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A group G is abelian if

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    show that a group G is abelian if g^2 = 1 for all g in G. Give an example showing that converse is false.

    3. The attempt at a solution
    Suppose g^2 = 1.
    gg = 1,
    (g^-1)gg = g^-1
    g = g^-1 -- means self inverse. but I'm not sure how to show G is abelian..
    and I don't know how I find counterexample.
     
  2. jcsd
  3. Feb 22, 2009 #2
    Try taking 3 elements from G, say g, h, and gh. You should be able to show gh=hg pretty easily.
     
  4. Feb 22, 2009 #3
    Just to add one more thing to the above post, you've shown that g=g^(-1) for all g. In particular, gh=(gh)^(-1)=...?

    The converse statement is "G is abelian implies g^2=1 for all g in G". You should have seen enough examples of groups by now to find an abelian group where this doesn't hold.
     
  5. Feb 24, 2009 #4
    So.. Suppose that gg = 1, then ggh = hgg for all g,h in G, this clearly is that G is abelian.., right??
    And counterexample...
    G = {4,8,12,16}; multiplication in Z_(20)..
    because g^2 not = 1 for all g in G, correct??
    Also 1 means identity in this problem? or only number 1..?
    Thanks
     
  6. Feb 24, 2009 #5
    Use the above poster's advice: gh=(gh)^(-1) by your discovery that each element is it's own inverse in this group.
    Now compute (gh)^-1

    For the counterexample perhaps you should consider Z_5
     
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