Determining if two subgroups of a symmetric group are conjugate

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This is quite old, but since it is not answered:

One way to do this is do use the (built-in) version of GAP to test the conjugacy.

This has the function RepresentativeAction(g,g1,g2) which takes three arguments, g, g1, g2 and either returns fail if g1 and g2 are not conjugate in g, or returns a conjugating element if they are.

You need to turn everything into a GAP-object in order to do this.

 gap.RepresentativeAction(gap("SymmetricGroup(3)"), gap(G1), gap(G2))

This returns () which is the identity, indicating that the groups are not only conjugate, but identical.

EDIT: if you define

S = SymmetricGroup(3)
gen1 = Permutation('(1,2)(3)')
gen2 = Permutation('(1,3)(2)')
H1=S.subgroup([gen1])
H2=S.subgroup([gen2])

You'll see that you get

sage: gap.RepresentativeAction(gap(S),gap(H1),gap(H2))
(2,3)

so the subgroups are conjugate, but not identical. Useful to know how to get this info, though!

This seems to work fine in Sage 8.4, for instance as tried on SageMathCell at https://sagecell.sagemath.org/.

Here is how to check the version of Sage (a good thing to include when reporting errors).

sage: print(version())
SageMath version 8.4, Release Date: 2018-10-17

Testing if two conjugacy classes are equal works:

sage: S = SymmetricGroup(3)
sage: a = Permutation('(1, 2, 3)')
sage: b = Permutation('(1, 3, 2)')
sage: c = Permutation('(1, 2)')
sage: d = Permutation('(1, 3)')
sage: G = PermutationGroup([a, c])
sage: H = PermutationGroup([b, d])
sage: GG = ConjugacyClass(S, G)
sage: HH = ConjugacyClass(S, H)
sage: GG == HH
False

I changed the notation from that in the question, but copy-pasting the exact code from the question works equally well (try it in SageMathCell).