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?

**Answer**

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.

**Attribution***Source : Link , Question Author : Mikko Korhonen , Answer Author : Jonas Meyer*