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Thread: 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 $\displaystyle a \in \mathbb{F}$ and suppose $\displaystyle a^n = 1$.
    also $\displaystyle 0 < k<n$ and $\displaystyle k \mid n$. Prove that if $\displaystyle 0<k<n$ then $\displaystyle a^k \neq 1$.


    So far my proof:

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

    where do i go from here?
    Last edited by strider2; Apr 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 $\displaystyle 0 < k<n$ and $\displaystyle k \mid n$
    Can you clarify this line? Is it meant to say there exists a $\displaystyle k$ such that $\displaystyle k|n$?
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    Re: primitive roots problem

    Another clarification...is $\displaystyle \mathbb{F}$ a group $\displaystyle \mathbb{F}( \cdot )$? 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 $\displaystyle a \in \mathbb{F}$ and suppose $\displaystyle a^n = 1$.
    also $\displaystyle 0 < k<n$ and $\displaystyle k \mid n$. Prove that if $\displaystyle 0<k<n$ then $\displaystyle a^k \neq 1$.


    So far my proof:

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

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

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

    We have
    $\displaystyle a^4 = e$

    We have
    0 < 2 < 4, 2|4

    and
    $\displaystyle a^2 = e$

    -Dan
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