Inner automorphism to automorphism is right tight for normality

Statement

Property-theoretic statement

Consider the subgroup property of being a Normal subgroup (?). Then, the following function restriction expression for this subgroup property:

Inner automorphism $\to$ Automorphism

Statement with symbols

Suppose $G$ is a group and $\sigma:G \to G$ is a function. Consider the following property for $\sigma$:

There exists a group $K$ containing $G$ as a normal subgroup, and an inner automorphism $\alpha$ of $K$ such that the restriction of $\alpha$ of $G$.

This property is precisely equivalent to $\sigma$ being an automorphism. (Note that by definition of normality, any such $\sigma$ must be an automorphism, so the content of the theorem is that every automorphism can be realized this way).

Facts used

1. Every group is normal fully normalized in its holomorph: If $G$ is a group, there is a group $\operatorname{Hol}(G)$ (the holomorph of $G$) containing $G$ as a normal subgroup, with the property that every automorphism of $G$ extends to an inner automorphism in $\operatorname{Hol}(G)$.

Related facts

Other facts using this idea or similar proofs

• Left transiter of normal is characteristic: If $H \le K$ is such that whenever $K$ is normal in $G$, $H$ is also normal in $G$, then $H$ is characteristic in $K$. This fact follows from the statement on this page and the right tightness theorem.
• Every group is normal fully normalized in its holomorph
• Normal upper-hook fully normalized implies characteristic

Other related facts

If we drop the requirement that $G$ be normal in $K$, then the result is no longer true. In fact, we have:

Proof

Proof using the holomorph

The proof using the holomorph finds a single big group that works for all automorphisms.

Given: A group $G$, an automorphism $\sigma$ of $G$.

To prove: There exists a group $K$ containing $G$ as a normal subgroup, such that $\sigma$ extends to an inner automorphism of $K$.

Proof: Let $K$ be the holomorph of $G$. Fact (1) now completes the proof.

Proof using semidirect product with just that automorphism

This proof uses a more minimalistic construction, but the construction differs for every automorphism. This proof technique generalizes to proving that every injective endomorphism arises as the restriction of an inner automorphism.

Given: A group $G$, an automorphism $\sigma$ of $G$.

To prove: There exists a group $K$ containing $G$ as a normal subgroup, such that $\sigma$ extends to an inner automorphism of $K$.

Proof: Consider the group $K := G \rtimes \mathbb{Z}$, where the generator of $\mathbb{Z}$ acts on $G$ by $\sigma$. $G$ is normal in the semidirect product by definition, and the automorphism $\sigma$ extends to the inner automorphism of $K$ induced by conjugation by that generator of $\mathbb{Z}$.