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., semi-strongly 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 semi-strongly potentially characteristic subgroup of a group is a normal-extensible automorphism-invariant subgroup.

Definitions used

Semi-strongly potentially characteristic subgroup

Further information: Semi-strongly potentially characteristic subgroup

A subgroup $H\le G$ is semi-strongly 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 ${\sigma }^{\prime }$ 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

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

Proof

Given: $H\le G$. 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 ${\sigma }^{\prime }\in \mathrm{A}\mathrm{u}\mathrm{t}(K)$ such that the restriction of ${\sigma }^{\prime }$ to $G$ equals $\sigma$. Since $H$ is characteristic in $K$, ${\sigma }^{\prime }(H)=H$, and hence, $\sigma (H)=H$.