# isomorphism between divisible, totally ordered, abelian groups

Let $G$, $H$ be divisible, abelian, linearly ordered groups, whose cardinalities are equal and satisfy $\mu := |G|=|H|>\aleph_{0}$.

These are supposed to be (order!) isomorphic. And just about every text, that I have looked at, points this out, only without proof.

How does one demonstrate the isomorphism? (Obvious as groups they are isomorphic, as they form $\mathbb{Q}$-vector spaces with the same dimension, but here the order structure plays a pivotal role.)

I can at most show, there are $\mathcal{G}, \mathcal{H}\subseteq\mathcal{P}(\mu)$ ultrafilters and $\phi:G\to\prod_{\mu}H\ /\ \mathcal{H}$ and $\psi:H\to\prod_{\mu}G\ /\ \mathcal{G}$ monomorphisms (but not first-order embeddings). More at this stage, not. Can one somehow out of these construct an iso? Or is this a wrong way?

Thanks in advance!

EDIT: I had misread. The existence of an isomorphism refers just to divisible, totally ordered groups. The real claim is simply, that the groups are elementarily equivalent — and a proof eludes me in these books. Would someone kindly point in a right direction, how to prove this?

• About your edit: I believe divisible, totally ordered groups have quantifier elimination, and using that you can find a universal countable model (one such that each archimedean class is divisible and each the set of classes is order isomorphic to ${\bf Q}$), implying that the theory is complete. Mar 22 '14 at 11:55
• Thanks for the tipp! That completes my search. A key ingredient in constructing the above mentioned monomorphisms, involved showing $\exists$-elimination, from this I can demonstrate quantifier elimination. (P. S. there is no need to rely on archimedian-ness; that is a second order property.) Mar 25 '14 at 7:31
• I'm glad I could help. About the P.S., it might be a second order property, but it is monotone under embeddings, so must be put under consideration when looking for an universal model. Of course, this is not the only way to show completeness. Mar 25 '14 at 18:07
• A universal model for, say, all models of cardinality $\omega$ --- \emph{i. e.}, a model, intwo which all models of this size elementarily embed --- would be $\prod_{i\in I}(\mathbb{Q};\leq,+,0)~/~\mathcal{F}$, wobei $\mathcal{F}\in Ult(I)$ nonprinzipal and $|I|=\aleph_{0}^{|Ult(\omega)|}$. This „universal" model (for countable models) is not archimedean, as it contains the countable divisible l. o. group $({\langle\mathbb{Q}\cup\{h\}\rangle_{Group}};\leq,+,0)$, with the lexical ordering $q+ph<q’+p’h$ iff $q<q’$ or $q=q’$ and $p<p’$, which makes $h$ an infinitesimal and the Group nonarch. Mar 30 '14 at 7:27
• Why would that even be countable? And even if it was, how would that counter any claim that I made? Mar 30 '14 at 20:49

## 1 Answer

It is not true that any two divisible ordered abelian groups of the same cardinality $\gt \aleph_0$ are isomorphic as ordered groups. For there is a non-Archimedean such group of cardinality $c$, and there is an Archimedean one.

• Also from model theoretical point of view, any theory with an order is unstable and hence has $2^\lambda$ nonisomorphic models in an uncountable cardinality $\lambda$. Mar 21 '14 at 18:10
• Thanks. Yes, I was considering the reals and hyperreals as a counterexample, but tied myself in knots, thinking, this might nevertheless be possible, as perspective play an huge role: with respect to the (hyper) naturals the hyper reals are achimedean. But nevertheless, one has to work in a single framework and in such the two structure are clearly not order isomorphic. Mar 22 '14 at 8:14