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

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.

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

Then:

Then from:

we get:

and we see that:


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

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:

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:

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

By, $z \odot g \in G$.

Hence $\preccurlyeq$ is a directed ordering.