Inner Automorphisms form Normal Subgroup of Automorphism Group

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $G$ be a group.


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

$\Inn G \lhd \Aut G$


Proof

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

From Inner Automorphisms form Subgroup of Automorphism Group, $\Inn G$ forms a subgroup of $\Aut G$.

It remains to be shown that $\Inn G$ is normal in $\Aut G$.


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

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


Let $\phi \in \Aut G$.

If we can show that:

$\forall \phi \in \Aut G: \forall \kappa_x \in \Inn G: \phi \circ \kappa_x \circ \phi^{-1} \in \Inn G$

then by the Normal Subgroup Test:

$\Inn G \lhd \Aut G$

Fix $\kappa_x \in \Inn G$.

We claim $\phi \circ \kappa_x \circ \phi^{-1} = \kappa_{\map \phi x}$.

Since $\phi \in \Aut G$ then $\phi$ is, in particular, a homomorphism.

Therefore:

\(\ds \forall g \in G: \, \) \(\ds \map {\paren {\phi \circ \kappa_x \circ \phi^{-1} } } g\) \(=\) \(\ds \map {\paren {\phi \circ \kappa_x} } {\map {\phi^{-1} } g}\)
\(\ds \) \(=\) \(\ds \map \phi {x \phi^{-1} \paren g x^{-1} }\)
\(\ds \) \(=\) \(\ds \map \phi x \map \phi {\map {\phi^{-1} } g} \map \phi {x^{-1} }\)
\(\ds \) \(=\) \(\ds \map \phi x g \map \phi x^{-1}\)
\(\ds \) \(=\) \(\ds \map {\kappa_{\map \phi x} } g\)

Therefore:

$\phi \circ \kappa_g \circ \phi^{-1} = \kappa_{\map \phi g} \in \Inn G$

Since $\kappa_x \in \Inn G$ and $\phi \in \Aut G$ were arbitrary:

$\Inn G \lhd \Aut G$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Inner_Automorphisms_form_Normal_Subgroup_of_Automorphism_Group&oldid=565275"