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
If $\left|{Z \left({G}\right)}\right| = p^2$, then $\left|{G / Z \left({G}\right)}\right| = p$ and is therefore cyclic.

Thus $G$ is abelian from Cyclic Quotient Group of Center, and $\left|{Z \left({G}\right)}\right| = p^3$ from Group is Abelian iff Center Equals Group.

The result follows.