A direct proof of the non-existence of a MOL(6)

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.


Introduction
Recall Euler's problem of the 36 officers [16]: Can 36 officers, drawn from 6 different ranks and also from 6 different regiments, be arranged in a 6 × 6grid so that in each row (both horizontal and vertical) there are 6 officers of different rank and different regiments? The first proof of the impossibility was by Tarry [19], so it became known as the Euler/Tarry problem. It is equivalent to the existence of a pair of orthogonal Latin Squares of order 6. Further proofs for the non-existence were given, also very short ones, [4,10,18]. A computer proof with the program Orbiter [2] is instantaneous but does not give insight.
Besides the proof of the non-existence we are also interested in the structure of the Latin Squares of order 6 and the structure of a (hypothetical) MOL (6), see [6][7][8]. For notions as Latin Squares, orthogonal Latin Squares, pairs of mutual orthogonal Latin Squares (MOL), we refer the reader to [11,[13][14][15]. We will view a Latin Square and a (hypothetical) MOL as an incidence structure called a linear space. Therefore, in Sect. 2, we recall the notion of linear spaces. Our approach uses partitions into subsquares. In [6][7][8] we considered partitions of type 6 = 3 + 3. Here, we look at partitions of type 6 = 2 + 2 + 2. For a fixed partition we denote by Q ij the subsquare arising at the intersection of the i th row class and the j th column class. In the following figure we show the two partition types: . .
Theorem 1 describes how the 6 elements of a digit are distributed to the 9 subsquares.
The non-existence of an orthogonal square in Case A is proved in Theorem 4, for Case B in Theorem 5, and C is ruled out in Theorems 7 and 8 together. We get the non-existence of a MOL(6) without relying on the classification of the Latin Squares.
In an Appendix we collect some properties of the Latin Squares. This is for illustration purposes only. It is not part of the proof.

Linear spaces
consists of a set V of points (or variables), and a collection B of subsets (called blocks) such that each pair of different points lies on exactly one block and each block has at least two points.
In the following example of a transversal the six elements are encircled. We give the transversal a number, say 1, and write this number at the right side of the corresponding digit: Each transversal number is combined with each digital number and therefore each of the 36 pairs (d, t), d, t ∈ {1, 2, ..., 6} occurs exactly once in the (6 × 6)- In the last row we had to put two question marks since a MOL(6) does not exist. The existence of a MOL(6) would give a solution of Euler's Problem: take the digits as ranks and the transversals as regiments. Example: A Latin Square LSQ(6) is a regular linear space: each point is incident with one 6-block and 6 3-blocks. Also a MOL(6) is regular: each point is incident with one 6-block and 6 4-blocks.
Regular spaces were studied in [3].

Parity considerations
Consider a Latin Square LSQ(6) with a decomposition of type (2,2,2). For each of the nine 2 × 2-squares, we call the two positions on the main diagonal positive, and the two positions on the opposite diagonal negative. So, altogether we get 18 positive and 18 negative positions: Proof. Each digit appears once in every row and also once in every column. Denote by Q i,j the 2 × 2-subsquare in row i and column j (i, j = 1, 2, 3) of the decomposition. Up to permutation of the blocks, there are 3 possibilities for how a digit, say 1, appears in the subsquares: We claim that it is possible to rearrange rows and columns of the LSQ within the decompositions in such a way that all digits lie in positive positions.
To see that we consider the possible types, one-by-one. In case {2, 2, 2} we interchange columns 1 and 2, columns 3 and 4, columns 5 and 6, if necessary. In the second case {2, 1, 1, 1, 1} we consider the four quadrangles Q 22 , Q 23 , Q 32 , Q 33 . If the element in Q 22 has a negative position, we may interchange the two rows or two columns to get

Assuming a pair of permutation type (2, 2, 2)
If there is a pair of rows (or of columns or of digits) with permutation type (2, 2, 2) then we may assume by isomorphism (Sect. 2) that the pair is in the Latin Square R ∪ C ∪ T , and the pair consists of two transversals of type (2, 2, 2). We then try to find an orthogonal mate R ∪ C ∪ D. We show that up to isomorphism we can choose 1 ∈ {f 1 , f 2 }: We take a reflection with respect the diagonal, then we interchange the 2 × 2-squares Q 22 and Q 33 and after this we interchange digits 3 and 5, and digits 4 and 6. Then we get 1 ∈ {f 1 , f 2 } and therefore (f 1 , We call a tripel of 2 × 2-squares tactical if they arise in the shape of a permutation matrix.

