# Is $x + y \ne y+nx$ for $x \ne 0$ and $n \ge 2$ (in an ordered group)?

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.

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Are you writing the group additively, even though it is not abelian? – Tobias Kildetoft Oct 28 '13 at 10:45
Yes, I am. This is common in additive theory (e.g., see Ruzsa's survey Sumsets and structure). – Salvo Tringali Oct 28 '13 at 10:52
This is true only if $A$ is abelian. Otherwise, there exist two distinct conjugate elements $x,z$; since your order is total, we can suppose $x\preceq z$. So there exists $y$ such that $xy=yz$ (I avoid your additive notation which is confusing to me). – YCor Oct 28 '13 at 11:58
Right, I was too hasty (and optimistic). But then, what if $z = 2x$? Actually, this is the case I am interested in (I edited the OP according to your comment). – Salvo Tringali Oct 28 '13 at 13:12

As I pointed out in the comments, Q1 has a negative answer in general and actually has a positive answer only for abelian groups. Your second question (in the version of October 28 2013, 14:54 GMT) also has a negative answer. Indeed, the Baumslag-Solitar group $G=\langle x,t\mid txt^{-1}=x^n\rangle=\mathbf{Z}[1/n]\rtimes_n\mathbf{Z}$ is bi-orderable if $n\ge 1$ (i.e. has a left and right invariant total order, what is sometimes called orderable, the latter sometimes also meaning left-orderable). Indeed, if you consider the set $P$ of $(y,m)\in\mathbf{Z}[1/n]\rtimes\mathbf{Z}$ such that either $m>0$ or $m=0$ and $y\ge 0$, then $P$ is invariant under conjugation (because $n\ge 1$), is stable by multiplication, and $P\cup P^{-1}=G$, $P\cap P^{-1}=\{1\}$. Hence $P$ defines a bi-invariant order (with $P\smallsetminus\{1\}$ as set of positive elements. But the element $x=(1,0)$ is conjugate to $x^n$, so this is a counterexample to your expectation as soon as $n\ge 2$.
Thank you, Yves, this is (very) helpful. I must admit I'm (happily) surprised to learn that ${\rm BS}(1,n)$ is bi-orderable for $n \ge 1$ (the case $n = 1$ being trivial). – Salvo Tringali Oct 30 '13 at 21:31