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.
order of "truncated" braid groups
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0@SteveD: well, don't spend too much time on it. I was just curious. – 2012-04-27
2 Answers
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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0Right - the discussion there seems to resolve the finiteness question completely. – 2012-04-27