How can I show the trefoil group is torsion-free

Question

I would like to show the trefoil group is torsion-free. The trefoil group has the presentation \begin{equation} G = \langle a, b \mid a^3 = b^2\rangle. \end{equation} I tried to map this to a simpler torsion-free group, for instance, if $h: G\to \mathbb{Z}$ by \begin{equation} a\to 2, b\to 3, \end{equation} then the torsion of $G$ must be in the kernel of $h$. However, the kernel is still pretty complicated.

Any ideas will be greatly appreciated!

Answer 1

I'll try to rewrite user1729's comment in more detail.

The theorem used is the following and can be found for example in Serre's Trees (Thm 8)

Every finite subgroup of $G=G_1\ast_A G_2$ is contained in a conjugate of $G_1$ or $G_2$.

Now, the trefoil group can be writen in the form $G=\mathbb{Z}\ast_A \mathbb{Z}$ where $A=\mathbb{Z}=\langle a \rangle$ and the two homomorphisms $f_2:A\to \mathbb{Z}=\langle x \rangle,\ f_3:A\to \mathbb{Z}=\langle y \rangle$ are $f_2(a)=x^2,\ f_3(a)=y^3$.

Therefore, if $g\in G$ is a torsion element then it should be an element of one of the two $\mathbb{Z}$'s. Since $\mathbb{Z}$ has no torsion elements it has to be $g=1$.