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