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YCor
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Updated after a comment killing Q1
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Salvo Tringali
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Does $x \prec z$ imply Is $x + y \ne y+z$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 $x \prec y$.

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.

Does $x \prec z$ imply $x + y \ne y+z$ (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 $x \prec y$.

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

Some observations: The answer is yes if (i) $y$ is in the commutator of either of $x$ or $z$, (ii) $x \preceq y \preceq z$, (iii) $x+y \preceq y+x$, (iv) $z+y \preceq y + z$, or (v) the subgroup, $S$, generated by $x$, $y$ and $z$ is Archimedean (this follows from a 1902 theorem by O. Hölder).

My feeling is that the answer is positive in general (here is an informal reasoning): For any $n \in \mathbb{Z}$ we have $-y+nx+y = nz$. In my fantasy, this shoud be in contradiction to the fact that for $x \prec z$ the "gap" between $nx$ and $nz$ "grows" larger and larger as $n \to \infty$, while the "distance" between $nx$ and $-y+nx+y$ is "uniformly bounded". The point is that I don't see how to turn this 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 Q while working at the proof of some estimates on sumsets in ordered semigroups / monoids.

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.

Removed an apparently useless remark
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Salvo Tringali
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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 $x \prec y$.

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

Some observations: The answer is yes if (i) $y$ is in the commutator of either of $x$ or $z$, (ii) $x \preceq y \preceq z$, (iii) $x+y \preceq y+x$, (iv) $z+y \preceq y + z$, or (v) the subgroup, $S$, generated by $x$, $y$ and $z$ is Archimedean (this follows from a 1902 theorem by O. Hölder).

My feeling is that the answer is positive in general (here is an informal reasoning): For any $n \in \mathbb{Z}$ we have $-y+nx+y = nz$, so the smallest subgroup, say $T$, containing $z$ and $-y+x+y$ is torsion-free, abelian, and finitely generated. In my fantasy, this shoud be in contradiction to the fact that for $x \prec z$ the "gap" between $nx$ and $nz$ "grows" larger and larger as $n \to \infty$, while the "distance" between $nx$ and $-y+nx+y$ is "uniformly bounded". The point is that I don't see how to turn this 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 $T$$S$, but it didn't help much.

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

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 $x \prec y$.

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

Some observations: The answer is yes if (i) $y$ is in the commutator of either of $x$ or $z$, (ii) $x \preceq y \preceq z$, (iii) $x+y \preceq y+x$, (iv) $z+y \preceq y + z$, or (v) the subgroup, $S$, generated by $x$, $y$ and $z$ is Archimedean (this follows from a 1902 theorem by O. Hölder).

My feeling is that the answer is positive in general (here is an informal reasoning): For any $n \in \mathbb{Z}$ we have $-y+nx+y = nz$, so the smallest subgroup, say $T$, containing $z$ and $-y+x+y$ is torsion-free, abelian, and finitely generated. In my fantasy, this shoud be in contradiction to the fact that for $x \prec z$ the "gap" between $nx$ and $nz$ "grows" larger and larger as $n \to \infty$, while the "distance" between $nx$ and $-y+nx+y$ is "uniformly bounded". The point is that I don't see how to turn this 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 $T$, but it didn't help much.

Motivation. I was led to Q while working at the proof of some estimates on sumsets in ordered monoids.

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 $x \prec y$.

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

Some observations: The answer is yes if (i) $y$ is in the commutator of either of $x$ or $z$, (ii) $x \preceq y \preceq z$, (iii) $x+y \preceq y+x$, (iv) $z+y \preceq y + z$, or (v) the subgroup, $S$, generated by $x$, $y$ and $z$ is Archimedean (this follows from a 1902 theorem by O. Hölder).

My feeling is that the answer is positive in general (here is an informal reasoning): For any $n \in \mathbb{Z}$ we have $-y+nx+y = nz$. In my fantasy, this shoud be in contradiction to the fact that for $x \prec z$ the "gap" between $nx$ and $nz$ "grows" larger and larger as $n \to \infty$, while the "distance" between $nx$ and $-y+nx+y$ is "uniformly bounded". The point is that I don't see how to turn this 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 Q while working at the proof of some estimates on sumsets in ordered semigroups / monoids.

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Salvo Tringali
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