Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 1

Theorem

Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.

Let $x \in G$.

Then the following equivalences hold:

$\ds \forall x \in G: \, $

$\ds x \preccurlyeq e$

$\iff$

$\ds e \preccurlyeq x^{-1}$

$\ds e \preccurlyeq x$

$\iff$

$\ds x^{-1} \preccurlyeq e$

$\ds x \prec e$

$\iff$

$\ds e \prec x^{-1}$

$\ds e \prec x$

$\iff$

$\ds x^{-1} \prec e$

$\ds \forall x \in G: \, $

$\ds x \preccurlyeq e$

$\iff$

$\ds e^{-1} \preccurlyeq x^{-1}$

$\ds e \preccurlyeq x$

$\iff$

$\ds x^{-1} \preccurlyeq e^{-1}$

$\ds x \prec e$

$\iff$

$\ds e^{-1} \prec x^{-1}$

$\ds e \prec x$

$\iff$

$\ds x^{-1} \prec e^{-1}$

Since $e^{-1} = e$, the theorem holds.

$\blacksquare$