Conjugacy-closedness is transitive

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)
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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

• Conjugacy-closed normality is transitive
• Central factor is transitive

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 $H$.