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GAP has the command ConjugacyClassesSubgroups which gives a list of the conjugacy classes of a finite group $G$. Is there a way I can specify further what types of subgroups GAP reports? For instance, can I ask GAP to only list conjugacy classes of subgroups of a certain order or isomorphism type?

My question is about subgroups in symmetric group isomorphic to others symmetric groups.

Thanks for your answers.

For instance, I have defined $S_3$ and $S_5$ and i would like to know the number of subgroups in $S_5$ isomorphic to $S_3$.

So the function IsomorphicSubgroups(s5,s3) enable to see 2 types of classes of subgroups in $S_5$ isomorphic to $S_3$. But how see the size of these classes?

If i put: 

emb :=  IsomorphicSubgroups(s5,s3);

And i ask: 

Size(emb[1]);

GAP returns: "Error no method found!"

Thanks for any answer to help me!

Olexandr Konovalov
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Karev
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  • Questions like this are on-topic here iff they contain some on-topic mathematical content, which this question does not. – Bill Dubuque Jun 28 '17 at 16:59
  • As my answer to https://math.stackexchange.com/q/1569349/ says, LatticeByCyclicExtension and SubgroupsSolvableGroup accept optional arguments which allow to put restrictions on computed subgroups. – Olexandr Konovalov Jun 28 '17 at 22:24
  • Thanks for your answer, but what are these arguments? How solve my problem? I don't understand how manipulate these functions could help me – Karev Jun 28 '17 at 22:27
  • And as for IsomorphicSubgroups, it returns monomorphisms, and you have to ask Size(Image(emb[1])); if you intend to see 6. – Olexandr Konovalov Jun 28 '17 at 22:28
  • I don't have time to write a complete answer now - see documentation here and here. Also, they may be less helpful if you actually want isomorphic subgroups, so you're interested in a very hard restriction. – Olexandr Konovalov Jun 28 '17 at 22:31
  • And welcome to MSE! I've edited the question to show how to use some formatting - see diffs here. In particular, enclose short code fragments in backticks and indent code lines by 4 spaces. – Olexandr Konovalov Jun 28 '17 at 23:15

1 Answers1

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IsomorphicSubgroups returns homomorphisms. The Image will be the subgroup that is isomorphic.

If you want the total number of subgroups that are isomorphic, the normalizer indices will give this

Sum(emb,x->Index(s5,Normalizer(s5,Image(x))));

However to understand the pattern, you might want to look at the image groups first.

Similarly for other groups replace S5, e.g.

gap> s6:=SymmetricGroup(6);;
gap> emb:=IsomorphicSubgroups(s6,s3);
[ [ (1,2,3), (1,2) ] -> [ (1,2,3), (1,2) ],
  [ (1,2,3), (1,2) ] -> [ (3,4,5), (1,2)(3,4) ],
  [ (1,2,3), (1,2) ] -> [ (1,2,5)(3,4,6), (1,2)(3,4) ],
  [ (1,2,3), (1,2) ] -> [ (1,3,5)(2,6,4), (1,2)(3,4)(5,6) ] ]
gap> Sum(emb,x->Index(s6,Normalizer(s6,Image(x))));
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ahulpke
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  • Thanks for your answer ! Yes it works for the case s3 in s5. But not for s3 in s6, or s3 in s7 for instance. Indeed I need to understand the pattern. I'm a beginner in GAP (wonderful software!), and I'm not very good with Normalizer notion.. So is there a formula in the case of s3 in sX with GAP to have the total number of subgroups that are isomorphic to s3 in sX ? – Karev Jun 28 '17 at 22:46
  • @Karev You will have to replace S5 with S6 and calculate new embeddings, of course. As for a formula, you will need to deal with regular $S_3$ and the natural $S_3$ and diagonals thereof, as well as further orbits of length 2. Its not horribly hard, but might take a day or two to sort out details. Basically you need to consider partitions of $n$ into cells of size $1,2,3,6$. – ahulpke Jun 28 '17 at 22:53
  • Oh sorry !! I made a mistake, of course that works for all these cases! Thanks very much for your help ahulpke !! (and thanks for your works in GAP !) – Karev Jun 28 '17 at 22:57