Inner Automorphisms form Subgroup of Automorphism Group

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Theorem

Let $G$ be a group.


Then the set $\Inn G$ of all inner automorphisms of $G$ forms a normal subgroup of the automorphism group $\Aut G$ of $G$:

$\Inn G \le \Aut G$


Proof

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

Let $\kappa_x: G \to G$ be the inner automorphism defined as:

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

We see that:

$\Inn G \ne \O$

as $\kappa_x$ is defined for all $x \in G$.


We show that:

$\kappa_x, \kappa_y \in \Inn G: \kappa_x \circ \paren {\kappa_y}^{-1} \in \Inn G$


So:

\(\ds \forall g \in G: \, \) \(\ds \map {\paren {\kappa_x \circ \paren {\kappa_y}^{-1} } } g\) \(=\) \(\ds \map {\kappa_x} {\map {\paren {\kappa_y}^{-1} } g}\)
\(\ds \) \(=\) \(\ds \map {\kappa_x} {\map {\kappa_{y^{-1} } } g}\) Inverse of Inner Automorphism
\(\ds \) \(=\) \(\ds \map {\kappa_x} {y^{-1} g y}\)
\(\ds \) \(=\) \(\ds x y^{-1} g y x^{-1}\)
\(\ds \) \(=\) \(\ds \paren {x y^{-1} } g \paren {x y^{-1} }^{-1}\) Inverse of Product
\(\ds \) \(=\) \(\ds \map {\kappa_{x y^{-1} } } g\)

As $x y^{-1} \in G$, it follows that:

$\kappa_{x y^{-1} } \in \Inn G$

By the One-Step Subgroup Test:

$\Inn G \le \Aut G$

$\blacksquare$


Sources

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