[GAP Forum] IdGroup and GroupHomomorphism

[t***@math.bu.edu]

Dear Forum members,

I am trying to construct an isomorphism between two regular subgroups
of S_12 which I know are isomorphic by IdGroup().

However, I am encountering the following:

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.

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.

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.

Thanks.

-T

[Hulpke,Alexander]

Dear Forum, Dear Tim Kohl,

Post by t***@math.bu.edu
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.

Post by t***@math.bu.edu
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.

Post by t***@math.bu.edu
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: ***@colostate.edu, Phone: ++1-970-4914288
http://www.math.colostate.edu/~hulpke

[t***@math.bu.edu]

Dear Alexander,

I cannot thank you enough,

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

is exactly what I need.

-Tim

Post by Hulpke,Alexander
Dear Forum, Dear Tim Kohl,

Post by t***@math.bu.edu
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.

Post by t***@math.bu.edu
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.

Post by t***@math.bu.edu
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
http://www.math.colostate.edu/~hulpke

--