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.

  • $\begingroup$ 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? $\endgroup$ – verret Nov 11 '19 at 20:50

No there is no such simple group.

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.

  • 2
    $\begingroup$ 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. $\endgroup$ – YCor Nov 12 '19 at 8:08
  • $\begingroup$ @ Professor Holt. Thank you so much for your complete and short proof. It is great. $\endgroup$ – H.Shahsavari Nov 13 '19 at 18:01

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