I found this theorem in Dummit and Foote, and there was no proof of it there. It looks difficult to prove and I also could not find any resources online to help me out with this theorem. So here is the theorem in full:
The finite group $G$ is solvable if and only if for every divisor $n$ of $|G|$such that $(n,|G|/n) = 1$, $G$ has a subgroup of order $n$.
I would be grateful for the proof.