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Groups whose elements have order 2

  1. Oct 18, 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
    DOES THIS ANSWER THE QUESTION????:

    Notice first that x2 = 1 is equivalent to x = x−1. Since every element of G
    has an inverse, we can distribute the elements of G into subsets {x, x−1}. Since
    the inverse of x−1 is x, these sets are all disjoint. All of them have either two
    elements (if x = x−1) or one element (if x = x−1). Let k be the number of
    two-element sets, and let j be the number of one-element sets. Then the total
    number of elements of G is 2k + j.
    Now notice that j > 0, since there is at least one element such that x = x−1,
    namely x = 1. So we know that 2k + j is even and that j > 0. It follows that
    j ≥ 2, so there must exist at least one more element x such that x = x−1. That’s
    the element we were looking for.
     
  2. jcsd
  3. Oct 18, 2009 #2

    Dick

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    No, that doesn't show anything. You assumption is that every nonidentity element has order two. That mean EVERY element is its own inverse. Try this. What you want to show is that ab=ba for every two nonidentity elements of G. If ab is not the identity then (ab)(ab)=1, that's your assumption. What's (ab)(ba)?
     
  4. Oct 19, 2009 #3

    to be honest, I am lost now. I'm not sure I understand your response. Does (ab)(ba)=-1???
     
  5. Oct 19, 2009 #4

    Dick

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    I'll be honest in return. What does (ab)(ba)=(-1) mean? Are you sure you understand multiplicative group notation? There's no such thing as (-1).
     
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