Inner Automorphism is Automorphism

Theorem
Let $G$ be a group.

Let $x \in G$.

Let $\kappa_x$ be the inner automorphism of $x$ in $G$.

Then $\kappa_x$ is an automorphism of $G$.

Proof
By definition, $\kappa_x: G \to G$ is a mapping defined as:
 * $\forall g \in G: \kappa_x \left({g}\right) = x g x^{-1}$

We need to show that $\kappa_x$ is an automorphism.


 * First we show $\kappa_x$ is a homomorphism.

Thus the morphism property is demonstrated.


 * Next we show that $\kappa_x$ is injective.

So $\kappa_x$ is injective.


 * Finally we show that $\kappa_x$ is surjective.

Note that $\forall h \in G: x^{-1} h x \in G$ from fact that $G$ is a group and therefore closed. So:

Thus every element of $G$ is the image of some element of $G$ under $\kappa_x$ (that is, of $x^{-1} h x$), and surjectivity is proved.

Comment
Note that performing an inner automorphism of a subgroup results in a conjugate subgroup.