Order-conjugate not implies order-dominating

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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.