Ordering Compatible with Group Operation is Strongly Compatible/Corollary/Proof 1

Theorem

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

Let $\prec$ be the reflexive reduction of $\preceq$.

Let $x, y \in G$.

Then the following equivalences hold:

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

$\ds x \preccurlyeq y$

$\iff$

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

$\ds x \preccurlyeq y$

$\iff$

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

$\ds x \preccurlyeq y$

$\iff$

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

$\ds x \preccurlyeq y$

$\iff$

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

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

$\ds x \prec y$

$\iff$

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

$\ds x \prec y$

$\iff$

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

$\ds x \prec y$

$\iff$

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

$\ds x \prec y$

$\iff$

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

By the definition of an ordered group, $\preccurlyeq$ is a relation compatible with $\circ$.

$\blacksquare$