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
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 7.2$. Some lemmas on homomorphisms: Example $134$
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Problem $\text{AA}$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $25$