most recent 30 from http://mathoverflow.net 2013-05-25T12:50:35Z http://mathoverflow.net/feeds/question/112198 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112198/solvable-groups liobei 2012-11-12T17:24:51Z 2013-01-06T07:23:11Z

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

http://mathoverflow.net/questions/112198/solvable-groups/112202#112202 Geoff Robinson 2012-11-12T17:51:39Z 2013-01-06T07:23:11Z

<p>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.</p>

http://mathoverflow.net/questions/112198/solvable-groups/118187#118187 majid arezoomand 2013-01-06T07:21:36Z 2013-01-06T07:21:36Z

<p>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)</p>