Proposition 4.
If for a tactical tripel of 2 × 2-squares two squares contain one element of the six elements of a digit, and the third square none of them, then the distribution is the cyclic distribution {1, 1, 1, 1, 1, 1}.
Proof. Up to permutation of rows or columns the tactical tripel has the form If one of the 6 empty squares contains at least two digits, then they must lie diagonally. This gives two digits 1 in a row or in a column, a contradiction. Therefore each square contains at most one digit and we get the type {1, 1, 1, 1, 1, 1}.
Proof. The partial structure is up to isomorphism 1 3 4 2 3 5 6 4 5 1 2 6 All 6 digits fulfill the condition of Proposition 4 and have type {1, 1, 1, 1, 1, 1}. Therefore we know for each digit the distribution to the 2 × 2-subsquares Q i,j : We now determine for each Q i,j the exact positions of the 4 digits it contains. We begin with the square Q 12 : Since in Q 22 the digits 5 and 6 are on a diagonal they have to be in different columns in Q 12 , and since the digits 1 and 2 in Q 11 are on a diagonal, they have to be in Q 12  We abbreviate these 8 cases by the notation Similarly one finds that in all nine 2 × 2-squares the two diagonals are made of two of the pairs {1, 2}, {3, 4} or {5, 6} (next figure): For each 2 × 2-square we call the main diagonal positive and the opposite diagonal negative. By Theorem 1, the number of positive positions of the 6 digits 1 is even, therefore the number of the positive diagonals {1, 2} is even as well. Similarly, the number of positive diagonals {3, 4} and {5, 6} is even and the sum of these three even numbers is also even. But there are 9 2 × 2squares which have 9 positive diagonals, and these 9 positive diagonals cannot be covered by an even number of positive diagonals, a contradiction.
This means that the Latin Square in R ∪ C ∪ T with pair (2, 2, 2), we started with has no orthogonal mate.

Definition 6.
We define the two Latin Squares LQS (6): The first square is the multiplication table of the group C 2 × C 3 (isomorphic to the cyclic group C 6 ) hence the name. Switching the digits in one of the subsquares of order two yields the Latin Square (C 2 × C 3 ) σ on the right. We first show that it extends to The notation 2 ↔ 5 means 2 5 or 5 2.
To prove this we use Proposition 4 and get the distributions of the 6 digits to the subsquares Q ij : We determine for each Q i,j the exact positions of the 4 digits it contains. We begin with the square Q 12 : Since in Q 11 there are the digits 2 and 5 in the second row, the digits 2 and 5 in Q 12 have to be in the first row, either as 2 5 or as 5 2, which we indicate by the notation 2 ↔ 5. Since on position 44 (i. e. row 4, column 4) there is a digit 4, the digit 4 in Q 12 has to be in column 3, and the positions of the four digits in the square Q 12 follow. Similarly we find the positions of digits in the other 2 × 2-squares and it also follows f 1 = 1, f 2 = 6, g 1 = 2 and g 2 = 4.
After a suitable permutation of rows we obtain Here we consider the 6 × 6-array as an array of 2 × 2-squares with parities By suitable inversions of columns and rows we may assume that the array becomes . We assume 4, 3, or 2 minus signs at the remaining positions. Then we take suitable inversions and get   Proof. From the assumption follows that that each pair of rows has either the cycle type (3,3) or (6). Take an arbitrary triple of rows. Then not all three pairs can have the type (6) since the permutation (6) is odd. It follows that there is at least one pair of rows having type (3,3).

Proposition 8. If a Latin
Square LSQ (6) has no subsquare of order two then it has a subsquare of order 3.
Proof. It suffices to prove that the triple b) in Proposition 7 has no extension to an LSQ(6) without a sub-LSQ (2). Then the first three columns of a) are a Latin Square of order 3.
We start with b): In the fourth row and column 1 we may take the digit 4. Then 43:5 or 43:6 since we avoid an LSQ(2) in the column pair 1,3. In both cases we get a unique fourth row: To prove this we find in the first case uniquely 42:1, then 45:3, then 46:2 and 44:6. In the second case there is 46:1, since digits 2 or 3 would give an LSQ(2). Row 4 is then detgermined uniquely.

Proposition 9.
If an LSQ (6)  In the squares right above and left down there are digits 4,5,6 and in the square right side down there are digits 1,2,3. Such a LSQ (6) has no transversal at all: A transversal has in one of the four squares at least two elements, say in the square left above. Then it has also two elements right side down, and therefore the transversal has four elements paired with digits from {1, 2, 3}, a contradiction. It follows that there exists no orthogonal mate.

