The Chromatic Number of Finite Group Cayley Tables

The chromatic number of a latin square $L$, denoted $\chi(L)$, is the minimum number of partial transversals needed to cover all of its cells. It has been conjectured that every latin square satisfies $\chi(L) \leq |L|+2$. If true, this would resolve a longstanding conjecture---commonly attributed to Brualdi---that every latin square has a partial transversal of size $|L|-1$. Restricting our attention to Cayley tables of finite groups, we prove two main results. First, we resolve the chromatic number question for Cayley tables of finite Abelian groups: the Cayley table of an Abelian group $G$ has chromatic number $|G|$ or $|G|+2$, with the latter case occurring if and only if $G$ has nontrivial cyclic Sylow 2-subgroups. Second, we give an upper bound for the chromatic number of Cayley tables of arbitrary finite groups. For $|G|\geq 3$, this improves the best-known general upper bound from $2|G|$ to $\frac{3}{2}|G|$, while yielding an even stronger result in infinitely many cases.


Introduction and preliminaries
Let n be a positive integer, let [n] := {0, 1, 2, . . . , n − 1}, and let L be a latin square of order n, which we define as an n × n array in which each row and each column is a permutation of some set of n symbols indexed by [n]. Let L be a latin square of order n. We define a partial transversal of L as a collection of cells which intersects each row, each column, and each symbol class at most once. A transversal of L is a partial transversal of size n, and a near transversal is a partial transversal of size n − 1. It is well known that L possesses an orthogonal mate if and only if it can be partitioned into transversals. But when L does not have an orthogonal mate, can we still efficiently partition its cells into partial transversals?  This question can be restated in terms of graph coloring. Associated with every latin square L is a strongly regular graph Γ(L) defined on vertex set {(r, c, L r,c ) : r, c ∈ [n]} with (r 1 , c 1 , s 1 ) ∼ (r 2 , c 2 , s 2 ) if and only if exactly one of r 1 = r 2 , c 1 = c 2 , or s 1 = s 2 holds (see Figure 1). It is straightforward to check that partial transversals of L correspond to independent sets in Γ(L). Thus, the graph chromatic number χ(Γ(L)) is the minimum number of partial transversals needed to cover all of the cells in L.
To see that Conjecture 1.2 implies Conjecture 1.3.1, suppose there exists a latin square L in which every partial transversal has size at most n − 2. Any set of n + 2 partial transversals in L could cover at most (n + 2)(n − 2) = n 2 − 4 of L's n 2 cells. Thus, L has no proper (n + 2)-coloring. A similar argument shows that Conjecture 1.2 implies Conjecture 1. 3

.2.
There is a growing body of evidence for Conjecture 1.2. As noted in [1], the conjecture has been verified for n ≤ 8. It is also known to hold in an asymptotic sense. Using the canonical representation of latin squares as 3-uniform hypergraphs, the following is an immediate corollary of a powerful theorem due to Pippenger and Spencer [10]. Furthermore, there are several families for which Conjecture 1.2 is known to hold. Cavenagh and Kuhl [3] showed that Conjecture 1.2 holds for circulant latin squares (i.e. Cayley tables of cyclic groups) when n ≡ 6 (mod 12). This result was confirmed and extended in [1], where Conjecture 1.2 was established for circulant latin squares of every order.
The present paper continues the work towards resolving Conjecture 1.2 in the special case where L is the Cayley table of a finite group. Observe that the chromatic number of Γ := Γ(L) is not affected by relabelling the rows, columns, or symbol classes of L, nor is it affected by applying a fixed permutation to each of the triples (r, c, s) ∈ V (Γ). Thus, χ(L) is a main class invariant, and it makes sense in this context to speak of the Cayley table of a group G, which we denote by L(G). In a slight abuse of notation, we write χ(G) for the chromatic number of L(G) and Γ(G) for the latin square graph Γ(L(G)).
Given a group G, let Syl 2 (G) denote the isomorphism class of its Sylow 2-subgroups. The groups for which χ(G) = n were recently characterized by Bray, Evans, and Wilcox [5,16], resolving a 50 year old conjecture due to Hall and Paige [7]. Theorem 1.5. Let G be a group of order n. Then the following are equivalent: 4. Syl 2 (G) is either trivial or non-cyclic.
In light of this, verifying Conjecture 1.2 amounts to showing that every group with nontrivial, cyclic Sylow 2-subgroups has an (n + 2)-coloring. In Section 2 we give such a construction under the additional assumption that G is Abelian, yielding our first main result. Theorem 1.6. Let G be an Abelian group of order n. Then In Section 3 we turn to the case of general (i.e. not necessarily Abelian) groups. We begin by showing that χ(G) is submultiplicative, thereby generalizing a classical result due to Hall and Paige. This allows us to establish an upper bound for χ(G) which depends only upon the largest power of 2 dividing |G|: every group G of order n = 2 l m ≥ 3 satisfies Previously, the best known general upper bound for χ(G) was due to Wanless; it follows directly from his work in [14] that every finite group satisfies χ(G) ≤ 2n. This bound is improved upon by (2) except when n ≡ 2 (mod 4). Dealing directly with this final case, we obtain our second main result.
Theorem 1.7. Let G be a group of order n ≥ 3. Then The condition n ≥ 3 is necessary because Γ(Z 2 ) ∼ = K 4 . It is worth noting that Theorem 1.7 is still somewhat far from the best possible. Indeed, if Conjecture 1.2 is true, then the bound (1) holds for every finite group.

