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.