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- 1. Can every torsion-free nilpotent group be ordered? 2. 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?

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.