Obstructions for a group to be the multiplicative group of a field 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.
 A: 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 $\ell$.)
A: There are various constraints. 
One of those is: the $\mathbf{Q}$-rank is either 0 or infinite. (Thus $F^*$ cannot be isomorphic to $\mathbf{Z}\times \mathbf{Q}/\mathbf{Z}$, among others).
Indeed, if $F$ has characteristic zero, then the primes form an infinite $\mathbf{Z}$-free family. If $F$ has characteristic $\ell>0$ and $F^*$ is not torsion, then there exists a transcendental element $x$, and then the $(x^p-1)$, where $p$ ranges over primes $\neq\ell$, form a $\mathbf{Z}$-free family. Indeed otherwise we could find a positive integer $k$, a non-negative integer $n$, and primes $p_1,\dots,p_k,q_1,\dots,q_n$, all distinct and distinct from $\ell$, and positive integers $t_1,\dots,t_k,m_1,\dots,m_n$, such that $\prod_i (x^{p_i}-1)^{t_i}=\prod_j (x^{q_j}-1)^{m_j}$. Since $x$ is transcendental, this equality still holds in the polynomial ring $\mathbf{F}_p[x]$. Evaluation at a primitive $p_1$-root of unity (in a suitable extension of $\mathbf{F}_\ell$) then yields a contradiction. 
