# Maximal subgroups that force solvability.

For which finite groups $$M$$ is it the case that every finite group $$G$$ with $$M$$ as a maximal subgroup solvable?

If $$M$$ satisfies this condition then $$M$$ is solvable. Also, if $$M$$ is abelian then $$M$$ satisfies this condition. Futhermore, I believe that if $$M$$ is nilpotent and if all 2-subgroups of $$M$$ are normal subgroups of $$M$$ (if Sylow 2-subgroups of $$M$$ are abelian or quaternion, for example) then $$M$$ satisfies this condition (proof below).

More specific questions:

1) Is there a non-nilpotent group that satisfies this condition?

2) Which 2-groups satisfy this condition?

Apparently, the dihedral group of order 8 satisfies this condition (see Mikko Korhonen's comment on this post). Also, if $$M\times N$$ satisfies this condition then $$M$$ and $$N$$ both satisfy this condition.

(This proof is adapted from j.p.'s answer to the linked question). Let $$G$$ be minimal such that $$G$$ is not solvable and such that $$G$$ contains a maximal subgroup $$M$$ that is nilpotent and whose 2-subgroups are normal. If $$M$$ contains a nontrivial normal subgroup $$N$$ of $$G$$ then $$G/N$$ contradicts the minimality of $$G$$. Thus, $$M$$ does not contain nontrivial normal subgroups of $$G$$. In particular, $$N_G(P)=M$$ for all Sylow $$p$$-subgroups $$P$$ of $$M$$. Then $$P$$ is a Sylow $$p$$-subgroup of $$N_G(P)$$ so $$P$$ is a Sylow $$p$$-subgroup of $$G$$. This shows that $$M$$ is a Hall subgroup of $$G$$.

If $$P$$ is a Sylow $$p$$-subgroup of $$M$$ and if $$Q$$ is a nontrivial normal subgroup of $$P$$ then $$N_G(Q)=M$$ which has a normal $$p$$-complement. For $$p=2$$, Frobenius' normal $$p$$-complement theorem gives that $$G$$ has a normal $$p$$-complement. For $$p\geq3$$, Thompson's normal $$p$$-complement theorem or Glauberman's normal $$p$$-complement theorem gives that $$G$$ has a normal $$p$$-complement (since you only have to consider characteristic $$p$$-subgroups).

Thus, for each prime $$p$$ dividing the order of $$M$$, $$G$$ has a normal $$p$$-complement. Then $$M$$ has a normal complement $$N$$ in $$G$$. Since $$M$$ is solvable but $$G$$ is not solvable, $$N$$ is not solvable. In particular, $$N$$ does not admit a fixed-point-free automorphism of prime order. If $$m\in Z(M)$$ has prime order then $$C_N(m)$$ is nontrivial. Then $$C_N(m)M$$ is a subgroup of $$G$$ that properly contains $$M$$ so $$C_N(m)M=G$$ by the maximality of $$M$$. Comparing cardinalities shows that $$C_N(m)=N$$ so $$m\in Z(G)$$. Then $$\langle m\rangle$$ is a nontrivial normal subgroup of $$G$$ contained in $$M$$ which is a contradiction.

• A famous example here is when $M$ is a $p$-group for an odd prime $p$. – Steve D Sep 21 '18 at 6:41
• I wonder if there is a reduction to the case where $G$ is almost simple? That is, could we show that $M$ "forces solvability as a maximal subgroup" if and only if $M$ is solvable and $M$ is not a maximal subgroup of any almost simple group? If the answer is yes, then your questions are answered by a paper of Li and Zhang from 2011 (Proceedings of the LMS), who classify the solvable maximal subgroups of almost simple groups. – Mikko Korhonen Sep 27 '18 at 1:50
• About 2), just to give an example: if $p = 2^n - 1$ is a Mersenne prime ($n > 3$), then $PSL(2,p)$ has the dihedral group of order $2^{n-1}$ as a maximal subgroup. – Mikko Korhonen Sep 27 '18 at 1:52
• Interesting. I computed maximal subgroups of a number of simple groups a while ago. It looked like all small nontrivial semi-direct products of two cyclic groups occurred with the exception of the dihedral group of order 8. I still don't know whether the dihedral group of order 8 forces solvability. – Thomas Browning Sep 27 '18 at 2:21
• @ThomasBrowning: In "A condition for the solvability of a finite group" (1961) Deskins shows that if $G$ has a nilpotent maximal subgroup of nilpotency class $\leq 2$, then $G$ is solvable. In particular, the dihedral group of order $8$ forces solvability as a maximal subgroup – Mikko Korhonen Sep 27 '18 at 3:46