Order-conjugate not implies order-dominating

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-dominating subgroup)
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Definition

We can have a group $G$ with an order-conjugate subgroup $H$ (i.e., a subgroup $H$ that is conjugate to any other subgroup of the same order) that is not an order-dominating subgroup: in other words, there exists a subgroup $K$ of $G$ whose order divides the order of $H$ , but such that $K$ is not contained in any conjugate of $H$ .

Related facts

Proof

Example of the alternating group of degree five

Further information: alternating group:A5

Suppose $G$ is the alternating group on the set $S = {1, 2, 3, 4, 5}$ . Suppose $H$ is the subgroup of $G$ that is the alternating group on ${1, 2, 3, 4}$ . In other words, $H$ is the stabilizer of the point ${5}$ . Then, $H$ is order-conjugate in $G$ : the subgroups of the same order as $H$ are precisely the stabilizers of points in $S$ , and these are conjugate to $H$ by suitable $5$ -cycles.

On the other hand, consider the subgroup $K$ :

$K : = {(), (1, 2, 3), (1, 3, 2), (1, 2) (4, 5), (2, 3) (4, 5), (1, 3) (4, 5)}$ .

$K$ is a group of order six, isomorphic to the symmetric group of degree three. However, $K$ is not contained in any conjugate of $H$ , because any conjugate of $H$ stabilizes some element, and $K$ does not stabilize any element.