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Proof about cycle with odd length

  1. Nov 23, 2017 #1
    1. The problem statement, all variables and given/known data
    In the following problems let ##\alpha## be a cycle of length ##s##, and say
    ##\alpha = (a_1a_2 . . . a_s)##.

    5) If ##s## is odd, ##\alpha## is the square of some cycle of length s. (Find it. Hint: Show ##\alpha = \alpha^{s+1}##)
    2. Relevant equations


    3. The attempt at a solution
    I know ##(12345)(12345) = (12345)## and ##(1234)(1234) = (13)(24)##. And I think I can prove this is true for any integer n, such that if a cycle ##\alpha## has an odd length, then squaring it produces another cycle of odd length, and ##\alpha## has an even length, s, then its square is the product of two disjoint cycles, each of length s/2. I guess i've pretty much restated the problem... i'm stuck.

    For the hint.. I know ##\alpha## of length ##s## has ##\alpha## distinct powers but I'm not sure how to prove it.

    Sorry this is so bare bones, thank you for any help
     
  2. jcsd
  3. Nov 24, 2017 #2

    andrewkirk

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    That doesn't look right. I get ##(12345)(12345)=(13524)##.
    The 'square root' of (12345) is (13524) because (13524)(13524)=(12345).

    The hint is a good one. Have you tried proving it?
    What is the position in the cycle to which the ##k##th element ##a_k## is mapped by one application of ##\alpha## (in other words, find an expression in terms of ##k## for the index ##j## such that ##\alpha a_k=a_j##).

    Then what about two applications of ##\alpha##. What about ##s## applications? What about ##s+1##?
     
  4. Nov 24, 2017 #3
    Thanks for the help and correction, I agree that (12345)12345) = (13524)

    So we can write ##\alpha^na_i = a_j## where ##j = (i + n) \mod s##
    Therefore ##\alpha a_i = a_{((i + 1) \text{mod s})}##
    and ##\alpha^{s+1}a_i = a_{((s+1+i) \text{mod s})}##
    Since (i + 1) = (s + 1 + i) (mod s)
    we have ##\alpha a_i = \alpha^{s+1}a_i##
    ##\alpha a_ia_i^{-1} = \alpha^{s+1}a_ia^{-1}##
    ##\alpha e = \alpha^{s+1}e##
    ##\alpha = \alpha^{s+1}##
    []

    ##\alpha^sa_i = a_{i + s} = a_i##
    ##\alpha^{s+1}a_i = a_{i + s + 1} = a_{i + 1}##

    I'm not sure how to apply this to the original problem but ill keep thinking.. I also thought of this which I think is a proof in response to the "Find it" part of the hint.

    Proof: Suppose ##\alpha## is a cycle of length ##s## where ##s## is an odd integer. Then we can write ##\alpha## as ##(a_1a_3...a_sa_2a_4...a_{s-1}).## But ##(a_1a_3...a_sa_2a_4...a_{s-1}) = (a_1a_2...a_s)(a_1a_2...a_s).## Let ##\beta = (a_1a_2..a_s).## Note that ##\beta## is a cycle of length s. Then ##\alpha = \beta^2##. This concludes the proof.
     
  5. Nov 24, 2017 #4

    andrewkirk

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    Give that ##\alpha=\alpha^{s+1}## and ##s## is odd, how can you write ##\alpha## as the square of an integer power of itself?
     
  6. Nov 24, 2017 #5

    s is odd, so we can write s = 2k + 1, where k is an integer.
    We note that ##\alpha^n## is a cycle of length s for all integers n.
    Then ##\alpha = \alpha^{s+1} = \alpha^{2(k+1)} = \alpha^{k+1}\alpha^{k+1}##.
    so ##\alpha## is the square of ##\alpha^{k+1}##.
     
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