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Let $G$ be a non-abelian group of order $pq^2$. Is it true that $G'\simeq \mathbb Z_q\times \mathbb Z_q$?

From a previous question, I learned that the only non-abelian group of order $pq^2$ is of the type $(\mathbb Z_q\times \mathbb Z_q)\rtimes \mathbb Z_p$ no matter which divisibility relation we use.

Thus $G/(\mathbb Z_q\times \mathbb Z_q)\simeq \mathbb Z_p$ and this implies that $ G' \subset \mathbb Z_q\times \mathbb Z_q$.

muser17
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  • Are you sure that all non-abelian group of order $pq^2$ are of the given type? I mean for $p=3$ and $q=2$ we have three non-abelian groups of order $12$. Two of them doesn't have a normal Sylow $2-$subgroups and also for one of them the Sylow $2-$ subgroup is $\mathbb{Z}_4$, while for the other it's $\mathbb{Z}_2 \times \mathbb{Z}_2$ – Stefan4024 Sep 13 '17 at 13:23

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No, the commutator subgroup of such a group could have order $q$. As an example, take $p=2$ and $q=3$. Let $Q$ be the additive group $\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$ and let $P$ be a cyclic group of order 2 with generator $x$. Finally, form the semi direct product of $Q$ by $P$ with action of $P$ on $Q$ given by $(a,b)^{x}=(b,a)$; call this group $G$ (so $G$ is isomorphic to the wreath product of $\mathbb{Z}/3\mathbb{Z}$ by $\mathbb{Z}/2\mathbb{Z}$).

You can check that the subgroups of $Q$ generated by $(1,1)$ and $(1,-1)$ are both normal in $G$, and that $P$ centralizes the former. It follows that $G'$ is equal to the latter subgroup of $Q$.

The situation is different if you add the assumption that $C_{G}(x)\leq Q$ for all nonidentity $x\in Q$. Such groups are known as Frobenius groups.

John McHugh
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