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Let $H$ be a subgroup of lattice-ordered group $G$. $H$ is a lattice-ordered group as a induced partial order from $G$ but $H$ is not a lattice-subgroup of $G$. For $a, b\in H$, let $ c=inf(a, b) \in H$ and let $d= inf(a, b) \in G$. Is it necessary to be $c = d$ or not? Thanks

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Isn't it true by definition? A lattice ordered group is an algebra with 4 operations: $\cdot, ^{-1}, \wedge, \vee$, so a lattice ordered subgroup is a subalgebra with respect to these 4 operations. – Mark Sapir Feb 18 '13 at 3:50
Sometimes a lattice ordered group is defined as a partially ordered group were the partial order happens to be a lattice, i.e., any two elements have infimum and supremum. In this case the answer to this question is not so obvious. – Stefan Geschke Feb 18 '13 at 12:56
@ Mark Sapir and Stefan Geschke, Thanks. I am trying to show $c=d$ but still I am not – Rajnish Feb 18 '13 at 21:24
As I understand, @Rajnish asks about a subgroup (not a lattice subgroup) which happens to be a lattice w.r. to the induced partial order, but which is NOT a sublattice of the whole group because the lattice operations in the subgroup are not the same as in the whole group. There are other algebraic structures (instead of a group) where this kind of a situation is common. – Włodzimierz Holsztyński Feb 19 '13 at 1:12
@ Wlodzimierz Thank you very much. I am completely agree with the "a subgroup (not a lattice subgroup) which happens to be a lattice w.r. to the induced partial order". I am going to correct my question. – Rajnish Feb 19 '13 at 5:21
up vote 2 down vote accepted

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 $G$.

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@Fred, Thank you. I did not get the idea used for subgroup $H$. Why is that on third compoent $(x + y)\vee (u + v) = x\vee u + y\vee u$. – Rajnish Feb 20 '13 at 21:18
Dear Rajnish, I do not understand your question. Please clarify. – Fred Rohrer Feb 20 '13 at 21:41
@Fred,Thank you very much. I did not get the result of third component in this line $sup_{H}((x,y,x+y), (u,v, u + v)) = (sup(x,u), sup(y,v), sup(x,y) + sup(y,v))$. – Rajnish Feb 20 '13 at 22:11
First, $\sup(x,u)+\sup(y,v)$ is greater than $x+y$ and than $u+v$. Second, if $(a,b,a+b)\in H$ is greater than $(x,y,x+y)$ and $(u,v,u+v)$, then $\sup(x,u)$ is smaller than $a$ and $\sup(y,v)$ is smaller than $b$. Hence, $\sup(x,u)+\sup(y,v)$ is smaller than $a+b$. This yields the claim. (Note that the third component needs to be the sum of the first and the second in order for the triple to be an element of $H$.) – Fred Rohrer Feb 20 '13 at 22:59
@ Fred, Thank you. I got it. – Rajnish Feb 20 '13 at 23:55

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