Ordering of Inverses

Theorem

Let $\left({S, \circ, \preceq}\right)$ be an ordered monoid whose identity is $e$.

Let $x, y \in S$ be invertible.

Then $x \prec y \iff y^{-1} \prec x^{-1}$.

Proof

• First, to show that $x \prec y \implies y^{-1} \prec x^{-1}$:

	\(\displaystyle \)	\(\displaystyle x\)	\(\prec\)	\(\displaystyle y\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle e\)	\(=\)	\(\displaystyle x^{-1} \circ x \prec x^{-1} \circ y\)	\(\displaystyle \)	Cancellability in Ordered Semigroup
	\(\displaystyle \implies\)	\(\displaystyle y^{-1}\)	\(=\)	\(\displaystyle e \circ y^{-1} \prec x^{-1} \circ y \circ y^{-1} = x^{-1}\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle y^{-1}\)	\(\prec\)	\(\displaystyle x^{-1}\)	\(\displaystyle \)

• Next, to show that $y^{-1} \prec x^{-1} \implies x \prec y$:

	\(\displaystyle \)	\(\displaystyle y^{-1}\)	\(\prec\)	\(\displaystyle x^{-1}\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle x\)	\(=\)	\(\displaystyle \left({x^{-1} }\right)^{-1} \prec \left({y^{-1} }\right)^{-1} = y\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle x\)	\(\prec\)	\(\displaystyle y\)	\(\displaystyle \)

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 15$: Theorem $15.2$