Almost-associative Latin squares

A group can be interpreted as a set with an associative operation whose Cayley table is a Latin square, as seen in the secret Simon Tatham puzzle called Group. (Ignore infinite groups for now.)

What if we only want it to be almost associative, i.e. as associative as possible without actually being associative?

I claim that any non-associative bibijective¹ operation ○ must have at least eight non-associative triples. The proof is as follows:

For any B and C, the functions x -> x ○ (B ○ C) and (x ○ B) ○ C are both bijective, so they must differ in at least two inputs. Therefore, for any A, B, and C such that A ○ (B ○ C) is unequal to (A ○ B) ○ C, there must be some other choice of A that does so. A similar line of reasoning proves that there is some other choice of B that does so, and some other choice of C that does so.

That means that for any non-associative triple, for any of its elements, there’s a different value that still makes it non-associative. Therefore, there are at least two different possible values for the first value in a non-associative triple, and for each of them there are at least two possible values for the second value, and similarly for the third, making at least 2 * 2 * 2 = 8 non-associative triples.

Now, an example of an almost-associative operation:

Notice how it differs from the Klein 4-group in the bottom left 2 by 2, which is A ○ B. For a triple to be non-associative, it has to involve a different parity of such pairs in the two calculation orders. It can be checked that A ○ A ○ B is the only such case, and thus that it and eight equivalent triples are the only non-associative triples of ○.

1. The first “bi” prefix in this word is used in the same sense as the word “bilinear”. ↩︎