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Math Help - Permutations

  1. #1
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    Permutations

    Write (43)(521)(51)(32)(1452)(23)(15)(125)(34) as the product of disjoint cycles. My answer is (12354), but I do not think this is correct. Will you explain?
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Permutations

    The answer i got was (1) (2345). For in this element,  1 \to 2 \to 3 \to 2 \to 1 = (1)
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  3. #3
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    Re: Permutations

    In working such a problem, you need to specify whether composition is left to right or right to left. That is, if p_1 and p_2 are permutations does p_1\compose p_2 mean first do p_1 and then p_2 or first do p_2 followed by p_1.

    Example (1 2)(1 2 3)=(1 3) for left to right but (1 2)(1 2 3)=(2 3) for right to left. I think probably most algebraists do composition left to right.

    Assuming left to right, I get (1 3 2 4)(5)

    Good case in point. The previous answer does composition right to left!
    Last edited by johng; February 7th 2013 at 07:13 PM.
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  4. #4
    Senior Member jakncoke's Avatar
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    Re: Permutations

    I've always assumed the standard used in university was to do it from right to left but john makes a very good point to keep in mind.
    Last edited by jakncoke; February 7th 2013 at 07:19 PM.
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  5. #5
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    Re: Permutations

    We are doing them from right to left.
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  6. #6
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    Re: Permutations

    it is a well-known fact that if τ is a cycle: say τ = (a1 a2 a3....ak),

    then στσ-1 = (σ(a1) σ(a2) σ(a3)....σ(ak))

    for example, if τ = (1 3 4 2), and σ = (1 2 3), so that:

    σ(1) = 2
    σ(2) = 3
    σ(3) = 1
    σ(4) = 4

    then στσ-1 = (σ(1) σ(3) σ(4) σ(2)) = (2 1 4 3) (this assumes we are working "right-to-left" so στ(a) = σ(τ(a)), as with functional composition).

    in the permutation to be evaluated at hand we have the 4-cycle:

    (1 4 5 2) which is conjugated by the transposition (2 3), that is:

    (3 2)(1 4 5 2)(2 3) = (2 3)(1 4 5 2)(2 3) = (2 3)(1 4 5 2)(2 3)-1 (transpositions are of order 2, and thus are their own inverses).

    the net effect of this is to replace the "2" in (1 4 5 2) with a 3:

    (2 3)(1 4 5 2)(2 3)-1 = (1 4 5 3).

    we then have this conjugated by (1 5) = (5 1), which has the effect of making the 1 and 5 switch places, giving:

    (5 1)(1 4 5 3)(1 5) = (1 5)(1 4 5 3)(1 5) = (1 5)(1 4 5 3)(1 5)-1 = (5 4 1 3) = (1 3 5 4).

    we then have THIS conjugated by (1 5 2) (note that (1 5 2) = (5 2 1)). so in (1 3 5 4) we replace 1 by the image of 1 under (1 5 2), which is 5; the 3 stays, since 3 is not moved by (1 5 2), the 5 is replaced by 2 (2 is the image of 5 under (1 5 2)), and finally the 4 is unaffected:

    (5 2 1)(1 3 5 4)(1 2 5) = (1 5 2)(1 3 5 4)(1 2 5) = (1 5 2)(1 3 5 4)(1 5 2)-1 = (5 3 2 4) = (2 4 5 3).

    again, this is conjugated by the transposition (3 4) = (4 3), so the 2 and 5 in (2 4 5 3) are unaffected, and the 3 and 4 change places, giving:

    (2 3 5 4) (it is customary, but not required, to put the least element of a cycle first). there is no reason to include 1-cycles in a disjoint cycle decomposition, so they are typically omitted.

    one can also do this "the long way" calculating each successive permutation in turn (starting with (3 4) as the first permutation applied, followed by (1 2 5) and so forth):

    1-->1-->2-->2-->3-->3-->2-->2-->1-->1
    2-->2-->5-->1-->1-->4-->4-->4-->4-->3
    3-->4-->4-->4-->4-->5-->5-->1-->5-->5
    4-->3-->3-->3-->2-->1-->1-->5-->2-->2
    5-->5-->1-->5-->5-->2-->3-->3-->3-->4

    which also gives the 4-cycle (2 3 5 4).
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