Let $(G,+,0,<)$ be an ordered divisible group of uncountable dimension. Consider the subset $G^{<0}$ of $G$. 


Question: Are $G$ and $G^{<0}$ isomorphic as ordered sets? Does there exist an order-preserving isomorphism of $G^{<0}$ onto $G$?
Intuitively I would say, this is true.


My ideas:
1. In the case where $(K,+, \cdot, 0,1,<)$ is an divisible ordered field we can give an explicit order-preserving isomorphism of $K$ onto $K^{<0}$, e.g. $f(x):= \begin{cases}
x-1 \quad \text{if} \; x \leq0, \\
-\frac{1}{x+1} \quad \text{else}.
\end{cases}
$


2. In the case where $G$ is in ordered set the statement is false.


How can we use the group structure and the divisibility of $G$ to construct an isomorphism or does anybody know a counterexample?



Uncountable divisible groups and the existence of order-preserving isomorphisms of their subsets
Let $G=\mathbb{Q}^{\omega\_1}$ with the lexicographic order where $\omega\_1$ is the first uncountable ordinal. So elements of $G$ have the form $(x\_\alpha)\_{\alpha<\omega\_1}$, and two elements are compared by looking at the first $\alpha$ on which the entries are not equal.


Then $G$ has a countable cofinal subset consisting of elements with $x\_0=n$ and $x\_{\alpha}=0$ for $\alpha\neq 0$ as $n$ ranges through $\mathbb{N}$, while $G^{<0}$ does not: if $A$ were a countable cofinal subset of $G^{<0}$ then extracting the first non-zero entry of each element of $A$ would give a countable cofinal subset of the ordinals less than $\omega\_1$ which is impossible. Therefore $G^{<0}$ and $G$ are not isomorphic as ordered sets.


