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So I was considering the following idea. Let a generalized Ring $gR$ be a set equipped with two operations $$u_1, u_2$$

Such that $gR$ is a group with the operation $u_1$ and the operation $u_2$ distributes over $u_1$ (given by):

$$ u_2(a,u_1(b,c)) = u_1(u_2(a,b), u_2(a,c)) $$

Further, $u_2$ is associative, closed, and has an identity element $s$ that isn't the same as the identity $t$ of $u_1$ (this is a bit artificial of a requirement now that I think about it).

I wrote some code to begin automatically finding groups, and then finding generalized rings that exist over these groups:

https://github.com/frogeyedpeas/RingWarrior/blob/master/findingRings.py

But... only abelian groups appear to support any type of ring structure (with some of them supporting an explosion of different rings for the same underlying group).

Why is this the case? I know the original definition for rings is that the group over which it is built is abelian. But my computer program appears to suggest that abelian groups ARE the only type of structure that could possibly support a ring like object over them.

Is there a way to prove that?

To make it clear what I wish to prove:

Given a set $X$ such that there are two binary operations

$$ u_1, u_2 $$

Defined and closed over elements in the set. Such that for any $a,b,c \in X$

$$ u_1(u_1(a,b),c) = u_1(a,u_1(b,c)) \ \text{associative}$$ $$ u_2(u_2(a,b),c) = u_2(a,u_2(b,c)) \ \text{associative}$$ $$ u_2(c,u_1(a,b)) = u_1(u_2(c,a),u_2(c,b)) \ \text{distributive}$$

And there exist identity elements $t,s$ such that $t \ne s$ and

$$ u_1(t,a) = a \forall a \in X$$ $$ u_2(s,a) = a \forall a \in X$$

And for each element $m \in X$ there exists a unique $n(m) \in X$ such that

$$ u_1(m,n(m)) = u_1(n(m),m) = t$$

I wish to show that this implies that $u_1$ is a commutative operation.

$$ u_1(a,b) = u_1(b,a) \forall a,b \in X$$

Since experimentally that is what appears to be suggested

Alex M.
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Sidharth Ghoshal
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    The reason we define rings to have abelian addition is that most rings that we are interested have abelian addition. Definitions do not come out of thin air but as abstractions of the objects we study. – Mariano Suárez-Álvarez Jul 02 '15 at 18:19
  • But what I'm getting at is more subtle than that. When I try to find rings that don't have abliean addition, my computer program verifies that no ring-like analogues exist. In a strict sense, having a second opeartion that distributes over the first, and carries its own unique identity, at least for small groups (I checked all groups up to order 6) yields no ring like objects with non abelian underlying groups – Sidharth Ghoshal Jul 02 '15 at 18:21
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    See http://math.stackexchange.com/questions/609364/why-is-ring-addition-commutative, by the way – Mariano Suárez-Álvarez Jul 02 '15 at 18:22
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    «All groups of order up to $6$» is an almost negligible amount of evidence! – Mariano Suárez-Álvarez Jul 02 '15 at 18:23
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    You might take a second look at your equation for the associativity of $u_2$; should it not read $u_2(u_2(a, b), c) =u_2(a, u_2(b, c))$? Cheers! – Robert Lewis Jul 02 '15 at 18:23
  • @MarianoSuárez-Alvarez, haha I don't disagree with that, but, my methods weren't most efficient. So I made do with up to that order and then tried to make a conjecture :) – Sidharth Ghoshal Jul 02 '15 at 18:24
  • @RobertLewis, Thanks! I made the correction – Sidharth Ghoshal Jul 02 '15 at 18:26
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    Also, what about $t$, $s$? Did you mean to type $t \ne s$? If not, what are $u$, $v$? – Robert Lewis Jul 02 '15 at 18:28
  • @RobertLewis thanks again! I was going to use u and v at variables but since i had u as a function i figured that wouldn't be too pretty, forget to edit it out. – Sidharth Ghoshal Jul 02 '15 at 18:31
  • @MarianoSuárez-Alvarez the question you posted was very helpful but the responses often resorted to an argument such as: (1+1)(x+y) = (1+1)x + (1+1)y = (1)(x+y) + (1)(x+y). It occurred to me what I am asking is weaker than that as the only think I can derive from my axioms is (1+1)(x+y) = (1+1)x + (1+1)y, and no statement can be made about the nature of (1)(x+y) + (1)(x+y), at least not obviously. So my version of the distributive property is in some sense, much weaker than classical distributive, if you use hte functional equations i have given above. – Sidharth Ghoshal Jul 02 '15 at 18:32
  • But given my tiny shred of experimental evidence it appears that doesn't really change anything at all. – Sidharth Ghoshal Jul 02 '15 at 18:34

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This is not an answer, but rather an extended comment.

In your comments, you state that the identity element $s$ is unique. I'm not sure how to prove this from the axioms, but OK. Anyway, I'm not convinced you've looked at enough examples. Why don't you do $D_8$ and $Q_8$?

If you take a group $G$, with $u_1$ the usual group multiplication, then your distributive law says that $u_2$ factors as a map $$u_2:G\times G\to\mathrm{Hom}(G,G)\times G\to G.$$

For abelian groups, any injective ring homomorphism $G\to\mathrm{Hom}(G,G)=\mathrm{Hom}_{\mathbb{Z}}(G,G)$ determines a generalized ring (there are $\mathrm{Aut}_{\mathrm{ring}}(G)$ many).

I'm not sure what is possible for non-abelian groups, and you may want to get some data on groups where $G/Z(G)$ is abelian. Also, you'll get more generalized rings if you relax the "s is unique" requirement.

David Hill
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