# 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 $G$.

## 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 $g$.