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Where does the abelian part come in?

  1. Nov 26, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove the set H of all solutions satisfying the equation x^n = e form a subgroup of an abelian group G with identity e.


    2. Relevant equations



    3. The attempt at a solution

    e^n = e so e belongs to H.

    x^n = e => x^(-n)= e => (x^(-1))^n= e so for any x belonging to H, x inverse belongs to H.

    if a^n=e and b^n= e => (ab)^n=e so ab belongs to H.

    where does the abelian part come in?
     
  2. jcsd
  3. Nov 26, 2007 #2
    (ab)^n = (ab)(ab)(ab)...(ab)(ab)(ab) =(<--here) (aa...aaa)(bb...bb) = a^nb^n = ee = e

    note that more is true, the set of all elements of finite order of an abelian group form a subgroup. the proof is the same as this except showing closure is slightly different, you could do it for practice if u wanted
     
    Last edited: Nov 26, 2007
  4. Nov 28, 2007 #3
    i'm going to do the practice problem just to make sure i've got it. if a^n=e and b^s=e then (a^n * b^s)=e. then by the group being abelian (ab)^s * a^(n-s)=e. we can assume n>s without loss of generality. therefore (ab)^s = a^s. if we take both sides to the power (n/s) then we have (ab)^n = e. thus ab belongs to G.
     
  5. Nov 28, 2007 #4
    by the way i still don't understand why G must be abelian in the first problem.
     
  6. Nov 28, 2007 #5

    HallsofIvy

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    That was answered in ircdan's first response. If G is not abelian, then (ab)n is NOT, in general, anbn so you cannot use the fact that an= bn= e.
     
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