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An equation with permutations, x^2 = sigma.

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data

    x^2 = sigma.
    The permutation sigma = (1 2 6 7 5 3 4), a cycle of length seven.

    Determine x!

    3. The attempt at a solution

    I have tried a few times but my attempts are totally wrong. I don't know where to start! :S

    I found a similar problem in the textbook. They could, by just lookin at the problem, say that there is no x for that (in the book not the one above) particular type of permutation. How? :S
     
  2. jcsd
  3. Oct 26, 2008 #2

    CompuChip

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    I don't know if this is the best way, but it definitely works:
    First determine what is the order of sigma. This can be used to solve the question. For example: suppose that the order is 3. Then sigma^3 = 1, so sigma^4 = sigma. From this, you can immediately find x.
    If the order would be 4, on the other hand, then you would get sigma^4 = 1 so sigma^5 = sigma. The left hand side is not a square of anything, so there is no solution.
     
  4. Oct 26, 2008 #3
    I'm not sure if i understood this correct. Lets see, with your reasoning...

    I know that sigma^7=1, therefore sigma^8=sigma. Then x=sigma^4. Right?
     
  5. Oct 26, 2008 #4

    HallsofIvy

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    Yes, that's exactly right. I didn't quite comprehend what CompuChip meant by "suppose that the order is 3" since the order was clearly 7! I'm glad you did.
     
  6. Oct 26, 2008 #5

    CompuChip

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    I tried not to give away the answer completely. Didn't work :smile:
     
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