Consider the braid group on n strands given in the usual Artin presentation. Then add extra relations: each Artin generator has order d. For example, if d=2, one recovers the symmetric group. I would like to know what the order of the group is for arbitrary n and d. Even knowing the name of such groups would be helpful, though, as my attempts to determine this by searching the literature have so far failed.

  • $\begingroup$ Is it clear that it has finite order? $\endgroup$ Apr 26, 2012 at 21:44
  • $\begingroup$ Is there a reason to believe a priori that the group will be finite? The negative solution to the Burnside problem says that there are finitely generated groups in which every element (not just the generators) has order $d$ that are infinite, for sufficiently large $d$. $\endgroup$ Apr 26, 2012 at 21:44
  • $\begingroup$ I don't really know that the group is finite. Perhaps I should have mentioned that explicitly in the question. $\endgroup$
    – StephenJ
    Apr 26, 2012 at 22:12
  • 2
    $\begingroup$ In particular, through Fox Calculus I can see that the commutator of that group has abelianization $\mathbb{Z}\times\mathbb{Z}$. $\endgroup$
    – user641
    Apr 26, 2012 at 23:18
  • 2
    $\begingroup$ A computer calculation shows that, if $G = G_4$ with $d=4$, then $G^{(4)}/G^{(5)}$ is infinite, where $G^{(i)}$ is the derived series of $G$. $\endgroup$
    – Derek Holt
    Apr 27, 2012 at 10:01

2 Answers 2


I thought I would add what is almost a complete answer which is outlined in the book

K. Murasugi & B. Kurpita, A Study of Braids, Kluwer Academic Publishers, 1999.

The following surprising theorem tells us when the truncated braid groups are finite, and the order of the groups when they are.

Theorem: Let $B_n(d)=B_n/\langle\sigma_i^d \rangle$. The group $B_n(d)$ is finite if and only if $d=2$ or $(n,d)$ is the type of one of the 5 platonic solids. For these cases, $$|B_n(d)|=\left(\frac{f(n,d)}{2}\right)^{n-1}n!$$ where $f(n,d)$ is the number of faces of the platonic solid of type $(n,d)$

The 5 platonic solids correspond to the pairs $(n,d)\in\{(3,3),(3,4),(4,3),(3,5),(5,3)\}$. This is equivalent to the pair $(n,d)$ being a solution to the inequality $$\frac{1}{n}+\frac{1}{d}>\frac{1}{2}.$$

For ease of calculation, we have the following table giving the number of faces of the corresponding platonic solids $$\begin{array}{|r|l|}\hline (n,d)&f(n,d)\\\hline (3,3)&4\\ (3,4)&8\\ (4,3)&6\\ (3,5)&20\\ (5,3)&12\\\hline \end{array}$$ and so we can calculate the table of group orders $$\begin{array}{|r|l|}\hline (n,d)&|B_n(d)|\\\hline (3,3)&24\\ (3,4)&96\\ (4,3)&648\\ (3,5)&600\\ (5,3)&155520\\\hline \end{array}$$

To me, this theorem and its application highlights one of the strangest links between two fairly weakly related areas of mathematics; finite groups arising from topological or combinatorial (pick you favourite description of the braid groups of the disk) considerations, and the geometric classification of regular solids in $\mathbb{R}^3$.


Let's consider just the 2-generator braid group, with added relations $a^d=b^d=1$. A computer coset enumeration shows that this is finite of order 6, 24, 96, and 600 for $d=2,3,4,5$.

If we now add the extra relation $(ab)^3=1$, giving

$G_d = \langle a,b \mid aba=bab, (ab)^3 = a^d = b^d = 1 \rangle.$

and peform a routine change of generator calculation with $x=ab$, $y=xa=aba$ using Tietze transformations, then we get the presentation

$\langle x,y \mid x^3 = y^2 = (xy)^d = 1 \rangle,$

a triangle group, which is well-known to be infinite for $d \ge 6$. So the 2-generator braid group with added relations is also infinite for $d \ge 6$.

With $d=3$, the 3- and 4-generator groups are finite of order 648 and 155520. I suspect that all other cases are infinite, but I don't known for sure.

This would be also be a reasonable question to ask on MathOverflow.


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