Groups of order p^2 are abelian

Proposition. Let $G$ be a finite group with $|G|=p^2$ for a prime $p$ . Then $G$ is abelian.

Proof

As a finite $p$ -group, $G$ has nontrivial center $Z(G)$ . If $|Z(G)|=p^2$ , then $Z(G)=G$ . Otherwise $|Z(G)|=p$ , so $G/Z(G)$ has order $p$ and is cyclic by prime-order implies cyclic. A group whose quotient by its center is cyclic is abelian: if $G/Z(G)$ is generated by $gZ(G)$ , write $x=g^a z_1$ and $y=g^b z_2$ with $z_1,z_2\in Z(G)$ ; then $xy=yx$ . Thus $G$ is abelian in either case.

Remarks

A common strategy for classifying groups of small order is to show that the center is nontrivial and then pass to the quotient by the center.