Property of Group Automorphism which Fixes Identity Only/Corollary 3

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Corollary to Property of Group Automorphism which Fixes Identity Only

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 $e$.


Let:

$\phi^2 = I_G$

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

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


Proof

Let $s, t \in G$.

Then:

\(\ds \map \phi s \, \map \phi t\) \(=\) \(\ds \map \phi {s t}\)
\(\ds \) \(=\) \(\ds \paren {s t}^{-1}\) Corollary 2
\(\ds \) \(=\) \(\ds t^{-1} s^{-1}\)
\(\ds \) \(=\) \(\ds t s\)

$\blacksquare$


Aiming for a contradiction, suppose $G$ is of even order.

Then from Even Order Group has Order 2 Element:

\(\ds \exists y \in G: \, \) \(\ds y^2\) \(=\) \(\ds e\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds y^{-1}\)
\(\ds \leadsto \ \ \) \(\ds \map \phi y\) \(=\) \(\ds y\)


But this contradicts the condition on $\phi$:

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

Hence there is no such element of $G$ of order $2$.

Thus $G$ must be of odd order.

$\blacksquare$


Sources

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