[GAP Forum] Matrix Generators of a Group Represented as a Permutation

[Hulpke,Alexander]

Dear Forum, Dear David Musyoka,

[it turns out the question is to find a particular irreducible matrix representation.]

> I have a permutation representation of a group G, and my intention is to get the generators of the group in Matrix form over GF(3),

What you can do generically is to use `IrreducibleRepresentations` to calculate the irreducible representations over a given field up to equivalence. For example, say I want a 42 dimensional representation of the given group:

gap> g:=Group( (1,22,15,23)(2,14,18,7)(3,6)(4,24)(5,16,17,9)(8,13,20,11)(10,12)(19,21),
>   (1,8)(2,3)(4,11)(5,13)(6,24)(7,12)(9,23)(10,16)(14,15)(17,20)(18,19)(21,22));;
gap> StructureDescription(g);
"(C2 x C2 x C2 x C2 x C2 x C2) : (PSL(3,2) x S3)”
gap> reps:=IrreducibleRepresentations(g,GF(3));;
gap> List(reps,x->DegreeOfMatrixGroup(Range(x)));
[ 1, 1, 6, 6, 6, 6, 7, 7, 21, 21, 21, 21, 42, 42, 63, 63, 63, 63, 126, 126 ]
gap> myrep:=First(reps,x->DegreeOfMatrixGroup(Range(x))=42);
[ (1,22,15,23)(2,14,18,7)(3,6)(4,24)(5,16,17,9)(8,13,20,11)(10,12)(19,21),
  (1,8)(2,3)(4,11)(5,13)(6,24)(7,12)(9,23)(10,16)(14,15)(17,20)(18,19)(21,22) ] ->
[ < immutable compressed matrix 42x42 over GF(3) >, < immutable compressed matrix 42x42 over GF(3) > ]
gap> matg:=Group(MappingGeneratorsImages(myrep)[2]);
<matrix group with 2 generators>
gap> Size(matg);
64512

This is clearly limited by the group order and the dimensions of the representations — for larger groups this quickly becomes a research level problem. Also make sure you use GAP 4.11, as this has a better algorithm implemented.

Best,

  Alexander Hulpke

-- Colorado State University, Department of Mathematics,
Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA
email: hulpke at colostate.edu, 
http://www.math.colostate.edu/~hulpke