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Math Help - Number theory, order of an integer

  1. #1
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    Exclamation Number theory, order of an integer

    Hi, do you guys know how to do this question? It's one of my assignment questions, pls help.......

    Q: Show that if n is a positive integer, and a and b are integers relatively prime to n such that (ordna, ordnb)=1, then ordn(ab)=ordna x ordnb

    Thanks alot!!!
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  2. #2
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    Quote Originally Posted by suedenation View Post
    Hi, do you guys know how to do this question? It's one of my assignment questions, pls help.......

    Q: Show that if n is a positive integer, and a and b are integers relatively prime to n such that (ordna, ordnb)=1, then ordn(ab)=ordna x ordnb

    Thanks alot!!!
    Let \gcd(a,n)=\gcd(b,n)=1
    Then let k,j be the orders of these integers respectively.
    That is,
    a^k\equiv 1(\mbox{ mod }n)
    b^j\equiv 1(\mbox{ mod }n)
    Where, k,j are minimal.

    Now, \gcd(ab,n)=1 so the order of ab exists. We note that, (ab)^{kj}\equiv 1 (\mbox{ mod }n)
    Because, (ab)^{kj}=(a^k)^j\cdot (b^j)^k
    So the order of ab is d and must be a divisor of kj. Since \gcd(k,j)=1 we must have by Euclid's Lemma that d|k, d|j. Thus, the order of k,j is actually smaller, that is, \frac{k}{d},\frac{j}{d} unless d=1 which is this case here.
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