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

edited Nov 13 at 5:35

Guntram
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asked Nov 12 at 17:24

liobei
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mathoverflow.net/howtoask – Yemon Choi Nov 12 at 17:38

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@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 at 10:17

Why is this not getting closed? – Igor Rivin Jan 7 at 0:04

Many thanks for your useful comments and statements. – liobei Jan 13 at 20:50

Geoff Robinson · Accepted Answer · 2013-01-06 07:23:11Z UTC

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Yes. Let $M$ be the cyclic maximal subgroup (actually the proof work 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 $m \in M.$ Hen $G$ is a Frobenius group with Frobeius 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.

edited Jan 6 at 7:23

answered Nov 12 at 17:51

Geoff Robinson
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	thank you so much! – liobei Nov 12 at 18:10
	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 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 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 at 7:51
	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. – Geoff Robinson Jan 6 at 9:31

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majid arezoomand · Answer 2 · 2013-01-06 07:21:36Z UTC

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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)

answered Jan 6 at 7:21

majid arezoomand
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solvable groups

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