Every Group of Order p^2 is Abelian

# Every Group of Order p^2 is Abelian

Recall from the Burnside's Theorem for p-Groups page that Burnside's Theorem for p-Groups states that if $G$ is a $p$-group then $Z(G)$ is nontrivial. The following proposition is an application of this theorem.

 Proposition 1: Let $p$ be a prime and let $G$ be a group of order $p^2$. Then $G$ is abelian.
• Proof: Let $G$ be a group of order $p^2$ for some prime $p$. Suppose instead that $G$ is nonabelian. Then $G \neq Z(G)$.
• Let $g \in G \setminus Z(G)$. Consider the centralizer:
(1)
\begin{align} \quad C_G(g) = \{ h \in G : gh = hg \} \end{align}
(2)
• Since $Z(G)$ is a subgroup of $G$, by Lagrange's Theorem we have that $|Z(G)| \mid |G| = p^2$. So $|Z(G)| \in \{ 1, p, p^2 \}$. Furthermore, since $G$ is a $p$-group, by Burnside's Theorem for $p$-Groups we have that $Z(G)$ is nontrivial, and so $|Z(G)| \neq 1$. Also, by the assumption that $G$ is nonabelian we have that $|Z(G)| \neq p^2$. So $|Z(G)| = p$.
• By $(*)$, since $Z(G)$ is group of order $p$ that is a proper subgroup of $C_G(g)$, and since $C_G(g)$ is a subgroup of $G$, we must have that $|C_G(g)| = p^2$, i.e., $C_G(g) = G$. So for all $h \in G$ we have that $gh = hg$, i.e., $g \in Z(G)$ - a contradiction.
• Thus the assumption that $G \neq Z(G)$ is false. Thus $G = Z(G)$, i.e., $G$ is abelian. $\blacksquare$