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Math Help - conjugacy question abstract algebra

  1. #1
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    conjugacy question abstract algebra

    If x~_c y in a group G, show that x and y have infinite order or o(x)=o(y)


    (Note: the relation x~_c y is called conjugacy, and as for any equivalence relation, the set G is partitioned by its equivalence classes)
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  2. #2
    Junior Member Nehushtan's Avatar
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    Re: Conjugacy question in abstract algebra

    Hint: Show that \left(gxg^{-1}\right)^n=gx^ng^{-1} for all g\in G and n\in\mathbb Z^+. It will follow that \left(gxg^{-1}\right)^n=e\Leftrightarrow x^n=e.
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  3. #3
    Senior Member jakncoke's Avatar
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    Re: conjugacy question abstract algebra

    This stuff becomes easy if you realize  (gxg^{-1})^{n} = gx^ng^{-1} for  (gxg^{-1})^{n} = gxg^{-1}gxg^{-1}gxg^{-1}....= gx^ng^{-1}

    If |x| =  \infty .
    if |y| = k ( a finite number)
    then y^k = e
    so since x ~ y, there exists a  g \in G s.t gxg^{-1} = y then  (gxg^{-1})^{k} = y^k = e
    so  gx^{k}g^{-1} = e
    or  gx^k = g
    or  x^k = e which cannot be since order of x is \infty .

    Thus is x has order \infty so does y.

    If |x| = n (finite)
    then (gxg^{-1})^{n} = y^n so  e = y^n , so either |y| = n, so |y| is less than n. if |y| = k < n  y^k = (gxg^{-1})^{k}
    so  e = gx^kg^{-1} or  x^k = e contradiction so n = k.
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