Groups of order $p^2$

Center is nontrivial. By the class equation, $|Z(G)|\equiv|G|\equiv0\pmod p$, so $p$ divides $|Z(G)|$.

Case 1: $Z(G)=G$. Done, $G$ abelian.

Case 2: $|Z(G)|=p$. Then $|G/Z(G)|=p$, so $G/Z(G)$ is cyclic (any group of prime order is cyclic). But if $G/Z(G)$ is cyclic, then $G$ is abelian (a standard lemma: if $G/Z(G)$ is cyclic, $G$ is abelian).

Contradiction. But $G$ abelian implies $Z(G)=G$, contradicting $|Z(G)|=p$.

Conclusion. Case 2 is impossible, so $G$ is abelian.