# Is there a non-right-orderable torsion-free quotient group of the braid group on 3 strands?

The braid group on 3 strands has the presentation $$\langle x,y \;|\; xyx=yxy\rangle$$. A group $$G$$ is called right-orderable if there is a total order $$<$$ on the set $$G$$ such that if $$a then $$ac for all $$c\in G$$. It is known that braid groups are right-orderable.

Is there a non-right-orderable torsion-free quotient group of the braid group on 3 strands?

• What do you mean by a factor? – HJRW Mar 15 '15 at 21:06
• @HJRW: factor means quotient. – Alireza Abdollahi Mar 15 '15 at 21:19

Yes, there are many such examples. The braid group is isomorphic to the fundamental group of the trefoil knot complement. The trefoil knot $T$ admits many Dehn fillings, parameterized by $r\in \mathbb{Q} \cup \{\infty\}$. If $|r|\geq 1$, then the Dehn filling $S^3_r(T)$ is an L-space. Moreover, it is usually Seifert-fibered and has torsion-free fundamental group. By Theorem 4 of Boyer-Gordon-Watson, these Dehn fillings do not have orderable fundamental group (the notion of L-space is actually not relevant to the answer to this question, it just gave the quickest way to cite the literature).
• What is $\hat{\mathbb{Q}}$? – Qiaochu Yuan Mar 16 '15 at 4:55
• $\mathbb{QP}^1$ - I changed the notation, since this is probably only used by 3-manifold topologists. – Ian Agol Mar 16 '15 at 5:00
• @IanAgol: Is it known a group presentation (not necessarily finite) for $\pi_1(S_r^3(T))$ for some $r$ with $|r|\geq 1$? – Alireza Abdollahi May 26 '15 at 20:03