Abstract

It is well known that a Latin Square of order 6 has no orthogonal mate. Many proofs are known, some of which are very short, see for instance (Betten in Unterricht 36:449–453, 1983; Beth et al. in Design theory, Bibliographisches Institut Mannheim Wien, Zürich, 1985; Tarry in Comptes Rendus Ass Franc Sci Nat 1900(2), 170–203, 1901; Stinson in J Comb Theory A 36:373–376, 1984). This paper provides a short proof of this fact which avoids a case distinction on the isomorphism types of the Latin Square. We observe that any Latin Square of order 6 falls within exactly one of three categories. Either it has two rows which form a permutation whose cycle type is three transpositions. Or it has no subsquare of order 2, or it is a single Latin Square with symmetry group the rotation group of the cube of order 24. In each case, the nonexistence of an orthogonal mate can be seen quickly.

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