Prove that every finite subgroup of Q/Z is cyclic.
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 , 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 in lowest terms, then where .
So suppose we have a finite subgroup 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 and . Then I claim . 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 is our identity.
Definitely some work to do, but I feel convinced.
Below is my proof so far.
Let H be a finite subgroup (order q) of Q/Z.
Then, , where (i=1,2,..,q-1) is an irreducible fraction and .
We then replace each with , where n is the least common denominator for all rational numbers 's.
Since nH is a subgroup of Z, nH is cyclic and q = n. We shall show that is a generator of H, where . Since , has a maximal order, which is n. Our choice of generates ,where , for some .Thus H is cyclic.
Any advice?