I need to show that the Baumslag-Solitar group $BS(1,2)$ is torsion-free. It is given by
$$BS(1,2) = \langle a,t \mid tat^{-1} = a^2\rangle$$
I have taken a non-trivial element $g$ from $BS(1,2)$. I assumed it is of finite order and tried to play with it, in order to get contradiction, with no luck.
${\rm BS}(1,2)$ has a normal subgroup $N = \langle a^G \rangle$ with $G/N$ infinite cyclic, so a torsion element would be contained in $N$. But $N$ is isomorphic to the additive subgroup $\{ a2^n : a,n \in {\mathbb Z} \}$ of ${\mathbb Q}$ and so is torsion-free.