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

$\displaystyle s-t>1$ 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?