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 $G$:


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

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

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

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

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:
 * $\forall \phi \in \operatorname{Aut} \left({G}\right): \forall \kappa_x \in \operatorname {Inn} \left({G}\right): \phi \circ \kappa_x \circ \phi^{-1} \in \operatorname{Inn} \left({G}\right)$

then by the Normal Subgroup Test:
 * $\operatorname{Inn} \left({G}\right) \lhd \operatorname{Aut} \left({G}\right)$

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

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

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

Therefore:

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

Since $\kappa_x \in \operatorname{Inn} \left({G}\right)$ and $\phi \in \operatorname{Aut} \left({G}\right)$ were arbitrary:
 * $\operatorname{Inn} \left({G}\right) \lhd \operatorname{Aut} \left({G}\right)$