Let $H$ be a subgroup of a [lattice-ordered group](https://www.encyclopediaofmath.org/index.php/Lattice-ordered_group) $G$. Suppose that $H$ with the induced order is a lattice (but a priori not a sublattice), so that $H$ is a lattice-ordered group too. For $a, b\in H$, let $ c=\inf\_H(a, b) \in H$ and let $d= \inf\_G(a, b) \in G$. Is it necessary to be $c = d$ or not? In other words, is $H$ an l-subgroup of $G$?



Subgroup of lattice-ordered group
No. A counterexample (essentially from Bourbaki's *Algèbre* VI.1 Exercice 12 a)) is the following.


We furnish $\mathbb{Z}$ with its usual structure of ordered group and consider the product of ordered groups $G=\mathbb{Z}^3$. This is a lattice, and for $(x,y,z),(u,v,w)\in G$ we have $$\textstyle\sup\_G((x,y,z),(u,v,w))=(\sup(x,u),\sup(y,v),\sup(z,w)).$$ Now we consider the subgroup $H=\{(x,y,z)\in G\mid z=x+y\}$ of $G$, furnished with its induced structure of ordered group. This is also a lattice, as one readily checks that for $(x,y,x+y),(u,v,u+v)\in H$ we have $$\textstyle\sup\_H((x,y,x+y),(u,v,u+v))=(\sup(x,u),\sup(y,v),\sup(x,u)+\sup(y,v)).$$ However, since $$\textstyle\sup\_G((0,1,1),(1,0,1))=(1,1,1)\neq(1,1,2)=\sup\_H((0,1,1),(1,0,1))$$ we see that $H$ is *not* a sublattice of $G$.


