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Math Help - Help in Mapping..

  1. #1
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    Help in Mapping..

    I need help in the mapping problems..
    Please help!

    Q1) A) Write (a1 a2 a3.....ak) ^ (-1) in cyclic notation (without the symbol
    for inverse)?

    B) For which values of k will every k-cycle be its own inverse?

    Here: ^ (-1) means inverse of the set.


    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Vedicmaths View Post
    I need help in the mapping problems..
    Please help!

    Q1) A) Write (a1 a2 a3.....ak) ^ (-1) in cyclic notation (without the symbol
    for inverse)?

    B) For which values of k will every k-cycle be its own inverse?

    Here: ^ (-1) means inverse of the set.


    Thanks in advance!
    Hint: Show that (a_1,a_2,...,a_k)(a_k,a_{k-1},...,a_1) = \text{id}.
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  3. #3
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    I am sorry I could not understand what do you mean?
    Are you saying that we have to show that the product of these cyclic notation you have mentioned above is equal to the identity mapping?
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  4. #4
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    Quote Originally Posted by Vedicmaths View Post
    I am sorry I could not understand what do you mean?
    Are you saying that we have to show that the product of these cyclic notation you have mentioned above is equal to the identity mapping?
    Yes. Show the product of these two is the indentity, and that will be the inverse.
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  5. #5
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    This is what I got...is this fine??


    The inverse for the k-cycle

    ( a[1] a[2] ... a[k] )

    is

    ( a[k] a[k-1] a[k-2] ... a[2] a[1] )

    = ( a[1] a[k] a[k-1] a[k-2] ... a[2] )

    Therefore a k-cycle is its own inverse if and only if

    - If there is no a[2], ( a[1] ) = identity permutation is obviously its own inverse. k = 1.
    - Otherwise, there is an a[2], and we must have a[2] = a[k] → k = 2 since a number may only appear once in a cycle.

    So, k = 1 or 2.
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  6. #6
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    Quote Originally Posted by Vedicmaths View Post
    So, k = 1 or 2.
    Yes.
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