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 $g^{-1}Qg \le P$.

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 $n=p^rm$ 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 $P$.