Is every group the unit group of some ring? Let the functor $F\colon\bf Ring\rightarrow\bf Grp$ send the ring $A$ to its group of units $A^\times,$ and the ring homomorphism $f\colon A\rightarrow B$ to the group homomorphism $f^\times\colon A^\times\rightarrow B^\times:a\mapsto f(a)$.
I was curious about this functor, and in particular, whether it is essentially surjective. That is, for any group $G$ (not just finite,) is there a ring $A$ such that $G\cong A^\times$? If not, what groups $G$ satisfies this? A similar question was asked in this question, but what can be said for infinite groups, or groups in general? Calling groups that satisfy this condition R-groups, I have proved that any finitely generated abelian group is an R-group.
I have no idea what to do from now, but I  have a conjecture that the unit group of the group ring $\mathbb F_2[G]$ is isomorphic to $G.$ If this is true, certainly, all groups are R-groups, and the functor $F$ is essentially surjective, but I am having trouble proving it. Can anyone help me?

Your statement about $\mathbb{F}_2[G]$ is incorrect. Consider when $G = \mathbb{Z}_5$, generated by some element $a$ with $a^5 = e$. Then,
$$(e + a^2 + a^3)(e + a + a^4) = (e + a^2 + a^3) + (a + a^3 + a^4) + (a^4 + a + a^2) = e + (a+a) + (a^2+a^2) + (a^3+a^3) + (a^4+a^4) = e$$
So, the unit group of $\mathbb{F}_2[G]$ includes the natural inclusion of $G$, but it also includes $e + a^2 + a^3$, as shown above.
For reference on how I found this example: I can call the "weight" of an element in $\mathbb{F}_2[G]$ the number of nonzero coefficients, so both of the elements above have weight 3, while $e+a$ has weight 2. Clearly weights multiply, so if we want to end up with an odd weight element like $e$, we must start with two odd-weight elements; and we don't want to use elements of weight 1. So we need $|G|$ at least 3. With $G = \mathbb{Z}_3$, there is only one element with odd weight more than 1, and it doesn't square to $e$. So I jumped to $\mathbb{Z}_5$ and it worked.
