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Problem concerning permutation groups

  1. Sep 27, 2006 #1
    Here is the problem concerning permutation groups:

    u =
    1 2 3 4
    -------
    3 4 2 1

    Show that there is no p such that p^2 (the second permutation) = u


    I've tried just substituting values for p1, p2, p3 and p4 in:
    1 2 3 4
    ------------
    p1 p2 p3 p4

    p1 = 1 doesn't work because 1 would never go to 3 the second time

    p1 = 2, p2 = 3, p3 = 4, p4 = 2, which doesn't work because 4 doesn't go to 1 the second time, it goes to 3.

    p1 = 3 doesn't work because then p3 would also have to be 3, so 3 wouldn't go to 2 the second time

    p1 = 4, p2 = 1, p3 = 1, p4 = 3, which doesn't work because 3 goes to 4 instead of the required 2.

    Is this sufficient to show what's being asked?
     
    Last edited: Sep 27, 2006
  2. jcsd
  3. Sep 27, 2006 #2

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    What about the other permutations for p1=2,4?
     
  4. Sep 27, 2006 #3
    I wrote those in the order I would do the second permutation. If it doesn't work in that order, it isn't going to work in any other order.
     
  5. Sep 27, 2006 #4

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    I'm saying, if you want that to count as a proof, you have to clearly explain why the other permutations won't work.
     
  6. Sep 27, 2006 #5
    If p1 = 2, then p2 has to equal 3 for p1 to go to 3 the second time. That means p3 = 4 for 2 to go to 4 the second time. For those to hold true, p4 = 2 so 3 goes to 2, but then 4 goes to 2 which goes to 3, so it doesn't hold.

    Similarly, if p1 = 4, then p4 = 3. 3 has to go to 1 for p4 to go to 1 on the second permutation, so p3 = 1. If p3 = 1, then p1 = 2 so 3 goes to 2 on the second permutation. Also, p2 = 1 because 1 goes to 4 and 2 has to go to 4 on the second permutation, but this means 3 goes to 1 which goes to 4, so 3 goes to 4 on the second permutation, which doesn't hold for u.
     
  7. Sep 27, 2006 #6

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    That works. Alternatively, do you know about the sign of permutations? This will give a one line proof.
     
  8. Sep 27, 2006 #7
    I know about the sign, but I don't see where you're going. We call it the signature in our book. It would be great to know how to prove it without brute force. I'm an engineering major, so I'm not great at proofs.
     
    Last edited: Sep 27, 2006
  9. Sep 27, 2006 #8

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    Well if the sign of permutation p1 is e1 (which is either +/-1) and the sign of p2 is e2, then the sign of p1 p2 (ie, p1 followed by p2) is e1*e2. So what is the sign of p1^2?
     
  10. Sep 27, 2006 #9
    It would have to be 1. I see now, thanks. I should've thought of that.
     
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