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No cycles in permutation N how to calculate sgn(N^2)?

  1. Feb 14, 2015 #1
    N is a 2 x n matrix:

    N =

    1 2 3 4 ... n-1 n
    n n-1 .... 4 3 2 1

    then N^2 =

    1 2 3 4 ... n-1 n
    1 2 3 4 ... n-1 n


    You COULD use the theorem: sgn(N^2) = sgn(N)sgn(N)

    however, im asked to find sgn(N^2) by the traditional method: sgn(N^2) = (-1)^([L1 - 1] + [L2 - 1].....) where L represents the respective lengths of each cycle. However, N^2 has no cycles, so im confused.




    How would I go about this? thanks.
     
    Last edited: Feb 14, 2015
  2. jcsd
  3. Feb 14, 2015 #2
    I think I know:

    since there are no cycles, you just have sgn(N^2) = (-1)^0 = 1.

    can anyone confirm if my method is correct? thanks
     
  4. Feb 15, 2015 #3

    ChrisVer

    User Avatar
    Gold Member

    If that's the N^2 you have, the cycles are 1-cycles... So their parity is + ...
    In particular you have the identity element of Sn:
    [itex]e=(1)(2)...(n)[/itex]
    and [itex]W(e)= (1-1)_{first} + (1-1)_{second} + ... + (1-1)_{n-th} =0 [/itex] and so the parity:

    [itex] \delta_P(e) = (-1)^{W(e)} = +1 [/itex]
     
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