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:

\(\ds \forall g \in G: \, \)

\(\ds \map {\kappa_x \circ \kappa_{x^{-1} } } g\)

\(=\)

\(\ds \map {\kappa_x} {\map {\kappa_{x^{-1} } } g}\)

Definition of Composition of Mappings

\(\ds \)

\(=\)

\(\ds \map {\kappa_x } {x^{-1} g x}\)

Definition of $\kappa_{x^{-1} }$

\(\ds \)

\(=\)

\(\ds x \paren {x^{-1} g x} x^{-1}\)

Definition of $\kappa_x$

\(\ds \)

\(=\)

\(\ds g\)

Group Properties

\(\ds \)

\(=\)

\(\ds x^{-1} \paren {x g x^{-1} } x\)

Group Properties

\(\ds \)

\(=\)

\(\ds \map {\kappa_{x^{-1} } } {x g x^{-1} }\)

Definition of $\kappa_{x^{-1} }$

\(\ds \)

\(=\)

\(\ds \map {\kappa_{x^{-1} } } {\map {\kappa_x} g}\)

Definition of $\kappa_x$

\(\ds \)

\(=\)

\(\ds \map {\kappa_{x^{-1} } \circ \kappa_x} g\)

Definition of Composition of Mappings

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.

$\blacksquare$