# How many representations of a non-commutative groups from $n$-dimensional Latin Squares

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.

• 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. – Dietrich Burde Jan 9 at 16:19
• 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. – kub0x Jan 9 at 16:27
• Then why aren't you just taking the known non-abelian groups, e.g., all finite simple non-abelian groups? – Dietrich Burde Jan 9 at 16:30

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.