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Is it possible to have a finitely generated lattice - ordered group has rational rank bigger than the number of generators of the group?

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Is the group abelian? Otherwise I don´t know the definition of rational rank. –  Giovanni De Gaetano Feb 7 '12 at 9:27

1 Answer 1

This answer suppose, given the tag, that we are dealing with an Abelian ordered group $G$. (Actually the fact that $G$ is ordered does not come into play once supposed the abelianity)

Then we can suppose, without loss of generality, $G$ to be torsion free. Indeed if $T(G)$ is the torsion group of $G$ then $r(G)=r(G/T(G))$. So we can suppose $G=G/T(G)$ which has equal rank and maybe a smaller set of generators.

But then $G$ is abelian, finitely generated and torsion free, so it is an abelian free group with $n$ generators and it has rank $n$.

In conclusion it is impossible to have such a group.

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Thank you Student 73. The group here is abelian group. Because of lattice -ordered group, the group is torsion free. I found the people using the statement as "finitely generated lattice-ordred abelian group of finite rational rank". This statement makes me curious is that, if the group is finitely generated why do we need to worry about finite rational rank, this is by defult should be finite rational rank but I could not convinced myself and I posted this question. In general the definition of rational rank is maximal number of rationally independent elements of the torsion free group G. –  Rajesh Feb 7 '12 at 14:25
    
Ok! It´s nice to know for the next time, thank you! If you are satisfied with the answer can I ask you to accept it? And, by the way, welcome to Math.SE! –  Giovanni De Gaetano Feb 7 '12 at 14:30

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