On the page https://en.wikipedia.org/wiki/Hyperoctahedral_group, under the subgroups section is stated:
"Viewed as a wreath product, there are two natural maps from the hyperoctahedral group to the cyclic group of order 2: one map coming from "multiply the signs of all the elements" (in the n copies of $\{\pm 1\}) $ $…$ The kernel of the first map is the Coxeter group $D_n$."
It is stated later that the kernal of the determinant map to $\{\pm 1\}$ is a distinct index 2 subgroup.
However on the page https://en.wikipedia.org/wiki/Generalized_permutation_matrix#Signed_permutation_group, under Signed permutation group(isomorphic to Hyperoctahedral group) section, is stated:
"Its index 2 subgroup of matrices with determinant 1 is the Coxeter group $D_{n}$"
Am I missing something or are these statements contradictory?
Best Answer
I think the statement on the second Wikipedia page, "Its index 2 subgroup of matrices with determinant 1 is the Coxeter group $D_n$" is a mistake. The first Wikipedia page, which states that the kernel of "multiply the signs of all the elements" map is $D_n$ is correct, and there is a natural isomorphism from that kernel to $D_n$.
In theory, both statements could be correct if we interpret the imprecise "is $D_n$" as meaning "is isomorphic to $D_n$".
I did some computer calculations for small $n$ ($n \le 10$) and, interestingly, it appears that the kernel of the determinant map is isomorphic to $D_n$ when $n$ is odd, but not when $n$ is even. I have not thought about why that should be true, but there must be a reason!