# Is $\mathbb{Z}^2$ endowed with the square of the strict order, a lattice-ordered group?

I was looking some lattice-ordered group structure. I have kind of difficulty to figure out about the group $$\mathbb{Z}^{2}$$ with positive cone is $$\mathbb{N}_{>0} \times \mathbb{N}_{>0} \cup \{(0,0)\}$$ is lattice-ordered group or not.

Added: Recall that a lattice-ordered group is a group $$(G,\cdot)$$ endowed with a poset structure $$(G,\le)$$ such that the partial order $$\le$$ is left and right invariant ($$x\le y$$ implies $$axb\le ayb$$), and the underlying order is a lattice. In a group, every left-invariant order is determined by its positive cone $$\{g:g\ge e_G\}$$.

• I edited your latex and made the notation a little more standard. – David Roberts Mar 13 '12 at 1:01
• [Updated comment (after Aaron answer)] Let us call P to your proposal for positive cone. Since P is closed under addition in $\mathbb{Z}^2$ and it contains the neutral element, it is obvious that we have a partially ordered group under the following order definition: $x \leq y$ iff $x - y \in P$. Thus, it is enough to check that this order is a lattice, i.e., every two elements have an infimum and a supremum. I thought this could be straightforwardly checked, but as Aaron says in his answer this is not the case; e.g., there is no infimum of $(a,b)$ and $(c,d)$ when $a<c$ and $b>d$. – boumol Mar 13 '12 at 15:25

This is not a lattice-ordered group. As mentioned by boumol, it is (partially-)ordered. A simple characterization of $(G,G_+)$ being lattice-ordered (where $G$ is an ordered group with positive cone $G_+$) is the following: every intersection of two translates of $G_+$ is itself a translate of $G_+$, i.e. for any $x,y \in G$, there exists $z \in G$ such that

$(x + G_+) \cap (y + G_+) = z + G_+.$

(This is equivalent to being lattice-ordered, since this says that $z$ is the supremum of $x$ and $y$.)

In your case, use $x=(0,1)$ and $y=(1,0)$ to show that your ordered group is not lattice-ordered.

• I don't have sufficient reputation to leave comments, so this comment is out of place. For clarity, I would point out why $(a-1,d-1)$ is not the infimum of $(a,b)$ and $(c,d)$ when $a < c$ and $b > d$: we have, for example, $(a-1,d-2) \leq (a,b), (c,d)$ yet $(a-1,d-2) \not\leq (a-1,d-1)$. – Aaron Tikuisis Mar 13 '12 at 14:00
• Alternatively, you could just check directly that $(0,1)$ and $(1,0)$ have no least upper bound. They have upper bounds $(2,1)$ and $(1,2)$ but no upper bound below both of these. – Andreas Blass Mar 13 '12 at 14:28
• @Aaron: Thanks for the clarification why my previous suggestion is wrong. @Andreas: I am afraid that in the partially ordered group proposed in the question it not true that $(2,1)$ is greater or equal than $(0,1)$ (because the difference is $(2,0)$, and hence not in the proposed positive cone). – boumol Mar 13 '12 at 15:21
• @Andreas: You're right (as long as you replace $(2,1)$ and $(1,2)$ by $(3,2)$ and $(2,3)$), this is a more elementary proof. I would like to nonetheless point out that the characterization that I gave can make it easier (at least conceptually) to check if other examples are lattice-ordered. – Aaron Tikuisis Mar 13 '12 at 16:31
• @Aaron: As you say in the last remark your characterization is very useful from a conceptual point of view. Let me add some clarification. For arbitrary partial orders it is false that the existence of supremum of two elements also gives the existence of infimum of two elements. However, using the group structure it is possible to prove that this is the case (because the infimum of $x$ and $y$ can be defined as $- \sup \{-x,-y\}$). – boumol Mar 13 '12 at 16:51