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[Abstract Algebra] Permutations and shuffling cards

  1. Jun 23, 2013 #1
    It's been a while since I've posted. This is a problem I had for a homework assignment a few weeks ago but I completely figure out. Any help appreciated.

    1. The problem statement, all variables and given/known data
    "A card-shuffling machine always rearranges cards in the same way relative to the order
    in which they were given to it. All of the hearts arranged in order
    from ace to king were put into the machine, and then the shuffled
    cards were put into the machine again to be shuffled. If the cards
    emerged in the order 10, 9, Q, 8, K, 3, 4, A, 5, J, 6, 2, 7, in what
    order were the cards after the first shuffle?"

    2. Relevant equations
    \

    3. The attempt at a solution

    We have a permutation
    [itex]\alpha^2 = \left[ \begin{array}{cc}
    A & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & J & Q & K \\
    10 & 9 & Q & 8 & K & 3 & 4 & A & 5 & J & 6 & 2 & 7 \end{array} \right] [/itex] which can be written as a single 13-cycle [itex]\alpha^2 = (A,10,J,6,3,Q,2,9,5,K,7,4,8)[/itex].

    The goal is to find out how where each card is mapped to under [itex]\alpha[/itex].

    And I'm not sure how to go about this.
     
  2. jcsd
  3. Jun 23, 2013 #2

    haruspex

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    Think about raising α2 to various powers.
     
  4. Jun 23, 2013 #3
    From what I recalled on my first attempt.

    We know that [itex](\alpha^2)^{13} = \alpha^{26} = e[/itex] since [itex]\alpha^2[/itex] is a 13-single cycle. So, [itex]|\alpha^{2}| \mbox{ divides } 13[/itex].

    That is, [itex]|\alpha^2| = 1 \mbox{ or } 13[/itex]. Clearly, [itex]|\alpha^2| \neq 1[/itex] since [itex]\alpha^2 \neq e[/itex]. Thus, [itex]|\alpha^2| = 13[/itex].

    Now that I thought about it.

    Using the formula [itex]|g| = |g^k|gcd(k,|g|)[/itex], we see that [itex]|\alpha| = |\alpha^{2}| gcd(2,|\alpha|) = 13 gcd(2,|\alpha|) = 26 \mbox{ or } 13[/itex].

    Since we're "working" with the symmetric group [itex]S_{13}[/itex] and [itex]26 = 2 \times 13[/itex], we know that no element (by product of cycles) in [itex]S_{13}[/itex] has order 26. So, [itex]|\alpha|=13[/itex].

    Hence, [itex]\alpha = \alpha^{13} \alpha = \alpha^{14} = (\alpha^{2})^{7}[/itex]. Therefore, starting with A, we find [itex]\alpha = (A,9,10,5,J,K,6,7,3,4,Q,8,2)[/itex].
     
  5. Jun 24, 2013 #4

    haruspex

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    Quite so - well done.
     
  6. Oct 26, 2016 #5
    So does this imply that the order of the cards after the first shuffle were 9 A 4 Q J 7 3 2 10 5 K 8 6?
     
  7. Oct 26, 2016 #6

    haruspex

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    Yes.
     
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