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?
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.