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Math Help - Prove [TEX] a^k \neq 1[/TEX] if [TEX]k \mid n[/TEX] and [TEX]k<n[TEX]

  1. #1
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    primitive roots problem

    I'm trying to figure this problem out and need a little help.


    So suppose n is an integer and  a \in \mathbb{F} and suppose  a^n = 1.
    also 0 < k<n and k \mid n. Prove that if 0<k<n then a^k \neq 1.


    So far my proof:

    Suppose a contradiction.
    Let a^k =1, then gcd(k,n) = d.
    I write the linear combination d = sk + tn for some integers t,n.
    then a^d = 1

    where do i go from here?
    Last edited by strider2; April 7th 2013 at 07:35 PM.
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  2. #2
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    Re: primitive roots problem

    Quote Originally Posted by strider2 View Post
    also 0 < k<n and k \mid n
    Can you clarify this line? Is it meant to say there exists a k such that k|n?
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    Re: primitive roots problem

    Another clarification...is \mathbb{F} a group \mathbb{F}( \cdot )? So I can presume that 1 is the multiplicative identity?

    -Dan
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    Forum Admin topsquark's Avatar
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    Re: primitive roots problem

    Quote Originally Posted by strider2 View Post
    I'm trying to figure this problem out and need a little help.


    So suppose n is an integer and  a \in \mathbb{F} and suppose  a^n = 1.
    also 0 < k<n and k \mid n. Prove that if 0<k<n then a^k \neq 1.


    So far my proof:

    Suppose a contradiction.
    Let a^k =1, then gcd(k,n) = d.
    I write the linear combination d = sk + tn for some integers t,n.
    then a^d = 1

    where do i go from here?
    There is something wrong with the problem statement as I understand it.

    Counter-example. Let \mathbb{F} be the multiplicative group of two elements e and a.

    We have
    a^4 = e

    We have
    0 < 2 < 4, 2|4

    and
    a^2 = e

    -Dan
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