# Relationship between Archimedean and Divisible ordered groups

Let $$(G,+,\leq)$$ be a linearly ordered abelian group (i.e. the order is total and compatible with the sum) and $$n\cdot x$$ denote the classical action of $$\mathbb{Z}$$ over $$G$$ (i.e. $$0$$ for $$n=0$$, sum of $$|n|$$ copies of $$x$$ if $$n>0$$ and of $$-x$$ if $$n<0$$). I was interested in the relationship between the following properties:

Archimedean property: $$\forall a,b\in G\:\:\:\exists n\in \mathbb{Z}|a\leq n\cdot b$$

Divisible group: $$\forall n\in \mathbb{Z}, a\in G\:\:\:\exists b\in G|n\cdot b =a$$

At a first glance, to me they look pretty similar. In fact, so far, I've encountered examples that satisfies both of them or neither of them but I stil didn't find anything that satisfies exactly one of them. I've read about hyperreal and surreal numbers which are not Archimedean but are divisible however they are proper classes and to me this seems a bit cheating so I'd prefer only group lying on a set rather then a proper class.

EDIT: Apparently the examples were pretty simple so I'll change my original question to: under which condition are they equal?

• If an linearly ordered [Abelian] group is Dedekind-complete and dense, then it is Archimedean and divisible. I hope I've remembered that right! I'll try to dig up a proper reference, if someone else hasn't posted an answer to the revised question by then. – Calum Gilhooley Mar 13 at 22:24

$$\mathbb{Z}$$ itself is Archimedean, but it is not divisible: for example, there is no $$b \in \mathbb{Z}$$ such that $$2b = 1$$.

The ring $$Q[x]$$, with the total ordering in which a polynomial in $$x$$ with rational coefficients is positive if and only if its trailing coefficient is positive, is divisible, but it is not Archimedean: for example, $$0 < nx < 1$$ for every positive integer $$n$$, because $$1 - nx > 0$$.

Further to my comment above, in response to the revised question, these extracts are from F. Loonstra, "Ordered groups", Proc. Nederl. Akad. Wetensch. 49 (1946), pp. 41-46 (PDF here):

If an ordered set $$O$$ is dense in itself, then generally $$O$$ need not be dense-ordered. (A set $$O$$ is called dense-ordered if for each pair of elements $$a$$ and $$b$$ of $$O$$ there is at least one element $$c$$ "between" $$a$$ and $$b$$.) For ordered groups however we prove:

If $$G$$ is an ordered group dense in itself, then $$G$$ is densely ordered.

If there was namely between two elements $$a$$ and $$b$$ no further element, then every element had an immediate successor and predecessor and thus $$G$$ was not dense in itself. Conclusion: An ordered group $$G$$ is dense in itself if and only if $$G$$ is densely ordered. If on the contrary one element is isolated, then every element is necessarily isolated; consequently:

An ordered group $$G$$ is either a discretely ordered group, which means that every element has an immediate successor and predecessor, or a densely ordered group. The property of an ordered group as being discrete resp. dense is as a topological property invariant for a mapping preserving the order.

If we divide an ordered set $$N$$ into two subsets $$N_1$$ and $$N_2$$ with the conditions:

I. Every element of $$N$$ belongs to one, but only one of the sets $$N_1$$ and $$N_2$$;

II. Neither $$N_1$$ nor $$N_2$$ is empty;

III. All elements of $$N_1$$ precede those of $$N_2$$.

then we call this division of $$N$$ a Dedekind-cut $$N_1 | N_2$$ in $$N$$. There are only the following cases that exclude each other:

$$1$$. $$N_1$$ has a last, $$N_2$$ a first element; then we call the cut $$N_1 | N_2$$ a sault in $$N$$.

$$2^a$$. $$N_1$$ has a last, $$N_2$$ no first element;

$$2^b$$. $$N_1$$ has no last, $$N_2$$ has a first element; we call the cut in $$2^a$$ and $$2^b$$ a continuous cut.

$$3$$. $$N_1$$ has no last. $$N_2$$ has no first element; the cut is called a lacuna in the set $$N$$.

The ordered set $$N$$ is called continuously ordered, if no cut in $$N$$ defines a sault or a lacuna.

A non-archimedean ordered group is not continuously ordered.

If the ordering was continuous then of every cut $$A | B$$ either $$A$$ had a last, $$B$$ no first element, or just the reverse. We shall try to prove, however, that in every non-archimedean ordered group there is to define a non-continuous cut. Suppose $$e < a < b$$ and $$a$$ non-archimedean with regard to $$b$$. Let the class $$A$$ contain all elements $$\leqslant a^n$$ for a positive integer $$n$$, $$B$$ the remaining elements; $$B$$ is not empty, as $$b \in B$$. $$A$$ has no last element, nor has $$B$$ a first element, for $$B$$ also contains the elements $$ba^{-1} > ba^{-2} > \cdots$$ elements that precede $$b$$. Therefore a non-archimedean ordered group is never a continuous ordered set or directly: continuously ordered groups are necessarily archimedean ordered groups.

A densely ordered group $$G$$ has the property that to every element $$a > e$$ there exists an element $$b > e$$, so that $$b^2 < a$$.

[A proof of that is provided, but ...]

Without a proof we pronounce the following theorem:

In a continuous ordered group there exists to every element $$a > e$$ one element $$b > e$$ with $$b^n = a$$.

[I was just typing this out, making edits and corrections, and joining it all up, and didn't notice that he hadn't provided a proof of this part! I think I have a proof in my own handwritten notes from years ago, but it might not be easy to find, and may be wrong. Let's just hope the result is obvious ... er ... an exercise for the reader?] :)

Noah Schweber's answer to the following question may help to fill in the, er, gap I've just left: Is $$\mathbb{R}$$ the only complete ordered Abelian group?.

• Sometimes the answer is so under your nose you don't manage to see it... I'll improve my answer. – Renato Faraone Mar 13 at 22:15
• It's just happened to me, too. :) – Calum Gilhooley Mar 13 at 22:17
• Glad I'm not the only one. Also, hope my question is more interesting now. – Renato Faraone Mar 13 at 22:21