
group homomorphism
Let $\displaystyle G\text{ and }H$ be additive subgroups of $\displaystyle \mathbb{R}$ both containing 1. Let $\displaystyle \eta:G\rightarrow H$ be an order preserving group homomorphism such that $\displaystyle \eta(1)=1$. Show that $\displaystyle \eta(g)=g$ for all $\displaystyle g\in G$.
The idea that I am working with is to let $\displaystyle s,t\in G$ such that
$\displaystyle st>1$ then there exists an integer $\displaystyle m$ such that
$\displaystyle s>m>t$.
now apply $\displaystyle \eta$ and get the following
$\displaystyle \eta(s)\eta(t)>1$ and $\displaystyle \eta(s)>\eta(m)>\eta(t)$
and suppose that $\displaystyle \eta(g)>g$ for all $\displaystyle g\in G$. In this way I hope to get some contradiction
Or is there another way?

This problem is still bugging me... does anybody have any ideas?