# Math Help - Finite subgroup of Q/Z

1. ## Finite subgroup of Q/Z

Prove that every finite subgroup of Q/Z is cyclic.

2. ## 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.

3. 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.