Overlapping Latin subsquares and full products. (English) Zbl 1224.05061

Summary: We derive necessary and sufficient conditions for the existence of a Latin square of order \(n\) containing two subsquares of order \(a\) and \(b\) that intersect in a subsquare of order \(c\). We also solve the case of two disjoint subsquares. We use these results to show the following.
A Latin square of order \(n\) cannot have more than \(\frac nm{n\choose h}/{m\choose h}\) subsquares of order \(m\), where \(h=\lceil (m+1)/2\rceil \). Indeed, the number of subsquares of order \(m\) is bounded by a polynomial of degree at most \(\sqrt {2m}+2\) in \(n\).
For all \(n\geq 5\), there exists a loop of order \(n\) in which every element can be obtained as a product of all \(n\) elements in some order and with some bracketing.


05B15 Orthogonal arrays, Latin squares, Room squares
20N05 Loops, quasigroups
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