Center of Group of Order Prime Cubed

Theorem
Let $G$ be a group of order $p^3$, where $p$ is a prime.

Let $Z \left({G}\right)$ be the center of $G$.

Then $\left|{Z \left({G}\right)}\right| \ne p^2$.

Proof
Suppose $\left|{Z \left({G}\right)}\right| = p^2$.

Then $\left|{G / Z \left({G}\right)}\right| = p$.

From Prime Group is Cyclic it follows that $G / Z \left({G}\right)$ is cyclic.

From Quotient of Group by Center Cyclic implies Abelian it follows that $G$ is abelian.

From Group equals Center iff Abelian it follows that $\left|{Z \left({G}\right)}\right| = p^3$.

The result follows.