Inversion Mapping on Ordered Group is Dual Order-Isomorphism

Theorem

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

Let $\iota: G \to G$ be the inversion mapping, defined by $\map \phi x = x^{-1}$.

Then $\iota$ is a dual order-isomorphism.

Proof

By Inversion Mapping is Involution and Involution is Permutation, $\iota$ is a permutation and so by definition bijective.

Let $x, y \in G$ such that $x \prec y$.

Then $y^{-1} \prec x^{-1}$ by Inversion Mapping Reverses Ordering in Ordered Group.

Thus $\map \iota y \prec \map \iota x$.

Since this holds for all $x$ and $y$ with $x \prec y$, $\iota$ is strictly decreasing.

If $\map \iota x \prec \map \iota y$, then $\map \iota {\map \iota y} \prec \map \iota {\map \iota x}$ by the above.

Thus by Inverse of Group Inverse: $y \prec x$.

Therefore, $\iota$ reverses ordering in both directions, and is thus a dual isomorphism.

$\blacksquare$