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Using the fact that G is abelian in this abstract algebra problem

  1. Oct 8, 2011 #1
    I'll post the problem and my attempt at solution all in one picture:

    jdjd.jpg



    In the red step, I'm using commutative multiplication. Am I allowed to do this? I'm not sure, because the subset of G might not be a subgroup, so I don't know if its necessarily abelian like G is. Or does the fact that G is abelian override this somehow?
     
  2. jcsd
  3. Oct 8, 2011 #2
    Any subgroup of an abelian group is abelian.
     
  4. Oct 8, 2011 #3
    i'm trying to PROVE it's a subgroup. That's the question itself.
     
  5. Oct 8, 2011 #4

    micromass

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    That looks good. I don't see the confusion. In an abelian group, it always holds that

    [tex](ab)^n=a^nb^n[/tex]

    You don't need any subgroups to prove this.
     
  6. Oct 9, 2011 #5
    Sorry any subset of an abelian group is abelian.
     
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