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 $H$, $K\leq H$.
2. For any group $L$ containing $G$, $H$ is weakly closed in $G$ relative to $L$.

Related facts

• Same order iff potentially conjugate
• Isomorphic iff potentially conjugate
• Left transiter of normal is characteristic
• Inner automorphism to automorphism is right tight for normality
• Every injective endomorphism arises as the restriction of an inner automorphism
• Subisomorph-containing iff strongly closed in any ambient group

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 $gHg^{-1}\leq G$, we have $gHg^{-1}\leq H$.

Proof: Suppose $g\in L$ is such that $gHg^{-1}\leq G$. Then, $gHg^{-1}$ is isomorphic to $H$, because conjugation by $g$ is an automorphism. In particular, $gHg^{-1}\leq H$ 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 $K\leq H$.

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 $gHg^{-1}=K\leq G$. Since $H$ is weakly closed in $G$ relative to $L$, we get $gHg^{-1}\leq H$, forcing $K\leq H$.