Show that every finite \(p\)-group is nilpotent.
Solution
Let \(G\) be a finite \(p\)-group. We prove by induction on \(|G|\) that its upper central series reaches \(G\).
1. Every nontrivial finite \(p\)-group has nontrivial center.
The class equation is \[ |G|=|Z(G)|+\sum_i [G:C_G(x_i)], \] where the sum runs over representatives of noncentral conjugacy classes. Each index in the sum is a nontrivial power of \(p\), hence divisible by \(p\). Since \(|G|\) is divisible by \(p\), so is \(|Z(G)|\). Thus \(|Z(G)|\ge p\).
2. The quotient \(G/Z(G)\) is a strictly smaller finite \(p\)-group.
By 1, \(|Z(G)|\ge p\), so \[ |G/Z(G)|=|G|/|Z(G)|<|G|. \]
3. The upper central series of \(G\) reaches \(G\).
Induct on \(|G|\). The trivial group is nilpotent. For nontrivial \(G\), the quotient \(G/Z(G)\) is nilpotent by induction. If its upper central series reaches \(G/Z(G)\) in \(c\) steps, then by the definition \[ Z_{i+1}(G)/Z(G)=Z_i(G/Z(G)) \] for the corresponding shifted series, so \(Z_{c+1}(G)=G\). Hence \(G\) is nilpotent.
Therefore every finite \(p\)-group is nilpotent.