Proof that groups of order p squared are abelian

Proof

We show that the centre is the whole group. Let $e$ be the identity of $G$, with $\abs{G} = p^2$.

1. Each positive divisor of $p^2$ is $1$, $p$ or $p^2$: a prime factor of a divisor divides $p$ by Euclid's lemma, so it equals $p$.
2. Let $g \in G$. The order of $g$ divides $\abs{G}$, so it is $1$, $p$ or $p^2$ by step 1. So $G$ is a $p$-group, and $G \neq \set{e}$.
3. By the nontrivial centre theorem, the centre $Z(G)$ is not $\set{e}$.
4. The centre is a subgroup of $G$, by the subgroup criterion. Indeed $e \in Z(G)$. If $z, w \in Z(G)$ and $g \in G$, then $w g = g w$ gives $w^{-1} g = g w^{-1}$, so $z w^{-1} g = z g w^{-1} = g z w^{-1}$. By Lagrange's theorem and step 1, $\abs{Z(G)}$ is $1$, $p$ or $p^2$. By step 3, it is $p$ or $p^2$.
5. Suppose $\abs{Z(G)} = p$. For $z \in Z(G)$ and $g \in G$, $g z g^{-1} = z g g^{-1} = z$. So the centre is a normal subgroup, and the quotient group $G/Z(G)$ exists. Its order is the index $[G : Z(G)]$.
6. By the index formula, $[G : Z(G)] = p^2 / p = p$. So $G/Z(G)$ has prime order $p$, and it is cyclic.
7. Since $G/Z(G)$ is cyclic, $G$ is abelian. Then each element commutes with all of $G$, so $Z(G) = G$ and $\abs{Z(G)} = p^2 \neq p$. This contradicts the assumption of step 5.
8. Hence $\abs{Z(G)} = p^2 = \abs{G}$, so $Z(G) = G$. Every element commutes with every other, and $G$ is abelian.