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
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.