Inner Automorphisms form Normal Subgroup of Automorphism Group

Theorem
Let $G$ be a group.

Then the set $\Inn G$ of all inner automorphisms of $G$ is a normal subgroup of the group of all automorphisms $\Aut G$ of $G$:


 * $\Inn G \lhd \Aut G$

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

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

We see that:
 * $\Inn G \ne \O$

as $\kappa_x$ is defined for all $x \in G$.

We show that:
 * $\kappa_x, \kappa_y \in \Inn G: \kappa_x \circ \paren {\kappa_y}^{-1} \in \Inn G$

So:

As $x y^{-1} \in G$, it follows that:
 * $\kappa_{x y^{-1}} \in \Inn G$

By the One-Step Subgroup Test:
 * $\Inn G \le \Aut G$

Now we need to show that $\Inn G$ is normal in $\Aut G$.

Let $\phi \in \Aut G$.

If we can show that:
 * $\forall \phi \in \Aut G: \forall \kappa_x \in \Inn G: \phi \circ \kappa_x \circ \phi^{-1} \in \Inn G$

then by the Normal Subgroup Test:
 * $\Inn G \lhd \Aut G$

Fix $\kappa_x \in \Inn G$.

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

Since $\phi \in \Aut G$ then $\phi$ is, in particular, a homomorphism.

Therefore:

Therefore:
 * $\phi \circ \kappa_g \circ \phi^{-1} = \kappa_{\phi \paren g} \in \Inn G$

Since $\kappa_x \in \Inn G$ and $\phi \in \Aut G$ were arbitrary:
 * $\Inn G \lhd \Aut G$