Does $\Bbb Q/ \Bbb Z$ have a proper subgroup that is not finite? 
Does $\Bbb Q/ \Bbb Z$ have a proper subgroup that is not finite?

I suspect it does not. However since we could take a subgroup of all $p$ sets $\{\frac{1}{p} + \Bbb Z\}$ if we consider $p$ to be arbitrarily large is this the same thing as a non-finite subgroup?
 A: The set of elements of the form $$\left\{\frac{k}{2^n} + \mathbb{Z} : k, n \in \mathbb{Z}\right\}$$
gives a proper subgroup which is certainly infinite. Replace $2$ by any prime for a similar result.
A: As mentionned here, $\mathbb{Q}/\mathbb{Z}$ is isomorphic to $\bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z}[p^{\infty}]$, so you can find infinitely many infinite proper subgroups. In particular, T. Bongers' answer generalizes as $$\left\{ \frac{k}{p^n} + \mathbb{Z} \mid k,n \in \mathbb{Z} \right\}$$ for any prime $p$ (ie. the image of $\mathbb{Z}[p^{\infty}]$ in $\mathbb{Q}/ \mathbb{Z}$).
A: To list all infinite subgroups: Let $P$ be proper subset of the set of prime numbers and let $f\colon P\to \mathbb N$ be a map. Consider
$$\left\{\,\frac ab+\mathbb Z\biggm|\forall p\in P\colon p^{f(k)}\not\mid b\,\right\}.$$
This is a subgroup because the divisibility constraint for denominators also holds for their lcm. It is infinite because $P$ is proper. There are no other subgroups because for $H\le G$ and primes $p$ either the powers occuring in denominators are bounded or (for infinite $H$: at last once) not and by suitable addition we find that all prime factorizations allowed by these bounds occur.
