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We see that on the cyclic additive group $(\mathbb Z,+) $ we can define a "natural" product induced by the sum. Let $(G,+) $ be a group, I wonder if there exists such a way I can define a function $\cdot : G\times G \rightarrow G $ such that $(G,+,\cdot ) $ is a ring (and the product is not the null function $a\cdot b=0\ \forall a,b\in G $). Or also, are there any conditions the group G has to satisfy in order to be given the structure of ring?

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    Consider the [Prüfer group](https://en.wikipedia.org/wiki/Prüfer_group) where the only ring structure is the trivial one (no identity, all products are zero). – 2017-01-12
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    a little addendum: What you can do is that if you have an *abelian* group $(G, +)$ (such as $\mathbb{Z}$) you can define a "$\mathbb{Z}$-multiplication onto" $G$, namely $$\cdot \colon \mathbb{Z} \times G \to G \\ (n,g) \mapsto n \cdot g$$, where $n \cdot g := g + \dots + g$. This makes $G$ a $\mathbb{Z}$-module. – 2017-01-12
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    The product on $\mathbb{Z}$ is not really "natural" given only its group structure: $1$ and $-1$ are both perfectly good generators of $\mathbb{Z}$ as a group, and there's no canonical way to choose one of them to be the unit of the ring. – 2017-01-12
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    @EricWofsey however the two rings are isomorphics. – 2017-01-12
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    @M.U. ok, but it does not give him a proper ring structure. Is a couple (R-module, R) isomorphic to a ring or even has it a kind of ring structure (I don't sincerely know it)? – 2017-01-12
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    @egreg where can I find a proof that there is no way to define a non trivial product? – 2017-01-13
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    @Blumer Suppose you're given a product. Take $a\ne0$ and $b$; then there is $n>0$ such that $nb=0$ and there is $c$ with $a=nc$; now $ab=ncb=c(nb)=c0=0$. – 2017-01-13

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