Property of Group Automorphism which Fixes Identity Only

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Theorem

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

Let $\phi: G \to G$ be a group automorphism.

Let $\phi$ have the property that:

$\forall g \in G \setminus \set e: \map \phi t \ne t$

That is, the only fixed element of $\phi$ is $e$.


Then:

$\forall x, y \in G: x^{-1} \, \map \phi x = y^{-1} \, \map \phi y \implies x = y$


Corollary $1$

$\forall x \in G: \exists g \in G: x = g^{-1} \, \map \phi g$


Corollary $2$

Let:

$\phi^2 = I_G$

where $I_G$ denotes the identity mapping on $G$.

Then:

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


Corollary $3$

Let:

$\phi^2 = I_G$

where $I_G$ denotes the identity mapping on $G$.

Then $G$ is an abelian group of odd order.


Proof

\(\ds x^{-1} \, \map \phi x\) \(=\) \(\ds y^{-1} \, \map \phi y\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(=\) \(\ds x y^{-1} \, \map \phi y\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x \paren {\map \phi y}^{-1}\) \(=\) \(\ds x y^{-1}\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x \, \map \phi {y^{-1} }\) \(=\) \(\ds x y^{-1}\)
\(\ds \leadsto \ \ \) \(\ds \map \phi {x y^{-1} }\) \(=\) \(\ds x y^{-1}\)
\(\ds \leadsto \ \ \) \(\ds x y^{-1}\) \(=\) \(\ds e\) Definition of $\phi$
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds y\)

$\blacksquare$


Sources

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