Groups of order p squared are abelian

Statement

Let $p$ be a prime number and let $G$ be a group of order $p^2$. Then $G$ is abelian.

Remarks

By the nontrivial centre theorem and Lagrange's theorem, $|Z(G)|$ is $p$ or $p^2$. If $|Z(G)| = p$, then $G/Z(G)$ has order $p$, so it is cyclic, and this criterion gives a contradiction. Thus $Z(G) = G$, and $G$ is isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$ or to $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$. For $p^3$ the result is false: the dihedral group $D_4$ and the quaternion group have order $8$ and are not abelian.

Explanations

L1 human

A group of order $p^2$ has a centre with more than one element. If the centre were not the whole group, the quotient by the centre would be cyclic, and that is impossible. So all elements commute.