Show that every cyclic group is solvable.
Solution
Every cyclic group is abelian. Hence its commutator subgroup is trivial: \[ [G,G]=1. \] Therefore the derived series reaches the identity after one step, so \(G\) is solvable.
E-AMD-KGEXBKIW
Show that every cyclic group is solvable.
Every cyclic group is abelian. Hence its commutator subgroup is trivial: \[ [G,G]=1. \] Therefore the derived series reaches the identity after one step, so \(G\) is solvable.