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I would like to know if there is a method to determine how many $n \times n$ Latin square define the same non-commutative group $G$. This is how many representations of $G$ can be obtained from $n$ dimensional Latin squares.

It results that I'm interested on building examples of non-abelian groups to toy with and test how many representations do they have.

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    $\begingroup$ Symmetric groups appear everywhere (so are ideal for toy examples). Any group table is a Latin square, see for example here. For a given Latin square we have its automorphism group. See also the other posts at this site, e.g., this one and its links. $\endgroup$ – Dietrich Burde Jan 9 at 16:19
  • $\begingroup$ Concretely the table can be built using the principle from the action of $f: G \times G \to G$ so $f(a,b)=c$ If I'm correct such representation is the core of Cayley's theorem as the group would be embedded into a permutation group (if it's action is faithful). My concern here is to build arbitrary non-abelian groups of order $n$ and to obtain the different number of Latin squares that define the same non-abelian group. I should point that on my question. $\endgroup$ – kub0x Jan 9 at 16:27
  • $\begingroup$ Then why aren't you just taking the known non-abelian groups, e.g., all finite simple non-abelian groups? $\endgroup$ – Dietrich Burde Jan 9 at 16:30
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(I'm going to assume that the elements of the Latin Square are the elements of the group. If that's not what you want, then make your question clearer.)

For a given group, once you've chosen how you label the rows and columns, the Cayley table is uniquely determined. So if the group has order $n$, it has $(n!)^2$ tables.

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