Exercise 3B.4 on page 84 of Isaacs's Finite Group Theory asks one to prove a precursor of the Sylow theorems for $π$-subgroups, and I only know how to do it using the Sylow theorems, so I have missed the point. I actually needed a similar (false) statement in my research, so I thought it would be wise to get the point.
Suppose $G$ is a finite group, and $N$ is a normal subgroup whose index and order are coprime. Let $U$ be a subgroup of $G$ whose order is coprime to the order of $N$. If either N or $U$ is solvable, then show that $U$ is contained in some subgroup $H$ whose order is the index of $N$ in $G$.
Available tools: We already have Schur–Zassenhaus, and so we know that $G$ has a subgroup whose order is the index of $N$.
In other words, if $G$ has a normal Hall $π$-subgroup, then every $π′$-subgroup is contained in some Hall $π′$-subgroup. However Hall subgroups are the next section, and containment and $π$-separability are even in the section after that, so appealing to Hall's generalization of Sylow's theorem will miss the point.
Attempt: Let $M$ be a maximal subgroup of $G$ containing $U$. If $N$ is not contained in $M$, then $M∩N$ is a normal subgroup of $M$ whose order is coprime to its index, and so we can find $H ≤ M$ with $U ≤ H$ and $|H| = [M:M∩N]=[G:N]$, and we are done.
If $N$ is contained in every maximal subgroup of $G$ containing $U$, then $\ldots$ . It'd be nice if $N$ was in the Frattini, but that's not true. Perhaps at this point it would be wise to use solvability?