Normal-potentially characteristic implies normal-extensible automorphism-invariant

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal-potentially characteristic subgroup) must also satisfy the second subgroup property (i.e., normal-extensible automorphism-invariant subgroup)
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Statement

Verbal statement

Any normal-potentially characteristic subgroup of a group is a normal-extensible automorphism-invariant subgroup.

Definitions used

Normal-potentially characteristic subgroup

Further information: Normal-potentially characteristic subgroup

A subgroup $H\leq G$ is normal-potentially characteristic if there exists a group $K$ containing $G$ as a normal subgroup such that $H$ is a characteristic subgroup of $K$.

Normal-extensible automorphism-invariant subgroup

Further information: Normal-extensible automorphism-invariant subgroup

An automorphism $\sigma$ of a group $G$ is termed a normal-extensible automorphism if, whenever $K$ is a group containing $G$ as a normal subgroup, there exists an automorphism $\sigma '$ of $K$ whose restriction to $G$ equals $\sigma$. A normal-extensible automorphism-invariant subgroup is a subgroup invariant under all normal-extensible automorphisms.

Intermediate properties

• Normal-potentially relatively characteristic subgroup: For full proof, refer: Normal-potentially characteristic implies normal-potentially relatively characteristic, normal-potentially relatively characteristic implies normal-extensible automorphism-invariant

Proof

Given: $H\leq G$. There exists a group $K$ such that $G$ is normal in $K$ and $H$ is characteristic in $K$. $\sigma$ is a normal-extensible automorphism of $G$.

To prove: $\sigma (H)=H$.

Proof: Since $\sigma$ is normal-extensible, there exists $\sigma '\in \operatorname {Aut} (K)$ such that the restriction of $\sigma '$ to $G$ equals $\sigma$. Since $H$ is characteristic in $K$, $\sigma '(H)=H$, and hence, $\sigma (H)=H$.