7
$\begingroup$

Here's how to show all subgroups of $S_4$ in GAP:

gap> AllSubgroups(SymmetricGroup(4));
[ Group(()), Group([ (1,2)(3,4) ]), Group([ (1,3)(2,4) ]), Group([ (1,4)(2,3) ]), Group([ (3,4) ]), Group([ (2,3) ]),
  Group([ (2,4) ]), Group([ (1,2) ]), Group([ (1,3) ]), Group([ (1,4) ]), Group([ (2,4,3) ]), Group([ (1,3,2) ]),
  Group([ (1,4,2) ]), Group([ (1,4,3) ]), Group([ (1,4)(2,3), (1,3)(2,4) ]), Group([ (3,4), (1,2)(3,4) ]),
  Group([ (1,4), (1,4)(2,3) ]), Group([ (2,4), (1,3)(2,4) ]), Group([ (1,3,2,4), (1,2)(3,4) ]), Group([ (1,4,3,2), (1,
   3)(2,4) ]), Group([ (1,2,4,3), (1,4)(2,3) ]), Group([ (3,4), (2,4,3) ]), Group([ (1,4), (1,4,3) ]),
  Group([ (2,3), (1,3,2) ]), Group([ (1,2), (1,4,2) ]), Group([ (1,4)(2,3), (1,3)(2,4), (3,4) ]), Group([ (1,2)
  (3,4), (1,3)(2,4), (1,4) ]), Group([ (1,2)(3,4), (1,4)(2,3), (2,4) ]), Group([ (1,4)(2,3), (1,3)(2,4), (2,4,3) ]),
  Group([ (1,4)(2,3), (1,3)(2,4), (2,4,3), (3,4) ]) ]

In Mathematica I can display the elements of $S_4$:

In[2]:= GroupElements[SymmetricGroup[4]]

Out[2]= {Cycles[{}],Cycles[{{3,4}}],Cycles[{{2,3}}],Cycles[{{2,3,4}}],Cycles[{{2,4,3}}],Cycles[{{2,4}}],Cycles[{{1,2}}],Cycles[{{1,2},{3,4}}],Cycles[{{1,2,3}}],Cycles[{{1,2,3,4}}],Cycles[{{1,2,4,3}}],Cycles[{{1,2,4}}],Cycles[{{1,3,2}}],Cycles[{{1,3,4,2}}],Cycles[{{1,3}}],Cycles[{{1,3,4}}],Cycles[{{1,3},{2,4}}],Cycles[{{1,3,2,4}}],Cycles[{{1,4,3,2}}],Cycles[{{1,4,2}}],Cycles[{{1,4,3}}],Cycles[{{1,4}}],Cycles[{{1,4,2,3}}],Cycles[{{1,4},{2,3}}]}

What's a good approach for listing the subgroups in Mathematica?

$\endgroup$
2
  • 5
    $\begingroup$ Unfortunately, the ClearAll["Global`*"]; FiniteGroupData[{"SymmetricGroup", 4}, "Subgroups"] command fails, outputting Missing["NotAvailable"]. $\endgroup$
    – user64494
    May 10, 2018 at 20:14
  • 1
    $\begingroup$ The command FiniteGroupData[{"SymmetricGroup", 4}, "Information"] says, in particular, "The symmetric group S_n of degree n is the group of all permutations on n symbols. S_n is therefore a permutation group of order n! and contains as subgroups every group of order n". Hope this would be useful. $\endgroup$
    – user64494
    May 24, 2022 at 14:36

2 Answers 2

3
$\begingroup$

My solution is too too too slow, which is so slow that I can't stand it. But it does work when the group orders less than $20$. Such as the $D_8$ of order $16$

 DeleteDuplicatesBy[
  PermutationGroup /@ Subsets[GroupElements[DihedralGroup[8]]], 
  GroupElements] // Timing

