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Math Help - Permutations: maximal order and odd/even

  1. #1
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    Permutations: maximal order and odd/even

    Problem number one goes like this: determine, in cycle form (eng?), a σ є S10 with maximal order, in other words so that the order of σ is greater than the order of any other permutation in S10. State this order, too.

    Is there any systematic method of doing this or is trying different combinations the only way?

    The second problem: Let π denote that permutation on the set {1, 2, 3, 4, 5, 6, 7} which can be described with the product

    π = (1 2 3 4 5) (1 7 6 5) (1 3 5 7)

    a) Write π as a product of disjunct cycles.

    b) Is π an odd or even function?

    I would go about solving a) like this: write the cycles of π in one-row form (eng?) from right to left. Then, write [1 2 3 4 5 6 7] at the top and the final result at the bottom
    (which is [4 3 2 5 6 1 7]) and figure out the transpositions. But the solution is completely different: (1 4 5 6) (2 3) (7). How did they get that answer? Also, I figure that I haven't understood what a disjunct cycle is. Is it when you write the cycle form like (a b c) (d e)? But isn't that always the way to do it?

    For b) the solution is what I tried to do in a) and count the number of transpositions. The answer they state is (1 6) (1 5) (1 4) (2 3). Is there a simple way to get this answer or do I have to write a long list with a bubble sort from top to bottom?

    Grateful for answers!
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  2. #2
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    Re: Permutations: maximal order and odd/even

    If a permutation is expressed as a product of disjoint cycles then the order of the permutation is the lcm of the lengths of the cycles.

    So the order of (1 2 3)(7 8) is 6.

    For S_{10}

    10 = 5 + 3 + 2.

    5 x 3 x 2 = 30
    Last edited by a tutor; July 31st 2012 at 01:09 PM. Reason: Clarity.
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  3. #3
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    Re: Permutations: maximal order and odd/even

    For your permutation \pi you are working from the wrong end. These product are worked out left to right.

    So
    1->2
    2->3->5
    3->4
    4->5->1->3
    5->1->7->1
    6->5->7
    7->6

    So you end up with

    1 2 3 4 5 6 7
    2 5 4 3 1 7 6
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  4. #4
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    Re: Permutations: maximal order and odd/even

    I may be exposing my ignorance now but I worked out a set of transpositions like this:

    Permutations: maximal order and odd/even-transpositions.jpg

    Resulting in (1 2)(1 5)(3 4)(6 7).

    I hope this is all correct.
    Thanks from Leibniz111
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