[GAP Forum] Recombining irreducible representations

[Max Horn]

Dear Jerry,

> On 18 Jun 2017, at 15:05, Jerry Swan <dr.jerry.swan at gmail.com> wrote:
> 
> Dear all,
> 
> For some element g of a group G for which irr :=
> IrreducibleRepresentations(G) have been obtained, is it possible to recover
> g from images := List(irr,r->Image(r,g)) ?

The representations are homomorphisms, and as such, you can compute preimages -- which in general are of course not unique, but rather cosets of the kernel. But for a finite group, the intersection of the kernel of all irreducible representations is trivial, so you can recover g like this:

pre:=Intersection(List([1..Length(irr)], i -> PreImagesElm(irr[i], images[i])));

Applied to a concrete example:

gap> G:=SymmetricGroup(5);;
gap> irr:=IrreducibleRepresentations(G);;
gap> g:=Random(G);
(1,4,3)
gap> images := List(irr,r->Image(r,g));;
gap> pre:=Intersection(List([1..Length(irr)], i -> PreImagesElm(irr[i], images[i])));
[ (1,4,3) ]


In a later email, you clarified that your group G is always a symmetric group S_n. In that case, at least for n>=5, most irreducible representations are actually faithful, the exception being the trivial and the sign representation. In that case, you can simply take a preimage of one of the faithful representations, like so:

gap> List(irr, IsInjective);
[ false, true, true, true, true, true, false ]
gap> irr[1];   # this is the sign representation
[ (1,2,3,4,5), (1,2) ] -> [ [ [ 1 ] ], [ [ -1 ] ] ]
gap> irr[7];   # this is the trivial representation
[ (1,2,3,4,5), (1,2) ] -> [ [ [ 1 ] ], [ [ 1 ] ] ]
gap> PreImage(irr[2], images[2]);
(1,4,3)
gap> PreImage(irr[3], images[3]);
(1,4,3)


Hope that helps,
Max