I am looking to define the Weyl group of type $D_n$ as a subgroup of Weyl group of type $B_n$ in the software GAP. In general, one can define these groups separately. For example, let's say G:=CoxeterGroup("B",4) and H:=CoxeterGroup("D",4) defines the Weyl group of type $B_4$ and $D_4$ respectively. Under this definition, I assume that $H$ is not recognized as a normal subgroup of $G$.

The main thing I want to achieve is to visualize the conjugacy classes in these groups. For example: cc:=ConjugacyClasses(H); will return the list of conjugacy classes in $H$. Now, a command as: ConjugacyClass(G,cc[1]); is not making sense here although theoretically it makes sense. I hope I have been able to explain the problem properly. Thank you in advance for any kind of help.

  • $\begingroup$ CoxeterGroup is not a command of GAP4. Are you using any package? Which ones? $\endgroup$
    – ahulpke
    Jan 1, 2022 at 22:24
  • $\begingroup$ I am using GAP 3 which has CHEVIE as a package. I couldn't get a way to work with CHEVIE in Gap 4. These are finite imprimitive complex reflection groups which are defined in CHEVIE. $\endgroup$
    – Riju
    Jan 2, 2022 at 10:00

1 Answer 1


GAP3 is a 20 year old version of the system, but it is easy to create the Coxeter/Weyl groups in GAP4 from the presentation. (The next release of GAP4 also will contain a function WeylGroupFp that does this directly.)

Basically, two different finitely presented groups do not have any connection. You will have to create the second group thus as a subgroup of the first.

Thus, let's start with the group of type B:


We now could look for normal subgroups of the right order, but (if I recall correctly), the $D_n$ in $B_n$ is generated by the generators $2,\ldots,n$ and a conjugate of the second by the inverse of the first.


(The ordering here gives the generators in correspondence to the $D_4$ presentation.)

We verify that the subgroup is D_4:

# returns:
[ F1^2, F2^2, F3^2, F4^2, (F4*F1)^2, (F2*F1)^2, (F4*F2)^2, (F3*F1)^3,  (F4*F3)^3, (F3*F2)^3 ]

Thus now you have $D_4$ as a subgroup of $B_4$.


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