Counting transversals in Latin squares over groups of even order
Abstract
The number of transversals in a Latin square L(G) whose rows and columns are indexed by a finite group G has been determined computationally for all G of order at most $23$ and for C25. However, computation of the number of transversals of L(G) has until now been considered intractable for all non-cyclic groups G of order $32$. We show, for a group G containing an order $2$ normal subgroup N, that the image of a transversal in L(G) under the quotient mapping from G to G/N is a $2$-plex in L(G/N). We then derive linear algebraic conditions that determine the number of transversals in L(G) whose image under the quotient mapping is a given $2$-plex in L(G/N). This allows us to count transversals in L(G) by examining all $2$-plexes in L(G/N). Although we can simplify the counting of $2$-plexes in L(G/N) by taking advantage of their symmetry group, the linear algebraic conditions can vary over a symmetry class of $2$-plexes. However, in the case that G takes the splitting form C2 × H, the linear algebraic conditions become invariant over each symmetry class and so the counting of transversals in L(C2 × H) reduces to counting the $2$-plexes in L(H) under the action of their symmetry group. We illustrate the power of these techniques by computing the exact number of transversals in L(C25).
Related
- Latin Squares with Few Transversals
- Counting Near-Spanning Matchings in Latin Squares and Steiner Triple Systems
- Symmetric Quantum Latin Squares with Maximal Cardinality
- Orthogonal Quantum Latin Squares with Maximal Cardinality from Full-Rank Factorizations
- Resolvable quantum Latin squares with maximal cardinality
BibTeX
@misc{davis2026counting,
title = {{Counting transversals in Latin squares over groups of even order}},
author = {James A. Davis and Jonathan Jedwab and Ian M. Wanless and Amy Wiebe},
howpublished = {arXiv},
year = {2026},
month = oct,
doi = {10.48550/arxiv.2610.35611},
eprint = {2610.35611},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2610.35611},
note = {CC BY 4.0},
}
RIS
TY - UNPB
TI - Counting transversals in Latin squares over groups of even order
AU - James A. Davis
AU - Jonathan Jedwab
AU - Ian M. Wanless
AU - Amy Wiebe
PY - 2026
DA - 2026/10/26
PB - arXiv
DO - 10.48550/arxiv.2610.35611
UR - https://arxiv.org/abs/2610.35611
AB - The number of transversals in a Latin square L(G) whose rows and columns are indexed by a finite group G has been determined computationally for all G of order at most $23$ and for C_25. However, computation of the number of transversals of L(G) has until now been considered intractable for all non-cyclic groups G of order $32$. We show, for a group G containing an order $2$ normal subgroup N, that the image of a transversal in L(G) under the quotient mapping from G to G/N is a $2$-plex in L(G/N). We then derive linear algebraic conditions that determine the number of transversals in L(G) whose image under the quotient mapping is a given $2$-plex in L(G/N). This allows us to count transversals in L(G) by examining all $2$-plexes in L(G/N). Although we can simplify the counting of $2$-plexes in L(G/N) by taking advantage of their symmetry group, the linear algebraic conditions can vary over a symmetry class of $2$-plexes. However, in the case that G takes the splitting form C_2 × H, the linear algebraic conditions become invariant over each symmetry class and so the counting of transversals in L(C_2 × H) reduces to counting the $2$-plexes in L(H) under the action of their symmetry group. We illustrate the power of these techniques by computing the exact number of transversals in L(C_2^5).
ER -