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 $x \prec y$.
QQ1. If $x,y,z \in A$ and $x \preceq z$, is it true that $x+y \ne y + z$ unless $x = z$?
Some observations: TheAs pointed out by Yves Cornulier in the comments below, the answer is yes if (i) $y$ is iniff the commutator of either of $x$ or $z$, (ii) $x \preceq y \preceq z$,ambient group is abelian. This means that I was too hasty (iiiand optimistic) $x+y \preceq y+x$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 (ivi) $z+y \preceq y + z$$x+y\preceq y+x$, or (vii) the subgroup, $S$, generated by $x$, $y$ and $z$$y$ is Archimedean (this follows from a 1902 theorem by O. Hölder).
MyHere is an feelinginformal is that the answer is positive in general (here is an informal reasoning) in support of a positive answer: For anyAssume to the contrary that $n \in \mathbb{Z}$ we have$x+y=y+nx$ for some $-y+nx+y = nz$$n \in \mathbb{N}^+$. In my fantasyThen, this$-ky+x+ky = n^kx$ for all $k \in \mathbb{N}$. This shoud be in contradiction to the fact that for $x \prec z$$n \ge 2$ and $x \ne 0$ the "gap" between $nx$$n^kx$ and $nz$$-ky+x+ky$ "grows" larger and larger as $n \to \infty$$k \to \infty$, while the "distance" between $nx$$x$ and $-y+nx+y$$-ky+x+ky$ is "uniformly bounded". The point is that I don't see how to turn thisthe 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 QQ2 while working at the proof of some estimates on sumsets in ordered semigroups / monoids.