# Schmidt group and maximal subgroups

Let $G$ be a Schmidt group, a minimal non-nilpotent group, so that $G$ is not nilpotent, but every proper subgroup of $G$ is nilpotent. I want to prove $G$ has precisely two classes of maximal subgroups: $\langle a^{q}\rangle \times P$ and $\{ P' \times Q^{x} ∣ x \in G‎ \}$ where $G = Q \ltimes P$, $Q=\langle a\rangle$, $Q$ is a Sylow $q$-subgroup, and $P$ is a Sylow $p$-subgroup.

Such groups are covered in Huppert's Endliche Gruppen Kapitel III §5 page 280, and Robinson's Course in the Theory of Groups 9.1.9 page 251.

• I cleaned up the tex and gave some definitions and references. Derek Holt mentioned he would help, and gave you a hint to work on: a maximal subgroup should either contain P or a conjugate of Q. – Jack Schmidt Jun 8 '13 at 22:47
• I'm not sure I fully understand: if $\,Q=\langle a\rangle\,$ is a cyclic Sylow $\,q$-subgroup then $\,a^q=1\,$ , so in fact $\,\langle a^q\rangle\times P\cong P\,$ ...is this what was meant? Also, is $\;P'\;$ meant to be the derived subgroup of $\,P\,$ ? – DonAntonio Jun 8 '13 at 23:31
• @DonAntonio: $Q$ could have order greater than $q^1$, so $\langle a^q\rangle$ is just a maximal subgroup of $Q$. $Q$ has to act trivially on every $Q$ invariant-subgroup of $P$, so by minimality it has to act irreducibly on $P/\Phi(P)$. A maximal $Q$-subgroup of $P$ is thus $\Phi(P)$. Huppert mentions that when $p$ is odd, $P$ has exponent $p$, and $\Phi(P)=P'$. When $p=2$, I'm not sure why $P'$ would be maximal. Huppert also mentions $\Phi(P)\Phi(Q) \leq Z(G)$. – Jack Schmidt Jun 9 '13 at 6:10