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.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.13$