Nice! To make this familiar to the unwashed masses, write [a,b] = ab^{-1}.

Yeah for the rest of the year, I'm only going to think about division! Goodbye, multiplication and composition.

Form a ring with one more operator.

I suspect there's nothing interesting to do for the ring multiplication, because this definition really relies on having inverses. But one thing you'd want to do anyway is express the notion of "abelian group". The best I can do is ‣ [a,b]=[[e,b],[e,a]]

It looks like you can reduce the axioms further to the structure of the division map [ , ] subject to the axioms that [a,b] = [[a,c],[b,c]] and that all [a,a] are equal to a single element.

I don't know how you'd get out of insisting [a,[a,a]]=a. But otherwise it's fine to do this… that's how it is presented in Hall's book actually. twitter.com/solifine/statu…

@johncarlosbaez can mention of the identity element be dropped with these three equations? [a,b]=[[a,c],[b,c]] [a,[b,b]]=a [a,a]=[b,b] What do we get with just [a,b]=[[a,c],[b,c]] [a,[b,b]]=a ?

Yes, you can drop mention of the identity as mentioned. In group theory there's a famous easy proof that if ax = a for all a and ya = a for all a then x = y, so if there are right units and left units they must all be equal.

What notation is this?

Easy to show on a Rubik's cube. Any combination of moves is a group element. The comma is performing the first operand, then _reverting_ the second operand.