# Inversion Mapping is Isomorphism from Ordered Abelian Group to its Dual

## 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 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.

$\blacksquare$