I just want to mention that, if $M$ be an abelian maximal subgroup of finte group $G$ then $G$ is solvable and its drived length is at most 3.(Exercise 3.4.7 of Permutation Groups, Dixon and Mortimer)