Totally Ordered Abelian Group Isomorphism

Theorem
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 $\N' \setminus \set {0'}$.

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

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

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:

Also:

Thus $\N'$ also satisfies.

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

Therefore $g$ is a surjection.

Now to 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 $\struct {\Z, +, \le}$ to $\struct {\Z', +', \le'}$.

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