Let \(G\) be a \(p\)-group of order \(p^4\), and suppose \[ |Z(G)|=p^2. \] Determine the number of conjugacy classes of \(G\).
Solution
1. Every noncentral element of \(G\) has conjugacy class of size \(p\).
Let \(x\in G\sm Z(G)\). Since both \(Z(G)\) and \(x\) lie in the centralizer \(C_G(x)\), the subgroup generated by them satisfies \[ \langle Z(G),x\rangle\subseteq C_G(x). \] Because \(x\notin Z(G)\) and \(\abs{Z(G)}=p^2\), this subgroup has order strictly larger than \(p^2\), hence at least \(p^3\).
The centralizer cannot be all of \(G\), since that would make \(x\) central. As \(\abs{G}=p^4\), it follows that \[ \abs{C_G(x)}=p^3. \] Therefore the conjugacy class of \(x\) has size \[ \abs{x^G} = [G:C_G(x)] = p. \]
2. The number of conjugacy classes of \(G\) is\[ \boxed{p^3+p^2-p}. \]
Each of the \(p^2\) elements of \(Z(G)\) forms a singleton conjugacy class. The remaining \[ p^4-p^2 \] elements are noncentral, and by step 1 they occur in conjugacy classes of size \(p\). Hence the number of noncentral conjugacy classes is \[ \frac{p^4-p^2}{p} = p^3-p. \] Adding the \(p^2\) central classes gives \[ p^2+(p^3-p) = p^3+p^2-p. \]
3. Q.E.D.
Step 2 gives the required number of conjugacy classes.