\beginsolution Let $|G| = p^2$. The center $Z(G)$ is nontrivial by the class equation (since $|G| = |Z(G)| + \sum |G:C_G(g_i)|$, each term divisible by $p$). So $|Z(G)| = p$ or $p^2$.
\subsection*Exercise 4.7.14 \textitProve that if $G$ is a group of order $p^2$ where $p$ is prime, then $G$ is abelian. Dummit And Foote Solutions Chapter 4 Overleaf High Quality
\subsection*Exercise 4.1.1 \textitProve that every cyclic group is abelian. \beginsolution Let $|G| = p^2$