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My question is simple, though it proves to be much more difficult than it sounds. Suppose I want to find a binary operation to add extra structure to a multiplicative group (so it becomes a ring). This binary operation must be associative, and also distributive over multiplication by definition.

My question is whether or not there exists a binary operation that is both associative and distributive over multiplication. Keep in mind that I am just an amateur mathematician, not an advanced ring theorist, so it would be kind for you to keep things simple.

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    A ring has a absorption element (zero) whereas a nontrivial group cannot have such an element. However sometimes it is possible to find an $R$ such that $R^\times=G$ for given groups $G$. But there will generally be more than one possible way to put a $+$ structure on $G\cup{0}$ (for example $\Bbb F_q^\times$ has many choices of generators), and I doubt there is a generic way to do it (I am not even sure if every group is a group of units). [It is true though that $G\subseteq R^\times$ is always solvable in $R$; just form a group ring.] – anon Dec 30 '13 at 04:34
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    Perhaps you are looking for http://en.wikipedia.org/wiki/Group_ring ? – Prahlad Vaidyanathan Dec 30 '13 at 04:47
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    You can only make it a ring if the multiplicative group is abelian. The distributivity of the binary operation somehow demands that. See http://math.stackexchange.com/q/609364/75923 and for a short answer to that question see http://math.stackexchange.com/a/609371/75923. – drhab Dec 30 '13 at 10:11
  • @user118228 just to be absolutely clear, you're looking for an operation $\ast$ on a group $G$ such that $a\ast(bc)=(a\ast b)(a\ast c)$ and similarly for the right handed distributive law? I just have some lurking suspicions that you might not have expressed what you actually meant... – rschwieb Dec 30 '13 at 14:32
  • @anon: http://math.stackexchange.com/questions/384422/which-finite-groups-are-the-group-of-units-of-some-ring is a smaller version of your question – Jack Schmidt Dec 30 '13 at 17:45
  • @rschwieb Yes, that is what I mean. I need some (non-trivial) binary operation $$ such that $(ab)c=a(bc)$ and $a(bc)=(ab)(ac)$. –  Dec 30 '13 at 20:26
  • @user118228 Distributive over one side or both? – rschwieb Dec 30 '13 at 20:40
  • @rschwieb Preferably both. –  Dec 30 '13 at 22:14

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