# 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 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

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.3$