The chromatic number of Abelian groups
Let G = Z n1 × Z n2 × · · · × Z n k be a finite Abelian group of order n. We say that G = {g 0 , g 1 , . . . , g n−1 } is ordered lexicographically if g i = (i 1 , i 2 , . . . , i k ) precedes g j = (j 1 , j 2 , . . . , j k ) (i.e. i < j) if and only if there is some l ∈ {1, 2, . . . , k} for which i l < j l and i m = j m for every positive integer m < l. In the statement of the following technical lemma, indices are expressed modulo n.
. . , g n−1 } be a lexicographically ordered Abelian group of odd order n. If gcd(s + 1, n) = 1 for some positive integer s, then the map φ : G → G given by Proof. Suppose G = Z n is cyclic and consider g, h ∈ G such that φ(g) = φ(h). We may treat g and h as integers in the set [n], in which case the group operation is simply addition modulo n, and We then have (s + 1)g ≡ (s + 1)h (mod n). But gcd(s + 1, n) = 1 tells us that s + 1 is a generator of G = Z n , and therefore g ≡ h (mod n), as desired. Now, we may assume G = Z n1 × H, where H = Z n2 × · · · × Z n k is a nontrivial Abelian group of odd order m := n/n 1 . If H = {h 0 , h 1 , . . . , h m−1 } is ordered lexicographically, then for every i ∈ [n] and, defining the map ψ : Consider i, j ∈ [n] such that φ(g i ) = φ(g j ). In this case we have ψ h i (mod m) = ψ h j (mod m) . It then follows by induction on |G| that i ≡ j (mod m). Thus, there is some r ∈ [n 1 ] such that j = i + rm, and We then have (s + 1)r ≡ 0 (mod n 1 ). But r ∈ [n 1 ] and gcd(s + 1, n 1 ) = 1, forcing us to conclude that r = 0, in which case i = j.
The Möbius ladder of order 2n, denoted M n , is the cubic graph formed from a cycle of length 2n by adding n edges, one between each pair of vertices at distance n in the initial cycle. We refer to this initial cycle as the rim of M n , and refer to the edges between opposite vertices in the rim as rungs. A pair {u, v} ⊆ V (M n ) is called near-antipodal if the shortest path from u to v along the rim of M n has length n − 1 (see Figure 2). There is a strong sense in which Möbius ladders are "nearly" bipartite.  Let G = {g 0 , g 1 , . . . , g n−1 } be a group satisfying χ(G) > n. We construct an (n + 2)-coloring of Γ = Γ(G) in two steps. First, we show that Γ can be partitioned into n 2 induced copies of M n . Second, we find a bipartite induced subgraph Λ ⊆ Γ which contains a near-antipodal pair from each copy of M n . By Proposition 2.2, we then have a partition of Γ into n 2 + 1 bipartite induced subgraphs. Using a disjoint pair of colors for each of these subgraphs, we obtain an (n + 2)-coloring of Γ.
Before formally presenting our construction, we introduce some notation which will be utilized both here and in Section 3. Letting L = L(G) and fixing an integer d ∈ [n], we define the dth right diagonal of L as the set where indices are expressed modulo n. When it is clear which latin square we are discussing, we drop the superscript and simply write T d . We also define the maps R, C : L → [n] and S : L → G by These maps send a cell of L to its row index, its column index, and its symbol, respectively. We then extend these functions to sets of cells. For A ⊆ L, let R(A) = {R(a) : a ∈ A} be the multiset containing the row-index of every cell in A (counted with multiplicity), and define C(A) and S(A) similarly.

