Totally Ordered Abelian Group Isomorphism

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Theorem

Let $\left({\Z', +', \le'}\right)$ be a totally ordered abelian group.

Let $0'$ be the identity of $\left({\Z', +', \le'}\right)$.

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


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:

$\forall n \in \Z: g \left({n}\right) = \left({+'}\right)^n 1'$

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


Proof

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:

$z \in \N' \setminus \left\{{0'}\right\}$

or:

$-z \in \N' \setminus \left\{{0'}\right\}$

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

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

Call this minimal element $1'$.


It is clear that $\N'$ is an ordered semigroup satisfying $(NO 1)$, $(NO 2)$ and $(NO 4)$ of the naturally ordered semigroup axioms.

Also:

\(\displaystyle \) \(\) \(\displaystyle 0' \le' x \le' y\)
\(\displaystyle \) \(\implies\) \(\displaystyle 0' \le' y - x\)
\(\displaystyle \) \(\implies\) \(\displaystyle y - x \in \N' \land x +' \left({y - x}\right) = y\)


Thus $\N'$ also satisfies $(NO 3)$ of the naturally ordered semigroup axioms.


So $\left({\N', +', \le'}\right)$ is a naturally ordered semigroup.

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

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


Next it is established that $g$ is surjective.

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

\(\displaystyle \) \(\) \(\displaystyle y <' 0'\)
\(\displaystyle \) \(\implies\) \(\displaystyle -y >' 0'\)
\(\displaystyle \) \(\implies\) \(\displaystyle \exists n \in \N: -y = g \left({n}\right)\)
\(\displaystyle \) \(\implies\) \(\displaystyle y = - g \left({n}\right) = g \left({-n}\right)\) Homomorphism with Identity Preserves Inverses


Therefore $g$ is a surjection.


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

Let $n < m$.

\(\displaystyle \) \(\) \(\displaystyle n < m\)
\(\displaystyle \) \(\implies\) \(\displaystyle m - n \in \N_{>0}\)
\(\displaystyle \) \(\implies\) \(\displaystyle g \left({m}\right) - g \left({n}\right) - g \left({m - n}\right) \in \N' \setminus \left\{ {0'}\right\}\)
\(\displaystyle \) \(\implies\) \(\displaystyle g \left({n}\right) <' \left({g \left({m}\right) - g \left({n}\right)}\right) +' g \left({n}\right) = g \left({m}\right)\) 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 $\left({\Z, +, \le}\right)$ to $\left({\Z', +', \le'}\right)$.


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

$\blacksquare$


Sources