For what it’s worth, I now see one way to deal with H = S_{n/2} wr S_2, completing my earlier sketch. The proof of the upper bound n!^n/e^(n^2 + o(n^2)) for the total number of Latin squares proceeds purely row-by-row: it bounds the number of choices for each subsequent row uniformly over all choices of previous rows. So we can bound the number of Latin squares all of whose rows lie in H by saying that there are at most |H|^k ways to fill out the first k rows, then proceeding with the previous method to bound the number of choices for the remaining n-k rows. By taking k to be n/10 or so, this results in a bound which is smaller than n!^n/e^(n^2 + o(n^2)) by a factor of e^(cn^2) for some constant c.

]]>I am sure that the “second row” conjecture is just the tip of the iceberg. There should be a general conjecture that says that the entries in any set of cells of a random Latin square are close to random subject to the obvious constraints, provided the set is not too large. But I don’t have a precise formulation of this. ]]>

Your conjecture about the distribution of the second row of a Latin square with first row the identity is interesting and challenging, but it seems like it might be overkill for your problem. We should be able to prove directly that almost all loops have multiplication group the symmetric group by arguing as follows. Fix a transitive subgroup H of S_n that we might be worried about. The number of Latin squares all of whose rows are in H is bounded by |H|^n = n!^n / [S_n : H]^n. On the other hand the total number of Latin squares is n!^n / e^(n^2 + o(n^2)) (a theorem of van Lint and Wilson I think?). Thus we need only worry about transitive subgroups H of index at most e^(n + o(n)), i.e., only H = A_n and H = S_{n/2} wr S_2. Admittedly I don’t immediately see how to deal with H = S_{n/2} wr S_2. Can the theorem of Häggkvist and Janssen you mention (which I know nothing about) be adapted somehow to different subgroups H?

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