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

4$\begingroup$ mathoverflow.net/howtoask $\endgroup$ – Yemon Choi Nov 12 '12 at 17:38

5$\begingroup$ @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). $\endgroup$ – Benoît Kloeckner Nov 13 '12 at 10:17

$\begingroup$ Why is this not getting closed? $\endgroup$ – Igor Rivin Jan 7 '13 at 0:04

$\begingroup$ Many thanks for your useful comments and statements. $\endgroup$ – 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 nontrivial normal subgroup of $G.$ Then for each nonidentity 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 nonidentity $m \in M.$ 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.

$\begingroup$ 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. $\endgroup$ – majid arezoomand Jan 6 '13 at 7:03

1$\begingroup$ 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 $q.$ $\endgroup$ – Geoff Robinson Jan 6 '13 at 7:29

$\begingroup$ 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. $\endgroup$ – majid arezoomand Jan 6 '13 at 7:51

$\begingroup$ Yes, OK, but you still need to deal with Frobenius groups to do the general case ( or use transfer, which will work too when the maximal subgroup is Abelian), which is the hard part of the proof. The only real saving in the proof you quote is to notice that the Frobenius kernel is elementary Abelian, which I had not noticed or made use of. $\endgroup$ – Geoff Robinson Jan 6 '13 at 9:31

$\begingroup$ Certainly, your answer is very nice and I added only a reference for the question. Thanks for your previous comments. $\endgroup$ – majid arezoomand Jan 6 '13 at 15:22
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)
Let $H<G$ be maximal. If $H$ is abelian, then $G$ is solvable (Herstein). If $H$ is nilpotent of odd order, then $G$ is solvable (Thompson). If $H=P\times F$ is nilpotent and $P\in\text{Syl}_2(H)$ is abelian, then $G$ is solvable (this follows from the previous two results). If $H$ is nilpotent nonSylow subgroup of $G$ and all Sylow subgroups of $H$ are nonnormal in $G$, then $G$ is solvable (Wielandt). All the mentioned results are well knowm.