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?
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 $\Bbb Z/p\Bbb Z \times \Bbb Z/p\Bbb Z$.
Theorem~4.3.8 implies a non-trivial center which must have order $p$, else the group is either trivially abelian $\left|Z\left(P\right)\right|=p^{3}$ or, by Exercise~3.1.36, $\left|Z\left(P\right)\right|=p^{2}$ implies the group is abelian. On the other hand, $\left|Z\left(P\right)\right|=p$ and $P/Z\left(P\right)\cong Z_{p^{2}}$ also imply, via Exercise~3.1.36, that the group is abelian. There is only one possibility, $P/Z\left(P\right)\cong Z_{p}\times Z_{p}$.
Altenatively, $P$ is a non-abelian group of order $p^{3}$ so there are the standard constructions \begin{align*} \left\langle y\right\rangle\rtimes\left\langle x\right\rangle &=\left\langle\left. x,y\right|\ x^{p}=y^{p^{2}}=1,\ xyx^{-1}=y^{p+1}\right\rangle\\ \left(\left\langle a\right\rangle\times\left\langle b\right\rangle\right)\rtimes\left\langle x\right\rangle &=\left\langle\left. x,a,b\right|\ x^{p}=a^{p}=b^{p}=1,\ ab=ba,\ xax^{-1}=ab,\ xbx^{-1}=b\right\rangle \end{align*} and Theorem~6.1.1(2) implies $\left\langle y\right\rangle\cap Z\left(P\right)=\left\langle y^{p}\right\rangle\cong Z_{p}$ and $\left(\left\langle a\right\rangle\times\left\langle b\right\rangle\right)\cap Z\left(P\right)=\left\langle b\right\rangle\cong Z_{p}$, respectively. Then there are the homomorphisms \begin{align*} \phi_{1}:\left\langle y\right\rangle\rtimes\left\langle x\right\rangle &\rightarrow\left\langle y\right\rangle\rtimes\left\langle x\right\rangle /\left\langle y^{p}\right\rangle\cong\left\langle y^{p}\right\rangle\times\left\langle x\right\rangle\\ \phi_{2}:\left(\left\langle a\right\rangle\times\left\langle b\right\rangle\right)\rtimes\left\langle x\right\rangle &\rightarrow\left(\left\langle a\right\rangle\times\left\langle b\right\rangle\right)\rtimes\left\langle x\right\rangle /\left\langle b\right\rangle\cong\left\langle a\right\rangle\times\left\langle x\right\rangle \end{align*} In either case, $P/Z\left(P\right)\cong Z_{p}\times Z_{p}$.
