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

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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:

\(\ds e\) \(\preccurlyeq\) \(\ds x\) Definition of $G^+$
\(\ds e\) \(\preccurlyeq\) \(\ds y\) Definition of $G^+$
\(\ds \leadsto \ \ \) \(\ds e \circ y\) \(\preccurlyeq\) \(\ds x \circ y\) Definition of Ordered Group as $\preccurlyeq$ is compatible with $\circ$
\(\ds x \circ e\) \(\preccurlyeq\) \(\ds x \circ y\)
\(\ds \leadsto \ \ \) \(\ds e\) \(\preccurlyeq\) \(\ds x \circ y\) $\preccurlyeq$ is an ordering and therefore transitive
\(\ds \leadsto \ \ \) \(\ds x \circ y\) \(\in\) \(\ds G^+\) Definition of $G^+$

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

$\Box$


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

Then:

\(\ds x\) \(\preccurlyeq\) \(\ds e\) Definition of $G^-$
\(\ds y\) \(\preccurlyeq\) \(\ds e\) Definition of $G^-$
\(\ds \leadsto \ \ \) \(\ds x \circ y\) \(\preccurlyeq\) \(\ds x \circ e\) Definition of Ordered Group as $\preccurlyeq$ is compatible with $\circ$
\(\ds x \circ y\) \(\preccurlyeq\) \(\ds e \circ y\)
\(\ds \leadsto \ \ \) \(\ds x \circ y\) \(\preccurlyeq\) \(\ds e\) $\preccurlyeq$ is an ordering and therefore transitive
\(\ds \leadsto \ \ \) \(\ds x \circ y\) \(\in\) \(\ds G^-\) Definition of $G^-$

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

$\Box$


Then we establish that:

\(\ds \forall x, y \in G: \, \) \(\ds \map \iota {x \circ y}\) \(=\) \(\ds \paren {x \circ y}^{-1}\) Definition of Inversion Mapping
\(\ds \) \(=\) \(\ds y^{-1} \circ x^{-1}\) Inverse of Group Product
\(\ds \) \(=\) \(\ds x^{-1} \circ y^{-1}\) Definition of Abelian Group
\(\ds \) \(=\) \(\ds \map \iota x \circ \map \iota y\) Definition of Inversion Mapping

demonstrating that for all $\iota$ has the morphism property over the whole of $G$.


Then we note that:

\(\ds \forall x \in G^+: \, \) \(\ds e\) \(\preccurlyeq\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds e\) \(\succcurlyeq\) \(\ds x^{-1}\) Inversion Mapping Reverses Ordering in Ordered Group: Corollary
\(\ds \leadsto \ \ \) \(\ds e\) \(\succcurlyeq\) \(\ds \map \iota x\) Definition of Inversion Mapping
\(\ds \leadsto \ \ \) \(\ds \map \iota x\) \(\in\) \(\ds G^-\) Definition of $G^-$
\(\ds \leadsto \ \ \) \(\ds \iota \sqbrk {G^+}\) \(=\) \(\ds G^-\) Definition of Image of Subset under Mapping


Similarly:

\(\ds \forall x \in G^-: \, \) \(\ds e\) \(\succcurlyeq\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds e\) \(\preccurlyeq\) \(\ds x^{-1}\) Inversion Mapping Reverses Ordering in Ordered Group: Corollary
\(\ds \leadsto \ \ \) \(\ds e\) \(\preccurlyeq\) \(\ds \map \iota x\) Definition of Inversion Mapping
\(\ds \leadsto \ \ \) \(\ds \map \iota x\) \(\in\) \(\ds G^+\) Definition of $G^+$
\(\ds \leadsto \ \ \) \(\ds \iota \sqbrk {G^-}\) \(=\) \(\ds G^+\) Definition of Image of Subset under Mapping


Then we note that:

\(\ds \forall x, y \in G: \, \) \(\ds \map \iota x\) \(=\) \(\ds \map \iota y\)
\(\ds \leadsto \ \ \) \(\ds x^{-1}\) \(=\) \(\ds y^{-1}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds y\) Group Axiom $\text G 3$: Existence of Inverse Element

demonstrating that $\iota$ is an injection.


Let $x \in G^+$.

Then a priori:

$\exists x^{-1} \in G^-: \map \iota {x^{-1} } = x$

Similarly, let $x \in G^-$.

Then a priori:

$\exists x^{-1} \in G^+: \map \iota {x^{-1} } = x$

So $\iota: G^+ \to G^-$ and $\iota: G^- \to G^+$ are both surjections.

Hence $\iota: G^+ \to G^-$ and $\iota: G^- \to G^+$ are both bijections by definition.


Thus we have that:

$\iota: \struct {G^+, \circ} \to \struct {G^-, \circ}$ is a semigroup isomorphism

and

$\iota: \struct {G^-, \circ} \to \struct {G^+, \circ}$ is a semigroup isomorphism.

$\Box$


Let $x, y \in G^+$ be arbitrary, such that $x \preccurlyeq y$.

We have:

\(\ds x\) \(\preccurlyeq\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds x^{-1}\) \(\succcurlyeq\) \(\ds y^{-1}\) Inversion Mapping Reverses Ordering in Ordered Group
\(\ds \leadsto \ \ \) \(\ds \map \iota x\) \(\succcurlyeq\) \(\ds \map \iota y\) Definition of Inversion Mapping


Let $x, y \in G^-$ be arbitrary, such that $x \succcurlyeq y$.

We have:

\(\ds x\) \(\succcurlyeq\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds x^{-1}\) \(\preccurlyeq\) \(\ds y^{-1}\) Inversion Mapping Reverses Ordering in Ordered Group
\(\ds \leadsto \ \ \) \(\ds \map \iota x\) \(\preccurlyeq\) \(\ds \map \iota y\) Definition of Inversion Mapping


Thus we have demonstrated that $\iota: \struct {G^+, \preccurlyeq} \to \struct {G^-, \succcurlyeq}$ and $\iota: \struct {G^-, \succcurlyeq} \to \struct {G^+, \preccurlyeq}$ are order isomorphisms.

So we have that:

Thus we have that:

$\iota: \struct {G^+, \circ} \to \struct {G^-, \circ}$ is a semigroup isomorphism
$\iota: \struct {G^+, \preccurlyeq} \to \struct {G^-, \succcurlyeq}$ is an order isomorphism

and so:

$\iota: \struct {G^+, \circ, \preccurlyeq} \to \struct {G^-, \circ, \succcurlyeq}$ is an ordered semigroup isomorphism

and

$\iota: \struct {G^-, \circ} \to \struct {G^+, \circ}$ is a semigroup isomorphism
$\iota: \struct {G^-, \succcurlyeq} \to \struct {G^+, \preccurlyeq}$ is an order isomorphism

and so:

$\iota: \struct {G^-, \circ, \succcurlyeq} \to \struct {G^+, \circ, \preccurlyeq}$ is an ordered semigroup isomorphism

$\blacksquare$


Sources

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