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:
 * $\paren {\kappa_x}^{-1} = \kappa_{x^{-1} }$

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

Let $x \in G$.

Let $\kappa_x \in \Inn G$.

Then from the definition of inner automorphism:
 * $\forall g \in G: \map {\kappa_x} g = x g x^{-1}$

As $G$ is a group:
 * $x \in G \implies x^{-1} \in G$

So:
 * $\kappa_{x^{-1} } \in \Inn G$

and is defined as:
 * $\forall g \in G: \map {\kappa_{x^{-1} } } g = x^{-1} g \paren {x^{-1} }^{-1} = x^{-1} g x$

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:
 * $\forall g \in G: \map {\kappa_x \circ \kappa_{x^{-1} } } g = \map {I_G} g = \map {\kappa_{x^{-1} } \circ \kappa_x} g$

Hence the result.