()  0  1  2 

0  1  2  0 
1  0  1  2 
2  2  0  1 
The simple result which causes our interest in these algebraic forms is:
Theorem I.1.1.1  The multiplication table of a quasigroup is a Latin square.
Proof: Let a_{1} ,a_{2} ,...,a_{n} be the elements of the quasigroup and let its multiplication table be as below:
 a_{1} a_{2} ........ a_{s} ....... a_{n}  a_{1}  a_{11} a_{2}  .  a_{r}  a_{rs} .  .  a_{n}  a_{nn}Where the entry a_{rs} which occurs in the rth row and sth column is the product a_{r}a_{s} of the elements a_{r} and a_{s} . If the same entry occurred twice in the rth row, say in the sth and tth columns so that a_{rs} = a_{rt} = b say, we would have two solutions to the equation a_{r}x = b, in contradiction to the quasigroup axiom. Similarly, if the same entry occurred twice in the sth column, we would have two solutions to the equation ya_{s} = c for some c. We conclude that each element of the quasigroup occurs exactly once in each row and column, and so the unbordered multiplication table (which is an n×n array) is a latin square.
If the quasigroup is a loop and we let e = a_{1}, then the elements in the first row and first column of the table would be in the same (natural) order as the row and column headings and the Latin square would then be a reduced square.
It should be pointed out that the association between quasigroups and Latin squares is not unique. The elements that border the multiplication table of the quasigroup can be arranged in any order, so the same quasigroup can give rise to many Latin squares  but they are clearly isotopic. On the other hand, given a Latin square, one could border it in many ways, giving many different quasigroups. In order to straighten out the relationships, we make the following definition: Let (G,) and (H,*) be two quasigroups. An ordered triple (,, ) of bijections (oneone, onto maps) between the sets G and H is called an isotopism of (G,) upon (H,*) if
Theorem I.1.1.2  Two quasigroups are isotopic iff any multiplication table of one is isotopic (as Latin squares) to any multiplication table of the other.
Proof: (Left as a homework assignment).
Observe that of the permutations , , introduced in this definition, operates on the elements of the latin square, while and operate on the borders.
If = = then the isotopism is called an isomorphism. Notice that if (G,) and (H,*) are groups then a quasigroup isomorphism between them becomes a group isomorphism. Actually, a much stronger statement can be made in this situation, namely, if two groups (viewed as quasigroups) are isotopic then they are isomorphic. This follows immediately from the next result whose proof is omitted.
Theorem I.1.1.3  If a loop is isotopic to a group, then the loop is a group which is isomorphic to the given group.
Example: The underlined cells form a transversal in this 5x5 Latin square.
1  2  3  4  5 
2  1  4  5  3 
3  5  1  2  4 
4  3  5  1  2 
5  4  2  3  1 
The corresponding idea in quasigroups is called a Complete Mapping. By definition, a complete mapping of a quasigroup (G,) is a permutation of G such that the mapping x(x) where (x) = x(x) is also a permutation of G.
That these two ideas correspond is the substance of:
Theorem I.1.2.1  If Q is a quasigroup which possesses a complete mapping, then its multiplication table is a latin square with a transversal. Conversely if L is a latin square having a transversal, then at least one of the quasigroups which have L as multiplication table has a complete mapping.
Proof: Left as a homework exercise.
The notion of transversal was first introduced by Euler who called it a formule directrix. It has also gone under the names directrix, 1permutation and diagonal.
Theorem I.1.2.2  If L is a latin square of order n which is the multiplication table of a group and possesses at least one transversal, then L has a decomposition into n disjoint transversals.
Proof: Let G be the group whose multiplication table is L, and let the given transversal in L be formed by the symbol c_{1} in the first row, c_{2} in the second row,..., c_{n} in the nth row. It will follow from the group axioms that another transversal can be obtained by taking c_{1} g from the first row, c_{2} g from the second row, ..., c_{n} g from the nth row, where g is any fixed element of G. As g varies through the n elements of the group, we shall thus obtain n disjoint transversals.
To show that this procedure gives transversals we need only show that the c_{i} g's are all in different columns. To see this, suppose that c_{i} = g_{i} g_{M(i)} that is, c_{i} is in the ith row and M(i)th column. Since, the c_{i} 's form a transversal, M is a permutation  M(i)M(j) if i j. Then because G is a group,
The converse of this theorem is not true. The Latin square below can be decomposed into 10 disjoint transversals but it is not the multiplication table of any group.
