Show that every group of order \(p^2\) is abelian and classify all such groups.
Solution
Let \(|G|=p^2\). The class equation for a finite \(p\)-group gives \[ p\mid |Z(G)|, \] so \(|Z(G)|=p\) or \(p^2\).
If \(|Z(G)|=p\), then \(G/Z(G)\) has order \(p\) and is cyclic. If \(G/Z(G)\) is cyclic, then \(G\) is abelian: if \(G/Z(G)=\langle gZ(G)\rangle\), every element has the form \(g^az\) with \(z\in Z(G)\), and any two such elements commute. This would imply \(Z(G)=G\), contradicting \(|Z(G)|=p\). Hence \[ Z(G)=G, \] so \(G\) is abelian.
If \(G\) has an element of order \(p^2\), then it is cyclic: \[ G\cong C_{p^2}. \] Otherwise every nonidentity element has order \(p\). Choose \(x\ne1\) and \(y\notin\langle x\rangle\). Then \[ \langle x\rangle\cap\langle y\rangle=1, \] and, since \(G\) is abelian, \[ G=\langle x\rangle\times\langle y\rangle\cong C_p\times C_p. \] Thus the two isomorphism types are \[ C_{p^2}\quad\text{and}\quad C_p^2. \]