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.