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Operations and Cycle Notation

  1. Nov 2, 2014 #1
    Hello everyone,

    The problem I am working on asks me to write a Cayley table for ##D_3## using cycle notation. However, I am having difficult with determining how to operate on a pair of "cycles." Here is my work so far:

    ##R_0 = e = ()##

    ##R_{120} = (123)##

    ##R_{240} = (312)##

    ##F_T = (321)## (This transformation corresponds to holding the top corner fixed and flipping it)

    ##F_L = (132)##

    ##F_R = (213)##

    Now, I realize that ##R_{120} \circ R_{120} = R_{240}##; but how do I write this mapping in cycle notation?

    ##(123) \circ (123) = (312)##...

    I am trying to figure out the rule by which I form a new cycle.
     
    Last edited: Nov 2, 2014
  2. jcsd
  3. Nov 2, 2014 #2

    Dick

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    I don't think you have the members of the group written correctly in cycle notation. If you are asking how to multiply cycles try You'll find (123)(123)=(321). And, for example, (123)=(312) - they represent the same permutation.
     
  4. Nov 2, 2014 #3
    I don't quite understand what (1 5 2 4 6 8)(3 7) is. This is all my textbook says regarding cycle notation:

    "...we can denote the same counterclockwise rotation by (123), where this notation is interpreted to mean that each number goes to the number to its right, except the last number in the parenthesis, which goes to the first number in the parenthesis. This notation is called the cycle notation."

    I have searched the internet to learn more about this cycle notation; however, I did not find much information.
     
  5. Nov 2, 2014 #4

    RUber

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    So it sounds to me that your ##R_{120}## means to conduct just this operation.
    ##R_{120}\circ R_{120}= (123)(123)##
    Means, each number goes to the number on its right, except for the last number which goes to the first number.
    (123)(ABC)= (CAB)
    (123) sends A to the right, B to the right and C to the front.
     
  6. Nov 2, 2014 #5
    I find this cycle notation very disagreeable. I really don't understand how to use this cycle notation. Also, I can't figure out why by cycles are incorrect.

    How would I, for instance, compute ##(123) \circ (132)##, where ##(132)## is applied first? It should result in the identity.
     
  7. Nov 2, 2014 #6

    RUber

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    I think the video Dick posted explained this pretty well.
    Start with 1.
    (132) sends 1 to 3, (123) acts on 3 and sends it to 1.
    For 2:
    (132) sends 2 to 1, (123) acts on 1 and sends it to 2.
    For 3:
    (132) sends 3 to 2, (123) acts on 2 and sends it to 3.
     
  8. Nov 2, 2014 #7
    I understand how a single transformation works. I would like to understand how a composition of transformations works.
     
  9. Nov 2, 2014 #8

    Dick

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    Ruber was explaining how to compute the product (123)(132). That is a composition of the two permutations (132) and (123).
     
  10. Nov 7, 2014 #9
    Okay, after having watched the video again, I believe I am understanding what happens a little better. However, there is still one aspect of the calculation that confuses me: how do we know when to close the cycle? In the example given in the video, he closes the first cycle because 6 went to 1; but he closes the second cycle because 3 went to 7 and 7 and went to 3 (basically, no transformation took place). How do I distinguish the two cases?
     
  11. Nov 7, 2014 #10
    Also, have I correctly translated the transformations into cycle notation?

    ##R_0 := ()##

    ##R_{120} := (123)##

    ##R_{240} := (132)##

    ##F_T := (23)##

    ##F_L := (12)##

    ##F_R := (13)##
     
  12. Nov 7, 2014 #11

    Dick

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    You close a cycle when the last number maps back into the first number. And those look more plausible. Hard to say without a diagram of what they are.
     
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