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