Inversion Mapping Reverses Ordering in Ordered Group

Theorem

Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group.

Let $x, y \in G$.

Let $\prec$ be the reflexive reduction of $\preceq$.

Then the following equivalences hold:

\(\ds \forall x, y \in G: \, \)

\(\ds x \preccurlyeq y\)

\(\iff\)

\(\ds e \prec y^{-1} \preccurlyeq x^{-1}\)

\(\ds \forall x, y \in S: \, \)

\(\ds x \prec y\)

\(\iff\)

\(\ds y^{-1} \prec x^{-1}\)

Corollary

\(\ds \forall x \in G: \, \)

\(\ds x \preccurlyeq e\)

\(\iff\)

\(\ds e \preccurlyeq x^{-1}\)

\(\ds e \preccurlyeq x\)

\(\iff\)

\(\ds x^{-1} \preccurlyeq e\)

\(\ds x \prec e\)

\(\iff\)

\(\ds e \prec x^{-1}\)

\(\ds e \prec x\)

\(\iff\)

\(\ds x^{-1} \prec e\)

Proof

By the definition of an ordered group, $\preceq$ is a relation compatible with $\circ$.

Thus by Inverses of Elements Related by Compatible Relation, we obtain the first result:

$x \preccurlyeq y \iff y^{-1} \preccurlyeq x^{-1}$

By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is also compatible with $\circ$.

Thus by again Inverses of Elements Related by Compatible Relation, we obtain the second result:

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

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Theorem $15.3$