Show that if \(G\) is finite and solvable, then every composition factor of \(G\) has prime order.
Solution
Let \[ 1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_r=G \] be a composition series. Each factor \[ G_i/G_{i-1} \] is simple by definition of a composition series.
Because \(G\) is solvable, every subgroup and every quotient of \(G\) is solvable. Hence each composition factor is both simple and solvable.
A nontrivial simple solvable group must be abelian: if \(H\) is solvable and nonabelian, then its derived subgroup \(H'=[H,H]\) is a proper nontrivial normal subgroup, contradicting simplicity. Thus every composition factor is a finite simple abelian group.
A finite simple abelian group is cyclic of prime order. Indeed, if \(A\) is finite abelian and \(1\ne a\in A\), then \(\langle a\rangle\normal A\); simplicity forces \(A=\langle a\rangle\), so \(A\) is cyclic. A finite cyclic group is simple exactly when its order is prime.
Therefore every composition factor of \(G\) has prime order.