Associocommutativity

Question

One thing I've noticed is that addition and multiplication both form commutative groups over the reals, but subtraction, division, and exponentiation are neither associative nor commutative. Ignoring issues with closure for division and possibly exponentiation, all 5 have the property that $(a \star b) \star c = (a \star c) \star b$ (that I call "right associocommutativity" because the swapped operands are on the right). Both left and right associocommutativity are implied by the combination of associativity and commutativity. However, tetration ($\uparrow\uparrow$, repeated exponentiation) has neither left nor right associocommutativity.

Now, my questions: Is there a better name for this? What other operations that aren't both associative and commutative have this property? Why doesn't this work for tetration? Is there a similar property that all of these operations have?

Answer 1

First, note that, if $G$ is an abelian group acting on the right on a set $X$ (say we denote the action by $\cdot$), then we can get another right action $*$ of $G$ on $X$ by $x*g=x\cdot g^{-1}$.

Since addition and multiplication can be seen as abelian groups acting on themselves, this allows us to also view subtraction and division in this way.

Now, note that any right action $*$ of a commutative semigroup $(G,\cdot)$ on a set $X$ will have the property that you want: $(x*g)*h=x*(g\cdot h)=x*(h\cdot g)=(x*h)*g$.

This explains all your examples.

Answer 2

The algebras satisfying $(a\cdot b)\cdot c=(a\cdot c)\cdot b$ for all $a,b,c \in A$ have been studied in geometry. For a special class, see our paper here. Denoting the right multiplication by an element $x$ by $R(x)$, we can rewrite the identity as $$ [R(x),R(y)]=R(x)R(y)-R(y)R(x)=0. $$ So the right multiplications all commute. There are several $K$-algebras which are neither commutative nor associative and they arise naturally in many areas of mathematics and physics.