Every maximal subgroup of a finite solvable group has prime power index

Prove that every maximal subgroup of a finite solvable group has prime power index.


Let G be a finite solvable group. We will proceed by induction on the breadth k of G; that is, the number of prime factors in the factorization of |G|, including multiplicity.

For the base case, if k = 1, we have G \cong Z_p for some prime p. Clearly then the (unique) maximal subgroup of G, namely 1, has prime power index.

For the inductive step, suppose that every finite solvable group of breadth at most k satisfies the conclusion. Let G be a finite solvable group of breadth k+1.

If G is simple, then G^\prime = 1 (since G is solvable). Thus G is an abelian simple group, and we have G \cong Z_p for some prime p; this is a contradiction.

Suppose now that G is not simple, so that nontrivial normal subgroups exist. H \leq G be a maximal subgroup and let M \leq G be a minimal nontrivial normal subgroup. Now G/M is a finite solvable group of breadth at most k. There are two cases:

If M \leq H, then H/M \leq G/M is maximal by the Lattice Isomorphism Theorem. By the Third Isomorphism Theorem and the induction hypothesis, [G:H] = [G/M : H/M] is a prime power.

If M \not\leq H, then G = MH. Now [G:H] = [HM:H] = |HM|/|H| = (|H| \cdot |M|/|H \cap M|)/|H| = |M|/|H \cap M| is a prime power, since M is a p-group.

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