Almost all quasigroups have rank 2
The multiplication group \(\text{Mlt}(Q)\) of a quasigroup \(Q\) is the group of permutations of the underlying set of \(Q\) generated by the maps \(\lambda_ a: x\mapsto a\cdot x\) and \(\rho_ a: x\mapsto x\cdot a\). The question of whether \(\text{Mlt}(Q)\) is two-transitive is of importance in the study of the character theory of \(Q\). Here it is shown that almost all quasigroups have the property that the permutation group generated by merely the left mappings is either the symmetric group or the alternating group (almost all in the usual sense of the ratio of the number of quasigroups of size \(n\) with this property to the number of all quasigroups tending to 1 as \(n\) tends to \(\infty\)). The result is actually proved in terms of Latin squares; the multiplication table of a quasigroup is a Latin square and the group generated by the left mappings is just the group generated by the rows of the Latin square, regarded as permutations. The van der Waerden permanent conjecture implies that the number of Latin squares of order \(n\) is roughly \((n!)^ n/e^{n^ 2}\), whereas the number of Latin squares with row group distinct from \(S_ n\), \(A_ n\) is at most \(\sum| H|^ n\), where the summation is over the transitive subgroups \(H\) which are maximal in the sense that \(H < K < S_ n\) implies \(K = A_ n\). A straightforward argument considering the various possibilities for \(H\) establishes that this number is not too big. It is noteworthy that this result does not require the classification of the finite simple groups (though, of course, it does require another big gun, namely the proof of the van der Waerden permanent conjecture).
scientific article; zbMATH DE number 94515