0  4  1  7  2  9  8  3  6  5 
8  1  5  2  7  3  9  4  0  6 
9  8  2  6  3  7  4  5  1  0 
5  9  8  3  0  4  7  6  2  1 
7  6  9  8  4  1  5  0  3  2 
6  7  0  9  8  5  2  1  4  3 
3  0  7  1  9  8  6  2  5  4 
1  2  3  4  5  6  0  7  8  9 
2  3  4  5  6  0  1  8  9  7 
4  5  6  0  1  2  3  9  7  8 
(Homework: Prove those statements about this Latin square)
For groups then, we can get this decomposition into transversals if its multiplication table has one transversal, or equivalently if the group has a complete mapping. What groups then have complete mappings? This question has not been fully answered, but there are some important results.
Theorem I.1.2.3  If G is a group of odd order, then G has a complete mapping.
Proof: If G is a group of odd order, then every element of G has a unique square root in G. To prove this, let gG. Since G has odd order, and the order of g must divide it, the order of g is odd, say 2i  1. Then, the element h = g^{i} satisfies h^{2} = g^{2i} = g and so h is a square root of g. Now, suppose k G also satisfies k^{2} = g, we would then have k^{4i2} = e the identity of G. Since k necessarily has odd order, k^{2i1} = e. Then k = k^{2i} = g^{i} = h, so h is the unique square root of g.
It follows that, in a group G of odd order, g_{i}^{2} = g_{j}^{2} only if i = j. Consequently, the mapping (g_{i} ) = g_{i}^{2} ( i = 1,2,...,n) is a permutation of G. Thus, the identity mapping (g) = g gG satisfies the definition of a complete mapping of G.
It follows from the proof, that in the multiplication table of a group of odd order, the main left to right diagonal is a transversal.
(Homework: Show that in the above theorem, group can be replaced by commutative quasigroup.)
As for groups of even order, the situation is much more difficult. We will state the main results, but their proofs require a substantial amount of group theory and so we omit them.
Theorem I.1.2.4  (Hall & Paige) A finite group of order n which has a cyclic Sylow 2subgroup does not possess a complete mapping.
Corollary I.1.2.5  If G is an arbitrary group of order n = 4k + 2, then G has no complete mapping. In particular, the symmetric group on three letters has no complete mapping.
Theorem I.1.2.6  (Hall & Paige) A finite solvable group G whose Sylow 2subgroups are noncyclic has a complete mapping.
It has been conjectured that the above theorem is true without the solvable restriction, if this were true then question of which groups had complete mappings would be completely answered.
Let L be a Latin square of order n (with entries 1,2,...,n) having a transversal. To form a Latin square L' of order n+1 we need to introduce a new row and column and a new element (n+1). This can be done in the following way: For each cell in the transversal, the element in the cell is placed in the new column and same row as the cell, and also placed in the new row and same column as the cell. The element n+1 is placed in the cell itself. Finally, the new element n+1 is placed in the intersection of the new row and new column. It is easy to see that the properties of the transversal will insure that the new row and column have no repeated elements and do not destroy the Latiness of the square.
Consider the following example of this procedure. Starting with an order 3 Latin square with underlined transversal, we prolong it to an order 4 Latin square.


If L has another disjoint transversal, then after prolongation this transversal together with the n+1 element in the new row and column would form a transversal of the larger square L' and so prolongation could be repeated.
It should be noted that prolongations of the same Latin square using different transversals generally do not produce isotopic Latin squares. This is definitely the case if the Latin square is the multiplication table of a group with at least one element of order greater than two.
(Homework: The inverse operation (i.e., shrinking an n+1 square to an n square) is called contraction. Carefully describe the property that a Latin square must have in order to carry out a contraction on it.)
Some of the research papers concerning Complete Mappings and Transversals are:
Hall,M. and Paige,L.J., "Complete mappings of finite groups", Pacific Journal of Mathematics, 5(1955), pp. 541 549.
Mann,H.B., "The construction of orthogonal Latin squares", Annals of Mathematical Statistics, 13(1942), pp. 418423.
Paige,L.J., "Complete mappings of finite groups", Pacific Journal of Mathematics, 1(1951), pp. 111116.
And some papers dealing with prolongation and quasigroups are:
Bruck, R.H.,"Some results in the theory of quasigroups", Transactions of the American Mathematical Society, 55(1944), pp. 1952.
Denes,J. and Pasztor, E.,"A kvázicsoportok néhány problémájárol.", Magyar Tud. Akad. Mat.
Fiz. Oszt.Kozl. 13(1963), pp. 109118.
 This one could be a little tough going, it's in Hungarian.
An excellent survey article on Loops is:
Bruck, R.H., "What is a loop?", in Studies in Modern Algebra, ed. A.A.Albert, Mathematical Association of America, 1963. (this is vol. 2 in the MAA Studies in Mathematics series of books)
An expository article on quasigroups is:
Frink,O., "Symmetric and selfdistributive systems", American Mathematical Monthly, 62(1955), pp. 697707.