Prove that every maximal subgroup of a finite solvable group has prime power index.
Let be a finite solvable group. We will proceed by induction on the breadth of ; that is, the number of prime factors in the factorization of , including multiplicity.
For the base case, if , we have for some prime . Clearly then the (unique) maximal subgroup of , namely 1, has prime power index.
For the inductive step, suppose that every finite solvable group of breadth at most satisfies the conclusion. Let be a finite solvable group of breadth .
If is simple, then (since is solvable). Thus is an abelian simple group, and we have for some prime ; this is a contradiction.
Suppose now that is not simple, so that nontrivial normal subgroups exist. be a maximal subgroup and let be a minimal nontrivial normal subgroup. Now is a finite solvable group of breadth at most . There are two cases:
If , then is maximal by the Lattice Isomorphism Theorem. By the Third Isomorphism Theorem and the induction hypothesis, is a prime power.
If , then . Now is a prime power, since is a -group.