Inverse of Inner Automorphism

Theorem
Let $G$ be a group.

Let $x \in G$.

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

Then:
 * $\left({\kappa_x}\right)^{-1} = \kappa_{x^{-1}}$

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

Let $x \in G$.

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

Then from the definition of inner automorphism, $\forall g \in G: \kappa_x \left({g}\right) = x g x^{-1}$.

As $G$ is a group, $x \in G \implies x^{-1} \in G$.

So $\kappa_{x^{-1}} \in \operatorname{Inn} \left({G}\right)$ and is defined as:

$\forall g \in G: \kappa_{x^{-1}} \left({g}\right) = x^{-1} g \left({x^{-1}}\right)^{-1} = x^{-1} g x$.

Now we need to show that $\kappa_x \circ \kappa_{x^{-1}} = I_G = \kappa_{x^{-1}} \circ \kappa_x$, where $I_G: G \to G$ is the identity mapping.

So:

Thus:
 * $\forall g \in G: \kappa_x \circ \kappa_{x^{-1}} \left({g}\right) = I_G \left({g}\right) = \kappa_{x^{-1}} \circ \kappa_x \left({g}\right)$

and the proof is finished.