# Abstract formulation of associativity

Say we are given a binary operation $$f$$ on a set $$X$$, that is, $$f : X \times X \to X.$$ Denote by $$\text{Id}$$ the identity map on $$X$$. We say that $$f$$ is associative if, for all $$x, y, z \in X$$, we have $$f(f(x, y), z) = f(x, f(y, z)).$$ I was wondering if there is a more abstract way of formulating this relationship, i.e. in a coordinate-free way. After looking at a couple of commutative diagrams, I came up with the following: $$f \circ (f \otimes \text{Id}) = f \circ (\text{Id} \otimes f),$$ where $$\otimes$$ is defined through $$(f \otimes g)(x, y) = f(x)g(y).$$ Is this a senisble abstract definition of associativity? Can it be simplified somehow? My goal ultimatiely us to understand associativity as a "form of higher-order commutativity", if that makes sense. Am I onto something here?

• Re: associativity as a "form of higher-order commutativity" it may be interesting to note that as one goes from complex numbers to quaternions, commutivity is lost. When one goes from quaternions to octonions, associativity is lost.
– Matt
Commented Dec 4, 2023 at 14:47

I assume you meant $$(f\otimes g)(x,y)=\big(f(x),g(y)\big)$$. Yes, it is sensible. In fact this can be generalized to any category with products. And so we can even get rid of pointwise definition, as long as you know how to form products. In particular that's how group objects are defined over any category.
• I assume you meant $(f \otimes g)(x, y) = \big(f(x), g(y)\big)$. Commented Dec 4, 2023 at 1:51
Indeed, if $$L_x$$ is left "multiplication" with $$x$$ (that is, $$L_x(y) = f(x,y)$$) and $$R_y$$ is right multiplication with $$y$$, then $$f$$ is associative iff every $$L_x$$ comutes with every $$R_y$$.