# Conditions for a group to be lattice-ordered

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.)

-
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.

-
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 $(x\vee y)z=yz\ne xz=xz\vee yz$.

-
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