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Latin squares with maximal partial transversals of many l...
Anthony B. Evans, Adam Mammoliti, Ian Wanless · 2019-11-14 · via math.CO updates on arXiv.org

A partial transversal $T$ of a Latin square $L$ is a set of entries of $L$ in which each row, column and symbol is represented at most once. A partial transversal is maximal if it is not contained in a larger partial transversal. Any maximal partial transversal of a Latin square of order $n$ has size at least $\lceil\frac{n}{2}\rceil$ and at most $n$. We say that a Latin square is omniversal if it possesses a maximal partial transversal of all feasible sizes and is near-omniversal if it possesses a maximal partial transversal of all feasible sizes except one. Evans showed that omniversal Latin squares of order $n$ exist for any odd $n \neq 3$. By extending this result, we show that an omniversal Latin square of order $n$ exists if and only if $n\notin\{3,4\}$ and $n \not\equiv 2 \mod 4$. Furthermore, we show that near-omniversal Latin squares exist for all orders $n \equiv 2 \mod 4$. Finally, we show that no non-trivial group has an omniversal Cayley table, and only 15 groups have a near-omniversal Cayley table. In fact, as $n$ grows, Cayley tables of groups of order $n$ miss a constant fraction of the feasible sizes of maximal partial transversals. In the course of proving this, we are led to consider the following interesting problem in combinatorial group theory. Suppose that we have two subsets $R,C\subseteq G$ of a finite group $G$ such that $|\{rc:r\in R,c\in C\}|=m$. How large do $|R|$ and $|C|$ need to be (in terms of $m$) to be certain that $R\subseteq xH$ and $C\subseteq Hy$ for some subgroup $H$ of order $m$ in $G$, and $x,y\in G$.