This is kinds of extension of 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 $p$.

What i want to do is following

If $p$ is a prime number and $G$ is non-abelian group of order $p^3$, $G/Z(G) \cong Z_p \times Z_p$.

What i know is that $|Z(G)|=p$, and conclude the isomorphism to product of $Z_p$.

How to prove this?

  • $\begingroup$ Imho this is not an extension but simply a duplicate. $\endgroup$ – MooS Jun 9 '17 at 9:10
  • $\begingroup$ I know $|G/Z(G)|=p^2$, but how this shows that this is isomorphic to $\mathbb{Z}_p\times \mathbb{Z}_p$? $\endgroup$ – phy_math Jun 9 '17 at 9:11

If the center has order $p$, then $G/Z(G)$ has order $p^2$, so according to this question, it must be equal to either $\Bbb Z/p^2\Bbb Z $ or $\Bbb Z/p\Bbb Z \times \Bbb Z/p\Bbb Z$. Since it can't be cyclic has otherwise $G$ would be abelian, $G/Z(G)$ must be isomorphic to $\Bbb Z/p\Bbb Z \times \Bbb Z/p\Bbb Z$.


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