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Groups whose elements have order 2

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data

    suppose that G is a group in which every non-identity element has order two. Show that G is commutative.

    2. Relevant equations



    3. The attempt at a solution
    Is my answer correct?

    Suppose that a,b and ab all have order two. we will show that a and b commute. By assumption, e=(ab)^2
    =abab
    As a and b are their own inverses, multiplying on the left by a and then b,
    we get
    ba=ab.
     
  2. jcsd
  3. Oct 19, 2009 #2
    Looks good to me.
    e = abab implies
    b = ababb = aba which implies
    ba = abaa = ab
     
  4. Oct 19, 2009 #3

    Dick

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    Yes, that's right.
     
  5. Oct 20, 2009 #4
    Thanks! Is there another way i can show this?
     
  6. Oct 20, 2009 #5

    Dick

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    Why do you need another way? What did you have in mind?
     
  7. Oct 20, 2009 #6
    well i just don't want to have the same answer as someone else.
     
  8. Oct 20, 2009 #7

    Dick

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    It's a simple problem. There's a simple answer. There's a few different permutations on the expression of that answer, but they are all really the same. Wouldn't it be better to move on to the next problem?
     
  9. Oct 20, 2009 #8
    That's true. I'm having trouble showing that an element k is a generator of Zn if and only if k and n are relatively prime.
     
  10. Oct 20, 2009 #9

    Dick

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    The order 2 part of the problem has nothing to do with proving k is a generator of Zn if k and n are relatively prime. Try thinking about them independently.
     
  11. Oct 20, 2009 #10
    Sorry, i forgot to mention that this question has nothing to do with the order 2 problem. Here's the new question:

    Consider Zn={0,1,...,n-1}. show that an element k is a generator of Zn if and only if k and n are relatively prime.
     
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