$\mathbb{Q}/\mathbb{Z}$ has a unique subgroup of order $n$ for any positive integer $n$? Viewing $\mathbb{Z}$ and $\mathbb{Q}$ as additive groups, I have an idea to show that $\mathbb{Q}/\mathbb{Z}$ has a unique subgroup of order $n$ for any positive integer $n$. You can take $a/n+\mathbb{Z}$ where $(a,n)=1$, and this element has order $n$. 
Why would such an element exist in any subgroup $H$ of order $n$? If not, you could reduce every representative, and then every element would have order less than $n$, but where is the contradiction?

Find all elements of order a divisor of $n$ in $\mathbb Q/\mathbb Z$ (that is, all elements $g$ such that $ng=0$). There are exactly $n$ of them, and there is in fact at least one of them whose order is exactly $n$. 
Show that those elements are in fact the elements of a subgroup. 
