Hello,
thanks for your reply, but I'm afraid I don't really understand.
Perhaps it would indeed help if I'd explain what I really want to do:
#small permutation representation
gsmall:=AllPrimitiveGroups(DegreeOperation,40,Size,51840)[1];
#big permutation representation
g:=AllPrimitiveGroups(DegreeOperation,1120, Size, 9170703360)[2];
subset:=[ 1, 3, 13, 53, 106, 143, 149, 161, 233, 263, 268, 281, 284,
295, 333, 370,
378, 410, 421, 439, 459, 496, 541, 546, 565, 597, 628, 683, 695, 696,
698,
738, 795, 868, 881, 907, 934, 999, 1021, 1038 ];
gstab:=Action(Stabilizer(g,subset,OnSets),subset,OnPoints);
Size(gstab);DegreeAction(gstab);
(note: the stabilizer in g of subset does not act faithfully on that
subset, it has size 51840*486.)
Now suppose I have found (through a separate and lengthy computation)
that [1,3,5,6,10] is a peculiar subset of [1..40] with respect to the
first representation. How could I transform it into a corresponding
subset of
[ 1, 3, 13, 53, 106, 143, 149, 161, 233, 263, 268, 281, 284, 295, 333, 370,
378, 410, 421, 439, 459, 496, 541, 546, 565, 597, 628, 683, 695, 696,
698,
738, 795, 868, 881, 907, 934, 999, 1021, 1038 ]
?
Many thanks,
Kind regards,
Frédéric
Op 05/12/12 18:59, Alexander Hulpke schreef:
Dear Forum, Dear Drederic,
some time ago the command ActionHomomorphism was recommend to me. I was now
wondering, if I have two equivalent permutation groups, can I get a bijection
between the groups as well as a corresponding bijection between the sets on
which they act?
For instance:
g1:=Group((1,2),(3,4),(1,3));
g2:=Group((1,2),(2,3));
gaction1:=Action(Stabilizer(g1,1),[2..4],OnPoints);
gaction2:=Action(g2,[1..3],OnPoints);
Your syntax is a bit confused here. Basically
Action(G,set,op)=Image(ActionHomomorphism(G,set,op)).
The action homomorphism preserves the connection to the initial group, the
action is just the resulting permutation group. Thus
ActionHomomorphism(gaction1,gaction2);
does not make sense.
If you call `RepresentativeAction(symmetricgroup,g1,g2)' you get a permutation
that conjugates one group into the other. So it is a bijection between the
groups given by a bijection of the underlying domain.
However I suspect that you actually want to test for equivalence of actions,
i.e. you have a group G acting on two sets omega1 and omega2 and you want to
see whether there is a bijection between the sets that makes the actions
equivalent. In this case you need to find a permutation that maps the images of
the same generating set to each other. This is actually a cheaper test than the
one for mapping the groups.
For example, consider these two actions of S4 on sets of order 2 or on cosets
of a subgroup.
gap> G:=SymmetricGroup(4);
Sym( [ 1 .. 4 ] )
gap> c:=Combinations([1..4],2);
[ [ 1, 2 ], [ 1, 3 ], [ 1, 4 ], [ 2, 3 ], [ 2, 4 ], [ 3, 4 ] ]
gap> act1:=ActionHomomorphism(G,c,OnSets,"surjective");
<action epimorphism>
gap> U:=Subgroup(G,[(1,3),(2,4)]);
Group([ (1,3), (2,4) ])
gap> Index(G,U);
6
gap> T:=RightCosets(G,U);
[ RightCoset(Group( [ (1,3), (2,4) ] ),()),
RightCoset(Group( [ (1,3), (2,4) ] ),(3,4)),
RightCoset(Group( [ (1,3), (2,4) ] ),(2,3)),
RightCoset(Group( [ (1,3), (2,4) ] ),(1,2)),
RightCoset(Group( [ (1,3), (2,4) ] ),(1,2)(3,4)),
RightCoset(Group( [ (1,3), (2,4) ] ),(1,3,4,2)) ]
gap> act2:=ActionHomomorphism(G,T,OnRight,"surjective");
<action epimorphism>
gap> imgs1:=List(GeneratorsOfGroup(G),x->Image(act1,x));
[ (1,4,6,3)(2,5), (2,4)(3,5) ]
gap> imgs2:=List(GeneratorsOfGroup(G),x->Image(act2,x));
[ (1,5)(2,3,4,6), (1,4)(2,5) ]
Now search for an element that maps the elements of imgs1 to those of imgs2 in
the same arrangement:
gap> rep:=RepresentativeAction(SymmetricGroup(6),imgs1,imgs2,OnTuples);
(1,3,2)
Verify:
gap> List(imgs1,x->x^(1,3,2));
[ (1,5)(2,3,4,6), (1,4)(2,5) ]
So (1,3,2) is the permutation indicating how to reorder the elements of c to
get the same action.
If you can replace the symmetric group with something smaller (you might do a
prearrangement by hand so that for example both actions have the same blocks)
this will help in larger degrees.
(If this is not of help or not what you want feel free to send me the actual
problem you want to solve.)
Best,
Alexander
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