Does a nontrivial finite solvable group have a subgroup of prime power index for each prime divisor? It is well-known that every maximal subgroup of $G$ is of prime power index if $G$ is a nontrivial finite solvable group.
My question is: Can we prove that for each prime  $r\in\pi(G)$ there exists a maximal subgroup of $G$ of index a power of $r$?
I tried to prove it but I found that I made a mistake in my proof. Here is my attempt:
Define $$\pi^*:=\{r\in\pi(G)\mid~\mbox{There is no maximal subgroup }H\mbox{ of }G\mbox{ such that }|G:H|\mbox{ is a power of }r\}.$$
We claim that $\pi^*$ is an empty set.
Assume that $\pi^*$ is non-empty. Then the indices of the maximal subgroups are exactly powers of primes in $\pi(G)\setminus\pi^*$. Take a Sylow $q$-subgroup $S_q$ for each $q\in\pi(G)$. For $p\in\pi(G)\setminus\pi^*$, take an arbitrary maximal subgroup $M$ of $G$ such that $|G:M|$ is a power of $p$. We have
$$\left|\prod_{q\in\pi(G)\setminus\pi^*}S_q\right|_p=|G|_p>|M|_p.$$
It implies that $\prod\limits_{q\in\pi(G)\setminus\pi^*}S_q$ is not contained in any maximal subgroup of $G$. But $\prod\limits_{q\in\pi(G)\setminus\pi^*}S_q$ is properly contained in $G$, which is a contradiction.
My mistake: $\prod\limits_{q\in\pi(G)\setminus\pi^*}S_q$ is not necessarily a subgroup of $G$, so in fact I cannot get any contradiction.
Could you give me some ideas? I think maybe I should prove it in a different way. Any help is appreciated. Thanks!
 A: This is Hall's theorem on soluble groups. It states:
A finite group is soluble if and only if, for each $p\mid |G|$, there exists a $p'$-subgroup $H$ whose index is a power of $p$.
A subgroup $H$ such that $|H|$ and $|G:H|$ are coprime is called a Hall subgroup, and if $\pi$ is a set of primes such that $p\in \pi$ divides $|G|$ if and only if it divides $|H|$, then $H$ is a Hall $\pi$-subgroup.
Proving this without hints is a little bit of a challenge. You can either look it up in your favourite textbook, or follow the outline below for the one direction. Let $\pi$ be a set of primes, and we aim to prove the existence of a Hall $\pi$-subgroup in $G$.

*

*Let $K$ be a minimal normal subgroup of $G$. If $K$ is a $\pi'$-subgroup then everything is done.

*If $K$ is a $p$-subgroup for $p\in \pi$, then you can use the Schur-Zassenhaus theorem to the preimage of a Hall $\pi$-subgroup in $G/K$.

You can find a full proof here, p.28.
A: Yes, for every set of primes the finite solvable group contains a Hall subgroup whose order is divisible only by these primes and index is not dividible by any of them. Now take the set of all primes which divide the order of the group but one. A corresponding Hall subgroup is what you need.
https://en.m.wikipedia.org/wiki/Hall_subgroup
