$\mathbb{Q}/\mathbb{Z}$ 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?

We can approach the problem using elementary number theory.  Look first at the subgroup $K$ of $\mathbb{Q}/\mathbb{Z}$ generated by (the equivalence class of) $q/r$, where $q$ and $r$ are relatively prime.
Since $q$ and $r$ are relatively prime, there exist integers $x$ and $y$ such that $qx+ry=1$.  Divide both sides by $r$.  We find that
$$x\frac{q}{r}+y=\frac{1}{r}.$$
Since $y$ is an integer, it follows that $\frac{1}{r}$ is congruent, modulo $1$, to $x\frac{q}{r}$. It follows that (the equivalence class of) $1/r$ is in $K$, and therefore generates $K$.

Now let $H$ be a subgroup of $\mathbb{Q}/\mathbb{Z}$ of order $n$.  Let $h$ be an element of $H$.  If $h$ generates $H$, we are finished. Otherwise, $h$ generates a proper subgroup of $H$. By the result above, we can assume that $h$ is (the equivalence class of) some $1/r_1$, and that there is some $1/b$ in $H$ such that $b$ does not divide $r_1$.  Let $d=\gcd(r_1,b)$.
There are integers $x$ and $y$ such that $r_1x+by=d$.  Divide through by $r_1b$. We find that
$$x\frac{1}{b}+y\frac{1}{r_1}=\frac{d}{r_1b}.$$
It follows that (the equivalence class of) $d/(r_1b)$ is in $H$.  But $r_1b/d$ is the least common multiple of $r_1$ and $b$. Call this least common multiple $r_2$. Then since $r_1$ and $b$ both divide $r_2$, the subgroup of $H$ generated by (the equivalence class of) $1/r_2$ contains both $1/r_1$ and $1/b$. 
If $1/r_2$ generates all of $H$, we are finished.  Otherwise, there is a $1/b$ in $H$ such that $b$ does not divide $r_2$.  Continue.
