Let $G$ be a minimal nonnilpotent group with cyclic Sylow $p$-subgroup $P$ and normal Sylow $q$-subgroup $Q$ [see Huppert, Endlich Gruppen I]. If $Q$ is abelian and $q > 2$, then can we get that $G$ is a minimal nonabelian group? If not, can anyone give a counterexample?


The answer is yes, and you do not need the assumption $q>2$.

Let $P = \langle x \rangle$. Then, by one of the properties of minimal non-nilpotent groups (Huppert III, 5.2 b)), $x^p \in Z(G)$, so $P$ acts on $Q$ by inducing an automorphism of order $p$.

Recall that $\Omega_1(Q)$ is the subgroup of $Q$ consisting of all elements of order dividing $q$. There is a result (e.g. Gorenstein, Finite Groups, 5.2.4) that says that, if a $q'$-automorphism of $Q$ acts trivially on $\Omega_1(q)$ then it acts trivially on $Q$. So $P$ cannot act trivially on $\Omega_1(Q)$ and hence $\Omega_1(Q)P$ is not nilpotent, so $G=\Omega_1(Q)P$ and $Q=\Omega_1(Q)$. That is, $Q$ is elementary abelian.

Now, by Mashcke's Theorem, $P$ acts completely reducibly on $Q$, and by choosing $R$ to be a $P$-invariant subgroup of $Q$ on which $P$ acts nontrivially and irreducibly, the minimality of $G$ gives $G=PR$, $Q=R$. That is, $P$ acts irreducibly on $Q$.

Now let $H$ be a proper subgroup of $G$. Let $P_1 = \langle x^p \rangle$ be the unique maximal subgroup of $P$. Then $QP_1$ is the unique maximal subgroup of $G$ that contains $Q$. So if $HQ \ne G$ then $HQ \le HP_1$. But we observed above that $x^p$ acts trivially on $Q$, so $HP_1 = H \times P_1$ is abelian.

On the other hand, if $HQ = G$ then, since $P \cong HQ/Q$ acts irreducibly on $Q$, we have $H \cap Q = 1$ or $Q$. If $H \cap Q = 1$, then $H$ is a $2$-group and hence cyclic whereas, $H \cap Q = Q$ implies $H=G$, contrary to assumption. So $H$ is abelian, and $G$ is minimal nonabelian.


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