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:
- 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.
Also:
\(\ds \) | \(\) | \(\ds 0' \le' x \le' y\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds 0' \le' y - x\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds y - x \in \N' \land x +' \paren {y - x} = y\) |
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.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.14$