Ordered Group Equivalences

Theorem
Let $\struct {S, \circ, \preceq}$ be an ordered group whose identity is $e$.

Let $x, y, z, \in S$.

Then the following are all equivalent:


 * $(1): \quad x \prec y$
 * $(2): \quad x \circ z \prec y \circ z$
 * $(3): \quad z \circ x \prec z \circ y$
 * $(4): \quad y^{-1} \prec x^{-1}$
 * $(5): \quad e \prec y \circ x^{-1}$
 * $(6): \quad e \prec x^{-1} \circ y$

Proof
Since $\struct {S, \circ, \preceq}$ is an ordered group, we have, for all $x, y, z \in S$:


 * $(R): \quad x \prec y \implies x \circ z \prec y \circ z$
 * $(L): \quad x \prec y \implies z \circ x \prec z \circ y$

Using $(R)$ and $(L)$, let us demonstrate the equivalences.

$(1) \iff (2)$ and $(1) \iff (5)$
We have by applying $(R)$ twice:

This last identity comes down to $x \prec y$ by definition of inverse.

Selecting $z = x^{-1}$ proves $(1) \iff (5)$ as well.

$(1) \iff (3)$ and $(1) \iff (6)$
We have by applying $(L)$ twice:

This last identity comes down to $x \prec y$ by definition of inverse.

Selecting $z = x^{-1}$ proves $(1) \iff (6)$ as well.

$(4) \iff (5)$
Applying that we know $(1) \iff (2)$, we obtain:

since $y \circ y^{-1} = e$.

The remaining equivalences follow from Material Equivalence is Equivalence.

Also see

 * Properties of Ordered Group