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Math Help - Question about finite cyclic groups

  1. #1
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    Question about finite cyclic groups

    Is it true that for every finite cyclic group <G,*> with generator a each of the unique elements of the group may be presented as a^n where n is non-negative integer.

    My proof: If G is finite, there is some positive integer m for which a^m=e. Then a^{m+1}=a^1, a^{m+2}=a^2 etc...
    Moreover a^m=a^{-m}=e. Then a^m*a^1=a^{-m}*a^1. Therefore a^{m+1}=a^{-m+1}=a^1.
    If a^{m+n}=a^{-m+n} for some n<m then a^{m+n+1}=a^{-m+n+1} too.

    This must cover all inverses of the elements from a^1 to a^m.
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  2. #2
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    Re: Question about finite cyclic groups

    well, what you want to do, is show that a^{-1} can be expressed as a positive power of a. and you have almost done done this, which power of a is it?

    also, you need to flesh out the existence of m a bit more. all we're given to work with, is that <a> is finite.

    i would argue thusly: there are infinitely many positive powers of a, but since G = <a> is finite, for some r ≠ s we must have a^r = a^s.

    without loss of generality, we may assume that r > s. then a^ra^{-s} = a^{r-s} = e. this shows that for SOME positive integer n, namely n= r-s,

    we have a^n = e. we can then let m be the smallest positive integer such that a^m = e (because we know there is at least one).

    this gives us a set of DISTINCT positive powers of a \{a,a^2,\dots,a^m = e\}. your proof that ANY positive power of a is in this set consists mainly of the word: "etc."

    you can do better. write an arbitrary positive number n, as n = qm + r, where 0 ≤ r < m (why can we do this?). show that a^n = a^r.

    if a^{-1} is in this set, then so is a^{-k} = (a^{-1})^k (why?)
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    Re: Question about finite cyclic groups

    This forum seems to be infected with malware. I'll stop posting for a while until this is sorted out. I understand the proof of existence of m. n = qm + r comes from division algorithm. Thanks for the advice!
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