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Question : Suppose that $G$ is a non-abelian group of order $p^{3}$ where $p$ is prime and $Z (G) \neq \{e\}$. Prove that $|Z (G)| =p$.

Any useful hint to this question is appreciated.

Thanks in advance.

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1 Answer 1

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Since the center is non trivial his order can be $p,p^2$ or $p^3$. But $G$ is non abelian, so $|Z (G)|\neq p^3$.

Also if $|Z (G)|=p^2$, then $|G/Z|=p$, so $G/Z$ is cyclic, so $G$ is abelian (proof here).

Finally $|Z (G)|=p$.

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  • $\begingroup$ I never remember, how to construct a non abelian $|H| = p^2$ ? $\endgroup$
    – reuns
    Jul 14, 2016 at 11:25
  • $\begingroup$ What is $H$ ? ${}$ $\endgroup$
    – Bérénice
    Jul 14, 2016 at 11:26
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    $\begingroup$ Well all the groups of order $p^2$ are abelian, so you can't find such a group. $\endgroup$
    – Bérénice
    Jul 14, 2016 at 11:28
  • $\begingroup$ yes that's what I just thought, with the same argument you used. How to construct $|G| = p^3$ non abelian then ? $\endgroup$
    – reuns
    Jul 14, 2016 at 11:29
  • $\begingroup$ How does the fact that the center is non trivial lead us to deduce the possible order of the center as p, $p^{2}$ or $p^{3}$. $\endgroup$ Jul 14, 2016 at 11:33

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