Strict Ordering Preserved under Product with Invertible Element

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 $15.1: \ 2^\circ$