Artin group

Remarks

For $m_{ij} = 2$ the relation is $s_i s_j = s_j s_i$, and for $m_{ij} = 3$ it is $s_i s_j s_i = s_j s_i s_j$. The quotient of $A(M)$ by the relations $s_i^2 = 1$ is the Coxeter group $W(M)$. $A(M)$ is of finite type, or spherical, when $W(M)$ is finite.

Example: the braid group $B_n$ is the Artin group with $I = \set{1, \ldots, n-1}$, $m_{i, i+1} = 3$ and $m_{ij} = 2$ for $\abs{i - j} \ge 2$. These are exactly Artin's relations $\sigma_i \sigma_{i+1} \sigma_i = \sigma_{i+1} \sigma_i \sigma_{i+1}$ and $\sigma_i \sigma_j = \sigma_j \sigma_i$. If all $m_{ij} = \infty$, there are no relations and $A(M)$ is the free group on $I$; if all $m_{ij} = 2$ for $i \ne j$, then $A(M)$ is free abelian.

Non-example: the group $\Z/2$ is not an Artin group. In the abelianization of $A(M)$, a relation with $m_{ij}$ even becomes trivial, and a relation with $m_{ij}$ odd becomes $s_i = s_j$. So the abelianization of an Artin group is free abelian, but $\Z/2$ is abelian and not free abelian.