Inner Automorphism is Automorphism

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Theorem

Let $G$ be a group.

Let $x \in G$.

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


Then $\kappa_x$ is an automorphism of $G$.


Proof

By definition, $\kappa_x: G \to G$ is a mapping defined as:

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


We need to show that $\kappa_x$ is an automorphism.


First we show $\kappa_x$ is a homomorphism.

\(\ds \forall g, h \in G: \, \) \(\ds \map {\kappa_x} g \map {\kappa_x} h\) \(=\) \(\ds \paren {x g x^{-1} } \paren {x h x^{-1} }\) Definition of $\kappa_x$
\(\ds \) \(=\) \(\ds x g \paren {x^{-1} x} h x^{-1}\) Group Axiom $\text G 1$: Associativity
\(\ds \) \(=\) \(\ds x \paren {g e h} x^{-1}\) Group Axiom $\text G 3$: Existence of Inverse Element
\(\ds \) \(=\) \(\ds x \paren {g h} x^{-1}\) Group Axiom $\text G 2$: Existence of Identity Element
\(\ds \) \(=\) \(\ds \map {\kappa_x} {g h}\) Definition of $\kappa_x$


Thus the morphism property is demonstrated.


Next we show that $\kappa_x$ is injective.

\(\ds \map {\kappa_x} g\) \(=\) \(\ds \map {\kappa_x} h\)
\(\ds \leadsto \ \ \) \(\ds x g x^{-1}\) \(=\) \(\ds x h x^{-1}\) Definition of $\kappa_x$
\(\ds \leadsto \ \ \) \(\ds g\) \(=\) \(\ds h\) Cancellation Laws


So $\kappa_x$ is injective.


Finally we show that $\kappa_x$ is surjective.

Note that $\forall h \in G: x^{-1} h x \in G$ from fact that $G$ is a group and therefore closed. So:

\(\ds \forall h \in G: \, \) \(\ds \map {\kappa_x} {x^{-1} h x}\) \(=\) \(\ds x \paren {x^{-1} h x} x^{-1}\) Definition of $\kappa_x$
\(\ds \) \(=\) \(\ds \paren {x x^{-1} } h \paren {x x^{-1} }\) Group Axiom $\text G 1$: Associativity
\(\ds \) \(=\) \(\ds e h e\) Group Axiom $\text G 3$: Existence of Inverse Element
\(\ds \) \(=\) \(\ds h\) Group Axiom $\text G 2$: Existence of Identity Element


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Thus every element of $G$ is the image of some element of $G$ under $\kappa_x$ (that is, of $x^{-1} h x$), and surjectivity is proved.

Since $\kappa_x$ is injective and surjective, it is bijective.

Since $\kappa_x$ is a homomorphism that is bijective, it is a (group) isomorphism.

Hence, $\kappa_x: G \to G$ is a (group) isomorphism from $G$ to itself, which is by definition, an automorphism of $G$.

$\blacksquare$


Also see


Sources

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