1) $X1$ is a cayley graph and a is an automorphism (an element of Aut($X1$)) of $X1$. I'm trying to compute the quotient graph $X1$/a in GAP. I tired to use the command "QuotientGraph(X1,[a]);" in GAP but it is giving an error.
Can someone please help me with this question.
The steps of my computation is as below.
gap> Read("C:/Users/buddhini/Desktop/Test1/UndirectedGeneratingSets.gap");
gap> LoadPackage("grape");
───────────────────────────────────────────────────────────────────────────────────────────────────────────────────── Loading GRAPE 4.8 (GRaph Algorithms using PErmutation groups) by Leonard H. Soicher (http://www.maths.qmul.ac.uk/~lsoicher/). Homepage: http://www.maths.qmul.ac.uk/~lsoicher/grape/ ───────────────────────────────────────────────────────────────────────────────────────────────────────────────────── true
gap> Z3:=Group((1,2,3));
Group([ (1,2,3) ])
gap> Z3gen:=GeneratorsOfGroup(Z3);
[ (1,2,3) ]
gap> h:=Group((1,2,3,4,5));
Group([ (1,2,3,4,5) ])
gap> g:=DirectProduct(h,h);
Group([ (1,2,3,4,5), (6,7,8,9,10) ])
gap> Auts:=AutomorphismGroup(g);;
gap> AutsOfOrd3:=Filtered(Elements(Auts),elt->Order(elt)=3);;
gap> ExHom:=GroupHomomorphismByImages(Z3,Auts,[Z3gen[1]],[AutsOfOrd3[1]]);
[ (1,2,3) ] -> [ [ (1,2,3,4,5), (6,7,8,9,10) ] -> [ (1,5,4,3,2)(6,10,9,8,7),
(1,2,3,4,5) ] ]
gap> s:=SemidirectProduct(Z3,ExHom,g);
Group([ (2,6,25)(3,11,19)(4,16,13)(5,21,7)(8,10,20)(9,15,14)(12,24,22)
(17,18,23), (1,6,11,16,21)(2,7,12,17,22)
(3,8,13,18,23)(4,9,14,19,24)(5,10,15,20,25), (1,2,3,4,5)(6,7,8,9,10)
(11,12,13,14,15)(16,17,18,19,20)
(21,22,23,24,25) ])
gap> S:=IrredUndirGenSetsUpToAut(s);
[ [ (2,6,25)(3,11,19)(4,16,13)(5,21,7)(8,10,20)(9,15,14)(12,24,22)
(17,18,23), (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,
15)(16,17,18,19,20)(21,22,23,24,25) ],
[ (2,6,25)(3,11,19)(4,16,13)(5,21,7)(8,10,20)(9,15,14)(12,24,22)
(17,18,23), (1,2,7)(3,12,25)(4,17,19)(5,22,13)(6,21,
8)(9,11,20)(10,16,14)(18,24,23) ] ]
gap> S1:=S[1];
[ (2,6,25)(3,11,19)(4,16,13)(5,21,7)(8,10,20)(9,15,14)(12,24,22)(17,18,23),
(1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,
15)(16,17,18,19,20)(21,22,23,24,25) ]
gap> X1:=CayleyGraph(s,S1);;
gap> H:=AutomorphismGroup(X1);
<permutation group with 4 generators>
gap> AutOfOrd5:=Filtered(Elements(H),elt->Order(elt)=5);;
gap> a:=AutOfOrd5[1];
(1,4,7,10,13)(2,5,8,11,14)(3,6,9,12,15)(16,19,22,25,28)(17,20,23,26,29)
(18,21,24,27,30)(31,34,37,40,43)(32,35,38,41,
44)(33,36,39,42,45)(46,49,52,55,58)(47,50,53,56,59)(48,51,54,57,60)
(61,64,67,70,73)(62,65,68,71,74)(63,66,69,72,75)
gap> QuotientGraph(X1,a);
Error, usage: QuotientGraph( <Graph>, <List> ) at /proc/cygdrive/C/gap-
4.9.2/pkg/grape/lib/grape.g:2169 called from
<function "QuotientGraph">( <arguments> )
called from read-eval loop at *stdin*:41
you can 'quit;' to quit to outer loop, or
you can 'return;' to continue
Even if I use QuotientGraph(X1,[a]); it still gives an error.
2)I was able to read the vertices of the Cayley graph by identifying them as elements of the semidirect product group by using the commands below. Let V[2] be the vertex of Cayley graph I need to identify,
gap> V[2];
(2,6,25)(3,11,19)(4,16,13)(5,21,7)(8,10,20)(9,15,14)(12,24,22)(17,18,23)
gap> proj:=Projection(s);
[ (2,6,25)(3,11,19)(4,16,13)(5,21,7)(8,10,20)(9,15,14)(12,24,22)(17,18,23),
(1,6,11,16,21)(2,7,12,17,22)(3,8,13,18,
23)(4,9,14,19,24)(5,10,15,20,25), (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)
(16,17,18,19,20)(21,22,23,24,25) ] ->
[ (1,2,3), (), () ]
gap> e1:=Embedding(s,1);
[ (1,2,3) ] -> [ (2,6,25)(3,11,19)(4,16,13)(5,21,7)(8,10,20)(9,15,14)
(12,24,22)(17,18,23) ]
gap> e2:=Embedding(s,2);
[ (1,2,3,4,5), (6,7,8,9,10) ] -> [ (1,6,11,16,21)(2,7,12,17,22)
(3,8,13,18,23)(4,9,14,19,24)(5,10,15,20,25),
(1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25) ]
gap> a:=Image(proj,V[2]);
(1,2,3)
gap> b:=PreImagesRepresentative(e2,Image(e1,Image(proj,V[2]))^-1*V[2]);
()
Therefore the vertex is ((1,2,3),()).
When we consider the automorphism of $X1$ the vertices are again the elements of $s$, right?
So then is it possible to identify the elements of the automorphism $a$ by converting them to a form as above? I tried using the same codes but its not working.
(I just checked by taking $V$:= (1,4,7,10,13)(2,5,8,11,14)(3,6,9,12,15)(16,19,22,25,28)(17,20,23,26,29) (18,21,24,27,30)(31,34,37,40,43)(32,35,38,41, 44)(33,36,39,42,45)(46,49,52,55,58)(47,50,53,56,59)(48,51,54,57,60) (61,64,67,70,73)(62,65,68,71,74)(63,66,69,72,75) , the representative of $a$, by thinking that this will be one of the elements of the automorphism but I don't know hoe to compute the remaining elements of the automorphism. Will it be by taking $V*V$, $V*V*V$,... etc. )
Please help and guide me with these questions.
Thanks a lot in advance.