Results 1 to 3 of 3

Thread: Finite subgroup of Q/Z

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    394

    Finite subgroup of Q/Z

    Prove that every finite subgroup of Q/Z is cyclic.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517

    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 $\displaystyle \mathbb{Q}/\mathbb{Z}$, 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 $\displaystyle q\in\mathbb{Q} $ in lowest terms, then $\displaystyle \bar{q}=q-\lfloor q \rfloor$ where $\displaystyle \bar{q}\in\mathbb{Q}/\mathbb{Z}$.

    So suppose we have a finite subgroup $\displaystyle H \leq \mathbb{Q}/\mathbb{Z}$ 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 $\displaystyle r=gcd(r_1,r_2,...,r_n)$ and $\displaystyle s=lcm(s_1,s_2,...,s_n)$. Then I claim $\displaystyle H=\langle\frac{r}{s}\rangle$. 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 $\displaystyle s\frac{r}{s}\in \mathbb{Z}$ is our identity.

    Definitely some work to do, but I feel convinced.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Below is my proof so far.

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

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

    Any advice?
    Last edited by aliceinwonderland; Dec 8th 2008 at 11:35 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finite subgroup of an infinite Group
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Dec 13th 2011, 03:26 AM
  2. Prove that no proper nontrivial subgroup of Z is finite.
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jul 18th 2011, 12:11 AM
  3. proof that subgroup has finite index
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Mar 10th 2010, 10:33 PM
  4. Finite group with identity subgroup.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Feb 18th 2010, 09:17 AM
  5. finite subgroup of C*
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Dec 9th 2008, 04:07 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum