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Proving the order of a group element

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data

    The problem states that [itex]G[/itex] is a group, [itex]a \in G[/itex] s.t. [itex]o\left(a\right)=dk[/itex], and [itex]x \in G[/itex] s.t. [itex]x^{d}=a[/itex]. Prove that [itex]o\left(x\right)=d^{2}k[/itex] where [itex]d,k > 1[/itex].

    Note: [itex]o\left(x\right)[/itex] denotes the order of element [itex]x[/itex] in [itex]G[/itex].

    2. Relevant equations



    3. The attempt at a solution

    So, it's rather easy to show that [itex]x^{d^{2}k} = e[/itex] (where [itex]e[/itex] is the identity element of [itex]G[/itex]), but I'm having trouble proving that it is the order of [itex]x[/itex].

    My thoughts were to try to assume [itex]x^{n}=e[/itex], then divide [itex]n[/itex] by [itex]d[/itex] to get [itex]n=qd+r \Rightarrow x^{qd+r}=x^{qd}x^{r} = {x^{d}}^{q}x^{r} = a^{q}x^{r} = e[/itex]. If I could somehow prove [itex]r=0[/itex], I think I could finish the proof, however, I cannot see how to proceed. If someone could point me in the right direction, it would be much appreciated.
     
  2. jcsd
  3. Oct 31, 2011 #2

    Deveno

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    instead of dividing n by d, try dividing it by dk.....(the idea is to prove the order of x is a multiple of dk).

    then show that if xdkq = e, with 0 < q < d, you get a contradiction (now use what you know about a).
     
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