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Math Help - Abstract Algebra: Cyclic Groups and GCD

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    is up to his old tricks again! Jhevon's Avatar
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    Abstract Algebra: Cyclic Groups and GCD

    Hail Mathematicians,

    Problem:

    Prove that <[a]> = \mathbb{Z}_n iff \mbox{gcd}(a,n) = 1



    Stuff that might come in handy:

    Perhaps some notation will be of assistance.

    Let G be a group and a \in G. We denote the subgroup generated by a as <a>

    For anything else, Ask, and it shall be given unto you.



    What I tried:

    ( \Rightarrow). Assume <[a]> = \mathbb{Z}_n. Then we have <[a]> = \{ [0],[a],[a]^2,[a]^3, \cdots , [a]^{n - 1}\} = \{ [0], [a], [2a], \cdots [(n - 1)a] \} (this is right, right?)

    By this problem: 0 + a + 2a + \cdots + (n - 1)a = \frac a2n(n - 1) = \left \{ \begin{array}{lr} 0~\mbox{mod }n & \mbox{ if }n \mbox{ is odd} \\ & \\ \frac n2~\mbox{mod }n & \mbox{ if }n \mbox{ is even } \end{array} \right.

    Say n is odd. Then we have \frac a2n(n - 1) = nk for some k \in \mathbb{Z}.

    and i don't know where to go from there.

    Say n is even. Then \frac a2n(n - 1) - \frac n2 = mn for some m \in \mathbb{Z}.

    and i don't know where to go from there.


    I do know that i want to show that ak + nl = 1 for k,l \in \mathbb{Z}. Furthermore, I have to make sure 1 is the gcd by showing that 1|a, 1|n and if there is some b such that b|a and b|n then b|1.

    ( \Leftarrow) For the converse, assume (a,n) = 1. Then ak + nl = 1 for some k,l \in \mathbb{Z}.

    and i don't know where to go from there.

    Help


    Thanks guys and gals
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  2. #2
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    Quote Originally Posted by Jhevon View Post
    Hail Mathematicians,

    Problem:

    Prove that <[a]> = \mathbb{Z}_n iff \mbox{gcd}(a,n) = 1
    By your other problem \left< [a] \right> = \left< [d] \right> where d=\gcd(a,n). Say \left< [d] \right> = \mathbb{Z}_n then it means [1]\in \left< [d] \right> and so [1] = [kd] so 1 = kd + qn for some q and therefore it means d=1 because otherwise RHS is divisible by d and LHS is not. Try it the other way.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    By your other problem \left< [a] \right> = \left< [d] \right> where d=\gcd(a,n). ...
    what problem are you referring to here? if you are referring to 15.26 from one of my other posts, we can't use that.
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