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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asked Jul 14, 2016 at 11:08
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$\begingroup$ What have you tried? Do you think the order of the center can be $1?$ And can it be $p^2?$ $\endgroup$– awllowerCommented Jul 14, 2016 at 11:10
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2$\begingroup$ math.stackexchange.com/questions/443710/… $\endgroup$– Behrouz MalekiCommented Jul 14, 2016 at 11:12
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$\begingroup$ 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. $\endgroup$– MathematicingCommented Jul 14, 2016 at 11:18
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$\begingroup$ @Mathematicing This question has been asked several time. $\endgroup$– Behrouz MalekiCommented Jul 14, 2016 at 11:21
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$\begingroup$ Does this answer your question? Let G be a nonabelian group of order $p^3$, where $p$ is a prime number. Prove that the center of $G$ is of order $p$. $\endgroup$– AKP2002Commented Dec 2, 2021 at 10:45
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1 Answer
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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 $|H| = p^2$ ? $\endgroup$– reunsCommented Jul 14, 2016 at 11:25
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1$\begingroup$ Well all the groups of order $p^2$ are abelian, so you can't find such a group. $\endgroup$– BéréniceCommented Jul 14, 2016 at 11:28
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$\begingroup$ yes that's what I just thought, with the same argument you used. How to construct $|G| = p^3$ non abelian then ? $\endgroup$– reunsCommented Jul 14, 2016 at 11:29
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$\begingroup$ 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 $p^{3}$. $\endgroup$ Commented Jul 14, 2016 at 11:33