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Abelian group

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove the collection of all finite order elements in an abelian group, G, is a subgroup of G.

    3. The attempt at a solution
    Let H={x[itex]\in[/itex]G : x is finite} with a,b [itex]\in[/itex]H.
    Then a[itex]^{n}[/itex]=e and b[itex]^{m}[/itex]=e for some n,m.
    And b[itex]^{-1}[/itex][itex]\in[/itex]H. (Can I just say this?)
    Hence (ab[itex]^{-1}[/itex])[itex]^{mn}[/itex]=a[itex]^{mn}[/itex]b[itex]^{-mn}[/itex]=e[itex]^{m}[/itex]e[itex]^{n}[/itex]=e (Since G is abelian the powers can be distributed like that)

    So ab[itex]^{-1}[/itex][itex]\in[/itex]H, and H≤G.
     
    Last edited by a moderator: Feb 16, 2013
  2. jcsd
  3. Feb 16, 2013 #2

    jbunniii

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    If you're going to say it, you should justify it. But your proof below doesn't use this fact. (Indeed, it proves it!)
    This part is fine. Note that as a special case, this shows that ##b^{-1} \in H##. (Take ##a = e##.) You didn't need to stipulate ##b^{-1}\in H## in the previously quoted section.
     
  4. Feb 16, 2013 #3

    Evo

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    Staff: Mentor

    We don't delete threads once they have a response.

    Thank you jbunniii!
     
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