Totally Ordered Abelian Group Isomorphism

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 minimal element of $$\N' - \left\{{0'}\right\}$$.

Proof

 * First we establish that $$g$$ is a homomorphism.

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

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

Thus either $$z \in \N' - \left\{{0'}\right\}$$ or $$-z \in \N' - \left\{{0'}\right\}$$ and thus $$\N' - \left\{{0'}\right\}$$ is not empty.

Therefore $$\N' - \left\{{0'}\right\}$$ has a minimal element. Call this $$1'$$.

It is clear that $$\N'$$ is an ordered semigroup satisfying NO 1, NO 2 and NO 4 of Naturally Ordered Semigroup.

Also:

$$ $$ $$

Thus $$\N'$$ also satisfies NO 3 of Naturally Ordered Semigroup.

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

So, by Naturally Ordered Semigroups Isomorphism 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 we establish that $$g$$ is surjective.

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

$$ $$ $$ $$

Therefore $$g$$ is a surjection.


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

Let $$n < m$$.

$$ $$ $$ $$

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.