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This question got me wondering: suppose $S$ is a set of operators such that for all $x, y \in S$, $x(y) \in S$.

For any $x,y \in S$, we say that $x \equiv y$ iff for all $z \in S$: $$x(z) = y(z)$$

We say that $S$ is associative if, for any $x, y, z \in S$: $$x(y(z)) \equiv (x(y))(z)$$

Finally, we say $S$ is fully commutative iff, for any two ordered tuples $T_1 = (x_1,y_1,z_1,w_1,\cdots)$ and $T_2 = (x_2,y_2,z_2,w_2,\cdots)$, where all elements of both tuples $\in S$, and where $T_2$ is any reordering of $T_1$: $$x_1(y_1(z_1(w_1(\cdots)))) \equiv x_2(y_2(z_2(w_2(\cdots))))$$ For example, $a(b(c(c))) \equiv c(b(c(a))) \equiv b(c(c(a))) \equiv \cdots$ where $a,b,c \in S$

If we treat $S$ as a group isomorphic to $\mathbb{Z}$ acting multiplicatively, for example, we know $S$ is fully commutative based on the nice commutivity of integers, e.g. $$1(4(16(12))) = 16(12(1(4)))$$

My questions are as follows:

  1. Suppose $S$ is fully commutative. Does this imply $S$ is also associative?

  2. If the answer to (1) is 'yes', is there a proof accessible to somebody who knows absolutely nothing about category theory (except the fact that it seems to pop up in problems involving abstract algebra a lot)?

  3. If the answer is 'no', does a reasonably simple (i.e. non-graduate-level math) counterexample exist? What additional conditions on $S$ are needed to guarantee associativity?

Bernard
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COTO
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1 Answers1

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Yes, full commutativity implies associativity, and there is a simple proof. $$ \begin{align} x(y(z)) &=z(x(y))\tag{$T_1=(x,y,z),T_2=(z,x,y)$} \\&=z(w)\tag{Let $w=x(y)$} \\&=w(z)\tag{$T_1=(z,w),T_2=(w,z)$} \\&=(x(y))(z) \end{align} $$

Side note: Usually an operator refers to a function $x:V\to V$ for some set $V$. If $x$ and $y$ are two operators, then $x(y)=x\circ y$ is the composition of these functions. In this case, any set $S$ of operators is automatically associative, since function composition is associative; no total commutativity assumption is necessary.

Mike Earnest
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