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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Bérénice
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Mathematicing
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What have you tried? Do you think the order of the center can be $1?$ And can it be $p^2?$ – awllower Jul 14 '16 at 11:10
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2http://math.stackexchange.com/questions/443710/let-g-be-a-nonabelian-group-of-order-p3-where-p-is-a-prime-number-prove-t?rq=1 – Behrouz Maleki Jul 14 '16 at 11:12
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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. – Mathematicing Jul 14 '16 at 11:18
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@Mathematicing This question has been asked several time. – Behrouz Maleki Jul 14 '16 at 11:21
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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 $p$. – AKP2002 Dec 02 '21 at 10:45
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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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1Well all the groups of order $p^2$ are abelian, so you can't find such a group. – Bérénice Jul 14 '16 at 11:28
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yes that's what I just thought, with the same argument you used. How to construct $|G| = p^3$ non abelian then ? – reuns Jul 14 '16 at 11:29
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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 $p^{3}$. – Mathematicing Jul 14 '16 at 11:33
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@user1952009 You can try a semidirect product $;C_{p^2}\rtimes C_p;$, or even use matrices (google Heisenberg Group) – DonAntonio Jul 14 '16 at 11:33
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@Mathematicing Use Lagrange's theorem + what are the possible divisors of $;p^3;$ . – DonAntonio Jul 14 '16 at 11:33
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If $p=2$ you have $D_4$ and $Q_8$, if $p$ is odd you have $\mathbb{Z}_{p^2} \rtimes \mathbb{Z}_p$ and $(\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_p$ – Bérénice Jul 14 '16 at 11:34
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Factor group of a cyclic group is cyclic. But wait! How do we even know that G is cyclic? I do not have my full stock of theorem with me atm which is a source of great frustration. .. – Mathematicing Jul 14 '16 at 12:17
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What about G having order $p^{3}$ implying the center of a group not being $p^{3}$? – Mathematicing Jul 14 '16 at 12:38
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@Mathematicing the center is an abelian (sub)group. If the full group is non-abelian the center cannot be the full group. – quid Jul 14 '16 at 12:44