Dear Forum,
as I already replied to the author, the following would work,
although I do not know how efficient this is.
gap> a0:=AbelianPcpGroup(3,[0,0,0]);
Pcp-group with orders [ 0, 0, 0 ]
gap> a:=Subgroup(a0,[a0.1*a0.2^-1,a0.3]);
Pcp-group with orders [ 0, 0 ]
gap> b:=Subgroup(a0,[a.1*a.2^-1,a.1^-1*a.2^-1]);
Pcp-group with orders [ 0, 0 ]
gap> a/b;
Pcp-group with orders [ 2 ]
gap>
(here we use multiplicative notation for Abelian groups, which feels odd to me)
Best,
Dmitrii
On 7 March 2010 22:53, Lyosha Beshenov <a...@cadadr.org> wrote:
> Hello,
>
> Here is a typical computational task.
>
> Given bases of two abelian groups A_1 and A_2, A_2 \subset A_1,
> compute the structure of A_1/A_2.
>
>
> For instance, if A_1 has a basis {a - b, c} and A_2 has a basis
> {a - b - c, -a + b - c}, then A_1/A_2 is isomorphic to Z/2Z.
>
>
> I'm wondering, is there a straightforward formulation and solution
> of this problem in terms of the GAP system?
>
>
> Thank you.
>
> -- Lyosha
>
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>
--
Dmitrii Pasechnik
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