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Abelian p-group

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data

    How would I prove that gnhn = (gh)n, for g,h [itex]\in[/itex]G where G is a p-group, and all its elements have order p?

    3. The attempt at a solution

    My aim is to prove this in order to prove that G is Abelian, but I don't want to prove it using centres. I've supposed that gnhn = (gh)m for some m, and now I'm trying to prove that m must be congruent to n modulo p, as this is the only way that (gh)m = (gh)n since gh has order p. And this is where I strike a wall, so to speak. Is there a better way to prove that G is Abelian, if this isn't good enough?

    Thanks!!
     
  2. jcsd
  3. Sep 12, 2011 #2
    Think: generator.
     
  4. Sep 12, 2011 #3
    As in...gh generates a cyclic group?
     
  5. Sep 12, 2011 #4
    Well, yes, but don't think about gh. You know that if g is in G, then g has order p, correct?
     
  6. Sep 12, 2011 #5
    That is, can you think of a way to write any element of G in terms of g?
     
  7. Sep 12, 2011 #6

    micromass

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    Yeah, what you're trying to prove is false. A counterexample is given by

    [itex](\mathbb{Z}_3\rtimes \mathbb{Z}_3)\times \mathbb{Z}_3[/itex]

    All (non-identity) elements have order 3 and the group is non-abelian.
     
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