- Show that nilpotent groups have nontrivial centers.
Solution
Let \(G\neq1\) be nilpotent. Using the lower central series \[ \gamma_1(G)=G,\qquad \gamma_{i+1}(G)=[\gamma_i(G),G], \] choose the nilpotency class \(c\), so that \[ \gamma_c(G)\neq1,\qquad \gamma_{c+1}(G)=1. \] Then \[ [\gamma_c(G),G]=1, \] so every element of \(\gamma_c(G)\) commutes with every element of \(G\). Thus \[ 1\neq\gamma_c(G)\le Z(G), \] and therefore \(Z(G)\neq1\).