A totally ordered abelian group is an abelian group (G,+) with a total order $\leq$ such that for all $a,b,c \in G$ if $a \leq b$ then $a+c \leq b+c$.
We will say that an ordered abelian group is dense in itself if for all $a<b$ exist $c$ such that $a<c<b$.
I want to prove that, if for all $g>0$ exist $c$ such that $0<c<g$, then for all $g>0$ and for all $n \in \mathbb{N}$ exist $h$ such that $0<nh<g$.
I have tried this for induction on $n$, but I can't prove the $n+1$-case.
Any suggestion will be welcomed.