Suppose we are given a multiplication table of a binary operation "$*$" on $\{1,2,\dots,n\}$: $$\begin{array}{c|cccc} * & 1 & 2 & \dots & n \\\hline 1 & \dots & \dots & \dots & \dots \\ 2 & \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots & \dots \\ n & \dots & \dots & \dots & \dots \tag1 \end{array}$$ such that $(1)$ is a Latin square (every $i\in\{1,\dots,n\}$ appears once and only once in each row/column). I think I managed to prove that $(1)$ is associative (and hence the Cayley table of a group) if and only if, for every $i,j\in\{1,\dots,n\}$: $$\sigma_i\sigma_j=\sigma_{i*j}\tag2$$ where $\sigma_i\in S_n$ is defined by: $$\sigma_i(j):=i*j\tag3$$ Despite the runtime of the algorithm "$(2)$ after $(3)$" is no better than $O(n^3)$ (like the naive associativity algorithm), from the condition $(2)$ follows, in particular, that: $$\{\sigma_1,\dots,\sigma_n\}\le S_n \tag4$$
Q. Is the runtime of the algorithm "$(4)$ after $(3)$" less than $O(n^3)$? If so, this approach could serve as a faster preliminary test before going through (possibly unncessarily) any $O(n^3)$ runtime algorithm.