How to compute the pointwise stabilizer subgroup of a fixed-point subspace?

Question

Definitions: Let $W$ be a representation of a group $G$, $K$ a subgroup of $G$, and $X$ a subspace of $W$.
Let the fixed-point subspace $W^{K}:=\{w \in W \ \vert \ kw=w \ , \forall k \in K \}$.
Let the pointwise stabilizer subgroup $G_{(X)}:=\{ g \in G \ \vert \ gx=x \ , \forall x \in X \}$.

Let $G$ be a finite group, $H$ a subgroup and $V$ an irreducible complex representation of $G$.

Question: How to compute $G_{(V^H)}$?

Remark: It's for doing checks about this problem.

Answer 1

There are basically two steps, first computing $V^H$ and then computing the stabilizer.

$V^H$ is simply the common eigenspace of $K$ for eigenvalue 1. We get this by intersecting the eigenspaces of the generators. Thus (One could go to formal vector spaces, but staying in the matrix world is less typing):

eig:=List(GeneratorsOfGroup(K),x->NullspaceMat(x-x^0));
ise:=eig[1];
for i in eig do ise:=SumIntersectionMat(ise,i)[2];od;

Now ise is a basis for $V^H$. To stabilize the space, as Derek already wrote, I cannot think of something better in the generic case than a naive orbit-stabilizer calculation. Take the vectors in use, and stabilize them in turn:

stb:=G;
for i in ise do
  stb:=Stabilizer(G,i,OnRight);
od;

If you instead want to stabilize the space as a set, represent it by its RREF basis:

ise:=TriangulizedMat(ise);
stb:=Stabilizer(G,ise,OnSubspacesByCanonicalBasis);;

(This implicitly enumerates the orbit, so if your stabilizer has large index this is costly.)