The recognition of symmetric latin squares

✍ Scribed by Edwin C. Ihrig; Benjamin M. Ihrig

• Publisher: John Wiley and Sons
• Year: 2008
• Tongue: English
• Weight: 118 KB
• Volume: 16
• Category: Article
• ISSN: 1063-8539
• DOI: 10.1002/jcd.20151

✦ Synopsis

Abstract

A latin square S is isotopic to another latin square S′ if S′ can be obtained from S by permuting the row indices, the column indices and the symbols in S. Because the three permutations used above may all be different, a latin square which is isotopic to a symmetric latin square need not be symmetric. We call the problem of determining whether a latin square is isotopic to a symmetric latin square the symmetry recognition problem. It is the purpose of this article to give a solution to this problem. For this purpose we will introduce a cocycle corresponding to a latin square which transforms very simply under isotopy, and we show this cocycle contains all the information needed to determine whether a latin square is isotopic to a symmetric latin square. Our results relate to 1‐factorizations of the complete graph on n + 1 vertices, K~n + 1~. There is a well known construction which can be used to make an n × n latin square from a 1‐factorization on n + 1 vertices. The symmetric idempotent latin squares are exactly the latin squares that result from this construction. The idempotent recognition problem is simple for symmetric latin squares, so our results enable us to recognize exactly which latin squares arise from 1‐factorizations of K~n + 1~. As an example we show that the patterned starter 1‐factorization for the group G gives rise to a latin square which is in the main class of the Cayley latin square for G if and only if G is abelian. Hence, every non‐abelian group gives rise to two latin squares in different main classes. © 2007 Wiley Periodicals, Inc. J Combin Designs 16: 291–300, 2008