TITLE: Obstructions for a group to be the multiplicative group of a field
QUESTION [14 upvotes]: It is well known that every finite multiplicative subgroup of a field is cyclic.
I somehow got interested in a possible reverse implication:
Assume we have an abelian group $G$ whose every finite subgroup is cyclic. 
Is $G$ necessarily the multiplicative group of all nonzero elements of some field $F$?
(i.e I ask whether we can "add" to $G$ a "zero" element, and define addition operation $+$ making $G\cup\{0\}$ a field w.r.t $+$ and the multiplication in $G$).
Of course for finite cyclic groups the answer is trivial. (A group $G$ can be completed to a field iff $|G|=p^n-1$ for some prime $p$).
Next, I considered the infinite group of all complex roots of unity (of all orders). I once proved to myself that it cannot be completed into a field, but I do not remember exactly how I did this.
I wonder if there are "nice" necessary & sufficient conditions which ensure $G$ can be completed.

REPLY [8 votes]: Restricting to $\mathbb{Q}$-rank zero, there are more complicated torsion examples, such as $G=\mathbb{Z}[1/5]/\mathbb{Z}$. If $K^\times=G$, then $K$ must be algebraic over a finite field $k=\mathbb{F}_p$ for some prime $p>0$, since $G$ is torsion. But then $k^\times$ is a subgroup of $G$, and so we have that $\# k^\times$ is $5^n$ for some integer $n$. But we also have $\# k = p$, hence
$$
p = 5^n + 1,
$$
which is clearly impossible unless $p=2$. So assume $p=2$ now. In that case, $K$ contains an ascending chain of field extensions of $k=\mathbb{F}_2$, giving infinitely many positive integer solutions to 
$$
2^m=5^n+1,
$$
which is a contradiction, for instance with Catalan's conjecture (proven by Mihailescu), although a variety of more elementary arguments are possible (for which see the comments).
(With some more work, I think one can rule out $G=\bigoplus_\ell \mathbb{Z}[1/\ell]/\mathbb{Z}$, where the sum ranges over any finite set of primes $\ell$.)