Mapping from Group Element to Inner Automorphism is Homomorphism

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Theorem

Let $G$ be a group.

Let $\kappa: G \to \Aut G$ be the mapping from $G$ to the automorphism group of $G$ defined as:

$\forall x \in G: \map \kappa x := \kappa_x$

where $\kappa_x$ is the inner automorphism on $x$:

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


Then $\kappa$ is a homomorphism.


Proof

Let $x, y \in G$.

By definition of automorphism group, we have that:

$\map \kappa x \map \kappa y = \kappa_x \circ \kappa_y$

where $\circ$ denotes composition of mappings.


Then $\forall g \in G$:

\(\ds \map {\kappa_x \circ \kappa_y} g\) \(=\) \(\ds \map {\kappa_x} {\map {\kappa_y} g}\) for all $g \in G$
\(\ds \) \(=\) \(\ds \map {\kappa_x} {y g y^{-1} }\) Definition of $\kappa_y$
\(\ds \) \(=\) \(\ds x \paren {y g y^{-1} } x^{-1}\) Definition of $\kappa_x$
\(\ds \) \(=\) \(\ds \paren {x y} g \paren {y^{-1} x^{-1} }\) Group Axiom $\text G 1$: Associativity
\(\ds \) \(=\) \(\ds \paren {x y} g \paren {x y}^{-1}\) Inverse of Group Product
\(\ds \) \(=\) \(\ds \map {\kappa_{x y} } g\) Definition of $\kappa_{x y}$

And so:

$\forall g \in G: \map {\kappa_{x y} } g = \map {\kappa_x \circ \kappa_y} g$

Thus by definition of $\kappa$:

$\map \kappa x \map \kappa y = \map \kappa{x y}$

demonstrating that $\kappa$ is a homomorphism.

$\blacksquare$


Sources

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