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 Center of Abelian Group is Whole Group.

The result follows.