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Math Help - Order of an element and GCD

  1. #1
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    Order of an element and GCD

    I have the following problem which i can only solve half of it.

    Problem: Let G = <g> be a cyclic group of order n. Prove that the order of g^k is n/d, where d = gcd(n,k).

    Solution: Since G has order n, order of g is n. i.e. g^n=1. Also, d|k, giving da = k for some integer a. Hence, ({g^k})^{n/d} = g^{(nk)/d} = g^{(ndb)/d} = {(g^n)}^b = 1.

    However, i am unable to show that n/d is the order. The only theorem about order that i know is that: If r is the order of g and g^m =1, then r|k. is there any other theorems that i need to know to solve this problems. Thank You!.
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  2. #2
    MHF Contributor

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    Tejas
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    Re: Order of an element and GCD

    the trick is to first prove <g^k> = <g^d>.

    now k = db, so g^k = g^{db} = (g^d)^b therefore:

    <g^k> \subseteq <g^d>

    writing d = ns + kt (which we can by the euclidean algorithm), we have:

    g^d = g^{ns+kt} = (g^{ns})(g^{kt}) = (g^n)^s(g^k)^t = (g^k)^t

    so we also know that

    <g^d> \subseteq <g^k>

    so <g^k> = <g^d>. this means that |g^k| = |g^d|.

    now obviously (g^d)^{n/d} = e. suppose that (g^d)^u = e

    for some 0 < u < n/d. then 0 < du < n, but g^{du} = (g^d)^u = e

    contradicting the fact that the order of g is n. thus there can be no such u,

    so |g^k| = |g^d| = n/d.
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