- Show that \(G\) is solvable iff its derived series terminates.
Solution
Write \[ G^{(0)}=G,\qquad G^{(i+1)}=[G^{(i)},G^{(i)}] \] for the derived series.
1. If the derived series terminates at the identity, then \(G\) is solvable.
Suppose \[ G=G^{(0)}\trianglerighteq G^{(1)}\trianglerighteq\cdots\trianglerighteq G^{(r)}=1. \] Each \(G^{(i+1)}\) is characteristic in \(G^{(i)}\), hence normal there. Moreover, \[ G^{(i)}/G^{(i+1)} \] is abelian because the commutator subgroup of \(G^{(i)}\) is exactly \(G^{(i+1)}\). Thus the derived series itself is a finite normal series with abelian factors, so \(G\) is solvable.
2. Conversely, suppose \(G\) is solvable. Then its derived series terminates.
Choose a finite normal series \[ 1=G_r\trianglelefteq G_{r-1}\trianglelefteq\cdots\trianglelefteq G_0=G \] whose factors \(G_i/G_{i+1}\) are abelian. Since \(G_i/G_{i+1}\) is abelian, \[ [G_i,G_i]\subseteq G_{i+1}. \] We prove by induction that \[ G^{(i)}\subseteq G_i. \] It is true for \(i=0\). If \(G^{(i)}\subseteq G_i\), then \[ G^{(i+1)}=[G^{(i)},G^{(i)}]\subseteq [G_i,G_i]\subseteq G_{i+1}. \] Hence \(G^{(r)}\subseteq G_r=1\), so \(G^{(r)}=1\).
Therefore \(G\) is solvable if and only if its derived series reaches the identity after finitely many steps.