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

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- December 5th 2008, 11:23 PMaliceinwonderlandFinite subgroup of Q/Z
Prove that every finite subgroup of Q/Z is cyclic.

- December 6th 2008, 04:25 AMGammaVery 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 , 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. - December 8th 2008, 03:29 AMaliceinwonderland
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?