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Math Help - Problem with order of element

  1. #1
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    Problem with order of element

    Let a be in a group G with |a| = m. If n is relatively prime to m, show that a = b^n for some b in G.

    My Proof so far:

    Now the order of a is m, so I have a^m = e, the identity. n is relatively prime to m, so I have ng + r = m for some integers g and r, with m > r > n.

    umm... what now?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tttcomrader View Post
    Let a be in a group G with |a| = m. If n is relatively prime to m, show that a = b^n for some b in G.

    My Proof so far:

    Now the order of a is m, so I have a^m = e, the identity. n is relatively prime to m, so I have ng + r = m for some integers g and r, with m > r > n.

    umm... what now?
    I should think that this would be easier to do if you start from the idea that G contains the subset \{ e, a, a^2, ~ ... ~, a^{m -1} \}. Then show that one of these elements must be your "b" for a given n.

    Just a thought.

    -Dan
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  3. #3
    Super Member Rebesques's Avatar
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    n is relatively prime to m, so I have
    Well, everything goes well to that point Now by Euclidean division, there are integers g and r, such that gn+rm=1. So \alpha^{gn+rm}=\alpha\Rightarrow \alpha^{gn}\alpha^{rm}=\alpha\Rightarrow (\alpha^{g})^n(\alpha^{m})^{r}=\alpha\Rightarrow (\alpha^{g})^{n}=\alpha, that is \beta=\alpha^{g}.





    Ps. Topsq:
    Then show that one of these elements must be your "b" for a given n.
    Suppose \alpha=\beta^n=\alpha^i for some i<n, and prove you chose a bad start
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  4. #4
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    We haven't learn the Euclidean division yet, so I don't think I'm allow to use that. And how to you have the (a^m)^r?

    thanks
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  5. #5
    Super Member Rebesques's Avatar
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    We haven't learn the Euclidean division yet
    Really? Then what is this:

    so I have ng + r = m
    Use that rule and amaze everyone by showing gn+rm=1 for some g,r!


    And how to you have the (a^m)^r?
    a^m=e and e^r=e.
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  6. #6
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    Quote Originally Posted by tttcomrader View Post
    Let a be in a group G with |a| = m. If n is relatively prime to m, show that a = b^n for some b in G.
    Let the order the order of a be m. And consider the cyclic subgroup \left< a\right> = \{a,a^2,...,a^m\}. Now the order of this cyclic group is m. Thus, the generators of this group are a^k where k is relatively prime to m. In particular, k=n by hypothesis. That means \left< a^n \right> = \left< a \right>. Now a^n =\{ a^n,a^{2n},...,a^{mn} \} is a premutation of \{ a,a^2, ... ,a^m\} thus a is found among \{a^n,...,a^{nm}\} and hence a=a^{nj} = \left( a^j \right) ^n. Let b=a^j for some j. And the proof is complete.
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