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It was shown by van Rees \cite{vR} that a latin square of order $n$ has at most $n^2(n-1)/18$ latin subsquares of order $3$. He conjectured that this bound is only achieved if $n$ is a power of $3$. We show that it can only be achieved if $n\equiv3\bmod6$. We also state several conditions that are equivalent to achieving the van Rees bound. One of these is that the Cayley table of a loop achieves the van Rees bound if and only if every loop isotope has exponent $3$. We call such loops \emph{v...
For a finite triangulation of the plane with faces properly coloured white and black, let A be the abelian group constructed by labelling the vertices with commuting indeterminates and adding relations which say that the labels around each white triangle add to the identity. We show that A has free rank exactly two. Let A* be the torsion subgroup of A, and B* the corresponding group for the black triangles. We show that A* and B* have the same order, and conjecture that they are isomorphic. F...
Suppose that all groups of order $n$ are defined on the same set $G$ of cardinality $n$, and let the \emph{distance} of two groups of order $n$ be the number of pairs $(a,b)\in G\times G$ where the two group operations differ. Given a group $G(\circ)$ of order $n$, we find all groups of order $n$, up to isomorphism, that are closest to $G(\circ)$.
An autotopism of a Latin square is a triple $(\alpha,\beta,\gamma)$ of permutations such that the Latin square is mapped to itself by permuting its rows by $\alpha$, columns by $\beta$, and symbols by $\gamma$. Let $\mathrm{Atp}(n)$ be the set of all autotopisms of Latin squares of order $n$. Whether a triple $(\alpha,\beta,\gamma)$ of permutations belongs to $\mathrm{Atp}(n)$ depends only on the cycle structures of $\alpha$, $\beta$ and $\gamma$. We establish a number of necessary conditions...
We derive necessary and sufficient conditions for there to exist 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 that: (a) 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 degr...
A multi-latin square of order $n$ and index $k$ is an $n\times n$ array of multisets, each of cardinality $k$, such that each symbol from a fixed set of size $n$ occurs $k$ times in each row and $k$ times in each column. A multi-latin square of index $k$ is also referred to as a $k$-latin square. A $1$-latin square is equivalent to a latin square, so a multi-latin square can be thought of as a generalization of a latin square. In this note we show that any partially filled-in $k$-latin square...
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