Maximal subgroups of solvable groups have prime power index

[Barre]

I would like to ask if somebody can verify the solution I wrote up to an exercise in my book. It's kind of long, but I have no one else to check it for me :)

Homework Statement

If $H$ is a maximal proper subgroup of a finite solvable group $G$, then $[G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />]$ is a prime power.

Homework Equations

Lemma 7.13 that I refer to is basically this:
http://crazyproject.wordpress.com/2...-finite-solvable-group-is-elementary-abelian/

The Attempt at a Solution

Statement is true for abelian groups, so only nonabelian solvable groups are considered in the proof. All finite nonabelian solvable groups have at least one normal group (the commutator) and therefore contain a minimal normal subgroup. By Lemma 7.13 (iii), these subgroups have prime power order.
Assume there exists minimal normal subgroup $N$ such that $N \not\subseteq H$. Since $NH$ is a subgroup of $G$ properly containing $H$, $NH = G$. By Second Isomorphism Theorem, $G/N = HN/N \cong H/(N \cap H)$ and as all cardinalities are finite, $\frac{|G|}{|N|} = \frac{|H|}{|N \cap H|}$ which implies $\frac{|G|}{|H|} = [G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />] = \frac{|N|}{|N \cap H|}$, where the last one is a prime power.
Assume all minimal normal subgroups of $G$ are contained in $H$ and let $N$ be one such. We work by induction. If $|G| = 2$, then it is of prime order and the maximal proper subgroup $\langle e \rangle$ has index 2, certainly a prime power. In the quotient $G/N$, $H/N$ is maximal since if $H/N$ is properly contained in some subgroup $T/N$ then it follows that $H \subseteq T$, so $T = G$. Since order $|G/N|$ is strictly less than $|G|$, $[G/N<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />/N] = \frac{[G/N]}{[H/N]} = \frac{[G:N]}{[H:N]} = p^n$ by induction. Also the identity $[G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />][H:N] = [G:N]$ implies $[G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />] = \frac{[G:N]}{[H:N]}$ so $[G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />] = p^n$.

 

[micromass]

Seems ok.

I also note that your proof also works in the abelian case, so there is no need to only look at nonabelian groups.

 

[Barre]

Seems ok.

I also note that your proof also works in the abelian case, so there is no need to only look at nonabelian groups.

Thanks for reading it and for the tip. An abelian group certainly has a normal minimal subgroup as long as it is not simple, and that is probably the case I could have handled separately.