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When going through a list of elements, GroupOrbits usually recognises if the orbit of an element has been calculated already and will skip it in that case to avoid repetitions. For example,

GroupOrbits[SymmetricGroup[2],{1}] == GroupOrbits[SymmetricGroup[2],{1,2}]

will result in True (because 2 is in the orbit of 1).

However, if I define

a = {{1,1},{1,2}}
b = {{2,1},{2,2}}

the sets GroupOrbits[SymmetricGroup[2],{a}] and GroupOrbits[SymmetricGroup[2],{b}] are practically the same (as sets, in mathematical notation: {(1,1),(1,2)}={(1,2),(1,1)} etc.), but aren't treated as the same by Mathematica. How can I achieve that the output of GroupOrbits[SymmetricGroup[2],{a,b}] doesn't add one more orbit?

Secondary question: How does Mathematica handle an input of the form {...}? As a list, as a tuple, or as a set? And how can all these types be distinguished? For example, If I have a list or a tuple, how can I pass it to Mathematica as a set (and the other way round)?

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  • $\begingroup$ Is there a reason you're not doing just GroupOrbits[SymmetricGroup[2], a] (without the curly braces)? Because in that case the two yield the same result. $\endgroup$
    – march
    Commented May 16, 2022 at 16:42
  • $\begingroup$ Yes, I want the orbit of {{1,1},{1,2}} (as a set), not the respective orbits of the two elements {1,1} and {1,2}. $\endgroup$
    – Thrash
    Commented May 16, 2022 at 16:48
  • $\begingroup$ Instead of b, I could of course write ReverseSort@b, that is GroupOrbits[SymmetricGroup[2],{a,ReverseSort@b}], but this requires the knowledge of b. It would be nice to have that done automatically because the above example is just a minimal example. $\endgroup$
    – Thrash
    Commented May 16, 2022 at 17:01

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You have nested braces. The inner braces {...} does not mean sets, but vectors. Group is acting on the set of vectors (component-wise). For example,

GroupOrbits[CyclicGroup[3], {{1, 1}}]

(* {{{1, 1}, {2, 2}, {3, 3}}} *)

but

GroupOrbits[CyclicGroup[3], {{1, 10}}]

(* {{{1, 10}, {2, 10}, {3, 10}}} *)

In other words, the orbit is obtained by repeated application of /.{1->2,2->3,3->1}.

In your case, a = {{1, 1}, {1, 2}}; b = {{2, 1}, {2, 2}}; are interpreted as 2x2 matrices, and in this four dimensional space, orbit of a is

GroupOrbits[SymmetricGroup[2], {a}]

(* {{{{1, 1}, {1, 2}}, {{2, 2}, {2, 1}}}} *)

Note that the latter 2x2 matrix is obtained from the former by /.{1->2,2->1}. Similarly, the orbit of b is

GroupOrbits[SymmetricGroup[2], {b}]   

(* {{{{1, 2}, {1, 1}}, {{2, 1}, {2, 2}}}} *).

These two orbits are disjoint in the space of 2x2 matrices, so GroupOrbits[SymmetricGroup[2], {a, b}] returned disjoint union of these

(*{{{{1, 1}, {1, 2}}, {{2, 2}, {2, 1}}}, {{{1, 2}, {1, 1}}, {{2, 1}, {2, 2}}}} *)

as it should be.

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