How can we prove, using Cayley tables, that for any order $n$ there is some latin square with no orthogonal pair? What would this mean for the Cayley table regarding transversals? I am not entirely sure how Cayley tables work and hence am confused.
Cayley tables- for any order $n$ there is some latin square with no orthogonal pair
1 Answers
A Latin square can be regarded as the Cayley table of a quasigroup. Whether you talk about "quasigroups" or "Latin squares", is largely based on personal preference, and what tools you're planning on using.
A transversal is a selection on n cells in a Latin square that contains one representative from each row, column and symbol type. In a pair of orthogonal Latin squares, the copies of any given symbol in one Latin square must correspond to a transversal in the other.
Latin squares without orthogonal mates are known as bachelor Latin squares, which exist for all orders except 1 and 3. A slick proof was given by Judith Egan in:
J. Egan, Bachelor Latin squares with large indivisible plexes. J. Combin. Des. 19 (2011), no. 4, 304–312.
This is a non-trivial research problem, and prior work had been case-by-case. Egan's proof was described as a "proof from the book" by her supervisor at the time:
Wanless, Ian M. Transversals in latin squares: a survey. Surveys in combinatorics 2011, 403–437, London Math. Soc. Lecture Note Ser., 392, Cambridge Univ. Press, Cambridge, 2011.