[GAP Forum] polycyclic group problem

Stefan Kohl kohl at mathematik.uni-stuttgart.de
Fri Jul 8 10:43:25 BST 2005


Dear Forum,

R.N. Tsai wrote:

> I'm experimenting with the polycyclic package and ran into this problem.
> 
> This test program brings it out :
>  
> Test:=function(p)local collector,group,aut;
>  # define group with 3 generators using polycyclic collector
>  collector:=FromTheLeftCollector(3);
>  SetCommutator(collector,2,1,[3,1]);
>  SetRelativeOrder( collector, 1, p );
>  SetRelativeOrder( collector, 2, p );
>  SetRelativeOrder( collector, 3, p );
>  # get size of group
>  group:=PcpGroupByCollector(collector);
>  Print(Size(group),"\n");
>  # get size of automorphism group
>  aut:=AutomorphismGroup(group);
>  Print(Size(aut),"\n");
> end;

[ ... ]

> # with p=4 we get :
> gap> Test(4);
> 64
> Error, no method found!

The main point is here that the Polycyclic package currently does not
contain a method to compute automorphism groups of polycyclically
presented groups.

However, erraneously an `AutomorphismGroup' method which is not capable
of handling pcp groups is called, since its declaration claims that it
would be applicable for any finite solvable group.

This is a bug, and will be fixed as soon as possible.

An easy workaround for your case is the following:

Test:=function(p)local collector,group,aut;
  # define group with 3 generators using polycyclic collector
  collector:=FromTheLeftCollector(3);
  SetCommutator(collector,2,1,[3,1]);
  SetRelativeOrder( collector, 1, p );
  SetRelativeOrder( collector, 2, p );
  SetRelativeOrder( collector, 3, p );
  # get size of group
  group:=RefinedPcpGroup(PcpGroupByCollector(collector));
  Print(Size(group),"\n");
  # get size of automorphism group
  aut:=AutomorphismGroup(group);
  Print(Size(aut),"\n");
end;

In any case, note the difference between pc groups, which are always
finite and are supported by the GAP library, and pcp groups, which can
be infinite, and are implemented in the Polycyclic package.

Hope this is of some help,

     Stefan Kohl

(I would like to acknowledge hints by Bettina Eick.)





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