Finitely-presented torsionfree group whose abelianization has torsion In chat someone asked

Does anyone know of a torsion-free group (finitely presentable) whose abelianisation has torsion?

 A: $G=\pi_1(K)=\langle a,b\ |\ aba^{-1}=b^{-1}\rangle$ where $K$ is the Klein bottle is in fact left-orderable, which is quite a bit stronger than torsion-free (this is in fact true of any non-trivial fundamental group of a surface besides $\Bbb R \Bbb P^2$). To see this consider $G$ as the group of isometries of the plane generated by $a(x,y)=(1+x,-y)$ and $b(x,y)=(x,1+y)$. Now for $g \in G$, say $g>e$ if for $g(0,0)=(x,y)$ either $x>0$ or $x=0$ and $y>0$. It is quite easy to see that this is a left-order on $G$. As $H_1(K,\Bbb Z)=\Bbb Z \oplus \Bbb Z_2$, its abelianization has torsion. 
This proof is ripped from Rolfsen's talk here.
A: Denote by $H_3(\mathbb{Z})$ the integral Heisenberg group of unitriangular matrices $\mathbb{Z}$-entries. It has
$$x=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}, \quad y=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \quad z=\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} $$
in which $z$ commutes with $x$ and $y$, and $[x,y]=z$. In fact,
$$H_3(\mathbb{Z})=\langle x,y,z \mid [x,y]=z, [z,x],[y,x]\rangle $$
is a presentation for it. Verifying that $x,y,z$ generate $H_3(\mathbb{Z})$ and that they satisfy those relations should not be so hard, but I don't know a proof of why all relations come from those three. And yet it seems to be a well-known fact. The abelianization $H_3(\mathbb{Z})^\mathrm{ab}$ may be found by assuming $x,y,z$ commute and reconsidering the presentation: its generated by $x,y,z$ with single relation $z$. This is $\mathbb{Z}^2$.
Then consider $G=\big(H_3(\mathbb{Z})\times\mathbb{Z}\big)/\langle (z,-n)\rangle$. Writing $\mathbb{Z}=\langle w\rangle$, we can say that
$$G=\left\langle w,x,y,z ~\left| ~\begin{array}{c} [x,y]=z=w^n \\ [z,x],[z,y], \\ [w,x],[w,y],[w,z]\end{array}~~ \right. \right\rangle.$$
Evidently it has finite presentation. Suppose an element of $G$ were torsion. Its matrix part must have $0$ in its mid-upper and mid-right entries, so it is in $\langle \overline{z},\overline{w}\rangle\cong\mathbb{Z}^2/\langle(1,-n)\rangle$. If $\ell(a,b)=k(1,-n)$ for some $(a,b)\in\mathbb{Z}^2$ and nonzero $\ell,k\in\mathbb{Z}$ then $a=\frac{k}{\ell}$ is an integer so $(a,b)=\frac{k}{\ell}(1,-n)$. So $G$ is torsionfree.
The abelianization of $G$ is generated by $w,x,y,z$ with relations $z,w^n$, so $G^\mathrm{ab}\cong\mathbb{Z}^2\times\mathbb{Z}/n\mathbb{Z}$.
