# 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 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 $\mathbb{Z}$.