solvable groups Let $G$ be a finite group which has a cyclic maximal subgroup. Is $G$ solvable? 
 A: 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)
A: 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.
A: 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 non-Sylow 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.
