# Maximal subgroups of order $pq^2$ in finite simple groups

This question is moved from math stackexchange, seems like it is a more advanced question. Here the link from the original question: https://math.stackexchange.com/questions/3415338/maximal-subgroups-of-order-pq2-in-finite-simple-groups

It is a well-known result in elementary group theory that if $$q^{2}\mid (p-1)$$, then there are two non-isomorphic nonabelian groups of the form $$\mathbb{Z}_{q^{2}}\ltimes\mathbb{Z}_{p}$$. One has a cyclic subgroup of order $$pq$$ while in the other one the subgroup of order $$pq$$ is nonabelian. Now my question:

Is there any finite simple group $$G$$ with a maximal subgroup $$M\cong\mathbb{Z}_{q^{2}}\ltimes\mathbb{Z}_{p}$$ such that the subgroup of order $$pq$$ in $$M$$ is abelian?

As far as I searched in ATLAS there is not such an example but I need a proof.

• In response to your stackexchange question, I gave you a paper which has a list which you just have to go through to answer your question. Have you tried that? Nov 11, 2019 at 20:50

Suppose that $$G$$ is simple and has $$M$$ as a maximal subgroup. Let $$P \in {\rm Syl}_q(M)$$ and $$Z=Z(M)$$. So $$|P|=q^2$$ and $$|Z|=q$$ with $$Z < P$$. Since $$M$$ is maximal in $$G$$ and $$M$$ is not normal in $$G$$, we have $$M = N_G(Z)$$.
Now $$N_G(Z)$$ contains the centre of a Sylow $$q$$-subgroup $$R$$ of $$G$$ containing $$P$$, and since $$Z$$ is the only subgroup of $$P$$ of order $$q$$, we must have $$Z \le Z(R)$$, and so $$R=P$$.
Also, $$N_G(P) \le N_G(Z) = M$$, so $$N_G(P) = N_M(P) = P$$. But now by Burnside's Transfer Theorem $$G$$ has a normal $$q$$-complement in $$G$$, contradicting its simplicity.
• Nice. As the OP's assumptions are quite restricted, let me mention it more generally proves the following: if $M$ is finite group, $q$ is prime, $P$ is a $q$-Sylow of $M$ equal to its own normalizer in $M$, and $P$ has a unique subgroup $Z$ of order $q$, and $Z$ is normal in $M$, then $M$ is not isomorphic to a maximal subgroup in any finite simple group.