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Math Help - Permutations cycles

  1. #1
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    Cool Permutations cycles

    Hi,
    I know this is an easy problem, but I just can't understand how it is done. What happens when you take a cycle (permutation in disjoint cycle notation) to the power of something. I know that (1 2 3 4)^4 = (1 2 3 4)^28 = id. But in the case of something like (1 3 5 7)^2, or (3 4 5)^2, I don't know what to do.

    Thanks you
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alireza1989 View Post
    Hi,
    I know this is an easy problem, but I just can't understand how it is done. What happens when you take a cycle (permutation in disjoint cycle notation) to the power of something. I know that (1 2 3 4)^4 = (1 2 3 4)^28 = id. But in the case of something like (1 3 5 7)^2, or (3 4 5)^2, I don't know what to do.

    Thanks you
    just go twice as opposed to once for the cycle

    example:

    for (1 3 5 7)^2

    1 --> 3 --> 5, so 1 --> 5
    5 --> 7 --> 1, so 5 --> 1

    so one cycle is (1 5)

    then, 3 --> 5 --> 7, so 3--> 7
    and 7 --> 1 --> 3, so 7 --> 3

    so another cycle is (3 7)

    thus (1 3 5 7)^2 = (1 5)(3 7)
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