Inversion Mapping is Isomorphism from Ordered Abelian Group to its Dual

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Theorem

Let $\struct {G, \circ, \preccurlyeq}$ be an ordered Abelian group.

Let $\struct {G, \circ, \succcurlyeq}$ be the dual of $\struct {G, \circ, \preccurlyeq}$.


Let $\phi: \struct {G, \circ, \preccurlyeq} \to \struct {G, \circ, \succcurlyeq}$ be the inversion mapping from $\struct {G, \circ, \preccurlyeq}$ to $\struct {G, \circ, \succcurlyeq}$ defined as:

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

where $x^{-1}$ denotes the inverse of $x$ in $\struct {G, \circ}$.

Then $\phi$ is an isomorphism.


Proof

$\struct {G, \circ, \preccurlyeq}$ is a fortiori an ordered semigroup.

Hence from Dual of Ordered Semigroup is Ordered Semigroup, $\struct {G, \circ, \succcurlyeq}$ is also an ordered semigroup.

That is, $\succcurlyeq$ is compatible with $\circ$.


From Inversion Mapping is Permutation, a fortiori the inversion mapping is a bijection.

From Inversion Mapping is Automorphism iff Group is Abelian, $\phi: \struct {G, \circ} \to \struct {G, \circ}$ is an automorphism and so a fortiori an isomorphism.


It remains to be shown that $\phi: \struct {G, \preccurlyeq} \to \struct {G, \succcurlyeq}$ is an order isomorphism.

We note that from Inversion Mapping is Involution that:

$\phi = \phi^{-1}$


Let $x, y \in G$ be arbitrary.

We have:

\(\ds x\) \(\preccurlyeq\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds y\) \(\succcurlyeq\) \(\ds x\) Definition of Dual Ordering
\(\ds \leadsto \ \ \) \(\ds x^{-1}\) \(\succcurlyeq\) \(\ds y^{-1}\) Inversion Mapping Reverses Ordering in Ordered Group
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(\succcurlyeq\) \(\ds \map \phi y\)


Then we have:

\(\ds x\) \(\succcurlyeq\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds y\) \(\preccurlyeq\) \(\ds x\) Definition of Dual Ordering
\(\ds \leadsto \ \ \) \(\ds x^{-1}\) \(\preccurlyeq\) \(\ds y^{-1}\) Inversion Mapping Reverses Ordering in Ordered Group
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(\preccurlyeq\) \(\ds \map \phi y\)
\(\ds \leadsto \ \ \) \(\ds \map {\phi^{-1} } x\) \(\preccurlyeq\) \(\ds \map {\phi^{-1} } y\) as $\phi = \phi^{-1}$


Hence we have shown that $\phi$ is an order-preserving mapping in both directions.

Hence the result.

$\blacksquare$


Sources

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