Conjugacy-closedness is transitive

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This article gives the statement, and possibly proof, of a subgroup property (i.e., conjugacy-closed subgroup) satisfying a subgroup metaproperty (i.e., transitive subgroup property)
View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about conjugacy-closed subgroup |Get facts that use property satisfaction of conjugacy-closed subgroup | Get facts that use property satisfaction of conjugacy-closed subgroup|Get more facts about transitive subgroup property


Statement

Statement with symbols

Suppose H \le K \le G are groups such that H is conjugacy-closed in K and K is conjugacy-closed in G.

Related facts

Proof

Hands-on proof

Given: Groups H \le K \le G such that H is conjugacy-closed in K and K is conjugacy-closed in G.

To prove: H is conjugacy-closed in G.

Proof: We need to show that if a,b \in H are conjugate in G, then they are conjugate in H. First, observe that since  H \le K, a,b are elements of K conjugate in G. Since K is conjugacy-closed in G, a,b are conjugate in K.

Thus, a,b are elements of H that are conjugate in K. Since H is conjugacy-closed in K, the elements a,b are conjugate in H.