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Show that a Group (G, *) definied by a condition is Abelian

  1. Jan 3, 2009 #1

    JPC

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    1. The problem statement, all variables and given/known data

    (G, *) is a group (where * is a law)
    And for all 'i' belonging to {2, 3, 4}, for all (x, y) belonging to G2
    (x * y) ^ i = (x^i) * (y^i)
    (where ^ is the law : to the power of)

    Question : Show that G is an Abelian (commutative) group


    2. Relevant equations



    3. The attempt at a solution

    we have never done any questions of that sort yet, all i can say is that
    "(x * y) ^ i = (x^i) * (y^i)" shows that the law '^' (to the power of) is distributive over the law '*' for 'i' belonging to {2, 3, 4}
    But then i don't know where to go

    Any help or directions would be appreciated, thank you :)
     
  2. jcsd
  3. Jan 3, 2009 #2

    tiny-tim

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    Hi JPC! :smile:

    My inclination would be to start by saying …

    if it's not Abelian, then there are a and b with ab - ba ≠ 0,

    so let ab - ba = c, and … :smile:
     
  4. Jan 4, 2009 #3

    JPC

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    oh, i should have precised, here "*" is not necessarily the multiply law, it can be any law that respects the given conditions

    i tried with your method, putting to the square after, but then i end up working with 4 laws at the same time :D

    Thank you for the indication anyways, i have to find the little trick inside that exercise now
     
  5. Jan 5, 2009 #4
    Hello Jpc,

    In the case that i = 2,

    (x*y)*(x*y) = x*x*y*y

    by associativity:

    x*(y*x)*y = x*(x*y)*y

    and then just left and right multiply by x inverse and y inverse respectively.

    This proves that G is Abelian, since we have shown * to be commutative.

    If you want to show this for the more complicated cases where i is only allowed to take the value 3 or only allowed to take the value 4, then you would need to rephrase your question.
     
  6. Jan 7, 2009 #5

    JPC

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    yes thank you Sisplat, that works very well :)
    I was looking for very complicated things, and i did a little confusion, thinking that the "^i" was associated to the multiply law and not the * law

    And yes, you were right, there were too many data given, to confuse you probably

    Thanks again
     
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