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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.

StephenJ
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  • Is it clear that it has finite order? – Cheerful Parsnip Apr 26 '12 at 21:44
  • 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 $d$. – Arturo Magidin Apr 26 '12 at 21:44
  • I don't really know that the group is finite. Perhaps I should have mentioned that explicitly in the question. – StephenJ Apr 26 '12 at 22:12
  • I do not think the case of $n=3$ and $d=6$ is finite. –  Apr 26 '12 at 23:10
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    In particular, through Fox Calculus I can see that the commutator of that group has abelianization $\mathbb{Z}\times\mathbb{Z}$. –  Apr 26 '12 at 23:18
  • Thanks, Steve. That is quite helpful. The cases of most interest to me are d=4 and d=6. – StephenJ Apr 27 '12 at 00:16
  • @SteveD: can you give your argument in an answer? I'm curious to see it. – Cheerful Parsnip Apr 27 '12 at 02:09
  • @JimConant: Perhaps tomorrow? I think it will take a while to typeset. But I will say this for now: it seems the isomorphism type of $G'/G''$ only depends on $d\pmod{6}$, and not on $n$ (as long as $n,d>2$). For $d=6$, $G'/G''$ is $\mathbb{Z}\times\mathbb{Z}$, so certainly in that case $G$ is infinite. When $d=4$, $G'/G''$ has order $3$, so perhaps another attack will prove useful. –  Apr 27 '12 at 04:13
  • In fact, it is not too hard to show that if we call your group $G_n$ (for some fixed $d$), then $G_n'$ contains a copy of $G_{n-2}$. So if we can just prove one infinite we would be done. I believe that $G_4$ is always infinite, but the $d=4$ case is proving very difficult. For $d=4$, $G_3$ is finite (of order 96). –  Apr 27 '12 at 05:12
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    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 $G$. – Derek Holt Apr 27 '12 at 10:01
  • @SteveD: well, don't spend too much time on it. I was just curious. – Cheerful Parsnip Apr 27 '12 at 11:48

2 Answers2

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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.

Derek Holt
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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 $\mathbb{R}^3$.

Dan Rust
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