Theorem 2.3. Let G be an Abelian group of order n. If Syl 2 (G) is cyclic and nontrivial then
Proof. Because Syl 2 (G) is nontrivial, n is even and the constant q := n/2 is well-defined. We may assume n ≥ 4, as Γ(Z 2 ) ∼ = K 4 has chromatic number 4 = 2 + 2. Moreover, letting t := |Syl 2 (G)|, there is some integer l ≥ 1 such that t = 2 l . By the fundamental theorem of finite Abelian We then impose an ordering on G by setting Arrange the rows and columns of L = L(G) according to this ordering and, for every i ∈ [q], define the set Letting Γ := Γ(G), recall that V = V (Γ) corresponds to the set of cells in L, and that two cells are adjacent if they lie in the same row, they lie in the same column, or they contain the same symbol. By definition, two cells in a latin square can satisfy at most one of these conditions. We therefore have a natural partition of E = E(Γ) into the sets E R , E C , and E S , corresponding to "row-edges," "column-edges," and "symbol-edges," respectively.

Claim. The induced subgraph
Expressing indices modulo n, let A j := L j,j+2i and B j := L j,j+2i+1 for every j ∈ [n]. It is not hard to see that the only row-edges in Γ i are {A j B j : j ∈ [n]} (see Figure 3). Similarly, the only column-edges in corresponds to a Hamilton cycle in Γ i that uses all of the edges in E( To prove the claim, it is left to show that E(Γ i ) ∩ E S contains exactly n edges, each of which connects opposite vertices (i.e. vertices at distance q) in the cycle given by (5).
Given an integer z, let z be the corresponding residue modulo t. It follows from (4) that, for every j ∈ [n], Because t is even, for every j, k ∈ [n] the first coordinates of S(A j ) and S(B k ) have different values modulo 2. Thus, E S contains no edges of the form A j B k . Fix j ∈ [n] and suppose there is some nonzero x ∈ [n] such that S(A j ) = S(A j+x ). It follows from (6) that 2j + 2i ≡ 2j + 2x + 2i (mod t). Recalling that t = 2 l , we may conclude that x is divisible by 2 l−1 . On the other hand, (6) Because m is odd, the only nonzero integer in [n] which is divisible by both m and 2 l−1 is q = m2 l−1 .
Observing that S(A j ) = S(A j+q ) for every j ∈ [n], we see that each A j is incident to exactly one edge in E S ∩ E(Γ i ): the edge connecting it to the opposite vertex in the cycle given by (5). A similar argument shows that, for every j ∈ [n], the only edge in E S ∩ E(Γ i ) incident to B j is B j B j+q . This establishes the claim.
We complete the proof (of Theorem 2.3) by finding a pair of independent sets X, Y ⊆ V such that . Given such X and Y , we properly (n + 2)-color Γ using a distinct pair of colors for each of the Keeping in mind that indices are here considered modulo n, for each i ∈ [k] let Similarly, for every j ∈ [q − k], define Recall that the edges on the rim of Γ i are exactly E(Γ i ) ∩ (E R ∪ E C ). It follows from the definition of D i that the shortest path from x i to x i along the rim of Γ i has length n − 1. Similarly, the shortest path from y j to y j along the rim of Γ k+j has length n − 1. Thus, x i , x i and y j , y j are near-antipodal pairs for every i, j ∈ [k]. Proposition 2.2 then implies that Γ i is bipartite for every i ∈ [q].
