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More group theory!

  1. Mar 26, 2005 #1
    seeing lots of group theory here after a really long time...
    let G be a finite group of order n, where n is not divisible by 3. suppose
    (ab)^3 = a^3 b^3 ,for a, b in G . prove that G is abelian.
  2. jcsd
  3. Mar 26, 2005 #2
    i guess i must( or rather i am supposed to) add my thought process...
    since 3 does not divide n (n,3)=1 so we have un +3v=1...now for any x in G, x = x^(un+3v)= x^3v.
    so G ={y^3 / y is in G}...using this i think we're supposed to show that ab=ba for all a in G. here's where i'm stuck.
  4. Mar 26, 2005 #3


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    It's frequently useful to reverse what's inside an exponent. For example:

    b * (ab)^3 * a = b(ababab)a = (ba)^4
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