Prove that the centre of a group of order \(p^r\) (\(p\) prime) is not trivial.
Solution
Let \(|G|=p^r\). Under conjugation, the class equation is \[ |G|=|Z(G)|+\sum_i [G:C_G(x_i)], \] where the \(x_i\) represent the noncentral conjugacy classes.
1. Every noncentral conjugacy-class size is divisible by \(p\).
For noncentral \(x_i\), the centralizer \(C_G(x_i)\) is a proper subgroup of the \(p\)-group \(G\). Hence \[ [G:C_G(x_i)]>1 \] and, being a divisor of \(|G|=p^r\), it is a positive power of \(p\). Therefore it is divisible by \(p\).
2. The order \(|Z(G)|\) is divisible by \(p\).
Reducing the class equation modulo \(p\) and using 1 gives \[ 0\equiv |G|\equiv |Z(G)|\pmod p. \] Thus \(p\mid |Z(G)|\).
3. Therefore \(Z(G)\) is nontrivial.
Since \(p\mid |Z(G)|\), one has \(|Z(G)|\ge p>1\). Hence \(Z(G)\ne\{e\}\).