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Identity element of a group

  1. Sep 15, 2008 #1
    1. The problem statement, all variables and given/known data
    In S_3, show that there are four elements satisfying x^2=e and three elements satisfying y^3=e.

    3. The attempt at a solution
    I don't understand what the question is asking at all.....
  2. jcsd
  3. Sep 15, 2008 #2
    S_3 is the group of permutations of three symbols, ie {1,2,3}. You can write it out in terms of cyclic permutations:

    S_3={(1), (12), (13), (23), (123), (132)}

    The question is asking you to show that of those six elements in S_3, four have the property that when squared equal (1), and three have the property that when cubed equal (1).

    Before trying to prove anything, play around with the elements a bit, multiply them together, see what you get. Eg:


  4. Sep 15, 2008 #3
    I don't understand the multiplication you did.


    Can you explain this a bit further?
  5. Sep 16, 2008 #4
    You start with the rightmost permutation ie (23). What does this do to 1? It sends it to 1, since it does not appear in (23). Move to the left. (12) sends 1 to 2. So the product (12)(23) sends 1 to 2. Do this again for 2. (23) sends 2 to 3, (12) doesn't affect 3, to (12)(23) sends 2 to 3. Likewise 3 goes to 1. We write this as (123).
  6. Sep 16, 2008 #5
    Why wouldn't (23) send 2 to 2? Are we working with mod 6, or is this something different?
  7. Sep 16, 2008 #6


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    Staff Emeritus
    Science Advisor

    Where did you get this problem? You seem to be saying that you do not know anything about permutations and have no idea what S3 is. Did you accidently get your homework from the wrong class?

    (23) is the permutation that sends 2 to 3 and 3 to 2 (and sends any other numbers to themselves). No, permutations have nothing whatever to do with "modulo" arithmetic.
  8. Sep 16, 2008 #7
    Each cyclic permutation (abc) only acts once in any given product. I echo HallsofIvy's sentiments, you should make sure this question is required in your class.
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