Say I have a set $S$ with a group operation $\cdot$ and lattice ordering $\leq$. Suppose further that:
- $x\leq 1\implies xy\leq y$
- $x\geq 1\implies xy\geq y$
For all $x,y$.
Does it follow that $(S,\cdot,\leq)$ is a lattice ordered group? i.e. does it hold that $(x\vee y)z=xz\vee yz$ and $(x\wedge y)z=xz\wedge yz$ for all $x,y,z$?
I think this is true because I can't think of any counter examples, but I'm having trouble proving it. Any hints?
EDIT: This was also asked on Math Overflow
If it is a partially ordered group then it is also a lattice ordered group. The usual way is to prove $$a≤b⇒xay≤xby.$$
Edit: Your task is to prove that it is a partially ordered group. Your implications are a consequence from the definition of right-ordered groups $$x≤y ⇒ xz≤yz$$.
You have either more information about your structure or you are stuck with a right-ordered group. So your question reads: Is every right-lattice-ordered group already a $\ell$-group?
So looking at this survey we know that such a group can be embedded into an $\ell$-group. As it doesn't tell anything about isomorphy neither as fact nor as open question nor as recently answered question, I guess the anwer to your question is: No.
Maybe someone who is familiar with right-ordered groups can help you.