Sylow implies order-dominated

Statement with symbols
Suppose $$G$$ is a finite group and $$P$$ is a $$p$$-Sylow subgroup of $$G$$. Suppose $$H$$ is a subgroup of $$G$$ such that the order of $$H$$ is a multiple of the order of $$P$$ (equivalently, the index of $$H$$ is relatively prime to $$P$$). Then, there exists $$g \in G$$ such that $$gPg^{-1} \le H$$.

Facts used

 * 1) uses::Sylow subgroups exist
 * 2) uses::Sylow implies order-conjugate: Any two $$p$$-Sylow subgroups in a finite group are conjugate.

Proof
Given: A finite group $$G$$, a $$p$$-Sylow subgroup $$P$$ of $$G$$. A subgroup $$H$$ of $$G$$ whose order is a multiple of the order of $$P$$.

To prove: There exists $$g \in G$$ such that $$gPg^{-1} \le H$$.

Proof:


 * 1) $$H$$ has a $$p$$-Sylow subgroup, say $$Q$$: This follows from fact (1).
 * 2) $$Q$$ is also a $$p$$-Sylow subgroup of $$G$$: This follows from order considerations. Since the order of $$P$$ divides the order of $$H$$, the largest power of $$p$$ dividing the order of $$H$$ is the same as the largest power of $$p$$ dividing the order of $$G$$. Thus, a $$p$$-Sylow subgroup of $$H$$ has the correct order for being a $$p$$-Sylow subgroup of $$G$$.
 * 3) There exists $$g \in G$$ such that $$gPg^{-1} = Q$$. In particular, $$gPg^{-1} \le H$$: This follows from fact (2), and the previous step, which established that both $$P$$ and $$Q$$ are $$p$$-sylow in $$G$$.