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Group orbits

  1. Nov 29, 2011 #1
    I just started learning about group orbits, and wanted to look it up online, because I needed some more clarification. However, stumbling upon this Wolfram's Mathworld entry, I ended up even more confused, especially after reading that example for the permutation group G1. Could someone perhaps explain to me how exactly it is that if [itex]G_{1} = \{(1234), (2134), (1243), (2143)\}[/itex], the orbits of 1 and 2 are {1, 2}? To me it seems that 1 can get sent to either 2, 3 or 4, but not to itself, and 2 to 1, 3 and 4, but I also know that orbits are either disjoint or equal, so I must not be getting something.
  2. jcsd
  3. Nov 29, 2011 #2
    Each of those four elements of the group is a permutation. 1 and 2 have the same orbit (1,2) because those permutations only map them only each other.

    An orbit is an equivalence class (have you learned about equivalence relations or partitions?) Two objects are in the same orbit if the group can be used to turn one into another.

    An example that shows up in combinatorics is the following:

    Suppose you have a square and beads of three colors. How many distinguishable ways can you place the colored beads on the corners of the square? Two ways are indistinguishable if they are the same up to a rotation.

    Our group is the rotation group, and our set is all the different combinations of colors. Two combinations are in the same orbit if we can use the rotation group to turn one into another. That is, they are the same if they are indistinguishable, which is exactly why orbits are useful for combinatorics.
  4. Nov 29, 2011 #3
    Hmm, how? That's what I don't get. I mean, (2 1 4 3) clearly sends 1 to 4, doesn't it, and (1 2 3 4) sends 2 to 3. Or what am I missing here?
  5. Nov 30, 2011 #4


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    it's a notation problem.

    when the mathworld article says:

    G1 = {(1234), (2134), (1243), (2143)}, they are not using cycle notation but rather image notation, so (1234) is the identity mapping on {1,2,3,4}.

    in cycle notation, this group is: {e, (1 2), (3 4), (1 2)(3 4)}

    this is NOT a rotation group, rather it is isomorphic to the klein 4-group (and thus a reflection group).

    it should be clear that we have just two orbits: {1,2} and {3,4}.

    a rotation group would be: {e, (1 2 3 4), (1 3)(2 4), (1 4 3 2)}. this group is transitive on the set {1,2,3,4}, and has just one orbit. it should be clear that this is the same (up to isomorphism) action we get by letting Z4 act on itself by left-multiplication:

    0.x is the map x→0+x, which is clearly the identity map on {0,1,2,3}
    1.x is the map x→1+x (mod 4), which is the 4-cycle (1 2 3 4) (using "4" instead of "0").

    clearly k.x is the map x→k+x, which is the map x→1+x composed with itself k times.

    a more interesting kind of orbit is given by the orbits of the action g.x = gxg-1, or conjugation. these orbits are called conjugacy classes.

    for example, for the quaternion group Q8 = {1,-1,i,-i,j,-j,k,-k}, we have the orbits: {1}, {-1}, {i,-i}, {j,-j}, {k,-k}. the corresponding subgroup of S8, that this action on Q8 represents is:

    {e, (3 4)(5 6), (5 6)(7 8), (3 4)(7 8)}, which is also isomorphic to the klein 4-group, which is another way of showing that:

    Q8/Z(Q8) ≅ Z2 x Z2,

    because the kernel of this action is the center of Q8.
  6. Nov 30, 2011 #5
    Ah, I see, makes sense then. Why are they using this notation then, is this something that's widely accepted or is it just something only they are using? I would assume I wasn't the first, and am not going to be the last getting confused by it.
  7. Dec 1, 2011 #6


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    it's not that uncommon, especially with people who look at permutations as functions, rather than algebraic objects. it's common practice in much of mathematics to refer to a function f by its image f(x).
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