## 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.