Conditions for Ordering in Ordered Group to be Directed

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Theorem

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


Then:

$\preccurlyeq$ is a directed ordering

if and only if

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.

$\blacksquare$


Sources