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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?

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  • $\begingroup$ @Peter I accidentally made a typo. I never meant to say that either is associative or commutative. However, exponentiation has "right associocommutativity". $\endgroup$ – Solomon Ucko Mar 13 at 11:29
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    $\begingroup$ $a*b=a$ is not commutative, but has the property. $\endgroup$ – Gerry Myerson Mar 13 at 11:31
  • $\begingroup$ What do you mean with "right associative" ? The convention is that power towers are calculated from right, not left. But I would not call this "right associative". $\endgroup$ – Peter Mar 13 at 11:31
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    $\begingroup$ $a*b=a+2b$ is neither commutative nor associative, but has the property. $\endgroup$ – Gerry Myerson Mar 13 at 11:35
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    $\begingroup$ What is there to clarify, Peter? I'm defining the binary operation $*$ by for all $a$ and for all $b$, $a*b=a$. $\endgroup$ – Gerry Myerson Mar 13 at 11:36
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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.

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  • $\begingroup$ 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? $\endgroup$ – Solomon Ucko Mar 13 at 22:24
  • $\begingroup$ $(x^{-1}\cdot g^{-1})\cdot(x\cdot g) = x^{-1}\cdot x\cdot g^{-1}\cdot g = \mathrm{id} = (x\cdot g)^{-1}\cdot(x\cdot g)$. The $\cdot(x\cdot g)$s then cancel. $\endgroup$ – Solomon Ucko Mar 13 at 23:34
  • $\begingroup$ I'm still confused how this would apply to exponentiation. What would the relevant group be? Thanks. $\endgroup$ – Solomon Ucko Mar 15 at 11:35
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    $\begingroup$ 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.) $\endgroup$ – verret Mar 16 at 23:29
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    $\begingroup$ Note that there are two operations in the expression above: exponentiation (which is the action), and integer multiplication (which is the associative binary operation). $\endgroup$ – verret Mar 16 at 23:31
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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.

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  • $\begingroup$ 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. $\endgroup$ – Solomon Ucko Mar 13 at 17:40

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