Let $(A, +, \preceq)$ be an ordered group, namely $(A, +)$ is a group and $\preceq$ is a total order on $A$ such that $x + z \prec y + z$ and $z + x \prec z + y$ for all $x,y,z \in A$ with $x \prec y$.

Q1.If $x,y,z \in A$ and $x \preceq z$, is it true that $x+y \ne y + z$ unless $x = z$?

As pointed out by Yves Cornulier in the comments below, the answer is yes iff the ambient group is abelian. This means that I was too hasty (and optimistic) with Q1, since my original question would have been:

Q2.If $x,y \in A$ and $n$ is an integer $\ge 2$, is it true that $x+y \ne y + nx$ unless $x = 0$?

A couple of observations: The answer is yes if (i) $x+y\preceq y+x$, or (ii) the subgroup, $S$, generated by $x$ and $y$ is Archimedean (this follows from a 1902 theorem by O. Hölder).

Here is an *informal* reasoning in support of a positive answer: Assume to the contrary that $x+y=y+nx$ for some $n \in \mathbb{N}^+$. Then, $-ky+x+ky = n^kx$ for all $k \in \mathbb{N}$. This *shoud* be in contradiction to the fact that for $n \ge 2$ and $x \ne 0$ the "gap" between $n^kx$ and $-ky+x+ky$ "grows" larger and larger as $k \to \infty$, while the "distance" between $x$ and $-ky+x+ky$ is "uniformly bounded". The point is that I don't see how to turn the above into rigorous arguments, and particularly how to give a formal meaning to words or expressions like "gap", "grows", "distance", and "uniformly bounded". I had thought of introducing a *group* norm on $S$, but it didn't help much.

**Motivation.** I was led to Q2 while working at the proof of some estimates on sumsets in ordered semigroups / monoids.

Sumsets and structure). – Salvo Tringali Oct 28 '13 at 10:52thisis the case I am interested in (I edited the OP according to your comment). – Salvo Tringali Oct 28 '13 at 13:12