Strict Ordering Preserved under Product with Invertible Element
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Theorem
Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $z \in S$ be invertible.
Suppose that either $x \circ z \prec y \circ z$ or $z \circ x \prec z \circ y$.
Then $x \prec y$.
Proof
Suppose $x \circ z \prec y \circ z$.
By Invertible Element of Monoid is Cancellable, $z^{-1}$ is cancellable.
Then from Strict Ordering Preserved under Product with Cancellable Element:
- $x = \paren {x \circ z} \circ z^{-1} \prec \paren {y \circ z} \circ z^{-1} = y$
Likewise, if $z \circ x \prec z \circ y$:
- $x = z^{-1} \circ \paren {z \circ x} \prec z^{-1} \circ \paren {z \circ y} = y$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Theorem $15.1: \ 2^\circ$