Isomorph-conjugate subgroup

This article defines a subgroup property: a property that can be evaluated to true/false given a group and a subgroup thereof, invariant under subgroup equivalence. View a complete list of subgroup properties[SHOW MORE]

BEWARE! This term is nonstandard and is being used locally within the wiki. [SHOW MORE]

Definition

Symbol-free definition

A subgroup of a group is termed isomorph-conjugate if any subgroup isomorphic to that subgroup is conjugate to it in the whole group.

Definition with symbols

A subgroup $H$ of a group $G$ is termed isomorph-conjugate if whenever there is a subgroup $K$ isomorphic to $H$, $H$ and $K$ are conjugate subgroups in $G$.

In terms of the relation implication operator

The property of being isomorph-conjugate can be viewed in terms of the relation implication operator with the relation on the left being that of a subgroup pair being isomorphic and the relation on the right being that of a subgroup pair being conjugate in the whole group.

Relation with other properties

Conjunction with other properties

• Any normal isomorph-conjugate subgroup is isomorph-free. Thus, this subgroup property is normal-to-characteristic

Metaproperties

Transitivity

NO: This subgroup property is not transitive: a subgroup with this property in a subgroup with this property, need not have the property in the whole group
ABOUT THIS PROPERTY: View variations of this property that are transitive|View variations of this property that are not transitive
ABOUT TRANSITIVITY: View a complete list of subgroup properties that are not transitive|View facts related to transitivity of subgroup properties | View a survey article on disproving transitivity

An isomorph-conjugate subgroup of an isomorph-conjugate subgroup is not necessarily isomorph-conjugate. For full proof, refer: Isomorph-conjugacy is not transitive

Intersection-closedness

This subgroup property is not intersection-closed, viz., it is not true that an intersection of subgroups with this property must have this property.
Read an article on methods to prove that a subgroup property is not intersection-closed

An intersection of two isomorph-conjugate subgroups of a group need not be isomorph-conjugate. For full proof, refer: Isomorph-conjugacy is not intersection-closed

Testing

GAP code

One can write code to test this subgroup property in GAP (Groups, Algorithms and Programming), though there is no direct command for it.
View the GAP code for testing this subgroup property at: IsIsomorphConjugateSubgroup
View other GAP-codable subgroup properties | View subgroup properties with in-built commands
GAP-codable subgroup property

This subgroup property can be tested using GAP code, though there is no direct built-in GAP function for it. The GAP code is available at GAP:IsIsomorphConjugateSubgroup, and is invoked as follows:

IsIsomorphConjugateSubgroup(group,subgroup);