Let \(G\) be a group of order \(p^3\), where \(p\) is prime. Show that \(Z(G)\) is nontrivial, and hence that \(G\) has a nontrivial normal subgroup.
Solution
Let \(G\) act on itself by conjugation. The class equation is \[ |G|=|Z(G)|+\sum_i [G:C_G(x_i)], \] where the \(x_i\) represent the noncentral conjugacy classes.
For noncentral \(x_i\), the centralizer \(C_G(x_i)\) is a proper subgroup of the \(p\)-group \(G\), so \[ [G:C_G(x_i)] \] is divisible by \(p\). Hence every term in the sum is divisible by \(p\). Since \(|G|=p^3\) is divisible by \(p\), the class equation gives \[ |Z(G)|\equiv0\pmod p. \] But the identity lies in \(Z(G)\), so \(|Z(G)|\ge1\). Therefore \(|Z(G)|\ge p\), and \(Z(G)\) is nontrivial.
The center is characteristic in \(G\), hence normal. Thus \(G\) has a nontrivial normal subgroup.