[GAP Forum] subgroup of fp-group from coset table

[Alexander Hulpke]

Dear Forum, Dear Derek,

> I would be grateful if someone could help me with the technicalities of the
> following problem.
> 
> I have a homomorphism f: F -> G from a finitely presented group F to a
> transitive permutation group G. I want to find the inverse image under f
> of a point stabilizer in G

The most basic way in GAP would be to actually use the homomorphism, and take the pre-image under the homomorphism:

gap> F:=free/ParseRelators(free,"a2,b3,(ab)5");;
gap> G:=Group((1,2)(3,4),(1,3,5));;
gap> hom:=GroupHomomorphismByImages(F,G,GeneratorsOfGroup(F),GeneratorsOfGroup(G));
[ a, b ] -> [ (1,2)(3,4), (1,3,5) ]
gap> S:=PreImage(hom,Stabilizer(G,1));
Group(<fp, no generators known>)

The resulting subgroup is stored as being pre-image under a homomorphism (thus it does not yet know generators, but is very low cost), but one could ask for these or the coset table etc. if one wanted to (however they would not be required for element tests, thus this approach will work even for subgroups of huge index). Now everything should work for S, even if generators or coset table have not been computed. (For example one could ask for subgroup intersections, or for abelian quotients.) Please let me know if you find this subgroup in any way unsatisfactory.

gap> CosetTableInWholeGroup(S);
[ [ 2, 1, 5, 4, 3 ], [ 2, 1, 5, 4, 3 ], [ 3, 2, 4, 1, 5 ], [ 4, 2, 1, 3, 5 ] ]
gap> GeneratorsOfGroup(S);
[ a*b*a^-1, b^-1*a*b ]

> SubgroupOfWholeGroupByCosetTable
> 
This is a somewhat more technical function (I personally would not use outside library code). An example would be:

gap> table:=[[2,1,5,4,3],[2,1,5,4,3],[3,2,4,1,5],[4,2,1,3,5]];;
gap> SubgroupOfWholeGroupByCosetTable(FamilyObj(F),table);
Group(<fp, no generators known>)

(The reason for requiring the family is that it requires not just the group F, but also the particular generators, while H:=Subgroup(F,[F.1,F.1*F.2]) is mathematically the same group, it would not fit with the coset table.)

All the best,

   Alexander