Show that a finite group \(G\) is solvable if and only if all of its composition factors are cyclic groups of prime order.
Solution
Let \(G\) be finite.
Suppose first that \(G\) is solvable. Every quotient and subgroup of a solvable group is solvable, so every composition factor of \(G\) is a finite simple solvable group. If \(S\) is simple and solvable, then its commutator subgroup \([S,S]\) is a normal subgroup of \(S\). It cannot equal \(S\), since then the derived series would never descend, so \([S,S]=1\). Thus \(S\) is abelian. A finite simple abelian group is cyclic of prime order.
Conversely, suppose every composition factor of \(G\) is cyclic of prime order. A composition series \[ 1=G_0\trianglelefteq G_1\trianglelefteq\cdots\trianglelefteq G_r=G \] then has abelian successive quotients \(G_{i+1}/G_i\). Hence it is a subnormal series with abelian factors, so \(G\) is solvable.
Therefore a finite group is solvable exactly when all of its composition factors are cyclic of prime order.