# Order-conjugate not implies order-dominated

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., order-conjugate subgroup) need not satisfy the second subgroup property (i.e., order-dominated subgroup)
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## Statement

It is possible to have a finite subgroup $H$ of a group $G$ such that $H$ is conjugate to any subgroup of $G$ of the same order as $H$, but such that there exists a subgroup $K$ of $G$ whose order is a multiple of the order of $H$, but such that $H$ is not conjugate to any subgroup of $K$.

## Proof

### Example of the alternating group of degree five

Further information: alternating group:A5, subgroup structure of alternating group:A5

Suppose $G$ is the alternating group on the set $\{ 1,2,3,4,5 \}$. Suppose $H$ is the subgroup of $G$ given by: $H := \{ (), (1,2,3), (1,3,2), (1,2)(4,5), (2,3)(4,5), (1,3)(4,5) \}$. $H$ is conjugate to any other subgroup of $G$ of the same order.

On the other hand, suppose $K$ is the stabilizer of $\{ 5 \}$ in $G$. Then, $K$ is the alternating group on $\{ 1,2,3,4 \}$ and is isomorphic to the alternating group of degree four. The order of $K$ is $12$, which is a multiple of the order of $H$. However, no conjugate of $H$ is contained in $K$, since $H$ does not stabilize any element, but every conjugate of $K$ stabilizes some element.