Prove that for any n in $ \Bbb N $ there exists a unique cyclic subgroup $H_n$ $\subset$ G of order n. Where G is the group $\Bbb Q/$$\Bbb Z$.
Existence was not difficult to show: Let $H_n$ = < 1/n + $\Bbb Z$ >. Then, it is clear that if we add 1/n + $\Bbb Z$ n times we achieve n/n + $\Bbb Z$, which is an integer, the identity element in $\Bbb Q/$$\Bbb Z$. So, this cyclic subgroup has order n.
My problem was showing uniqueness. Let $H_2$ be another subgroup of G with order n. Now, assume that $H_2$ = < p/q + $\Bbb Z$ > is another subgroup of order n. Additionally, assume that p and q are co-prime, so that gcd (p,q) = 1. Now, by hypothesis, $H_2$ has order n, so that n * (p/q) = k, where k $\in $ $\Bbb Z$. We also know that for all l < n, l*(p/q) $\notin$ $\Bbb Z$. At this point in the proof I lose any ideas. One problem that I am having is where to arrive in the uniqueness proof. In other words, how do I prove that $H_n$ is unique? What should I demonstrate about $H_2$ in order to show that $H_n$ must be unique? Thanks kindly for your time, as always.
Let $<p/q + \mathbb Z>$ be ANY element of $H_2$ (using your notation). Then as you have shown, you can write $p/q = k/n$ for some $k$, which also means $<p/q + \mathbb Z> = <k/n + \mathbb Z>$. But all elements of this latter form are already in $H_n$, so you have shown that $H_2 \subset H_n$. And since they have the same finite cardinality, they must be equal.