Hall retract implies order-conjugate

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., Hall retract) must also satisfy the second subgroup property (i.e., order-conjugate subgroup)
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This article states and (possibly) proves a fact that involves two finite groups of relatively prime order, requiring the additional datum that at least one of them is solvable. Due to the Feit-Thompson theorem, we know that for two finite groups of relatively prime orders, one of them is solvable. Hence, the additional datum of solvability can be dropped. However, the proof of the Feit-Thompson theorem is considered heavy machinery.
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Name

This result is often stated as the conjugacy part of the Schur-Zassenhaus theorem.

Statement

Suppose $G$ is a finite group and $H$ is a Hall retract of $G$. In other words, $H$ is a Hall subgroup of $G$ that is also a retract: there exists a normal complement $N$ to $H$ in $G$. Note that $N$ is thus a normal Hall subgroup. Assume, further, that either $N$ or $H$ is a solvable group.

Then, if there is any subgroup $K$ of $G$ of the same order as $H$, $H$ and $K$ are conjugate subgroups.

Note that the assumption that either $N$ or $H$ is a solvable group is superfluous because, as a corollary of the odd-order theorem, given two groups of coprime order, one of them is solvable.

Proof

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