[GAP Forum] The outer automorphism of S6

[Alexander Hulpke]

Dear Gary McConnell, Dear Forum,

> I am new to GAP - I am using it inside SAGE ... I am trying to do a bunch of
> computations involving counting how certain elements of S6 behave under
> "the" outer automorphism. I have tried to do something from the manual pages
> If say I had a list of elements of S6, entered in the usual way as cycles,
> and if I chose a section s from Out(G) back to Aut(G), how would I get the
> image s(rho) of rho (if I write Out(G)=<1,rho> say) under that section,
> and/or how would I get the image of my elements under s(rho)?

This is how you would do it in GAP:

gap> G:=SymmetricGroup(6);
Sym( [ 1 .. 6 ] )
gap> A:=AutomorphismGroup(G);
<group with 3 generators>

Now, in your case you just want a generator of G outside the inner automorphisms:

gap> inn:=InnerAutomorphismsAutomorphismGroup(A);            
<group of size 720 with 2 generators>
gap> s:=First(GeneratorsOfGroup(A),x->not x in inn);
[ (5,6), (1,2,3,4,5) ] -> [ (1,2)(3,5)(4,6), (1,2,3,4,5) ]

To calculate element images:
gap> Image(s,(1,2,3));
(1,4,3)(2,6,5)

Here s is probably not the nicest possible
gap> Order(s);
10

Lets see what we can get:
gap> Collected(List(Elements(inn),x->Order(x*s)));
[ [ 2, 36 ], [ 4, 180 ], [ 8, 360 ], [ 10, 144 ] ]

So order 2 for an outer automorphism is possible. To find one of this kind either:

gap> news:=First(Elements(inn),x->Order(x*s)=2)*s;
[ (1,2,3,4,5,6), (1,2) ] -> [ (2,6)(3,5,4), (1,2)(3,4)(5,6) ]
gap> Order(news);
2
gap> news in inn;
false


or (less memory use, but more complicated -- use classes of A and find a representative, do so via permrep):
gap> hom:=IsomorphismPermGroup(A);
MappingByFunction( <group with 3 generators>, Group(
[ (1,2,3,4,5,6)(7,12,8)(9,11), (1,2)(7,8)(9,10)(11,12), 
  (1,7,4,10,2,8,5,11,3,9)(6,12) 
 ]), function( auto ) ... end, function( perm ) ... end )
gap> AP:=Image(hom);
Group([ (1,2,3,4,5,6)(7,12,8)(9,11), (1,2)(7,8)(9,10)(11,12), 
  (1,7,4,10,2,8,5,11,3,9)(6,12) ])
gap> innP:=Image(hom,inn);
Group([ (1,2,3,4,5,6)(7,12,8)(9,11), (1,2)(7,8)(9,10)(11,12) ])
gap> cl:=ConjugacyClasses(AP);;
gap> cl:=Filtered(cl,x->not Representative(x) in innP);;Length(cl);
5
gap> List(cl,x->[Order(Representative(x)),Size(x)]);
[ [ 8, 180 ], [ 4, 180 ], [ 2, 36 ], [ 8, 180 ], [ 10, 144 ] ]
gap> newsp:=Representative(cl[3]);      
(1,8)(2,10)(3,11)(4,7)(5,12)(6,9)
gap> PreImagesRepresentative(hom,newsp);
[ (1,2,3,4,5,6), (1,2) ] -> [ (1,4,5)(3,6), (1,4)(2,6)(3,5) ]

I hope this helps. Best wishes,

  Alexander Hulpke


-- Colorado State University, Department of Mathematics,
Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA
email: hulpke at math.colostate.edu, Phone: ++1-970-4914288
http://www.math.colostate.edu/~hulpke