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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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  • $\begingroup$ What have you tried? Do you think the order of the center can be $1?$ And can it be $p^2?$ $\endgroup$ – awllower Jul 14 '16 at 11:10
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    $\begingroup$ math.stackexchange.com/questions/443710/… $\endgroup$ – Behrouz Maleki Jul 14 '16 at 11:12
  • $\begingroup$ A fact is that the center of a group G is a subgroup of a group G. I'm trying to work from another fact that any grup of prime order is cyclic. $\endgroup$ – Mathematicing Jul 14 '16 at 11:18
  • $\begingroup$ @Mathematicing This question has been asked several time. $\endgroup$ – Behrouz Maleki Jul 14 '16 at 11:21
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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 '16 at 11:25
  • $\begingroup$ What is $H$ ? ${}$ $\endgroup$ – Jennifer Jul 14 '16 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. $\endgroup$ – Jennifer Jul 14 '16 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 '16 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 $p^{3}$. $\endgroup$ – Mathematicing Jul 14 '16 at 11:33

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