Weyl group of $B_n$ and $D_n$

Question

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.

Answer 1

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 $W(D_n)\simeq W(B_n)/\ker (f)$.