Condition for Group Endomorphism to Commute with All Inner Automorphisms

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Theorem

Let $G$ be a group.

Let $\phi: G \to G$ be an endomorphism on $G$.

Let $\phi$ be such that:

$\forall a \in G: \kappa_a \circ \phi = \phi \circ \kappa_a$

where:

$\kappa_a$ denotes the inner automorphism of $G$ given by $a$
$\circ$ denotes composition of mappings.


Then:

$H = \set {x \in G: \map \phi {\map \phi x} = \map \phi x}$

is a normal subgroup of $G$.

Also, the quotient group $G / H$ is an abelian group.


Proof

We have for all $a, g \in G$:

\(\ds a \map \phi g a^{-1}\) \(=\) \(\ds \map {\kappa_a} {\map \phi g}\) Definition of Inner Automorphism
\(\ds \) \(=\) \(\ds \map {\paren{\kappa_a \circ \phi} } g\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map {\paren{\phi \circ \kappa_a} } g\)
\(\ds \) \(=\) \(\ds \map \phi {\map {\kappa_a} g}\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map \phi {a g a^{-1} }\) Definition of Inner Automorphism

Let $x \in H$.

Then by the above:

\(\ds \map \phi {\map \phi {a x a^{-1} } }\) \(=\) \(\ds \map \phi {a \map \phi x a^{-1} }\)
\(\ds \) \(=\) \(\ds \map \phi a \map \phi {\map \phi x} \map \phi {a^{-1} }\) Definition of Group Endomorphism
\(\ds \) \(=\) \(\ds \map \phi a \map \phi x \map \phi {a^{-1} }\) $\map \phi x = \map \phi {\map \phi x}$
\(\ds \) \(=\) \(\ds \map \phi {a x a^{-1} }\) Definition of Group Endomorphism

Hence $a x a^{-1} \in H$, and thus $a H a^{-1} \subseteq H$.

By definition of a normal subgroup, $H$ is a normal subgroup of $G$.


It remains to show that the quotient group $G / H$ is an abelian group.

We need to show that for any $a, b \in G$, we have:

$a H b H = b H a H$

that is,

$a b H = b a H$

By Cosets are Equal iff Product with Inverse in Subgroup, this is equivalent to:

$\paren {b a}^{-1} a b = a^{-1} b^{-1} a b \in H$

We have:

\(\ds \map \phi {\map \phi {a^{-1} b^{-1} a b} }\) \(=\) \(\ds \map \phi {\map \phi {a^{-1} b^{-1} a} \map \phi b}\) Definition of Group Endomorphism
\(\ds \) \(=\) \(\ds \map \phi {a^{-1} \map \phi {b^{-1} } a \map \phi b}\) by commutativity condition above
\(\ds \) \(=\) \(\ds \map \phi {a^{-1} } \map \phi {\map \phi {b^{-1} } a \map \phi b}\) Definition of Group Endomorphism
\(\ds \) \(=\) \(\ds \map \phi {a^{-1} } \map \phi {\map \phi b^{-1} a \map \phi b}\) Group Homomorphism Preserves Inverses
\(\ds \) \(=\) \(\ds \map \phi {a^{-1} } \map \phi b^{-1} \map \phi a \map \phi b\) by commutativity condition above
\(\ds \) \(=\) \(\ds \map \phi {a^{-1} } \map \phi {b^{-1} } \map \phi a \map \phi b\) Group Homomorphism Preserves Inverses
\(\ds \) \(=\) \(\ds \map \phi {a^{-1} b^{-1} a b}\) Definition of Group Endomorphism

and thus $a^{-1} b^{-1} a b \in H$.

Hence the result.

$\blacksquare$


Sources

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