Inner Automorphisms form Normal Subgroup of Automorphism Group

Theorem
Let $G$ be a group.

Then the set $\operatorname {Inn} \left({G}\right)$ of all inner automorphisms of $G$ is a normal subgroup of the group of all automorphisms $\operatorname{Aut} \left({G}\right)$ of $G$:


 * $\operatorname{Inn} \left({G}\right) \triangleleft \operatorname{Aut} \left({G}\right)$

Proof
Let $G$ be a group whose identity is $e$.

Let $\kappa_x: G \to G$ be the inner automorphism defined such that $\forall g \in G: \kappa_x \left({g}\right) = x g x^{-1}$.

We see that $\operatorname{Inn} \left({G}\right) \ne \varnothing$ as $\kappa_x$ is defined for all $x \in G$.

We show that $\kappa_x, \kappa_y \in \operatorname{Inn} \left({G}\right): \kappa_x \circ \left({\kappa_y}\right)^{-1} \in \operatorname{Inn} \left({G}\right)$.

So:

As $x y^{-1} \in G$, it follows that $\kappa_{x y^{-1}} \in \operatorname {Inn} \left({G}\right)$.

By the One-Step Subgroup Test it follows that $\operatorname {Inn} \left({G}\right) \le \operatorname{Aut} \left({G}\right)$.

Now we need to show that $\operatorname {Inn} \left({G}\right)$ is normal in $\operatorname{Aut} \left({G}\right)$.

Let $\phi \in \operatorname{Aut} \left({G}\right)$.

If we can show that $\forall \phi \in \operatorname{Aut}(G): \forall \kappa_x \in \operatorname {Inn} \left({G}\right): \phi \circ \kappa_x \circ \phi^{-1} \in \operatorname{Inn} \left({G}\right)$, then by the Normal Subgroup Test, $\operatorname{Inn} \left({G}\right) \triangleleft \operatorname{Aut} \left({G}\right)$.

Fix $\kappa_x \in \operatorname{Inn} \left({G}\right)$.

We claim $\phi\circ\kappa_x\circ\phi^{-1} = \kappa_{\phi(x)}$.

Since $\phi \in \operatorname{Aut}(G)$ then $\phi$ is, in particular, a homomorphism.

Therefore:

Therefore $\phi\circ\kappa_g\circ\phi^{-1} = \kappa_{\phi(g)}\in \operatorname{Inn} \left({G}\right)$.

Since $\kappa_x\in \operatorname{Inn}(G)$ and $\phi \in \operatorname{Aut}(G)$ were arbitrary we have $\operatorname{Inn} \left({G}\right) \triangleleft \operatorname{Aut} \left({G}\right)$.