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Math Help - Finite subgroup of Q/Z

  1. #1
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    Finite subgroup of Q/Z

    Prove that every finite subgroup of Q/Z is cyclic.
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  2. #2
    Super Member Gamma's Avatar
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    Very Interesting

    Wow, now this problem when I first read it I thought, no way this can be true. I spent forever trying to come up with a counter example instead of actually just trying to prove it. I am not positive if this method holds up or not, but it was what came to mind, let me know what you think.

    So first of all what is \mathbb{Q}/\mathbb{Z}, the way I am thinking of it as basically just reducing the rationals mod 1 to get our coset representative. I guess formally it would be like let q\in\mathbb{Q} in lowest terms, then \bar{q}=q-\lfloor q \rfloor where \bar{q}\in\mathbb{Q}/\mathbb{Z}.

    So suppose we have a finite subgroup H \leq \mathbb{Q}/\mathbb{Z} then we can just look at the coset representative that is an element of [0,1) in lowest terms. Then H={0, \frac{r_1}{s_1},\frac{r_2}{s_2},...,\frac{r_n}{s_n }}. Let r=gcd(r_1,r_2,...,r_n) and s=lcm(s_1,s_2,...,s_n). Then I claim H=\langle\frac{r}{s}\rangle. It should follow from the euclidean algorithm (to get r to generate all the numerators) and the fact that this subgroup of rational number representatives must be closed under + and - and s\frac{r}{s}\in \mathbb{Z} is our identity.

    Definitely some work to do, but I feel convinced.
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  3. #3
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    Below is my proof so far.

    Let H be a finite subgroup (order q) of Q/Z.
    Then, H = \{ Z+0, Z+g_{1}, Z+g_{2}, ... , Z+g_{q-1} \}, where g_{i} (i=1,2,..,q-1) is an irreducible fraction and g_{q}=0.
    We then replace each g_{i} with \frac{m_{i}}{n}, where n is the least common denominator for all rational numbers g_{i}'s.

    Since nH is a subgroup of Z, nH is cyclic and q = n. We shall show that Z+ <\frac{m_{k}}{n}> is a generator of H, where (n, m_{k})=1. Since (n, m_{k})=1, Z + \frac{m_{k}}{n} has a maximal order, which is n. Our choice of  m_{k} generates  h_{i} = i \cdot \frac{m_{k}}{n} ,where h_{i}=g_{j}, for some i, j = 1,2,...,n.Thus H is cyclic.

    Any advice?
    Last edited by aliceinwonderland; December 8th 2008 at 12:35 PM.
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