# Schur–Zassenhaus exercise: coprime subgroup contained in a complement

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?

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Let $K$ be a complement to $N$ in $G$, which exists by Schur-Zassenhaus. Then $|K|=|G|/|N|$. Consider the subgroup $UN$. Since $K(UN)=G$, $|K\cap UN|=|K|\cdot |UN|/|G|=|K|\cdot|U|\cdot|N|/|G|=|U|$, since $K$ is a complement to $N$. Thus, $K\cap UN$ is a complement to $N$ in $UN$. Since either $U$ or $N$ is solvable, all complements to $N$ in $UN$ are conjugate, so $K\cap UN=U^g$. Thus, $K^{g^{-1}}$ contains $U$.