Let $H$ be a subgroup of a 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 $G$?
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$\begingroup$ 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. $\endgroup$– user6976Feb 18, 2013 at 3:50
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$\begingroup$ 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. $\endgroup$– Stefan GeschkeFeb 18, 2013 at 12:56
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$\begingroup$ @ Mark Sapir and Stefan Geschke, Thanks. I am trying to show $c=d$ but still I am not $\endgroup$– RajnishFeb 18, 2013 at 21:24
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$\begingroup$ 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. $\endgroup$– Włodzimierz HolsztyńskiFeb 19, 2013 at 1:12
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$\begingroup$ @ 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. $\endgroup$– RajnishFeb 19, 2013 at 5:21
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1 Answer
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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 $G$.
answered Feb 19, 2013 at 6:04
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$\begingroup$ @Fred, Thank you. I did not get the idea used for subgroup $H$. Why is that on third compoent $(x + y)\vee (u + v) = x\vee u + y\vee u$. $\endgroup$– RajnishFeb 20, 2013 at 21:18
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$\begingroup$ Dear Rajnish, I do not understand your question. Please clarify. $\endgroup$ Feb 20, 2013 at 21:41
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$\begingroup$ @Fred,Thank you very much. I did not get the result of third component in this line $sup_{H}((x,y,x+y), (u,v, u + v)) = (sup(x,u), sup(y,v), sup(x,y) + sup(y,v))$. $\endgroup$– RajnishFeb 20, 2013 at 22:11
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$\begingroup$ 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 $H$.) $\endgroup$ Feb 20, 2013 at 22:59