Does every Abelian group admit a ring structure? Given some Abelian group $(G, +)$, does there always exist a binary operation $*$ such that $(G, +, *)$ is a ring? That is, $*$ is associative and distributive:
\begin{align*}
&a * (b * c) = (a*b) * c \\
&a * (b + c) = a * b + a * c \\
&(a + b) * c = a * c + b * c \\
\end{align*}
We also might have multiplicative identity $1 \in G$, with $a * 1 = 1 * a = a$ for any $a \in G$. Multiplication may or may not be commutative.
Depending on the definition, the answer could be no in the case of the group with one element: then $1 = 0$. But the trivial ring is not a very interesting case. For cyclic groups the statement is certainly true, since $(\mathbb{Z}_n, +, \cdot)$ and $(\mathbb{Z}, +, \cdot)$ are both rings. What about in general? Is there some procedure to give arbitrary abelian groups ring structure?

Every finite Abelian group is a product of finite cyclic groups, so you get a ring for free. Similarly, every finitely generated abelian group is isomorphic to some copies of $\mathbb{Z}$ times a finite Abelian group, so you get a ring for free there as well. The only interesting case remaining would be a non-finitely-generated Abelian group. There are a few more steps that one can probably go, but I don't know how we can get a ring in general.
