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Counterexample so that (ab)^i=a^ib^i for two consecutive integers

  1. Dec 10, 2007 #1
    counterexample so that (ab)^i=a^ib^i for two consecutive integers for any a and b in a group G does not imply that G is abelian.

    this is a problem in herstein and i'm struggling to find an example. The previous problem to show that if (ab)^i=a^ib^i for 3 consecutive integers then G is abelian is a starred problem but seems to be easier.
     
  2. jcsd
  3. Dec 10, 2007 #2

    morphism

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    Hint: It's trivial if you pick the right two consecutive integers.
     
  4. Dec 10, 2007 #3
    im sorry but is i the square root of 1 here?
     
  5. Dec 10, 2007 #4
    no, i is an integer. It's a group theory problem
     
  6. Dec 14, 2007 #5

    morphism

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    I wonder if we can find an easy nontrivial counterexample (i.e. where we choose a pair of values for i other than 0 and 1). I scribbled down some stuff and managed to convince myself that the symmetric and dihedral groups aren't going to be good candidates, but I haven't had the energy to pursue this any further.

    Edit:
    I did some searching and found this. Apparently the quaternion group Q works with i=4 and i=5.
     
    Last edited: Dec 14, 2007
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