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

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.

• No, according to this MO answer by Geoff Robinson. – YCor Oct 30 '19 at 11:09

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.

• @ Mikko Korhonen. Many thanks for your helpful and complete answer. – H.Shahsavari Oct 30 '19 at 15:24