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Group Theory: g^n = e proof

  1. Apr 3, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    If G is a group with n elements and g ∈ G, show that g^n = e, where e is the identity element.

    2. Relevant equations


    3. The attempt at a solution

    I feel like there is missing information, but that cannot be.
    This seems too simple:
    The order of G is the smallest possible integer n such that g^n = e. If no such n exists, then G is of infinite order.

    From this definition of order can we simply state that since G is a group with 'n' elements then there must exist an n such that g^n = e ?

    order is denoted as °(g)
    So
    °(g) = n ==> g^n = e.
     
  2. jcsd
  3. Apr 3, 2016 #2

    vela

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    Aren't you mixing up G and g here? You're describing the definition of the order of the element g, not the order of the group G.

     
  4. Apr 4, 2016 #3

    HallsofIvy

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    Where did you get this: "The order of G is the smallest possible integer n such that g^n = e"? The definition of "order of a group" is simply the number of elements in the group (the cardinality of the underlying set). It looks to me like you are being asked to prove the statement you give.
     
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