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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\kappa_a \circ \phi} g\) | Definition of Composition of Mappings | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\phi \circ \kappa_a} g\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {\map {\kappa_a} g}\) | 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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.21$