Parallel classes on an 8-point set
In order to study triples of disjoint parallel classes we recall from [3]: A regular linear space (12|3, 8, 24) is defined on 12 points and has 3 disjoint 4-blocks (i. e., subsets with 4 points), 8 3-blocks and 24 2-blocks with the condition: each pair of points is on exactly one block. Proof. In the following figure a regular linear space is given by means of its incidence matrix (the 24 2-blocks deleted).
Taking the set of 3-blocks as 8-set, then we have on this 8-set three parallel classes (the rows of the incidence structure). If, conversely, a triple of parallel classes on an 8-point set is given, then we extend each 2-block to a 3-block (horizontally). We get 3 disjoint 4-blocks (vertically) and therefore a regular linear space on 12 points.
Each tripel has a row number r, a column number c and a digit number d. If we insert in an 4 × 4-array at row r and column c the digit d then we get the figure at the right hand side.This presents the triple as a substructure of an LSQ(6). . 6 4 The last two spaces have three pairs of type (4).

The remaining case
So far, we have treated all LSQ(6) which have a pair of cycle type (2, 2, 2). We have also shown that the Latin Square without a sub-LSQ(2) has no orthogonal mate. The remaining case is treated next.

Theorem 7.
Assume there is at least one sub-LSQ (2), but no pair with permutation type (2, 2, 2), then there is exactly one LSQ(6) with this property. Proof. We start with an LSQ(2), say in rows 1,2, columns 1,2 and digits 1 and 2. We have to put in the digits {3, 4, 5, 6} in row 1,2 and columns 3,4,5,6. In order to avoid a pair (2, 2, 2) we need there a 4-cycle. Similarly the digits {3, 4, 5, 6} in column 1,2 and rows {3, 4, 5, 6} have to form a 4-cycle. Then we have to insert the 4 digits 1 and 4 digits 2 into the square Q of rows 3 to 6 and columns 3 to 6. Also here we need a 4-cycle, say Then in Q there remain 8 empty positions which have to be filled with the digits 3, 3, 4, 4, 5, 5, 6, 6. All pairs rows/columns, rows/digits, columns/digits have the type (4) and by Theorem 6 there are up to isomorphism exactly two possibilities and we get for the subsquare Q the two cases We now try to extend these partial structures to a latin square of order 6. We fill in rows 1,2 and columns 3 to 6 (unique up to interchanging the two rows). Also columns 1,2 and rows 3 to 6 follow uniquely (up to interchanging column 1 and 2).  Next we show that this latin square has no orthogonal mate: the group is transitive on the set Z of 8 central (encircled) elements. No transversal is contained in the set of 28 outer elements (the elements in the complement of Z). Such a transversal would have at least two elements in, say, the square left above (rows 1 to 3, columns 1 to 3) and therefore also in the square right down. This would give a transversal with at least 4 elements from {1, 2, 3}, a contradiction. It follows that each transversal intersects the set Z in at least one element. Each automorphism maps transversals to transversals and pairs of disjoint transversals to pairs of disjoint transversals. Since the group is transitive on the set Z, all 6 transversals of an orthogonal square intersect Z in the same number a > 0. But a × 6 = 8, a contradiction.
Remark: This Latin Square has 8 transversals. One of them is given by (*) in the following figure, the other transversals follow by applying the automorphism group: The transversal (*) intersects the cube Z in 3 elements. Description: for the fixed vertex (2) in the figure take the three neighboring vertices 4 * , 2 * , 5 * on the cube. Since the cube has 8 vertices this gives 8 triples (which extend to 8 transversals of the Latin Square).

Appendix
In this appendix we give descriptions of some latin squares occuring before. In this computer printout of the 12 isomorphism types of latin squares LSQ(6) the order of the automorphism group is listed on top. Three of them have the same order 24. For these we also give the number l of subsquares LSQ(2), namely l = 15, l = 11, l = 4.
The two cases A1 and A2 in Theorems 2 and 3 can be described by the tactical decompositions: Proof. We choose the following substructure and apply an isomorphism.Then we describe it by its incidence matrix: Here we see a cube (3-blocks 1 to 8) which is extended by a parallel class (4 pairwise disjoint 3-blocks), denoted by 9 * , ..., 12 * . original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons. org/licenses/by/4.0/.
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