Maximal subgroups of order $p^3$ in finite simple groups

Question

Is there any finite simple group $G$ such that $G$ has a maximal subgroup of the form $\mathbb{Z}_{p}\ltimes\mathbb{Z}_{p^2}$, for some prime divisor $p$? In case the answer is positive please guide me to find a classification of such simple groups. Otherwise please give a proof for nonexistence of such maximal subgroups.

Answer 1

By a theorem of Thompson, if a finite group has a nilpotent maximal subgroup of odd order, it is solvable.

Janko and Deskins have shown that if a finite group $G$ has a maximal Sylow $2$-subgroup of class $\leq 2$, then $G$ is solvable. Proofs can also be found in Endliche Gruppen I by Huppert: Satz 7.4, p. 445.

It is immediate from these results that if a finite simple group has a maximal Sylow $p$-subgroup $P$, then $p = 2$ and $|P| \geq 16$. (Note that $\operatorname{PSL}(2,17)$ has a maximal Sylow $2$-subgroup of order $16$.)

References:

[1] J. Thompson, Finite groups with fixed-point-free automorphisms of prime order. Proc. Nat. Acad. Sci. U.S.A. 45 (1959) 578–581.

[2] Z. Janko, Finite groups with a nilpotent maximal subgroup. J. Austral. Math. Soc. 4 (1964) 449–451.

[3] W. E. Deskins, A condition for the solvability of a finite group. Illinois J. Math. 5 (1961) 306–313.

[4] J. S. Rose, On finite insoluble groups with nilpotent maximal subgroups. J. Algebra 48 (1977) no. 1, 182–196.

Answer 2

Just a quick comment if you want more.

In my paper with J. Chen, we classified all the pairs $(G,H)$ where $G$ is almost simple and $H$ a maximal subgroup of $G$ of prime-power order. See our Proposition 6.3.

It is a very short list (possibly finite, depending on whether there are finite many Mersenne/Fermat primes), and $H$ is dihedral except 2 cases.