Conditions for Ordering in Ordered Group to be Directed

Theorem

Let $\struct {G, \odot, \preccurlyeq}$ be an ordered group whose identity element is $e$.

Then:

$\preccurlyeq$ is a directed ordering
for every $x \in G$ there exist $y, z \in G$ such that $e \preccurlyeq y$, $e \preccurlyeq z$ and $x = y \odot z^{-1}$.

Proof

Sufficient Condition

Let $\preccurlyeq$ be a directed ordering.

By definition of directed ordering:

$\forall x, z \in G: \exists y \in G: x \preccurlyeq y$ and $z \preccurlyeq y$

Let $x \in G$ be arbitrary.

 $\ds \exists y \in G: \,$ $\ds x$ $\preccurlyeq$ $\ds y$ Definition of Directed Ordering $\, \ds \land \,$ $\ds e$ $\preccurlyeq$ $\ds y$ as it holds for all $z$, it definitely holds for $e$ $\ds \leadsto \ \$ $\ds x \odot y^{-1}$ $\preccurlyeq$ $\ds y \odot y^{-1}$ Definition of Relation Compatible with Operation $\, \ds \land \,$ $\ds e$ $\preccurlyeq$ $\ds y$ $\ds \leadsto \ \$ $\ds x \odot y^{-1} \odot z$ $\preccurlyeq$ $\ds e \odot z$ $\, \ds \land \,$ $\ds e$ $\preccurlyeq$ $\ds y$ $\ds \leadsto \ \$ $\ds \paren {y \odot x^{-1} }^{-1} \odot z$ $\preccurlyeq$ $\ds z$ Inverse of Group Product $\, \ds \land \,$ $\ds e$ $\preccurlyeq$ $\ds y$

Let $e = \paren {y \odot x^{-1} }^{-1} \odot z$.

Then:

 $\ds z$ $=$ $\ds y \odot x^{-1}$ $\ds \leadsto \ \$ $\ds x \odot z$ $=$ $\ds y$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds y \odot z^{-1}$

Then from:

 $\ds \paren {y \odot x^{-1} }^{-1} \odot z$ $\preccurlyeq$ $\ds z$ $\, \ds \land \,$ $\ds e$ $\preccurlyeq$ $\ds y$

we get:

 $\ds e$ $\preccurlyeq$ $\ds z$ Group Axiom $\text G 3$: Existence of Inverse Element $\, \ds \land \,$ $\ds e$ $\preccurlyeq$ $\ds y$

and we see that:

$\exists y, z \in G: e \preccurlyeq y, e \preccurlyeq z, x = y \odot z^{-1}$

$\blacksquare$

Necessary Condition

Let $\preccurlyeq$ be such that:

for every $x \in G$ there exist $y, z \in G$ such that $e \preccurlyeq y$, $e \preccurlyeq z$ and $x = y \odot z^{-1}$.

Let $x \in G$ be arbitrary.

By the hypothesis, there exist $z, w \in G$ such that $e \preccurlyeq z$, $e \preccurlyeq w$ and $x = z \odot w^{-1}$.

We have:

 $\ds e$ $\preccurlyeq$ $\ds w$ by hypothesis $\ds \leadsto \ \$ $\ds w^{-1}$ $\preccurlyeq$ $\ds e$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds z \odot w^{-1}$ $\ds$ $\preccurlyeq$ $\ds z \odot e$ Definition of Relation Compatible with Operation $\ds$ $=$ $\ds z$

Similarly, for another arbitrary $y \in G$, there exists some $g \in G$ such that $e \preccurlyeq g$ and $y \preccurlyeq g$.

Then we have:

 $\ds x$ $=$ $\ds x \odot e$ Definition of Identity Element $\ds$ $\preccurlyeq$ $\ds z \odot e$ Definition of Relation Compatible with Operation $\ds$ $\preccurlyeq$ $\ds z \odot g$ Definition of Relation Compatible with Operation $\ds y$ $=$ $\ds e \odot y$ Definition of Identity Element $\ds$ $\preccurlyeq$ $\ds z \odot y$ Definition of Relation Compatible with Operation $\ds$ $\preccurlyeq$ $\ds z \odot g$ Definition of Relation Compatible with Operation

This shows that $z \odot g$ is an upper bound of $\set {x, y}$.

By Group Axiom $\text G 0$: Closure, $z \odot g \in G$.

Hence $\preccurlyeq$ is a directed ordering.

$\blacksquare$