TITLE: Can every torsion-free nilpotent group be ordered?
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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.