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I know that the hyperoctahedral group $(\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$ has the presentation

$$\langle s_{\text{1}},\ldots,s_n\mid s_{\text{1}}^{\text{2}}=s_i^2=1, (s_1s_2)^4=(s_is_{i+1})^3=(s_is_j)^2=1~| ~2\leq i\leq n, i+2\leq j\leq n \rangle,$$

where the generator $s_1$ is the diagonal matrix with a $-1$ in the top left entry and $1$s on the rest of the diagonal entry and $s_i$ are the permutation matrix formed from the identity switching both the $i-1$st and $i$th rows and corresponding columns.

Where can I get a proof for this?

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  • 1
    $\begingroup$ You mean a proof if your representation is correct? $\endgroup$ – draks ... Feb 2 '16 at 12:07
  • $\begingroup$ yes.. this is the coxeter representation which can be obtained from the coxeter graph. I need an explicit proof for this without depending anything on root systems and simple systems. $\endgroup$ – Anupam Ah Feb 5 '16 at 9:16
  • $\begingroup$ What homomorphism defines the semidirect product? $\endgroup$ – Shaun Nov 29 '18 at 22:57

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