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

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, $$x \in G \Longrightarrow x^{-1} \in G$$.

So $$\kappa_{x^{-1}} \in \mathrm {Inn} \left({G}\right)$$ 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.