Let $V:=\Bbb R^\ell $, $\ell \geq 2$ with an orthonormal basis $\{e_i\}_{i=1}^\ell $. The set
$$\Phi = \{ \pm e_i \pm e_j \mid 1 \le i\neq j \le \ell\} \cup \{ \pm e_i \mid 1 \le i \le \ell\} $$ is known as a root system of types $B_\ell$. Let $$\Delta = \{\alpha_1:=e_1-e_2,\ldots,\alpha_{\ell-1}:=e_{\ell-1}-e_\ell,\alpha_{\ell}:=e_\ell\}$$ be a base of $\Phi$. I want to understand why the Weyl group $W$ of $\Phi$ is isomorphic to the semidirect product of an elementary abelian two group and a symmetric group i.e. $W \simeq (\Bbb Z/2\Bbb Z)^\ell \rtimes S_\ell$? I may need to prove the following:

(1) $W=NH$ where $N \simeq (\Bbb Z/2\Bbb Z)^\ell$ and $H\simeq S_\ell$.

Denote the reflections by the letter $r$. Set $N:=\langle r_{e_i}\mid1 \le i \le \ell\rangle$ and $H:=\langle r_{\alpha_i}\mid1 \le i \le \ell-1\rangle$ then $N \simeq (\Bbb Z/2\Bbb Z)^\ell$ and $H\simeq S_\ell$. But I cannot see why $W=NH$? My only concern is that $r_{e_\ell}$ and $r_{\alpha_{\ell-1}}$ do not commute.

(2) $N=\langle r_{e_i}\mid1 \le i \le \ell\rangle$ is a normal subgroup of $W$.

I don't know any appropriate criterion to prove $N$ is normal.

(3) $N \cap H =\{e\}$.

This part is done. Intuitively, elements of $H$ are permutations of $\{e_1,e_2,\ldots,e_{\ell}\}$ while elements of $N$ are sign changes. So their intersection necessarily consists of only the identity.

Any help would be much appreciated.

  • $\begingroup$ $W$ acts on the collection of pairs of vectors $\pm e_i, i=1,2,\ldots,\ell$. There are $\ell$ such pairs, so we have a homomorphism $W\to S_\ell$. The kernel of that homomorphism consists of the sign changes, and it is easy to check that all sign changes occur, because $r_{e_i}$ only changes the $i$th sign. $\endgroup$ – Jyrki Lahtonen Oct 27 '16 at 6:03
  • $\begingroup$ Which part of my post are you commenting to? Or you're using another method to prove the isomorphism? Can you give me the explicit definition of your homomorphism ,say $f: W \to S_\ell$ mentioned above? $\endgroup$ – user Oct 27 '16 at 6:12
  • $\begingroup$ $f:W\to S_\ell$ is defined as follows. Each $w\in W$ gives rise to a permutation $\sigma\in S_\ell$ defined by $w(e_i)=\pm e_{\sigma(i)}$ for all $i=1,2,\ldots,\ell$. We then define $f(w)=\sigma$. The homomorphism $f$ is split by $g:S_\ell\to H$ determined by $(i;i+1)\mapsto r_{\alpha_i}$. Therefore $W=\operatorname{ker}(f)\rtimes H$. Then we only need to prove that $N=\operatorname{ker}(f)$. $\endgroup$ – Jyrki Lahtonen Oct 27 '16 at 6:30
  • $\begingroup$ As far as I understand, $\ker(f)=\{w \in W \mid \sigma=id\}$. It's clear to me $N \subseteq \ker(f)$. To prove the reverse inclusion, let $w \in \ker(f)$ we have $w(e_i)=\pm e_i$. How can I deduce that $w \in N$? $\endgroup$ – user Oct 27 '16 at 6:47
  • $\begingroup$ The reflection $r_{e_i}$ maps $e_i\mapsto -e_i$ and $e_j\mapsto e_j$ for all $j\neq i$. Therefore any element $w$ such that $w(e_i)=\pm e_i$ for all $i$ is a product of reflections of some subset of the reflections $r_{e_i}, i\in\{1,2,\ldots,\ell\}$. $\endgroup$ – Jyrki Lahtonen Oct 27 '16 at 8:00

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