I was wondering if "rings" with noncommutative addition are studied at all? Of course, if a ring $R$ has a $1$, then for all $a, b\in R$, $a+a+b+b=(1+1)a+(1+1)b=(1+1)(a+b)=(a+b)+(a+b)=a+b+a+b$, from which it follows from cancellation that $a+b=b+a$. Thus, rings with $1$ automatically must have commutative addition.

So, are there interesting, necessarily non-unital, "rings" with noncommutative addition? Of course, you can take any non-abelian group and make such a "ring" by having all multiplications give the additive identity. Are there more interesting examples?

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    $\begingroup$ You are probably already aware of the concept of a near-ring, see en.wikipedia.org/wiki/Near-ring $\endgroup$ – Andreas Caranti Jun 13 '14 at 20:16
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    $\begingroup$ possible duplicate of Why is ring addition commutative? $\endgroup$ – Sir Jective Jun 13 '14 at 20:18
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    $\begingroup$ The enveloping near-ring of the endomorphisms of a non-abelian group is an interesting and useful such thing (with identity). Rather than lacking an identity, it lacks one of the distributive laws (but has the other). As a silly example: $x^{g+2h}$ is shorthand for $x^g (x^2)^h$, and $g+2h$ lives in that near-ring. Similarly $x^{-1+g} = x^{-1} x^g$ is shorthand for $[x,g]$, and it is nice to have $-1+g$ apart from the $x$. $-1+g$ is called $ad(g)$ in some areas, I guess. $\endgroup$ – Jack Schmidt Jun 13 '14 at 20:23
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    $\begingroup$ @silvascientist Not a duplicate. That question asks why rings are defined this way. Here the question is about interesting examples without commutativity. $\endgroup$ – user147263 Jun 13 '14 at 21:28

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