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A^n = e

  1. Nov 26, 2007 #1
    1. The problem statement, all variables and given/known data
    Show that if a belongs to G where G is finite with identity e, then there exists a positive n such that a^n = e


    2. Relevant equations



    3. The attempt at a solution

    Someone tell me if i am reasoning correctly.

    Suppose a^n is not equal to e.

    By finiteness of G, a^n = b for some b belonging to G. Then b^t = (a^n)^t is not equal to e for any positive t. so G is infinite? i dunno i am stuck.
     
  2. jcsd
  3. Nov 26, 2007 #2
    since a is in G, then the elements a, a^2, a^3, a^4, ... are also in G, but G is finite, so a^k = a^j for some j, k, say j < k, so a^(k-j) = e, set n = k - j, so a^n = e and n > 0.

    realize what you proved, you proved that every element of a finite group has finite order.
     
  4. Nov 26, 2007 #3

    Office_Shredder

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    In your attempt, you assume b^t =/= e for all t, which isn't a great assumption, considering you're supposed to prove that (in this case given b) there IS t for which b^t=e
     
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