{169.313,{PermutationGroup[{}],PermutationGroup[{Cycles[{{2,8},{3,7},{4,6}}]}],PermutationGroup[{Cycles[{{1,2},{3,8},{4,7},{5,6}}]}],PermutationGroup[{Cycles[{{1,2,3,4,5,6,7,8}}]}],PermutationGroup[{Cycles[{{1,3},{4,8},{5,7}}]}],PermutationGroup[{Cycles[{{1,3,5,7},{2,4,6,8}}]}],PermutationGroup[{Cycles[{{1,4},{2,3},{5,8},{6,7}}]}],PermutationGroup[{Cycles[{{1,5},{2,4},{6,8}}]}],PermutationGroup[{Cycles[{{1,5},{2,6},{3,7},{4,8}}]}],PermutationGroup[{Cycles[{{1,6},{2,5},{3,4},{7,8}}]}],PermutationGroup[{Cycles[{{1,7},{2,6},{3,5}}]}],PermutationGroup[{Cycles[{{1,8},{2,7},{3,6},{4,5}}]}],PermutationGroup[{Cycles[{{2,8},{3,7},{4,6}}],Cycles[{{1,2},{3,8},{4,7},{5,6}}]}],PermutationGroup[{Cycles[{{2,8},{3,7},{4,6}}],Cycles[{{1,3},{4,8},{5,7}}]}],PermutationGroup[{Cycles[{{2,8},{3,7},{4,6}}],Cycles[{{1,5},{2,4},{6,8}}]}],PermutationGroup[{Cycles[{{1,2},{3,8},{4,7},{5,6}}],Cycles[{{1,3,5,7},{2,4,6,8}}]}],PermutationGroup[{Cycles[{{1,2},{3,8},{4,7},{5,6}}],Cycles[{{1,5},{2,6},{3,7},{4,8}}]}],PermutationGroup[{Cycles[{{1,3},{4,8},{5,7}}],Cycles[{{1,5},{2,6},{3,7},{4,8}}]}],PermutationGroup[{Cycles[{{1,4},{2,3},{5,8},{6,7}}],Cycles[{{1,5},{2,6},{3,7},{4,8}}]}]}}

But this method obviously cannot get all subgroups of $S_4$, because I need to consider a combination of $2^{24}=16777216$ elements, which is unthinkable for me. There may be other specialized algorithms to do this.

$\endgroup$
1
+50
$\begingroup$

Some things change to better. In version 13 on Windows 10

FiniteGroupData[{"SymmetricGroup",  4},"Subgroups"]

{"Trivial", {"CyclicGroup", 2}, {"CyclicGroup", 3}, {"CyclicGroup", 4}, "Vierergruppe", {"SymmetricGroup", 3}, {"AlternatingGroup", 4}, {"SymmetricGroup", 4}}

$\endgroup$
8
  • 3
    $\begingroup$ Note that this is only a subset of the subgroups, not all of them. :) $\endgroup$
    – yode
    May 24, 2022 at 13:28
  • 1
    $\begingroup$ @yode: One may continue:FiniteGroupData[{"SymmetricGroup", 3}, "Subgroups"] which produces {"Trivial", {"CyclicGroup", 2}, {"CyclicGroup", 3}, {"SymmetricGroup", 3}} and FiniteGroupData["Vierergruppe", "Subgroups"] and FiniteGroupData[{"AlternatingGroup", 4}, "Subgroups"] and ... I leave the rest on your own. $\endgroup$
    – user64494
    May 24, 2022 at 13:38
  • $\begingroup$ Note all your subgroup of subgroups all in the subgroup list of $S_4$ $\endgroup$
    – yode
    May 24, 2022 at 16:27
  • $\begingroup$ @yode: No, it is not true : FiniteGroupData[{"AlternatingGroup", 4}, "Subgroups"] results in {"Trivial", {"CyclicGroup", 2}, {"CyclicGroup", 3}, {"AbelianGroup", {2, 2}}, {"AlternatingGroup", 4}}. $\endgroup$
    – user64494
    May 25, 2022 at 5:12
  • $\begingroup$ FiniteGroupData returns a non-homogeneous group, but even if it is non-homogeneous, $S_4$ has 11 instead of 8 as the maple $\endgroup$
    – yode
    May 28, 2022 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.