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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!

froyooo
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    In general, an open subset of $R^3$ has no torsion in the fundamental group, see here. Perhaps there is a more direct approach though. – Connor Malin Jul 19 '21 at 17:27
  • You could talk about the group elements in terms of words consisting of 'a's and 'b's and their inverses, and talk about their length after resolving a letter with its inverse, and frame the relation $aaa \to bb$ as a "rewrite rule" for those words, and prove that with that one rewrite rule there is now way to get a word of length one. ... This would all have to be made rigorous though, but I think approaching the task with that mindset helps. – Mike Pierce Jul 19 '21 at 17:29
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    @MikePierce I think you mean there no way to reduce a word to length zero. – Arthur Jul 19 '21 at 17:41
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    One way to see torsion-free-ness is to realise that this is a free product with amalgamation of two torsion-free groups (namely two copies of $\mathbb{Z}$). The result from here is pretty standard (e.g. via Bass-Serre theory, or I think in Magnus, Karrass and Solitar's book Combinatorial Group Theory). Alternatively, the group is a one-relator group whose relator is not a proper power, and so is torsion-free (this is definitely proven in Magnus, Karrass and Solitar, although better proofs exist now). – user1729 Jul 19 '21 at 17:51
  • The trefoil group is isomorphic to $ B_3 $, the braid group on 3 strands. And all braid groups are known to be torsion-free. Not self contained but maybe convincing for someone who thinks about braid groups a lot? – Ian Gershon Teixeira Jul 21 '23 at 18:23

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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 $g=1$.

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    +1 Nice profile picture! (Also, it is maybe worth pointing out that the proof of Serre's Theorem 8 is not difficult, and is a nice introductory proof on groups acting on trees.) – user1729 Jul 20 '21 at 08:59