Show that if \(G\) is finite and solvable, then every composition factor of \(G\) has prime order.
Solution
Let \(S\) be a composition factor of the finite solvable group \(G\).
1. The group \(S\) is simple and solvable.
A composition factor is simple by definition. Subgroups and quotients of solvable groups are solvable, so every composition factor of \(G\) is solvable.
2. A nontrivial simple solvable group is abelian.
The derived subgroup \([S,S]\) is characteristic, hence normal, in \(S\). Simplicity gives \([S,S]=1\) or \(S\). Solvability excludes \([S,S]=S\), so \([S,S]=1\) and \(S\) is abelian.
3. A finite simple abelian group has prime order.
Choose a prime \(p\mid |S|\). By Cauchy’s theorem there is \(x\in S\) of order \(p\). Since \(S\) is abelian, \(\langle x\rangle\) is normal; simplicity forces \(\langle x\rangle=S\). Hence \(|S|=p\).
Thus every composition factor is cyclic of prime order.