[GAP Forum] RightCosets and reconstructing a subgroup

[Derek Holt]

Dear Caroline Keil, GAP Forum,

I don't think that just knowing coset representatives is enough information
to determine the subgroup H uniquely. For example, there are two non-conjugate
subgroups of G of index 3, H1 = < B, S*B*S > and H2 = < B, S^-1*B*S, S*B*S^-1 >,
and for both of these subgroups, [ One(G), S, S^-1 ] is a right transversal.

So I believe that you will need some extra information in order to find
generators of H. For example, if for an arbitrary element g of G you were
able to determine the coset representative rep(g) of g from your list, then
you could find the Schreier generators of H;
i.e.  H is generated by { r_i * x * rep(r_i*x)^-1 | i in {1..n}, x in {B,S} }.

Derek Holt.

On Tue, Jun 06, 2006 at 06:13:17PM +0200, Caroline Keil wrote:
> 
> Hello,
> 
> I am doing some research with G=PSL(2,Z) and subgroups. Therefore, I have 
> constructed G as follows:
> 	# free group mit B=F.1, S=F.2
> 	F := FreeGroup( "B", "S");;
> 	# Relations
> 	G := F / [ F.1^2,F.2^3];;
> 	B:=G.1;;S:=G.2;;
> 
> For some subgroups H in G, I only know representatives r_1,...r_n given as 
> words in B and S of the right cosets H\G and with this information I want 
> to construct the group H.
> How can I get a proper coset table for H\G, or is there any other 
> possibily to gain the generators of H ?
> 
> I hope you can help me !
> 
> Thank you
> 
> Caroline Keil
> 
> 
> 
> > *********************************
> > Caroline Keil
> > Mathematisches Institut
> > Heinrich-Heine-Universitaet Duesseldorf
> > Universitaetsstrasse 1
> > 40225 Duesseldorf
> > ckeil at math.uni-duesseldorf.de
> > *********************************
> 
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