# Inner Automorphisms form Subgroup of Automorphism Group

## 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:

$\Inn G \le \Aut G$

$\blacksquare$