# Isomorph-containing iff weakly closed in any ambient group

This article gives a proof/explanation of the equivalence of multiple definitions for the term isomorph-containing subgroup
View a complete list of pages giving proofs of equivalence of definitions

## The definitions that we have to prove as equivalent

The following are equivalent for a subgroup $H$ of a group $G$:

1. $H$ is an isomorph-containing subgroup of $G$: For any subgroup $K$ of $G$ isomorphic to $H$, $K \le H$.
2. For any group $L$ containing $G$, $H$ is weakly closed in $G$ relative to $L$.

## Facts used

1. Isomorphic iff potentially conjugate

## Proof

### Isomorph-containing implies weakly closed in any ambient group ((1) implies (2))

Given: A group $G$, a subgroup $H$ that is isomorph-containing. A group $L$ containing $G$.

To prove: $H$ is weakly closed in $G$ relative to $L$: for any $g \in L$ such that $gHg^{-1} \le G$, we have $gHg^{-1} \le H$.

Proof: Suppose $g \in L$ is such that $gHg^{-1} \le G$. Then, $gHg^{-1}$ is isomorphic to $H$, because conjugation by $g$ is an automorphism. In particular, $gHg^{-1} \le H$ because $H$ is isomorph-containing in $G$, and we are done.

### Weakly closed in ambient group implies isomorph-containing ((2) implies (1))

Given: A group $G$. A subgroup $H$ of $G$ such that $H$ is weakly closed in $G$ relative to any group $L$ containing $G$.

To prove: If $K$ is a subgroup of $G$ isomorphic to $H$, then $K \le H$.

Proof: Let $\sigma:H \to K$ be an isomorphism. By fact (1), there exists a group $L$ containing $G$ and an element $g \in L$ such that conjugation by $g$ induces $\sigma$ on $H$. In particular, we get $gHg^{-1} = K \le G$. Since $H$ is weakly closed in $G$ relative to $L$, we get $gHg^{-1} \le H$, forcing $K \le H$.