# How to get Latin squares from a finite group and a subgroup

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.

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.
• Is it possible to construct one such set in particular for $G=S_5$ and $H=S_4$ the symmetric groups? – Anupam Ah Aug 23 '17 at 10:32
• In the special case of $S_{n-1}$ inside $S_n$, a loop transversal amounts to picking a set of permutations $X$ such that distinct elements $f,g \in X$ satisfy $\forall x: f(x) \neq g(x)$. There will be many such choices in general (the most obvious is to take the powers of an $n$-cycle), but it's not easy to tell which will give isomorphic loops. – Colin Reid Aug 23 '17 at 11:00
• Sure, but the loop you get in that case will just be the cyclic group $C_5$. What do you want to do with the resulting Latin square? – Colin Reid Aug 23 '17 at 11:56