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Is it true that the Weyl group $W(D_n)$ is also a quotient of the Weyl group $W(B_n)$? One can see that $W(D_n)$ is a normal subgroup of $W(B_n)$ irrespective of $n$ even or odd.

Frunobulax
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Nutan
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As abstract groups we have $W(D_n)\simeq (\mathbb{Z}/2)^{n-1}\ltimes S_n$ and $W(B_n)\simeq (\mathbb{Z}/2)^{n}\ltimes S_n$. Now try to find a surjetive morphism $f\colon W(B_n)\rightarrow W(D_n)$, which would give $W(D_n)\simeq W(B_n)/\ker (f)$.

Dietrich Burde
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  • @Burde: Actually I have just proved that there is no surjective map $W(B_n) \to W(D_n)$ if $n$ is even. There is one such if $n$ is odd. So $W(D_n)$ is a quotient of $W(B_n)$ only if $n$ is even. – Nutan Sep 09 '14 at 09:08