[Math] every maximal subgroup of a finite solvable group has prime power index.

Question

If $H$ is a maximal proper subgroup of a finite solvable group $G$, then $[\,G:H\,]$ is a prime power.

$G$ is solvable, so we can consider the minimal normal subgroup $N$ in $G$.

I got the hint:

Apply induction to the quotient group $G/N$ and consider separately the two cases $N\le H$ and $N\nleq H$

But I still have no idea for this hint to follow. Is there anyone can give me more direction to proceed? any suggestion will be appreciated.Thanks for considering my request.

Answer 1

Hint:

Proof by induction. Remark that $N$ is not trivial so the cardinal of $G/H$ is strictly inferior to the cardinal of $G$. Let $p:G\rightarrow G/N$ the canonical projection.

1. Suppose that $N\subset H$, $p(H)$ is a normal subgroup of $G/H$ and is maximal, so $[G:H]=[G/N:H/N]$ is prime, (see the reference)

if $N$ is not contained in $H$, remark that the cardinal of $N$ is prime, and $G$ is generated by $G$ and $n, n\in N$ we have $G=H\bigcup nH,...,n^{p-1}H$

Every minimal normal subgroup of a finite solvable group is elementary abelian