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I Proving a property when elements of a group commute

  1. Mar 29, 2017 #1
    By commutative, we know that ##ab = ba## for all a,b in G. Thus, why do we need to prove separately that ##a^n b^m = b^ma^n##? Isn't it the case that ##a^n## and ##b^m## are in fact elements of the group? So shouldn't the fact that they commute automatically be implied?
     
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  3. Mar 29, 2017 #2

    blue_leaf77

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    Yes but each of ##a^n## and ##b^m## might equal another element of the group and they need not commute if the group is non-Abelian.
     
  4. Mar 29, 2017 #3
    Oh, I see. I neglected to see that the group is not actually abelian
     
  5. Mar 30, 2017 #4

    mathman

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    The proof you want seems trivial.
     
  6. Apr 1, 2017 #5

    WWGD

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    Why do you believe you need to prove it separately?
     
  7. Apr 2, 2017 #6

    lavinia

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    If you know that the group is commutative then there is nothing to prove. ##a^nb^m = b^ma^n## by definition. If you do not know that the group is commutative then you need a proof.
     
    Last edited: Apr 2, 2017
  8. Apr 3, 2017 #7

    mathman

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    Proof: [tex]a^nb^m=a^{n-1}bab^{m-1}[/tex], etc.
     
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