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:


GAP returns: "Error no method found!"

Thanks for any answer to help me!


closed as off-topic by Bill Dubuque, Trevor Gunn, user370967, ccorn, BruceET Jun 28 '17 at 23:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – Bill Dubuque, Trevor Gunn, Community, ccorn
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Questions like this are on-topic here iff they contain some on-topic mathematical content, which this question does not. $\endgroup$ – Bill Dubuque Jun 28 '17 at 16:59
  • $\begingroup$ As my answer to math.stackexchange.com/q/1569349 says, LatticeByCyclicExtension and SubgroupsSolvableGroup accept optional arguments which allow to put restrictions on computed subgroups. $\endgroup$ – Alexander Konovalov Jun 28 '17 at 22:24
  • $\begingroup$ Thanks for your answer, but what are these arguments? How solve my problem? I don't understand how manipulate these functions could help me $\endgroup$ – Karev Jun 28 '17 at 22:27
  • $\begingroup$ And as for IsomorphicSubgroups, it returns monomorphisms, and you have to ask Size(Image(emb[1])); if you intend to see 6. $\endgroup$ – Alexander Konovalov Jun 28 '17 at 22:28
  • $\begingroup$ 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. $\endgroup$ – Alexander Konovalov Jun 28 '17 at 22:31

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


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))));
  • $\begingroup$ 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 ? $\endgroup$ – Karev Jun 28 '17 at 22:46
  • $\begingroup$ @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 $1,2,3,6$. $\endgroup$ – ahulpke Jun 28 '17 at 22:53
  • $\begingroup$ 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 !) $\endgroup$ – Karev Jun 28 '17 at 22:57

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