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I am sure this question was asked here before, maybe even multiple times. But I can't find it right now and it really bugs me:

Let $G$ be a group with $|G|=p^2$. Then $G$ is abelian.

For the proof we have 3 cases: $|Z(G)|\in\{1,p,p^2\}$, two of which I understand. Let $|Z(G)|=p$, then $G/Z(G)\cong\mathbb{Z}_p$, so $G/Z(G)$ is cyclical and therefore $G$ is abelian.

Question: This would mean that $Z(G)=G$, so there it's ultimately a contradiction although it proofs that $G$ is abelian. How do I interpret this proof?

Buh
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  • "But I can't find it right now" - I can't believe that you have searched; it comes up immediately, with multiple duplicates. – Dietrich Burde Nov 22 '17 at 11:14
  • I have found the one linked above. But my question is not being answered there. – Buh Nov 22 '17 at 11:17
  • That argument shows $G$ does not have a center of order $p$. It does not prove $G$ is abelian. As you wrote, the argument leads to a contradiction. Do not be fooled because it involves a concision you want. – KCd Nov 22 '17 at 11:17
  • See also https://math.stackexchange.com/questions/999247/if-g-zg-is-cyclic-then-g-is-abelian-what-is-the-point. – lhf Nov 22 '17 at 12:32

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As you have mentioned, $|Z(G)|=1,p \text{ or } p^2$. If $|Z(G)|=p$, then by this argument, $|G/Z(G)|=p$ which implies $G/Z(G)$ is cyclic. But as the proof mentions, this cannot be the case because of this result. Hence $|Z(G)|=p^2=|G|$, so that $Z(G)=G$ and $G$ is abelian.

creative
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