# Sylow implies order-dominated

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., Sylow subgroup) must also satisfy the second subgroup property (i.e., order-dominated subgroup)
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## Statement

### 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. Sylow subgroups exist
2. 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 $G$.