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Given a Latin square

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how would one tell if the operation is associative without trying every combination? Or is there something to look for that would at least limit the amount of combinations I have to try?

In this case I found $(bb)b \ne b(bb)$ but I struggled quite a bit.

Kan't
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DividedByZero
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  • See number 27 of Matousek's 33 Miniatures https://kam.mff.cuni.cz/~matousek/ – Angina Seng May 13 '18 at 02:44
  • @LordSharktheUnknown That looks very interesting! However I don't think an algorithm like that would be any quicker than the naive approach for most people :) – DividedByZero May 13 '18 at 02:57
  • That Miniature deals with the wider case of associativity of magmas, @AnginaSeng. The OP may have sharper solutions. – Kan't Oct 07 '25 at 05:15

2 Answers2

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In this particular case, it is easy to see that $e$ is an identity.

Lemma: Given a latin square with an identity, the operation is associative if and only if the latin square is the table of a group.

This is immediate. Indeed $\Leftarrow$ is obvious.

For $\Rightarrow$ associativity and identity are given, and it is immediate that latin square implies inverses.

\qed

For the given table it is trivial to see it is not coming from a group, as any group with 5 elements is cyclic thus commutative, and your square is not.

In general, given a larger latin square it may be hard to decide if it is the table of a group or not.

N. S.
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  • Yes, I agree that (given knowledge of Lagrange's theorem) this particular case would be trivial, but I was asking how to find an example of non-associativity. However your last sentence answers that question, thanks! – DividedByZero May 13 '18 at 02:46
  • It's almost as easy to see that a Latin square is associative if and only if it's the table of a group. Therefore, if there is no identity, it can't be associative. – user14111 Dec 18 '23 at 10:18
  • Of course there is an identity in the given table, @user14111. – Kan't Sep 20 '25 at 05:45
  • @Kan't, in the table, $cb=e$, but $bc=a\ne e$. That one observation is enough to show it's not a group. – Gerry Myerson Sep 20 '25 at 06:26
  • @GerryMyerson, unless you assume as known that a group of order 5 must be abelian, your point is good only once acknowledged that $e$ plays as identity, which seems to have been missed by user14111, to whom my comment was addressed. – Kan't Sep 20 '25 at 11:41
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Rename your Latin square into the isotopic one under the bijection: \begin{alignat}{1} &e\mapsto 1 \\ &a\mapsto 2 \\ &b\mapsto 3 \\ &c\mapsto 4 \\ &d\mapsto 5 \\ \end{alignat} Then you get to: $$\begin{array}{c|ccccc} * & 1 & 2 & 3 & 4 & 5 \\\hline 1 & 1 & 2 & 3 & 4 & 5 \\ 2 & 2 & 3 & 4 & 5 & 1 \\ 3 & 3 & 1 & 5 & 2 & 4 \\ 4 & 4 & 5 & 1 & 3 & 2 \\ 5 & 5 & 4 & 2 & 1 & 3 \\ \tag1 \end{array}$$ For every $i\in\{1,2,3,4,5\}$, the position: $$i*j=\sigma_i(j) \tag2$$ defines the permutation $\sigma_i\in S_5$. From $(1)$ and $(2)$: \begin{alignat}{1} &\sigma_1=() \\ &\sigma_2=(12345) \\ &\sigma_3=(13542) \\ &\sigma_4=(143)(25) \\ &\sigma_5=(15324) \\ \tag3 \end{alignat} Now to the associativity. From $(2)$: \begin{alignat}{1} &i*(j*k)=(i*j)*k &\iff \\ &\sigma_i(\sigma_j(k))=\sigma_{i*j}(k) &\iff \\ &\sigma_i\sigma_j=\sigma_{i*j} \\ \end{alignat}

Therefore, the Latin Square is associative only if $\Sigma:=\{\sigma_1, \dots, \sigma_n \}$ is a group, which is clearly not your case as, for example, $\sigma_4^2\notin\Sigma$.

Kan't
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