Sylow implies order-dominating

Property-theoretic statement
The subgroup property of being a Sylow subgroup is stronger than, or implies, the subgroup property of being an order-dominating subgroup: for any subgroup whose order divides the order of the Sylow subgroup, some conjugate of that subgroup is contained in the given Sylow subgroup.

Statement with symbols
Suppose $$G$$ is a finite group and $$p$$ is a prime number. Then, if $$P$$ is a $$p$$-Sylow subgroup and $$Q$$ is a $$p$$-subgroup of $$G$$, there exists $$g \in G$$ such that $$g^{-1}Qg \le P$$.

This statement is a part of Sylow's theorem.

Corollaries

 * Sylow implies order-conjugate: This follows from the fact that order-dominating implies order-conjugate in co-Hopfian: in particular, in a finite group, any order-dominating subgroup is order-conjugate. In other words, any two $$p$$-Sylow subgroups are conjugate.

Proof using coset spaces
Given: $$G$$ a finite group, $$P$$ a $$p$$-Sylow subgroup, and $$Q$$ a $$p$$-group. Suppose $$n=p^rm$$ where $$n$$ is the order of $$G$$ and $$m$$ is relatively prime to $$p$$ (So, $$|P| = p^r$$.

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

Proof: We prove this through a series of observations:
 * $$G$$ naturally acts (on the left) on the left coset space of $$P$$.
 * Since $$Q$$ is a subgroup of $$G$$, $$Q$$ also acts on the left coset space of $$P$$
 * The left coset space of $$P$$ has cardinality $$m$$, which is relatively prime to $$p$$. Hence, under the action of $$Q$$ (which is a $$p$$-group) on this set, there is at least one fixed point. Let the fixed point be $$gP$$.
 * We have $$QgP = gP$$. This gives us: $$(g^{-1}Qg)P = P$$, and hence $$g^{-1}Qg \subseteq P$$. Thus, a conjugate of $$Q$$ is contained inside $$P$$.