Showing that $|Z(G)| = p$ and $G/Z(G) \simeq \mathbb{Z}_p \times \mathbb{Z}_p$ for finite p-groups with order $|G|=p^3$

Question

I have a finite, non-abelian $p$-group $G$ with $|G|=p^3$. I want to show that $|Z(G)| = p$ and $G/Z(G) \simeq \mathbb{Z}_p \times \mathbb{Z}_p$, where $Z(G)$ is the center of $G$.

From the definitions, I know that for every $g \in G$, there's a natural number $\alpha_g$ such that ord$(g)=p^{\alpha_g}$, that $Z(G) = \{ z \in G \,|\, zg =gz \,\, \forall g \in G\}$ and that $G/Z(G) = \{gZ(G) \,|\, g \in G \}$. From a lemma, I know that $|Z(G)| > 1$. Also, I suspect that I need the class equation $|G| = |Z(G)| + \sum_i (G:C_G(\omega_i))$.

However, I don't see how these parts may fit together.

Answer 1

We will use following lemma:

Lemma If $G/Z(G)$ is cyclic then $G$ is abelian.

Since $|Z(G)|>1$ and $G$ is nonabelian, the cardinality of $Z(G)$ is $p$ or $p^2$. But if $|Z(G)|=p^2$ then $|G/Z(G)|=p$ so $G/Z(G)$ is cyclic. Therefore by lemma, $G$ is abelian, contradicting that $G$ is nonabelian. So $|Z(G)|=p$.

Since $|G/Z(G)|=p^2$, $G/Z(G)$ is isomorphic to $\Bbb{Z}/p^2\Bbb{Z}$ or $(\Bbb{Z}/p\Bbb{Z})^2$. But if $G/Z(G)$ is isomorphic to $\Bbb{Z}/p^2\Bbb{Z}$ then $G$ is abelian by Lemma, so $G/Z(G)$ is isomorphic to $(\Bbb{Z}/p\Bbb{Z})^2$.