It remains to show that X and Y are independent sets in Γ. We begin by showing that there are no row-edges and no column-edges between cells in X. Recalling the definitions in (3) and (8), we see that the multiset of row-indices of cells in X is But, having assumed n ≥ 4, we have k − 1 < q 0 and q 0 + k − 1 < n. It follows that R(X) is simple-it contains no repeated entries. Now, define Again looking to (8), we see that Because 3(k − 1) < n, both C( X) and C(X ) are simple sets. Thus, C(X) = C( X) ∪ C(X ) is simple unless C( X) ∩ C(X ) = ∅. Suppose there were some x ∈ C( X) ∩ C(X ). As x ∈ C( X), there is an i 0 ∈ [k] such that x = 3i 0 , which implies x ≡ 0 (mod 3). On the other hand, x ∈ C(X ) means x = q 1 + 3i 1 (mod n) for some i 1 ∈ [k]. We claim that this implies x ≡ 0 (mod 3), yielding a contradiction.
To prove this claim, first suppose that n is a multiple of 3. In this case q is also divisible by 3, and (7) tells us that q 1 = q + 1 ≡ 1 (mod 3). But then x ≡ q 1 + 3i 1 ≡ 1 (mod 3). So, we may assume n is not divisible by 3. In this case (7) tells us that q 1 = q ≡ 0 (mod 3). When q + 3i 1 < n this implies x = q + 3i 1 ≡ 0 (mod 3), while when q + 3i 1 ≥ n we have Now, observe that X and Y have the same "shape" in L in the sense that R(Y ) ⊆ R(X) and C(Y ) ⊆ {c + 2k : c ∈ C(X)}. Thus, having shown that R(X) and C(X) are simple, we may conclude that R(Y ) and C(Y ) are also simple. In other words, there are no row-edges or column-edges between cells in Y .
We next show that S(X) is a simple set. From here to the end of the proof indices are expressed modulo m. Recalling (7), observe that q 0 + q 1 ∈ {n − 1, n + 1}. Combinined with (4), (8), and the fact that t divides n, this implies the existence of some w ∈ {−1, 1} such that, for every i ∈ [k], Considering the parity of entires in the first coordinate, we immediately see that S( To see that S( X) is simple, consider x i , x j ∈ X such that S(x i ) = S(x j ). We then have h i +h 3i = h j +h 3j , and applying Lemma 2.1 with c = d = 0 and s = 3 tells us that i ≡ j (mod m). We also have 4i ≡ 4j (mod t). (10) Suppose t ≤ 4. In this case (10) is trivially satisfied. However, because 0 As distinct numbers are congruent modulo m only if their difference is at least m, we may conclude that i = j. So, recalling that t = 2 l for some integer l ≥ 1, we may assume t ≥ 8. It then follows from (10) that i − j ≡ 0 (mod 2 l−2 ). Because m is odd, gcd m, 2 l−2 = 1 and the Chinese Remainder Theorem tells us that x = 0 is the unique x ∈ [2 l−2 m] satisfying x ≡ 0 (mod 2 l−2 ) and x ≡ 0 (mod m). But we have just shown that |i − j| ≡ 0 (mod m) and |i − j| ≡ 0 (mod 2 l−1 ). Thus, as 0 ≤ |i − j| < k = 2 l−2 m, we have i = j. A similar argument shows that S(X ) is simple. Indeed, when S(x i ) = S(x j ), applying Lemma 2.1 with c = q 0 , d = q 1 , and s = 3 yields i ≡ j (mod m), while 4i + w ≡ 4j + w (mod t) implies 4i ≡ 4j (mod t). From here we may proceed exactly as above.
The proof that S(Y ) is simple is nearly identical. By (4) and (9), there is some w ∈ {−1, 1} such that S(y i ) = (4i + 2k, h i + h 3i+2k ) and S(y i ) = (4i + 2k + w, h i+q0 + h 3i+q1+2k ) for every i ∈ [q−k]. Considering the parity of entries in the first coordinate, we see S( Y )∩S(Y ) = ∅. We then check that S( Y ) and S(Y ) are both simple by applying Lemma 2.1 and noting that 4i + z ≡ 4j + z (mod t) if and only if 4i ≡ 4j (mod t) for every z ∈ Z.
Following directly from this esult is Theorem 1.6, which states that every finite Abelian group G of order n satisfies is either trivial or non-cyclic, n + 2 otherwise.

