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Let $G$ be a group of order $p^2q$ where $p$ and $q$ are distinct primes. Show that $G$ is abelian if there are elements $g,g'\in Z(G)$ such that $|g|=p$ and $|g'|=q$.

First of all, I want to prove the statement without using Sylow Theorems. Here is the link of the proof of statement for general case. However, I'm not sure how to start but at least I can say that order of element $gg'$ is $pq$ as $g,g'\in Z(G)$.

Fuat Ray
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2 Answers2

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As $gg'\in\textbf Z(G)$ has order $pq$, then by Lagrange's theorem $|G:\textbf Z(G)|\in\{1,p\}$. Since the index of the center can't be prime, it follows that $|G:\textbf Z(G)|=1$ and hence $G$ is abelian.

Deif
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  • Dear @Deif, we are stating that $|Z(G)| \geq pq$ i.e. $|Z(G)|$ is at least $pq$ because $gg' \in Z(G)$, correct? This implies that the index $|G : Z(G)|$ can be either $1$ or $p$. As you mentioned, the index of the center cannot be a prime, so the only possibility is $1$. – Fuat Ray Mar 14 '25 at 19:59
  • @FuatRay Yes, that's the reasoning. – Deif Mar 14 '25 at 22:55
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Your hypothesis implies $Z(G)$ would have order $pq$ or $p^2q$. The latter immediately implies $G$ is abelian, so assume for contradiction that $Z(G)$ has order $pq$. Then $G/Z(G)$ would have order $p$. This would make $G/Z(G)$ cyclic since it has prime order. But then this implies $G$ is abelian contradicting our assumption that $Z(G)$ only has order $pq$.

Matthew H.
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