Totally Ordered Abelian Group Isomorphism

From ProofWiki
Jump to navigation Jump to search


Let $\struct {\Z', +', \le'}$ be a totally ordered abelian group.

Let $0'$ be the identity of $\struct {\Z', +', \le'}$.

Let $\N' = \set {x \in \Z': x \ge' 0'}$.

Let $\Z'$ contain at least two elements.

Let $\N'$ be well-ordered for the ordering induced on $\N'$ by $\le'$.

Then the mapping $g: \Z \to \Z'$ defined by:

$\forall n \in \Z: \map g n = \paren {+'}^n 1'$

is an isomorphism from $\struct {\Z, +, \le}$ onto $\struct {\Z', +', \le'}$, where $1'$ is the smallest element of $\N' \setminus \set {0'}$.


First we establish that $g$ is a homomorphism.

Suppose $z \in \Z'$ such that $z \ne 0'$.

Then by Ordering of Inverses in Ordered Monoid, either $z >' 0'$ or $-z >' 0'$.

Thus either:

$z \in \N' \setminus \set {0'}$


$-z \in \N' \setminus \set {0'}$

and thus $\N' \setminus \set {0'}$ is not empty.

Therefore $\N' \setminus \set {0'}$ has a minimal element.

Call this minimal element $1'$.

It is clear that $\N'$ is an ordered semigroup satisfying:

Naturally Ordered Semigroup Axiom $\text {NO} 1$: Well-Ordered
Naturally Ordered Semigroup Axiom $\text {NO} 2$: Cancellability
Naturally Ordered Semigroup Axiom $\text {NO} 4$: Existence of Distinct Elements.


\(\ds \) \(\) \(\ds 0' \le' x \le' y\)
\(\ds \) \(\leadsto\) \(\ds 0' \le' y - x\)
\(\ds \) \(\leadsto\) \(\ds y - x \in \N' \land x +' \paren {y - x} = y\)

Thus $\N'$ also satisfies Naturally Ordered Semigroup Axiom $\text {NO} 3$: Existence of Product.

So $\struct {\N', +', \le'}$ is a naturally ordered semigroup.

So, by Naturally Ordered Semigroup is Unique, the restriction to $\N$ of $g$ is an isomorphism from $\struct {\N, +, \le}$ to $\struct {\N', +', \le'}$.

By Index Law for Sum of Indices, $g$ is a homomorphism from $\struct {\Z, +}$ into $\struct {\Z', +'}$.

Next it is established that $g$ is surjective.

Let $y \in \Z': y <' 0'$.

\(\ds \) \(\) \(\ds y <' 0'\)
\(\ds \) \(\leadsto\) \(\ds -y >' 0'\)
\(\ds \) \(\leadsto\) \(\ds \exists n \in \N: -y = \map g n\)
\(\ds \) \(\leadsto\) \(\ds y = -\map g n = \map g {-n}\) Homomorphism with Identity Preserves Inverses

Therefore $g$ is a surjection.

Now to show that $g$ is a monomorphism, that is, it is injective.

Let $n < m$.

\(\ds \) \(\) \(\ds n < m\)
\(\ds \) \(\leadsto\) \(\ds m - n \in \N_{>0}\)
\(\ds \) \(\leadsto\) \(\ds \map g m - \map g n - \map g {m - n} \in \N' \setminus \set {0'}\)
\(\ds \) \(\leadsto\) \(\ds \map g n <' \paren {\map g m - \map g n} +' \map g n = \map g m\) Strict Ordering Preserved under Product with Cancellable Element

Therefore it can be seen that $g$ is strictly increasing.

It follows from Monomorphism from Total Ordering that $g$ is a monomorphism from $\struct {\Z, +, \le}$ to $\struct {\Z', +', \le'}$.

A surjective monomorphism is an isomorphism, and the result follows.