Associativity test for a Magma

Solution 1:

In the absence of any further information then, yes, you need to check every triple. There is a theorem (due to G. Szasz) which asserts that on any set with at least four elements, there is a binary operation for which there is exactly one non-associative triple. (In fact, there are such operations on three-element sets also; $10$ of them, up to isomorphism.)

A reference for the Szasz theorem is:

@ARTICLE{Szasz1953,
AUTHOR = {G. Szasz},
TITLE = {{D}ie {U}nabh\"{a}ngigkeit der {A}ssoziativit\"{a}tsbedingungen},
JOURNAL = {Acta Sci. Math. Szeged},
VOLUME = {15},
YEAR = {1953},
PAGES = {20--28},
LANGUAGE = {German},
REVIEW = {\MR{56575 (15,95d) 09.1X}},
}

I should add that I've not actually seen this paper. (I've never found it online, and I don't read German anyway.) However, the proof is not difficult. Suppose you have a set $S$ with four distinct elements $a$, $u$, $v$ and $w$. Define the binary operation $\cdot$ on $S$ by putting $a\cdot a = u$, $a\cdot u = v$, and $x\cdot y = w$, for all pairs $(x,y)$ other than $(a,a)$ and $(a,u)$. Then it is easy to see that $(a\cdot a)\cdot a\neq a\cdot(a\cdot a)$. It is then tedious, but completely elementary to check (case by case, as it were) that every other triple does associate.

Solution 2:

In Rajagopalan and Schulman "Verification of Identities" (1997), an algorithm is given that probabilistically checks whether a given operation $\circ$ on a set $S$ is associative. If the operation is nonassociative, the test detects this fact with with any desired probability $\delta$ in $$O(\kappa n^2\log\delta^{-1})$$ time, where $n=|S|$ and$\kappa$ is the time to calculate $\circ$ for one pair of arguments. A variant of the algorithm will generate a specific triple $\langle a,b,c\rangle$ for which $(a\circ b)\circ c\ne a\circ (b\circ c)$, when one exists, in time $$O(\kappa n^2\log n\log \delta^{-1}).$$

This algorithm works for arbitrary operations. Unlike Light's test, it works even for "noncancellative" operations, where the equations $x\circ a = b$ and $a\circ x = b$ may not have solutions for all given $a,b$.

The paper also shows that if the operation $\circ$ is cancellative, one can compute a small ($O(\log n)$)set that generates it in time $O(n^2)$, and then apply Light's test to deterministically verify associativity in total time $O(\kappa n^2\log n)$.

Solution 3:

Generally, checking for associativity can be computationally very difficult. There are no easy visual criteria on the multiplication table to discern associativity.