# Associocommutativity

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?

• @Peter I accidentally made a typo. I never meant to say that either is associative or commutative. However, exponentiation has "right associocommutativity". Mar 13, 2019 at 11:29
• $a*b=a$ is not commutative, but has the property. Mar 13, 2019 at 11:31
• $a*b=a+2b$ is neither commutative nor associative, but has the property. Mar 13, 2019 at 11:35
• What is there to clarify, Peter? I'm defining the binary operation $*$ by for all $a$ and for all $b$, $a*b=a$. Mar 13, 2019 at 11:36
• mathworld.wolfram.com/LeftIdeal.html right and left handedness is possible peter. Also if parenthesized the exponents in a power tower can have any order.
– user645636
Mar 18, 2019 at 12:44

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.

• This is a bit too complicated for me at the moment, but I'll accept this at some point if there aren't any better answers. Thanks for the pointers, though. Mar 13, 2019 at 17:40

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: $$(x*g)*h=x*(g\cdot h)=x*(h\cdot g)=(x*h)*g$$.

• A more spelled-out version: $(x*g)*h = (x\cdot g^{-1})\cdot h^{-1} = x\cdot(g^{-1}\cdot h^{-1}) = x\cdot(g\cdot h)^{-1} = x*(g\cdot h) = x*(h\cdot g) = x\cdot(h\cdot g)^{-1} = x\cdot(h^{-1}\cdot g^{-1}) = (x\cdot h^{-1})\cdot g^{-1} = (x*g)*h$ (assuming $(x\cdot g)^{-1} = x^{-1}\cdot g^{-1}$, which I'm sure is (easily) provable). Also, how would this apply for exponentiation? Mar 13, 2019 at 22:24
• $(x^{-1}\cdot g^{-1})\cdot(x\cdot g) = x^{-1}\cdot x\cdot g^{-1}\cdot g = \mathrm{id} = (x\cdot g)^{-1}\cdot(x\cdot g)$. The $\cdot(x\cdot g)$s then cancel. Mar 13, 2019 at 23:34
• Are you familiar with the concept of an action? The integers with multiplication are a semigroup, and they act on $\mathbb{R}$ by exponentiation: $(x^a)^b=x^{a\cdot b}$. (This is exactly the definition of what an action is.) Mar 16, 2019 at 23:29