A general upper bound
Consider a finite group G and a normal subgroup H G. In [7], Hall and Paige showed that a sufficient condition for the existence of a transversal in L(G) is that both L(H) and L(G/H) possess transversals. This turns out to be a special case of a more general result concerning colorings of finite group Cayley tables.
Our proof of this fact relies upon a modification the mappings R, C, and S-introduced just before Theorem 2.3-which map sets of cells in a latin square to multisets of rows indices, column indices, and symbols, respectively. Given a multiset X, let Supp(X) be the underlying simple set. For every set of cells X ⊆ L(G), we set R (X) := Supp(R(X)), C (X) := Supp(C(X)), and S (X) := Supp(S(X)).

Lemma 3.1. Let G be a finite group and let H G be a normal subgroup. Then
Proof. Letting n := |G| and m := |H|, set k := n m . We begin by constructing a block representation Fixing arbitrary i, j ∈ [k], we define A ij as the unique m × m subsquare of L(G) satisfying To establish χ(A ij ) = χ(H), we provide a graph isomorphism between Γ(A ij ) and Γ(H). Indeed, for every v ab = (a, b, h a h ). It then follows from the definition of a group that the triples v ab = (a, b, h a h b ) and v cd = (r, s, h r h s ) match in exactly one coordinate if and only if the corresponding triples φ(v ab ) and φ(v rs ) match in exactly one coordinate.
Let K be the k × k array formed from (11)  To see that f is indeed a proper coloring, consider c, c ∈ L such that f (c) = f (c ). Because f ∞ is a proper coloring, c and c cannot lie in adjacent blocks of V (Γ(K)). They could lie in the same block, say A ij , but because f ij is also a proper coloring, it is nonetheless impossible for c and c to be adjacent in Γ(L).
Recall from Theorem 1.5 that, in determining an upper bound for the chromatic number of all finite groups, we need only consider groups whose Sylow 2-subgroups are nontrivial and cyclic. The following structural theorem for such groups was observed in [7] as a direct corollary of a classical result due to Burnside ([8] Theorem 14.3.1).    Proof. We may assume t ≥ 2 and Syl 2 (G) = Z t , as otherwise Theorem 1.5 implies χ(G) = n ≤ n + 2n t . Lemma 3.2 then tells us that G has a normal subgroup H of order m satisfying G/H ∼ = Z t . With χ(H) and χ(Z t ) determined by Theorem 1.5 and Theorem 2.3, respectively, it follows from Lemma 3.1 that Recall that the previously best known general upper bound was χ(G) ≤ 2n. Theorem 3.3 improves significantly upon this bound for general groups whose order is divisible by large powers of 2. Indeed, as t grows with respect to m, Theorem 3.3 approaches the conjectured best possible bound of χ(G) ≤ n + 2.
Given a (full) transversal T in a latin square L of order n, there is a unique bijection φ : [n] → [n], known as the index map of T , which sends the row index x column index of the unique cell in T with row index x, so that If L is the Cayley table of a group G = {g 0 , g 1 , . . . , g n−1 }, then φ ∈ S n is the index map of some transversal if and only if g i g φ(i) is an enumeration of G (or, in the language of [5], g i → g φ(i) is a complete mapping). We end by proving our second main result, Theorem 1.7, which states that every finite group G of order n satisfies Proof of Theorem 1.7. Let P be a Sylow 2-subgroup of G. We may assume P = p ∼ = Z 2 ; indeed, if P is trivial then Theorem 1.5 implies χ(G) = n ≤ 3 2 n, while if |P | ≥ 4 then it follows from Theorem 3.3 that Letting m := n 2 , Lemma 3.2 tells us that G has a normal subgroup H of order m satisfying G/H ∼ = P . As in the proof of Breaking L into four m × m subsquares, we obtain the block representation and note that T i ⊆ A 1 . To see that T i is a transversal of A 1 , note that φ i is a bijection and ph j h φi(j) is an enumeration of pH. It is then easy to check that {T i : i ∈ [m]} is an m-coloring of A 1 .

Observe that each
To find m-colorings for A 2 and A 3 , we use the fact that H is normal in G to define a permutation π ∈ S m for which h j p = ph π(j) for every j ∈ Observing that S(L j,ψi(j) ) = h j ph φi(π(j)) = ph π(j) h φi(π(j)) and S(L m+j,ψi(j) ) = h π(j) h φi(π(j)) , it is easy to check that Because X i is the union of four subsquare transversals, it contains exactly two cells from each row, column, and symbol class of L. Thus, the induced subgraph Γ i := Γ[X i ] is cubic. Noticing that {X i : i ∈ [m]} partitions L, if we can show that χ(Γ i ) ≤ 3 for every i ∈ [m], then we may conclude that χ(L) ≤ 3 2 n. Fixing an arbitrary i ∈ [m], Brooks' Theorem tells us that Γ i is 3-colorable unless it contains a connected component isomorphic to the complete graph K 4 .
Suppose we could find a connected component Λ ⊆ Γ i which is isomorphic to K 4 . As V (Γ i ) is the union of four independent sets-one corresponding to each subsquare A j -we may assume V (Λ) = {v j ∈ A j : j ∈ [4]}. Moreover, the row and column edges of Λ must form a 4-cycle. Thus, if R(v 0 ) = j, we must have C(v 0 ) = φ i (j) and R(v 2 ) = j. It then follows from the definition of Q i that C(v 2 ) = ψ i (j). But this implies C(v 3 ) = ψ i (j), so that R(v 3 ) = m + ψ −1 i (ψ i (j)) = m + j. We then have R(v 1 ) = m + j and C(v 1 ) = φ i+1 (j). If v 0 , v 2 , v 3 , v 1 is to form a 4-cycle, we must have C(v 0 ) = C(v 1 ). However, this would mean φ i (j) = φ i+1 (j), contradicting the fact that T i ∩ T i+1 = ∅.