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Order Problem

  1. Feb 17, 2007 #1
    Hey guys. Ive been stuck on the following thing for a little while now. Some help would be appreciated.

    If p divides o(G) (G an abelian group and p a prime), then show that
    G(p) = {g from G | o(g) = p^k for some k }

    I keep going round in circles.

    P.S. - this is not a homework question, just something I saw in an abstract algebra book that they stated without proof.
  2. jcsd
  3. Feb 17, 2007 #2
    What is G(p)? Are you sure this wasn't a definition?
  4. Feb 17, 2007 #3
    What about Cauchy's Theorem?
  5. Feb 18, 2007 #4

    matt grime

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    If you don't define G(p) for us we can't help. (The sylow p subgroup perhaps? but that is an easy exercise - all elements of order p^k lie in some sylow subgroup, and there is only one in an abelian group - so it must be something more difficult than that.)
    Last edited: Feb 18, 2007
  6. Feb 19, 2007 #5
    Did it:)

    Thanks to all who posted. My apologies for not defining G(p) properly. In any event, I solved the problem. The fundamental theorem of arithmetic did the trick (I underestimated the strength of the uniqueness of prime factorization in my earlier attempts!). Ciao
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