Subsets Greater Than and Less Than Identity of Ordered Abelian Group are Isomorphic Ordered Semigroups

Theorem
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered abelian group whose identity element is $e$.

Let $G^+$ and $G^-$ denote the subsets of $G$ defined as:
 * $G^+ = \set {x \in G: e \preccurlyeq x}$
 * $G^- = \set {x \in G: x \preccurlyeq e}$

Then $\struct {G^+, \circ, \preccurlyeq}$ and $\struct {G^-, \circ, \succcurlyeq}$ are subsemigroups of $G$ such that the inversion mapping $\iota: G \to G$ defined as:
 * $\forall g \in G: \map \iota g = g^{-1}$

is:
 * an isomorphism from $\struct {G^+, \circ, \preccurlyeq}$ to $\struct {G^-, \circ, \succcurlyeq}$

and:
 * an isomorphism from $\struct {G^-, \circ, \preccurlyeq}$ to $\struct {G^+, \circ, \succcurlyeq}$

Proof
Let $x, y \in G^+$ be arbitrary.

Then:

So by the Subsemigroup Closure Test $\struct {G^+, \circ}$ is a subsemigroup of $\struct {G, \circ}$.