Can every torsion-free nilpotent group be ordered? I know that a torsion-free abelian group can be ordered and have done two proofs for that too. But the next two question that popped up in my mind were-


*

*Can every  torsion-free nilpotent group be ordered?

*Can every  torsion-free solvable group be ordered?


After googling I got to know that answer to first is positive but I could not  find a proof. I am thinking on it. May be use induction on class of nilpotent group as it is usually useful with nilpotent groups. If anyone can guide me on that , I will be thankful. 
What is the answer for solvable, or is it unknown?
 A: Since I'm not sure what you call "orderable", I'll consider both bi-orderability and left orderability.
1) Every torsion-free nilpotent group is bi-orderable. It's true by an induction using: if $Z$ is central, torsion-free in $G$ and $G/Z$ is bi-orderable, then $G$ is bi-orderable (easy, can be checked directly).
2) Every poly-(torsion-free abelian) group is left-orderbale. It's true because being left-orderable is stable under taking extensions. This can be applied, for instance, for the $\pi_1$ of the Klein bottle $\langle x,y\mid x^2=y^2\rangle$ mentioned by Derek (which is not bi-orderable as it admits conjugate nontrivial inverse elements).
3) There exist torsion-free solvable (polycyclic) groups that are not left-orderable (and hence not bi-orderable). Any nontrivial torsion-free solvable finitely generated group with finite abelianization is such an example.
A: I know very little about ordered groups, but I believe that one of their properties is that they have unique roots; i.e. for all $n \ge 1$, $x^n=y^n \Rightarrow x=y$.
The group $G$ defined by the presentation $\langle x,y \mid x^2 = y^2 \rangle$ obviously does not have unique roots, so it cannot be orderable, but it is torsion-free and solvable.
To see that, note that the subgroup $Z = \langle x^2 \rangle$ is central, and $G/Z$ is infinite dihedral. So $G$ is solvable. The only torsion elements of $G/Z$ are the images of conjugates of $x$ and $y$, and these all square into $x^2$, so $G$ is torsion-free.
