Intervals in divisible ordered groups

Question

Is it true that if $(G,+,0,<)$ is a divisible ordered abelian group with at least two elements, then for $a,b >0 \in G$, there is an injective order preserving map from $[0;a)$ to $[0;b)$?

It is true if $G$ is archimedean because then there is a natural number $n$ such that $n.b \geq a$ and $x \mapsto \frac{1}{n}.x$ works fine.

It is false in general if $(G,0,<)$ is just a dense linear order with more than two elements, for instance if it is a copy of $(\mathbb{Q},<)$ followed by a copy of $(\mathbb{R},<)$.

If one could find a divisible ordered abelian group with a definable (with parameters) such map then this would yield the result since the theory of divisible ordered abelian group with several elements is complete. But I doubt one can.

Answer 1

It's not hard to cook up examples that fail for cardinality reasons like your linear order example.

For example, take $G = \mathbb{R}\times\mathbb{Q}$, ordered lexicographically (i.e. replace every element of $\mathbb{R}$ with a copy of $\mathbb{Q}$) and with the usual product group structure. Then the interval from $(0,0)$ to $(0,1)$ is countable, but the interval from $(0,0)$ to $(1,0)$ has size continuum.