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What are the explicit generators and relations for type $B_3$ Weyl group? Thank you very much.

Edit: type $B_3$ Weyl group $G$ is $(\mathbb{Z}/2\mathbb{Z})^{3} \rtimes S_3$, so the order of $G$ is $48$. Type $B_2$ Weyl group can be written as $$ \{ s_1, s_2 \mid s_1^2=s_2^2=1, (s_1s_2)^4=1 \}. $$ How can we write $G$ in this way?

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The Coxeter diagram tells you directly what the Coxeter presentation is. What's your background here? –  Qiaochu Yuan Jul 25 '12 at 14:21
    
@Qiaochu, given a weight $\Lambda=\lambda_1 \omega_1 + \lambda_2 \omega_2 + \lambda_3 \omega_3$, I want to compute the character $\chi(\Lambda)$ using Weyl character formula. So I need to know the weyl group of type $B_3$. –  LJR Jul 25 '12 at 15:47

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The type $B_3$ Weyl group is $$\left\langle s_1, s_2, s_3 : s_1^2 = s_2^2 = s_3^2 = 1, (s_1 s_2)^4 = (s_2 s_3)^3 = (s_1 s_3)^2 = 1 \right \rangle$$

In terms of signed permutation matrices:

$$s_1 =\begin{bmatrix} -1 & . & . \\ . & 1 & . \\ . & . & 1 \end{bmatrix}, \qquad s_2 = \begin{bmatrix} . & 1 & . \\ 1 & . & . \\ . & . & 1 \end{bmatrix}, \qquad s_3 = \begin{bmatrix} 1 & . & . \\ . & . & 1 \\ . & 1 & . \end{bmatrix}$$

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The type $B_3$ Weyl group is generated by three elements $s_0, s_1, s_2$ subject to the relations $s_0^2 = s_1^2 = s_2^2 = (s_0 s_1)^4 = (s_0 s_2)^2 = (s_1 s_2)^3 = 1$. There is a well-known systematic approach to Weyl groups that Qiaochu Yuan referred to in his comment which is covered in many sources. One book that I recommend as an introduction to this area is Humphreys' Reflection Groups and Coxeter Groups.

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