Hi Frédéric,
Good to see you on the GAP forum! I have the solution for you, and it is just
the last few lines to Alex's solution:
tau := RepresentativeAction(SymmetricGroup(40), gsmall, gstab);
peculiar := [1,3,5,6,10];
peculiartwo := subset{OnSets(peculiar,tau)};
That should give you the equivalent set for your group
Stabilizer(g,subset,OnSets) (before you took the permutation representation).
Cheers,
John.
On 06/12/2012, at 6:02 PM, Frederic Vanhove wrote:
> 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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