Sylow implies order-dominating

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a Finite group (?), every subgroup satisfying the first subgroup property (i.e., Sylow subgroup (?)) must also satisfy the second subgroup property (i.e., Order-dominating subgroup (?)). In other words, every Sylow subgroup of finite group is a order-dominating subgroup of finite group.
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Statement

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 $g^{- 1} Q g \leq P$ .

This statement is a part of Sylow's theorem.

Related facts

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

Proof using coset spaces

Given: $G$ a finite group, $P$ a $p$ -Sylow subgroup, and $Q$ a $p$ -group. Suppose $n = p^{r} m$ 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 $g^{- 1} Q g \leq P$ .

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 $g P$ .
We have $Q g P = g P$ . This gives us: $(g^{- 1} Q g) P = P$ , and hence $g^{- 1} Q g \subseteq P$ . Thus, a conjugate of $Q$ is contained inside $P$ .