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 leftcancellative; if $X$ is in fact a left transversal to every conjugate of $H$, then $*$ is also rightcancellative 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

$\begingroup$ Is it possible to construct one such set in particular for $G=S_5$ and $H=S_4$ the symmetric groups? $\endgroup$ – Anupam Ah Aug 23 '17 at 10:32

$\begingroup$ In the special case of $S_{n1}$ 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. $\endgroup$ – Colin Reid Aug 23 '17 at 11:00

$\begingroup$ I need just one set only. You are telling powers of (1,2,3,4,5) will work. $\endgroup$ – Anupam Ah Aug 23 '17 at 11:09

$\begingroup$ 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? $\endgroup$ – Colin Reid Aug 23 '17 at 11:56

$\begingroup$ I need a particular latin square which does not arise from a group. It exists only in order 5. see it in en.wikipedia.org/wiki/Latin_square. For that which I should choose? $\endgroup$ – Anupam Ah Aug 23 '17 at 12:41