Let G be a finite group and we know its group table is a Latin square of order |G|. Now let H be any subgroup of G of index n. Then we can form G/H which is a collection of left cosets. My question is, is there any natural construction to get a Latin square of order n from G/H? If H is normal in G it is clear since G/H itself is a group. What if H is not normal in G? Not only G/H, using any properties of G and H. Not restricted to coset set. But the order should be the index.



How to get Latin squares from a finite group and a subgroup
I don't know of any general construction. Perhaps the most 'natural' examples are loop transversals: Given a left transversal $X$ to $H$ in $G$ with $1 \in X$, define a binary operation on $X$ by setting $x\*y$ to be the unique element of $X \cap xyH$. Then $\*$ is left-cancellative; if $X$ is in fact a left transversal to every conjugate of $H$, then $\*$ is also right-cancellative and hence $X$ forms a loop.


I don't know who thought of this idea first, but I found a series of articles by E.A. Kuznetsov on the subject, for example:
<http://www.quasigroups.eu/contents/download/1999/6_1.pdf>


