# Maximal subgroups of order $pq^2$ 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.

• It's a big hammer, but you could have a look at "Li, Cai Heng; Zhang, Hua The finite primitive groups with soluble stabilizers, and the edge-primitive s-arc transitive graphs. Proc. Lond. Math. Soc. (3) 103 (2011), no. 3, 441–472." they have a list of all the soluble maximal subgroups of primitive groups (and in particular, of almost simple groups). Oct 31, 2019 at 2:15
• I deleted my answer since it was not correct, but here it is as a comment for the record. In $PSL(2,19)$ we have a maximal subgroup which is dihedral of order $2^2 \cdot 5$. So we can have maximal subgroups of order $p^2q$ with a cyclic subgroup of order $pq$. However in this example the $p$-Sylow is not cyclic, so this does not answer your question.
– spin
Oct 31, 2019 at 9:44
• Another example: in the alternating group $A_{19}$, there is a maximal subgroup $C_{q^2} \ltimes C_{p}$ for $p = 19$ and $q = 3$, but this does not have a cyclic subgroup of order $pq$.
– spin
Oct 31, 2019 at 10:08
• For what it's worth, I checked (by computer) for simple groups of order up to 16 million, and didn't find an example. I would guess this is true, and should be easy (but tedious) by using the reference above. OP, if you're really interested, I suggest you do this. (There might be an easier reason though, which doesn't rely on the classification.) Nov 1, 2019 at 1:20