order of "truncated" braid groups 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.
 A: 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.
A: 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 $\mathbb{R}^3$.
