Pi-dominating pi-subgroup implies pi-Hall

Statement
Suppose $$G$$ is a finite group, $$H$$ is a subgroup of $$G$$, and $$\pi$$ is a set of primes such that:


 * The set of prime divisors of the order of $$H$$ is in $$\pi$$.
 * Given any $$\pi$$-subgroup $$K$$ of $$G$$ (i.e., any subgroup for which all prime divisors of its order are in $$\pi$$), there exists $$g \in G$$ such that $$gKg^{-1} \le H$$.

Then, $$H$$ is a $$\pi$$-Hall subgroup: in particular, its index is relatively prime to $$\pi$$, or equivalently, its order is the unique largest $$\pi$$-number dividing the order of $$G$$.

Note that this also shows that a subgroup is $$\pi$$-dominating for a set of primes $$\pi$$ iff it is an order-dominating Hall subgroup.

Facts used

 * 1) uses::Sylow subgroups exist
 * 2) uses::Lagrange's theorem

Proof
Given: A finite group $$G$$, a $$\pi$$-dominating subgroup $$H$$.

To prove: $$H$$ is $$\pi$$-Hall.

Proof: It suffices to show that for every $$p \in \pi$$, the largest power of $$p$$ dividing the order of $$G$$ also divides the order of $$H$$.

For this, let $$P$$ be a $$p$$-Sylow subgroup (existence follows from fact (1)). The order of $$P$$ is the largest power of $$p$$ dividing the order of $$G$$. By the assumption, some conjugate $$gPg^{-1}$$ is in $$H$$. The order of $$gPg^{-1}$$ equals the order of $$P$$, and by fact (2), this divides the order of $$H$$. Thus, the order of $$H$$ is a multiple of the largest power of $$p$$ dividing the order of $$G$$, and this completes the proof.