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Abelian group proof

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data
    If G is an abelian group, then (ab)^2=a^2b^2 for all a, b in G. Give an example to show that abelian is necessary in the statement of the theorem.



    2. Relevant equations



    3. The attempt at a solution
    Abelian implies commutativity.
    a*b=?b*a
    a^2b^2=b^2a^2
    For example ab=ba must be true for the statement to work.
     
  2. jcsd
  3. Mar 30, 2009 #2
    take for example a^2 b^2 and expand it to aabb

    what can you do if ab is abelian?
     
  4. Mar 30, 2009 #3
    aabb
    Implies commutativity so aabb=bbaa
    I think there's something with inverse to make this work
     
  5. Mar 30, 2009 #4
    a(ab)b
     
  6. Mar 30, 2009 #5
    Ok so a(ab)b and b(ab)a
    so this implies ab=ba
     
  7. Mar 30, 2009 #6
    the definition of abelian is not axb = bxa so can't do that

    so instead we can swap the ab
     
  8. Mar 30, 2009 #7
    I don't really understand. You're saying we can swap the ab, but I don't gte what you mean by this.
    (ab)ab=(ab)ba
     
  9. Mar 30, 2009 #8

    matt grime

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    I don't know what waht it making you do. The point was to show that you can find G non-abelian where (ab)^2 is not equal to a^2b^2 for some a and b.
     
  10. Mar 30, 2009 #9
    Yeah and I wanted to prove that abelian implied (ab)^2=a^2b^2
    Is there a hint for an a and b to use where the 2 aren't equal?
     
  11. Mar 30, 2009 #10
    (ab)^2 = a^2 b^2 --->

    abab = aabb -------->

    cancel the leftmost a and the rightmost b:

    ba = ab
     
  12. Mar 30, 2009 #11
    well

    a^2b^2 = aabb = a(ab)b

    because they are abelian ab = ba

    so aabb = a(ba)b = (ab)(ab)
     
  13. Mar 30, 2009 #12
    But you are supposed to prove that they are Abelian.
     
  14. Mar 30, 2009 #13
    Ok I understnd how to prove it now.
    I'm having trouble with the example to show abelain is necessary.
     
  15. Mar 30, 2009 #14
    It is necessary because:

    So, we have that

    (ab)^2 = a^2 b^2 implies that ab = ba

    This is the same as saying:

    Not[ab = ba] implies Not[(ab)^2 = a^2 b^2]

    Or, in plain English, if you don't have the Abelian property, you don't have that (ab)^2 = a^2 b^2. So, the Abelian property is a necessary assumption for the identity to hold.
     
  16. Mar 30, 2009 #15
    Hmmm, I think I might see this better if I saw a way where it doesn't work. For example (ab)^2 not equaling a^2b^2. Therefore, not abelian
     
  17. Mar 30, 2009 #16
    But that's equivalent to Abelian ---> (ab)^2 =a^2b^2

    which is the trivial case.

    If you know that this is true (so Abelian implies the identity), then should you encounter a case where the identity is not satisfied, you know that the group cannot be Abelian.
     
  18. Mar 30, 2009 #17
    that's what I'm trying to do to find a case now where the identity is not satisfied to show then that the group is not abelian
     
  19. Mar 30, 2009 #18

    matt grime

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    Um, as count Iblis has pointed out *any* non-abelian group will provide a counter example.
     
  20. Mar 30, 2009 #19
    You can take some group of matrices. Matrix multiplication does not commute. So, it should be easy to find a particular example where the identity does not hold.

    But such a particular example does not show that the fact that the identity does not hold implies that the group is non-abelian. To show that, you need to prove that whenever the identity does not hold, you also have that the group is not abelian.
     
  21. Mar 30, 2009 #20
    Oh, that makes sense.
     
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