Dear Forum, Dear Minghui Liu,
>
> I thought it was because my groups are too complicated for GAP and thus I
> tried a much simpler version of groups:
>
> F2:=FreeGroup("a","b");
> B2:=[Comm(Comm(F2.1,F2.2),F2.1),Comm(Comm(F2.1,F2.2),F2.2)];
> K2:=F2/B2;
> A2:=Subgroup(K2,[K2.1*K2.2,Comm(K2.1,K2.2)])
> *
> *
> When I run the command AbelianInvariants(A2), the same error message
> appeared.
The algorithm for abelian invariants of a subgroup constructs a coset table and
rewrites relators. In your case A2 has infinite index in K2, so this will not
terminate.
I am not aware of any algorithm that would deal with such situations
automatically. What might be of help is to look at subgroups of finite index,
containing A2. For example:
gap> pq:=EpimorphismPGroup(K2,2,10);Size(Image(pq));
[ a, b ] -> [ a1, a2 ]
536870912
gap> A3:=PreImage(pq,Image(pq,A2));
Group(<fp, no generators known>)
gap> AbelianInvariants(A3);
[ 0, 0, 1024 ]
gap> pq:=EpimorphismPGroup(K2,3,5);Size(Image(pq));
[ a, b ] -> [ a1, a2 ]
4782969
gap> A3:=PreImage(pq,Image(pq,A2));
Group(<fp, no generators known>)
gap> AbelianInvariants(A3);
[ 0, 0, 243 ]
gap> pq:=EpimorphismPGroup(K2,5,3);Size(Image(pq));
[ a, b ] -> [ a1, a2 ]
390625
gap> A3:=PreImage(pq,Image(pq,A2));
Group(<fp, no generators known>)
gap> Index(K2,A3);
125
gap> AbelianInvariants(A3);
[ 0, 0, 125 ]
from which I would naively guess that the invariants of A2 might be [0,0,0],
but this is certainly no proof.
Best,
Alexander Hulpke
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