[GAP Forum] easy way to generate permutation groups?

[John Dixon]

    I don't know whether this answers the more general question which 
you recently posed on the GAP forum but it produces two permutations u, 
v on an appropriate interval [1..m] which generate the group which you 
describe.

gap> a := x -> [x[1],x[1]*x[2]];;
gap> b := x -> [x[2]*x[1],x[2]];; 
gap> g := SymmetricGroup(4);;
gap> omega := Cartesian(g,g) ;;
gap> u := PermListList(omega, List(omega,a));;
gap> v := PermListList(omega, List(omega, b));;
gap> h := Group(u,v);;
gap> Size(h);
2239488

    It is rather intriguing how the final group h varies with the input g.

          - John Dixon

On 2009-11-30 5:22 PM, R. Keith Dennis wrote:
> Hi, I'm interested in studying certain groups which arise as
> permutations of subsets of a given group.  As a simple example, let G
> be any finite group, and S = G x G, the set of pairs of elements of G.
> Define a to be the permutation of S generated by (x,y) --> (x,xy) and
> b to be the one generated by (x,y) --> (yx,y).  Is there a simple way
> to construct the subgroup of the permutation group on S generated by a
> and b?  In this case both a and b have order exp(G) & for example I'd
> like to be able to compute a presentation for the group of
> permutations they generate.  More generally I'd like to study other
> operations on certain subsets (or sequences) of elements derived from
> a fixed group G.
>
> As permutation groups seem to be given as permutations of sets of
> integers, it almost seems that I should (in essence) have to describe
> a one-to-one correspondence of S with a set of integers [1..m] and 
> describe a and b by explicitly computing via this correspondence.
> Is that sort of thing really necessary?  It usually seems that there
> are built-in operations in GAP to avoid such.
>
> Perhaps I'm missing something obvious.  Suggestions most welcome!
>
> Keith
>
>
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