All rings below are commutative with unity. For a ring $R$, let $U(R)$ denote its group of units, which is in particular abelian.
Now, let $R$ be a ring. Consider the dual group of $U(R)$ namely $\hat {U(R)} := Hom_{\mathbb Z} (U(R),\mathbb Z)$.
Under what conditions on $ U(R)$ , can we say that there exists a ring $S$ such that $U(S) \cong \hat {U(R)}$ ?
In other words : Given an abelian group $G$, which is the group of units of some ring, under what conditions on $G$, can we say that $U(S)\cong Hom_{\mathbb Z} (G,\mathbb Z)$, for some ring $S$ ?
My thoughts : Given any torsion free abelian group $G$, I can show that the group of units of the Group ring $(\mathbb Z/(2) )[G]$ is isomorphic to $G$ . Since the dual group of a torsion free abelian group is again torsion free, so I'm done in the case of torsion free abelian groups.
I don't know what happens if the group has a torsion.
Are there any other general sufficient or necessary conditions ?