Normalizer of isomorph-conjugate implies isomorph-dominating

Statement

Suppose $G$ is a group and $H$ is an Isomorph-conjugate subgroup (?) of $G$: any subgroup of $G$ isomorphic to $H$ is also conjugate to $H$ within $G$. Then, the Normalizer (?) $N_{G}(H)$ of $H$ in $G$ is isomorph-dominating in $G$: if $K\leq G$ is isomorphic to $N_{G}(H)$, there exists $g\in G$ such that $K\leq gN_{G}(H)g^{-1}$.

Related facts

Corollaries

• Isomorph-conjugacy is normalizer-closed in finite: In a finite group, any isomorph-dominating subgroup is also isomorph-conjugate, so the normalizer of any isomorph-conjugate subgroup is isomorph-conjugate.

Note that the normalizer of an isomorph-conjugate subgroup in an infinite group is not necessarily isomorph-conjugate. For instance, consider the group of integers. The normalizer of the trivial subgroup is the whole group, which is not isomorph-conjugate: there are proper subgroups of the group of integers isomorphic to it.

Proof

(This proof uses the left-action convention).

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

To prove: If $K\leq G$ is a subgroup such that $K\leq N_{G}(H)$, then there exists $g\in G$ such that $K\leq gN_{G}(H)g^{-1}$.

Proof: Let $\sigma :N_{G}(H)\to K$ be an isomorphism. Then, $H$ and $\sigma (H)$ are isomorphic groups. Thus, there exists $g\in G$ such that $gHg^{-1}=\sigma (H)$.

Since taking normalizer commutes with automorphisms, we get $gN_{G}(H)g^{-1}=N_{G}(\sigma (H))$. Now, since $H$ is normal in $N_{G}(H)$, $\sigma (H)$ is normal in $\sigma (N_{G}(H))=K$. In particular, $K$ is contained in the normalizer of $\sigma (H)$, so $K\leq N_{G}(\sigma (H))=gN_{G}(H)g^{-1}$.