Let \(G\) be a finite group. Prove that \(G\) is solvable if and only if all of its composition factors are of prime order (i.e. cyclic groups \(\mathbb{Z}_p\)).
Solution
Suppose first that \(G\) is finite and solvable. Every subgroup and quotient of a solvable group is solvable, so every composition factor of \(G\) is both simple and solvable. If \(S\) is simple and solvable, then \([S,S]\trianglelefteq S\) and solvability forces \([S,S]\ne S\); hence \([S,S]=1\), so \(S\) is abelian. A finite simple abelian group is cyclic of prime order. Thus every composition factor of \(G\) is \(C_p\) for some prime \(p\).
Conversely, if a composition series \[ 1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_r=G \] has factors \(G_{i+1}/G_i\) of prime order, then every factor is cyclic and hence abelian. Therefore this is a subnormal series with abelian factors, so \(G\) is solvable.
Thus a finite group is solvable if and only if all of its composition factors have prime order.