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Given a set $S$ with a group operation $\cdot$ and a lattice ordering $\leq$, I wish to know when we can say that $\cdot$ preserves $\leq$, i.e. $(x\vee y)z=xz\vee yz$ and similarly for meets.

Specifically, say we know this for $x\leq 1$ - i.e. $x\leq 1\implies xz\leq z$ and similarly for $x\geq 1$. Is this sufficient for us to state that $G=(S,\cdot,\leq)$ is a lattice ordered group?

(This was previously asked on MSE, but I didn't get any answers so cross-posting here.)

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Please explain me: 1) why $\{g\in G: g\not\parallel 1\}$ is a subgroup? 2) why $\{g\in G: g\not\parallel 1\}$ is lattice-ordered? It don't seem following from your condition. – Boris Novikov Sep 13 '13 at 19:47
@Boris: You are right! The group of functions $\mathbb R\to\mathbb R$ under addition with pointwise ordering is a counter example to this subset being closed. But the overall group still is lattice ordered. I will update the question. – Xodarap Sep 14 '13 at 11:56
I added the answer. – Boris Novikov Sep 14 '13 at 16:42
Is every right-ordered group with a lattice order already an ordered group? If not there are counterexamples as the implications $x≤1 ⇒ xz≤z$ and similarly for $x≥1$ is a conclusion of being right-ordered. Otherwise it would be sufficient to prove that it is a right-ordered group. – Tobias Schlemmer Sep 14 '13 at 17:15
Taking into account the comment of Tobias Schlemmer I corrected the answer. – Boris Novikov Sep 15 '13 at 13:21

Known facts

  • $(G,≤)$ is a lattice
  • $x≤1 ⇒ xy ≤ y$
  • $1≤x ⇒ y ≤ xy$
  • an $\ell$-group is defined to be a po-group with a lattice order (Копытов: Решеточно упорядоченные группы, chap. 2.1.1; English version published by Springer)
  • every po-group is a right partially ordered group
  • each right partially ordered group implies the two conditions $x≤1 ⇒ xy ≤ y$ and $1≤x ⇒ y ≤ xy$

Easy conclusions

  • $(G,∙,≤)$ is a po-group $⇒$ $(G,∙,≤)$ is an $\ell$-group (by definition).
  • If every right lattice ordered group is already a po-group then $(G,∙,≤)$ is a $\ell$-group.
  • If $(G,∙,≤)$ is a right partially ordered group then it fulfils Xodaraps conditions.

Provable facts

Statement 1

If $≤$ is a linear order then $(G,∙,≤)$ is a right ordered group.

Suppose that for some $x,y,z∈G$, $x≠y$ the conditions $x≤y$ and $xz\not≤yz$ are true. The latter can be rewritten to $yz≤xz$ as $≤$ is a linear order. Then we know that either $xy^{-1}≤1$ or $1≤xy^{-1}$ is true. Multiplying the latter with $y$ leads to $y≤x$ which is a contradiction to $x≠y$. Thus we know $xy^{-1}≤1$. Multiplying this inequality from the right side with $yz$ leads to $xz≤yz$. Thus $(G,∙,≤)$ is right ordered.

A remark to 2) from Boris' answer: An arbitrary order extension may change global properties, such as being a right partially ordered group. In fact a linear order defines that also those elements must be comparable with $1$ that have been incomparable. Thus, there are many ways to violate Xodaraps conditions.

Statement 2

If $≤$ is the intersection of linear orders that fulfil Xodaraps conditions then $(G,∙,≤)$ is a right partially ordered group.

By Statement 1 each linear order is a right ordered group. If $x≤y$ holds, then this condition is true in all of the other linear order extensions. Then in all of these extensions we get $xz≤yz$, thus this inequality is preserved by the intersection.

Other remarks

Searching the internet after “right lattice ordered groups” reveals some articles about “half lattice ordered groups”. This looks very promising in order to find an example of a group that fulfils the conditions but is not an $\ell$-group.

The terms “partially ordered” and “ordered” are used synonymous as well as “linearly ordered” and “ordered”, depending on the focus of the author and context.

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Thanks Tobias. Based on your suggestion of "half lattice ordered groups", I feel that a variant $\mathbb R^\times$ seems promising but I'm not able to find an explicit example yet. – Xodarap Sep 17 '13 at 0:17
I think you should contact one of the authors of the corresponding papers and ask them if they can help you. It is a very special branch of mathematics and propably of the contributers are not active participants of MO. – Tobias Schlemmer Sep 17 '13 at 18:22

Partially ordered group $G$ is a lattice ordered if and only if for every $a\in G$ there is a least upper bound $a\vee e$ in $G$. [L.Fuchs, Partially Ordered Algebraic Systems, 1963].

Addendum: Thanking to the comments of Mark Sapir and Tobias Schlemmer one can prove:

Proposition. Let $S(\cdot,\leq)$ be a group with a lattice order and the condition: $x\leq 1\implies xz\leq z, \ zx\leq z$, and similarly for $x\ge 1$.

1) If $S(\cdot,\leq)$ is a partially ordered group, then it is a lattice ordered group.

2) If $S(\cdot,\leq)$ is not a partially ordered group, then there is such a linear (hence lattice) exstension $\preceq$ of $\leq$ that $(x\vee y)z\ne xz\vee yz$ for some $x,y,z$.

Proof. 1) Since there is $a\vee e$, $S$ is lattice ordered (see above).

2) If $S(\cdot,\leq)$ is not a partially ordered, then $x< y$, but $xz\not< yz$ or $zx\not< zy$ for some $x,y,z$. Let, for example, $xz\not< yz$. By Szpilrajn theorem we can extend $\leq$ up to linear $\preceq$ such that $yz\prec xz$. Then $(x\vee y)z=yz\ne xz=xz\vee yz$.

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But does this answer the question? The OP already has a lattice order. My interpretation of the question is that what he wants to know is whether he has a partially ordered group or not. – Joel David Hamkins Sep 13 '13 at 12:35
@Joel is correct. The hard part is showing that $G$ is order-preserving - I already know it has an order. (I'm not sure if there's good terminology to distinguish "lattice ordered group" from "group with a lattice order that might not be preserved under the operation"...) – Xodarap Sep 13 '13 at 12:40
@Xodarap OK, you are right. I will think. – Boris Novikov Sep 13 '13 at 12:45
@Joel David Hamkins You are right. I will think. – Boris Novikov Sep 13 '13 at 12:46
@Xodarap: Every group has an linear order which "might not be preserved under the operation": just take any linear order on the set of elements of the group. – Mark Sapir Sep 13 '13 at 19:45

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