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 $\{g:g\ge e\_G\}$.



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


