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 exists an order-preserving ismorphism 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?