Show that every nilpotent group is solvable.
Solution
Suppose \(G\) is nilpotent. Then its lower central series \[ \gamma_1(G)=G, \qquad \gamma_{i+1}(G)=[\gamma_i(G),G] \] terminates: \[ \gamma_{c+1}(G)=1 \] for some \(c\).
Let \(G^{(i)}\) denote the derived series, \[ G^{(0)}=G, \qquad G^{(i+1)}=[G^{(i)},G^{(i)}]. \] We claim inductively that \[ G^{(i)}\le \gamma_{2^i}(G). \] For \(i=0\) this is clear. If it holds for \(i\), then \[ G^{(i+1)}=[G^{(i)},G^{(i)}] \le [\gamma_{2^i}(G),\gamma_{2^i}(G)] \le \gamma_{2^{i+1}}(G). \] Choose \(i\) with \(2^i>c\). Then \[ G^{(i)}\le \gamma_{2^i}(G)=1. \] Hence the derived series terminates, so \(G\) is solvable.