Associativity may fail by little?

Question

It is a well-known result on group theory that if a group has many pairs of commuting elements then it is abelian.

This motivated the following pseudo-conjecture.

If a (possibly infinite) set $S$ with a binary operation $\cdot$ is such that for many triples $a,b,c\in S$ it holds $(a\cdot b)\cdot c = a \cdot (b\cdot c)$, then $(S,\cdot)$ is a semigroup.

Exploring a little bit on this, a colleague told me he has read somewhere (he can't remember where) that the statement above is false.

If $S$ is a finite set, then there exists a binary operation $\star$ that satisfies $(a\star b)\star c = a\star(b\star c)$ for all but just one ordered triple $(a,b,c)\in S\times S\times S$.

Such a result implies that an algorithm that checks if a certain operation is associative must check indeed all triples of elements, which is kind of funny and rather unintuitive (to me, at least).

We couldn't find a proof of this last result, although playing with small sets it seems to be true.

My questions are the following:

1) Does anybody here know a reference for this result?

2) Is there a constructive proof for such an example?

Answer 1

The result you quoted appears in this reference: G. Szasz, Die Unabhängigkeit der Assoziativitätsbedingungen, Acta. Sci. Math. Szeged 15 (1953), 20-28.

The Szasz theorem requires that the set $S$ have at least four elements, though it is also true for sets of size $3$.

Szasz' proof is constructive and goes as follows. Assuming $a$, $u$, $v$ and $w$ are distinct members of $S$, define $a\cdot a = u$, $a\cdot u = v$ and $x\cdot y = w$, for any $(x,y)$ other than $(a,a)$ and $(a,u)$. Then a case by case verification shows that $(a\cdot a)\cdot a = w\neq v = a\cdot(a\cdot a)$ but that, for every other triple $(x,y,z)\in S^3$, we have $(x\cdot y)\cdot z = x\cdot(y\cdot z)$.

Direct enumeration shows that there are (up to isomorphism) $10$ magmas of order $3$ with exactly one non-associative triple. (There are $124$ of order $4$.)