[GAP Forum] IdGroup and GroupHomomorphism

[Hulpke,Alexander]

Dear Forum, Dear Tim Kohl,

> gap> A:=Group([ (1,2)(3,5)(4,6)(7,9)(8,10)(11,12), (1,3)(2,5)(4,7)(6,9)(8,11)(10,12), (1,4,8)(2,10,6)(3,7,11)(5,12,9) ]);
> Group([ (1,2)(3,5)(4,6)(7,9)(8,10)(11,12), (1,3)(2,5)(4,7)(6,9)(8,11)(10,12), (1,4,8)(2,10,6)(3,7,11)(5,12,9) ])
> gap> B:=Group([ (1,5)(2,3)(4,9)(6,7)(8,12)(10,11), (1,3)(2,5)(4,11)(6,12)(7,8)(9,10), (1,8,4)(2,6,10)(3,11,7)(5,9,12) ]);
> Group([ (1,5)(2,3)(4,9)(6,7)(8,12)(10,11), (1,3)(2,5)(4,11)(6,12)(7,8)(9,10), (1,8,4)(2,6,10)(3,11,7)(5,9,12) ])
> gap>  IdGroup(A);
> [ 12, 4 ]
> gap>  IdGroup(B);
> [ 12, 4 ]
> gap> GroupHomomorphismByImages(A,B);
> fail
> gap> GroupHomomorphismByImages(A,AllSmallGroups(12)[4]);
> [ (1,2)(3,5)(4,6)(7,9)(8,10)(11,12), (1,3)(2,5)(4,7)(6,9)(8,11)(10,12), (1,4,8)(2,10,6)(3,7,11)(5,12,9) ] -> [ f1, f2, f3 ]
> gap> GroupHomomorphismByImages(B,AllSmallGroups(12)[4]);
> fail
> 
> I guess my question is, how does IdGroup determine 
> that a given group is in an isomorphism class of one 
> of the groups in the SmallGroups library. 

Briefly, it first determines a number of isomorphism-invariant properties. If this does not leave a unique candidate it tries to find elements in a pcgs that correspond to the presentation in the library group.

> I realize that the generators of A satisfy the relations of the generators of AllSmallGroups(12)[4]
> which is why GroupHomomorphismByImages(A,AllSmallGroups(12)[4]) succeeds whereas those of B
> do not match which is why GroupHomomorphismByImages(B,AllSmallGroups(12)[4]) fails.
> 
> Is there a way to 'correct' the generating set of B so that I *can* construct a homomorphism from A to B
> (or from B to AllSmallGroups(12)[2]) which I could also use.

Generically, you could call 
iso:=IsomorphismGroups(A,B);
to find such an isomorphism.

> I am trying to take advantage of the ismomorphic = conjugate property of regular permutation
> groups of the same degree to construct an element of S_12 which conjugates A to B, but this
> homomorphism failure is getting in the way.

gap> RepresentativeAction(SymmetricGroup(12),A,B);
(2,3)(4,8)(6,7)(9,12)(10,11)

will find such a permutation.

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, Phone: ++1-970-4914288
http://www.math.colostate.edu/~hulpke