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
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