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?

Solutions Collecting From Web of "Does every Abelian group admit a ring structure?"

If your group has the property that every element has finite order, but there is no upper bound on the orders of the elements, then it is not the additive abelian group of a ring with identity. The reason is that if there were such a ring structure with an identity $1$, then $1$ would have finite additive order $k$, and then for all $a$ in your group, $k\cdot a=(k\cdot1)a=0a=0$, which forces $a$ to have order at most $k$.

For each prime $p$, the Prüfer $p$-group $\mathbb Z(p^\infty)$ is an example of such a group. The quotient group $\mathbb Q/\mathbb Z$ is another. Direct sums (but not direct products) of infinitely many finite cyclic groups of unbounded order would also be examples.

There are several canonical group structures that can be equipped naturally on a group ; I am going to list the ones that I know that are interesting, perhaps other people can add some to the list.

Finite abelian groups are isomorphic to product of finite cyclic groups so you can naturally form a ring from a direct product of finite rings. The same goes for finitely generated abelian groups ; you can deduce a ring structure from the products of the underlying natural rings.

One ring structure that is always there will be the trivial ring structure, i.e. just add a multiplication that means nothing ; it sends everything to the identity (of $G$). Clearly it respects all the ring axioms and it gives you a ring (without a ring identity, though). It is not very interesting in such, because it means “we do have a ring structure on $G$, but it is not telling us anything at all about $G$”.

Another ring structure which does not always have its underlying abelian group isomorphic to $G$ but is canonically defined by $G$ is the endomorphism ring of $G$, i.e. let
$$
R = \{ \phi : G \to G \, | \, \phi \text{ is a group endormorphism of } G \text{ to } G \}.
$$
Equip a ring structure on $R$ by saying that $(R,+, \circ)$ is a ring by defining the following : the endomorphism $\phi_1 + \phi_2$ is defined as
$$
(\phi_1 + \phi_2)(g) = \phi_1(g) + \phi_2(g)
$$
and
$$
(\phi_1 \circ \phi_2)(g) = \phi_1(\phi_2(g)).
$$
If $G$ is abelian, then you can readily see that the operation $+$ gives you an abelian group on $R$, because the inverse of the endomorphism $\phi$ will just be the endomorphism that sends $g$ to $-\phi(g)$ instead of $\phi(g)$, so that $\phi + (-\phi) = 0_R$. Commutativity follows from the fact that $G$ is abelian and the other properties follow naturally. The fact that the $\circ$ operation gives you a ring structure comes from the fact that endomorphisms are in particular homomorphisms, so that
$$
(\phi \circ (\psi_1 + \psi_2))(g) = \phi( \psi_1(g) + \psi_2(g)) = (\phi \circ \psi_1)(g) + (\phi \circ \psi_2)(g)
$$
In particular, the identity of the ring $R$ would be the identity endomorphism of $G$.

I know it is not exactly what you asked for (in this last part), but I didn’t know much about it at first when I saw it and I believed it was interesting, so I just thought you might wanna take a look.

Hope that helps,

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.