Automorph-conjugacy is normalizer-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., automorph-conjugate subgroup) satisfying a subgroup metaproperty (i.e., normalizer-closed subgroup property)
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Statement

Suppose $G$ is a group and $H$ is an automorph-conjugate subgroup of $G$, then $N_{G}(H)$ (the normalizer of $H$ in $G$) is also an automorph-conjugate subgroup of $G$.

Related facts

Generalizations

• Automorphism-based relation-implication-expressible implies normalizer-closed: This is a general formalism result stating that any subgroup property that can be expressed in a given format is closed under taking normalizers.

Proof

Proof idea

The key idea behind the proof is that taking the normalizer commutes with automorphisms. In other words, for any automorphism $\sigma$, we have $\sigma (N_{G}(H))=N_{G}(\sigma (H))$.

Proof details

Given: A group $G$, an automorph-conjugate subgroup $H$.

To prove: $N_{G}(H)$ is an automorph-conjugate subgroup: given any automorphism $\sigma$ of $G$, $N_{G}(H)$ and $\sigma (N_{G}(H))$ are conjugate.

Proof: $\sigma (N_{G}(H))=N_{G}(\sigma (H))$. By assumption, $H$ is automorph-conjugate, so there exists a $g\in G$ such that $\sigma (H)=gHg^{-1}$. Thus, $\sigma (N_{G}(H))=N_{G}(gHg^{-1})=gN_{G}(H)g^{-1}$, and thus, $N_{G}(H)$ and $\sigma (N_{G}(H))$ are conjugate by $g$.