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For $n\in\mathbb N$, the multiplication table of a quasigroup $Q=([n],*)$ induces a transitive subset $\Sigma=\{\sigma_1,\dots,\sigma_n\}\subseteq S_n$ via the position: $$\sigma_i(j):=i*j\tag1$$ and viceversa (just by moving in $(1)$ the colon "$:$" on the RHS). A necessary condition for $Q$ to be a group, is that $\Sigma$ is a transitive subgroup of $S_n$. In fact: \begin{alignat}{1} (i*j)*k= i*(j*k) &\iff \sigma_{\sigma_i(j)}(k)=(\sigma_i\sigma_j)(k) \\ &\iff \sigma_i\sigma_j=\sigma_{i*j} \\ &\color\red{\space\space\Longrightarrow}\Sigma\le S_n,\text{ transitive} \\ \end{alignat} I don't believe that this fact means that the other implication (sufficiency) holds true either, also because, if so, we'd have a deterministic algorithm able to test, in $O(n^2\log n)$ time, whether a given latin square is the Cayley table of a group. Hence my question:

Q. As a counterexample, I'm looking for of a transitive subgroup of order $n$, $\Sigma\le S_n$, such that the induced latin square is non-associative.

Kan't
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1 Answers1

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Here is a counterexample. Take $n=6$ and let $\Sigma$ be the cyclic subgroup generated by $s=(1,2,3,4,5,6)$. Now, take an enumeration $(\sigma_k)_{1\leq k\leq 6}$ of $\Sigma$ such that $\sigma_2=s^2,\sigma_4=s^4,\sigma_6=s^5$.

Then

$$ 2*(4*1)=2*(s^4(1))=2*5=s^2(5)=1 $$,

but on the other hand

$$ (2*4)*1=(s^2(4))*1=6*1=s^5(1)=6 $$

So this is not associative.

Ewan Delanoy
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  • Is it obvious that such an enumeration exists? – Kan't Sep 03 '25 at 13:34
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    Of course. $\Sigma$ consists of six elements, which you may enumerate any way you want. If you insist on a fully explicit example, say $\sigma_{k}=s^k$ for $1\leq k\leq 4$ and $\sigma_5=s^6$, $\sigma_6=s^5$. – Ewan Delanoy Sep 03 '25 at 14:10