Does a nontrivial finite solvable group have a subgroup of prime power index for each prime divisor?

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 $G$.

1. Let $K$ be a minimal normal subgroup of $G$. If $K$ is a $\pi'$-subgroup then everything is done.
2. 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.

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