Hall retract implies order-conjugate

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
We reduce the problem to basic cases by induction on the order.

Given: A finite group $$G$$. A normal $$\pi$$-Hall subgroup $$N$$ of $$G$$, and two $$\pi'$$Hall subgroups $$H,K$$ of $$G$$.

To prove: $$H$$ is conjugate to $$K$$.

Proof:


 * 1) Reduction to the case that $$N$$ is minimal normal in $$G$$: Suppose $$M$$ is a nontrivial normal subgroup of $$G$$ properly contained in $$N$$. Let $$\varphi:G \to G/M$$ be the quotient map. $$\varphi(N)$$ is normal $$\pi$$-Hall in $$G/N$$, and $$\varphi(H), \varphi(K)$$ are $$\pi'$$-Hall subgroups. By the inductive hypothesis, $$\varphi(H)$$ and $$\varphi(K)$$ are conjugate in $$G/N$$. Suppose $$g \in G$$ is such that $$\varphi(g)$$ conjugates $$\varphi(H)$$ to $$\varphi(K)$$. Then, $$g$$ conjugates $$H$$ to some $$\pi'$$-Hall subgroup inside $$\varphi^{-1}(\varphi(K)) = KM$$. Thus, it suffices to show that any $$\pi'$$-Hall subgroup of $$KM$$ is conjugate to $$K$$. This again follows by the inductive hypothesis on the order, since $$M$$ is normal $$\pi$$-Hall in $$KM$$.
 * 2) Reduction to the case that $$O_{\pi'}(G)$$ is trivial; in other words, $$G$$ has no nontrivial normal $$\pi'$$-subgroups: Suppose $$L$$ is a nontrivial normal $$\pi'$$-subgroup of $$G$$. Then, by the product formula, $$LH$$ and $$LK$$ are $$\pi'$$-subgroups, and since $$H,K$$ are Hall, this forces $$L \le H$$ and $$L \le K$$. Consider the quotient map $$\alpha:G \to G/L$$. Under this quotient map, $$\alpha(N)$$ is a normal $$\pi$$-Hall subgroup, and $$\alpha(H), \alpha(K)$$ are $$\pi'$$-Hall. The inductive hypothesis yields that $$\alpha(H)$$ and $$\alpha(K)$$ are conjugate in $$G/L$$. Since $$H$$ and $$K$$ both contain the kernel of $$\alpha$$, this yields that $$H$$ and $$K$$ are conjugate in $$G$$.
 * 3) Resolution of the case where $$H$$ is solvable:
 * 4) Resolution of the case where $$N$$ is solvable: