aa r X i v : . [ m a t h . C O ] A p r RAINBOW–FREE –COLORINGS OF ABELIAN GROUPS AMANDA MONTEJANO AND ORIOL SERRA
Abstract.
A 3–coloring of the elements of an abelian group is said to be rainbow–free if thereis no 3–term arithmetic progression with its members having pairwise distinct colors. We givea structural characterization of rainbow–free colorings of abelian groups. This characterizationproves a conjecture of Jungi´c et al. on the size of the smallest chromatic class of a rainbow–free3–coloring of cyclic groups. Introduction A k –coloring of a set X is a map c : X → [ k ] where [ k ] = { , , ...k } . A subset Y ⊂ X is rainbow under c if the coloring assigns pairwise distinct colors to the elements of Y . The study of theexistence of rainbow structures falls into the anti–Ramsey theory initiated by Erd˝os, Simonovitsand S´os [3]. Arithmetic versions of this theory were initiated by Jungi´c, Licht, Mahdian, Neˇsetˇriland Radoiˇci´c [5] where the authors study the existence of rainbow arithmetic progressions incolorings of cyclic groups and of intervals of integers.In the case of colorings of the integers, it was shown by Axenovich and Fon der Flaas [1] thatevery 3–coloring of the integer interval [1 , n ] such that each color class has cardinality at least( n + 4) / k –colorings with no rainbow arithmetic progressions of length k ≥
5. Conlon,Jungi´c and Radoiˇci´c [2] gave a construction of equinumerous 4–colorings with no rainbow 4–term arithmetic progressions. The canonical version of van der Waerden theorem by Erd˝osand Graham states that every coloring of the integers (with possibly infinitely many colors)contains either a monochromatic or a rainbow k –term arithmetic progression for each k . By thecelebrated theorem of Szemer´edi, if one of the color classes has positive density then one findsa monochromatic arithmetic progression of length k . In contrast, Jungi´c et al. [5] show thatthere are colorings with all color classes with positive density with no rainbow 3–term arithmeticprogressions.In the above mentioned reference of Jungi´cet al. [5] the authors also study the existence ofrainbow 3–term arithmetic progressions in 3–colorings of finite cyclic groups. The authorscharacterize all integers n such that every 3–coloring of the cyclic group Z /n Z contains a rainbow3-term arithmetic progression. Theorem 1 (Jungi´c et al. [5]) . For every integer n , there is a rainbow–free –coloring of Z /n Z with non–empty color classes, if and only if n does not satisfy any of the following conditions: (a) n is a power of . (b) n is a prime and the multiplicative order of is n − . (b) n is a prime, the multiplicative order of is ( n − / and ( n − / is odd. The above result motivates the following notation. We denote by P the set of primes p forwhich 2 has either multiplicative order p −
1, or multiplicative order ( p − / p − / P be the set of remaining primes.Following [5] we let m ( n ) denote the largest integer m for which there is a rainbow–free 3–coloring of Z /n Z such that the cardinality of the smallest color class is m . Among other results,the authors in [5] proved that if the smaller class in a 3–coloring of the cyclic group Z /n Z hassize greater than n/
6, then there exists a rainbow AP (3). For n divisible by 6 this condition istight, but for other values of n it is possible to obtain better bounds. Theorem 2 (Jungi´c et al. [5]) . Let n be not a power of , q be the smallest prime factor of n ,and r be the smallest odd prime factor of n , then: j n r k ≤ m ( n ) ≤ min ( n , nq )Motivated by the above result, Jungi´c, Neˇsetˇril and Radoiˇci´c [6] mention that “Computing theexact value of m ( n ) remains a challenge” and they formulate the following conjecture. Conjecture 1 (Jungi´c, Neˇsetˇril and Radoiˇci´c [6]) . Let n be an integer which is not a power of .Let p denote the smallest odd prime factor of n in P and let q be the smallest odd prime factorof n in P . Then the largest cardinality of the smallest color class in a rainbow–free –coloringof the cyclic group Z /n Z satisfies m ( n ) = (cid:22) n min { p, q } (cid:23) . In this paper we give a structural characterization of 3–colorings of finite abelian groups ofodd order with no rainbow 3–term arithmetic progressions. This characterization provides aproof of Conjecture 1 for general abelian groups of odd order. Combined with the study of theeven case, we complete the proof of Conjecture 1 for cyclic groups of even order as well. Notethat Conjecture 1 does not hold for general abelian groups of even order as illustrated by thefollowing counterexample. Let G := H ⊕ ( Z / Z ⊕ Z / Z ) where | H | is not a power of 2. Considerthe following 3–coloring of G : let the subgroup H be colored by A , color one of the threeremaining H –cosets of G by B and the two cosets left by C . This coloring —textcolorblackhasno rainbow 3–term arithmetic progressions, since such a progression (a triple ( x, y, z ) such that x − y + z = 0) must have two of its elements in the same H –coset. However the smaller colorclass has cardinality | H | = | G | / { | G | p , | G | q } according tothe choice of H .Our main result, , Theorem 3 below, identifies the three possible kinds of rainbow-free coloringsof an abelian group G of odd order which can be described as follows. There is a proper subgroup H of G such that, either the coloring of G is obtained by lifting a rainbow–free coloring with AINBOW–FREE 3–COLORINGS OF ABELIAN GROUPS 3 a color class of size one from the quotient group
G/H , or there is one coset of H which isbichromatic and G \ H is monochromatic, or a combination of the two possibilities above.In order to state the main result, let us introduce some notation. Let G be a finite abeliangroup. Recall that the Minkowski sum of two nonempty subsets X, Y ⊂ G is defined as X + Y = { x + y : x ∈ X, y ∈ Y } . The period (or stabilizer) of a subset S ⊆ G , denoted by P ( S ), is the subgroup of G defined by: P ( S ) = { g ∈ G : S + g = S } Thus, S is a union of cosets of P ( S ), and P ( S ) lies above any subgroup of G such that S is aunion of cosets. We say that a set S is H – periodic , where H is a subgroup of G , if S + H = S ,and S is periodic if P ( S ) is a nontrivial subgroup of G (i.e. P ( S ) = { } ). If P ( S ) = { } we saythat S is aperiodic .For a subset X ⊂ G we denote by 2 · X = { x : x ∈ X } and − X = {− x : x ∈ X } .A 3–term arithmetic progression is an ordered triple ( x, y, z ) with x, y, z ∈ G satisfying theequation x + y = 2 z . In the present setting we say that a 3–coloring of the elements of G is rainbow–free if there are no rainbow 3–term arithmetic progressions. Note that the property ofbeing rainbow–free is invariant by translations: c is a rainbow–free coloring of G if and only if,for each fixed g ∈ G , the coloring c ′ ( x ) := c ( x + g ) is also rainbow–free. We will often use thisremark without explicit reference. We identify a 3–coloring with the partition of G into its threecolor classes which we denote by { A, B, C } . Theorem 3.
Let G be a finite abelian group of odd order n and let c be a –coloring of G withnon–empty color classes A, B, C . Then c is rainbow–free if and only if, up to translation, thereis a proper subgroup H < G and a color class, say A , such that the following three conditionshold: (i) A ⊆ H , and the –coloring induced in H is rainbow–free, (ii) both e B = B \ H and e C = C \ H are H –periodic sets, and (iii) e B = − e B = 2 · e B and e C = − e C = 2 · e C . The ‘if’ part of Theorem 3 can be easily checked and its proof is detailed in Section 6, Proposi-tion 2. For the “only if” part we use results by Kneser [ ? ], Kemperman [7] and Grynkiewicz [8]which give the structure of sets with small sumset in an abelian group.The paper is organized as follows. In Section 2 we recall some results in Additive Combinatoricsand prove a simple Lemma which will be used in the remainder of the paper. Section 4 dealswith colorings in which one color class is either small or an arithmetic progression. In Section5 we consider colorings with structured color classes. The proof of Theorem 3 and the proofof Conjecture 1 for abelian groups of odd order is contained in Section 6. In Section 7 we givea structural characterization for the case of cyclic groups of even order (Theorem 7) which isanalogous to Theorem 3. With this version of the characterization one can complete the proofof Conjecture 1 for cyclic groups of even order. AMANDA MONTEJANO AND ORIOL SERRA Some tools from Additive Combinatorics
We shall use the following well–known result of Kneser (see e.g. [11, Theorem 5.5])
Theorem 4 (Kneser) . Let ( A, B ) be a pair of finite non-empty subsets of an abelian group G .Then, letting H := P ( A + B ) , we have: | A + B | ≥ | A + H | + | B + H | − | H | . Moreover, if | A + B | ≤ | A | + | B | − then we have equality. It follows from Kneser‘s Theorem that, if | A + B | ≤ | A | + | B | −
1, then either A + B is periodicor | A + B | = | A | + | B | −
1. We shall use this remark in the following sections.The structure of pairs of sets (
X, Y ) in an abelian group G verifying | X + Y | = | X | + | Y |− H = { } be a subgroup of G . A set S ⊂ G is said to be H – quasiperiodic if it admits a decomposition S = S ∪ S , where each of S and S can be empty, S is a maximal H –periodic subset of S and S is (properly) contained ina single coset of H . Note that every set S ⊂ G is quasiperiodic with S = ∅ and H = G . Theorem 5 (Kemperman [7]) . Let A and B be nonempty subsets of an abelian group G verifying | A + B | = | A | + | B | − ≤ | G | − . If A + B is aperiodic then one of the following holds: (i) min {| A | , | B |} = 1 . (ii) Both A and B are arithmetic progressions with the same common difference. (iii) Both A and B are H –quasiperiodic for some nontrivial proper subgroup H < G . We shall also use the following extension of KST, recently obtained by Grynkiewicz [8], whichdescribes the structure of pairs of sets (
X, Y ) in an abelian group G verifying | X + Y | = | X | + | Y | .Again we only need a simplified version of the full result. Theorem 6 (Grynkiewicz [8]) . Let A and B be nonempty subsets of an abelian group G of oddorder n verifying | A + B | = | A | + | B | ≤ | G | − . If A + B is aperiodic then one of the following holds: (i) min {| A | , | B |} = 2 or | A | = | B | = 3 . (ii) Both A and B are H –quasiperiodic for some nontrivial proper subgroup H < G . (iii) There are a, b ∈ G such that | A ′ + B ′ | = | A ′ | + | B ′ | − where A ′ = A ∪ { a } and B ′ = B ∪ { b } . It is well–known that, if A and B are subsets of a group G such that | A | + | B | > | G | then A + B = G . The following lemma characterizes the structure of sets A, B ⊂ G with | A | + | B | = | G | and A + B = G . We include here a short proof for the benefit of the reader. AINBOW–FREE 3–COLORINGS OF ABELIAN GROUPS 5
Lemma 1.
Let
A, B be subsets of a finite abelian group G . (i) If | A | + | B | > | G | then A + B = G . (ii) If | A | + | B | = | G | then either A + B = G or there is a subgroup H and a ∈ G such thatboth A and B are H –periodic and A + B = G \ ( a + H ) . Proof.
We only prove (ii). If | A + B | = | G | − H = { } . Supposethat | A + B | ≤ | G | − H = P ( A + B ) be the period of A + B . By Kneser’s Theorem | G | > | A + B | = | A + H | + | B + H | − | H | ≥ | A | + | B | − | H | = | G | − | H | . Since A + B is H –periodic, equality holds in the second inequality. It follows that A + B = G \ ( a + H ) for some a ∈ G and that A + H = A and B + H = B . ✷ One of the applications of Lemma 1 is the following result which will be often used. Let H bea proper subgroup of G . As usual we denote by G/H the quotient group. If X is a subset of G we write X/H for the image of X in G/H by the natural projection π : G → G/H . We saythat a triple (
X, Y, Z ) of H –cosets is in arithmetic progression if ( X/H ) + (
Y /H ) = 2 · ( Z/H ).For X an H –coset and U a color class of a coloring we write X U := X ∩ U . Lemma 2 (The 3-cosets Lemma) . Let { A, B, C } be a rainbow–free –coloring of an abeliangroup G with odd order n . Let H < G be a subgroup of G and let ( X, Y, Z ) be a triple of H –cosets in arithmetic progression.If each of X A , Y B and Z C is non–empty, then (1) max {| X A | + | Y B | , | X A | + | Z C | , | Z C | + | Y B |} ≤ | H | . In particular, none of the three cosets can be monochromatic.Moreover, if equality holds then there is a proper subgroup
K < H such that two of the sets X A , Y B , Z C are K –periodic (the two involved in the equality holding) and the third one is con-tained in a single coset of K .Proof. Since the coloring is rainbow–free and the three cosets are in arithmetic progression wehave X A + Y B ⊆ (2 · Z ) \ (2 · Z C ) . Hence | X A + Y B | < | H | which, by Lemma 1 (i), implies | X A | + | Y B | ≤ | H | . Similarly X A − (2 · Z C ) − Y and Y B − (2 · Z C ) − X imply | X A | + | Z C | ≤ | H | and | Y B | + | Z C | ≤ | H | respectively.This proves the first part of the statement.Suppose that | X A | + | Y B | = | H | . By Lemma 1 (ii) there is a subgroup K < H such that both X A and Y B are K –periodic and (2 · Z ) \ ( X A + Y B ) consists of a single K –coset, which contains2 · Z C . A symmetric argument applies if | X A | + | Z C | = | H | or | Y B | + | Z C | = | H | . ✷ AMANDA MONTEJANO AND ORIOL SERRA The prime case
The proof of Theorem 3 (in Section 6) is by induction on the number of (not necessarily distinct)primes dividing n = | G | . In this section we prove Theorem 3 for groups of prime order, whichis a direct consequence of Theorem 1 and Theorem refthm:n/q. Proposition 1.
Let G be a group of prime order p and let c be a –coloring of G with nonemptycolor classes A, B, C . Then c is rainbow–free if and only if, p ∈ P and, up to translation, thereis a color class, say A , such that: (i) A = { } , (ii) 2 · B = B = − B and · C = C = − C .Proof. Suppose first that c is rainbow–free. Since G ≃ Z /p Z , it follows from Theorem 2 that m ( p ) ≤ p ∈ P then there are no rainbow–free 3–coloring of Z /p Z with non–empty color classes. That is, if c is a rainbow–free 3–coloring with nonemptycolor classes, then necessarily p ∈ P and there is a color class, say A , such that | A | = 1. Wecan assume that A = { } satisfying (i).To prove (ii) note that, for every x ∈ B , since G is a cyclic group of prime order, then both − x, x ∈ { B, C } . Hence we must have − x, x ∈ B , otherwise we get a rainbow 3–term arithmeticprogression of the form ( − x, , x ) or (0 , x, x ). Thus 2 · B = B = − B and similarly 2 · C = C = − C .Reciprocally, if the coloring satisfies (i) and (ii), then any 3–term arithmetic progression con-taining 0 has its remaining members in the same color class, thus c is rainbow–free. ✷ Small color classes and color classes in arithmetic progresion
Throughout this section G denotes an abelian group of odd order n and c is a rainbow–free3–coloring of G with non–empty color classes { A, B, C } . The coloring is said to be H – regular if,up to transllation, it satisfies conditions (i), (ii) and (iii) of Theorem 3 for the subgroup H < G .We begin with the case when there is a color class with just one element.
Lemma 3. If | A | = 1 then the coloring is H –regular with H = { } .Proof. We may assume that A = { } . By choosing H = { } , parts (i) and (ii) of Theorem 3are satisfied.To prove (iii) note that,since G is a group of odd order, for every x ∈ B then both − x, x ∈{ B, C } . Hence we must have − x, x ∈ B , otherwise we get a rainbow 3–term arithmetic pro-gression of the form ( − x, , x ) or (0 , x, x ). Thus 2 · B = B = − B and similarly 2 · C = C = − C . ✷ Lemma 3 provides the description given in Theorem 3 when one of the colors has cardinalityone. The 3-cosets lemma (Lemma 2) can be used to show the analogous result if one of the colorclasses is contained in a single coset.
AINBOW–FREE 3–COLORINGS OF ABELIAN GROUPS 7
Lemma 4. If A is contained in a single coset of a proper subgroup H ′ < G and | A | > , thenthe coloring is H –regular for some proper subgroup H < G .Proof.
We may assume that 0 ∈ A . Let H be the the minimal proper subgroup of G whichcontains A . Suppose that Y = H is an H –coset which intersects the two remaining color classes B and C . Let Z be a third coset such that ( X = H, Y, Z ) are in arithmetic progression. Since A does not meet Y and Z , we have | Y B | + | Y C | + | Z B | + | Z C | = 2 | H | . It follows from Lemma 2 that | Y B | + | Z C | = | H | and | Y C | + | Z B | = | H | . Moreover, thereis a subgroup K ≤ H such that X A = A is contained in a single coset of K and each of Y B , Y C , Z B , Z C is K –periodic. By the minimality of H we have K = H contradicting theexistence of the bichromatic coset Y . Thus parts (i) and (ii) of Theorem 3 are satisfied.Now consider the 3–coloring of G/H with color classes A ′ = { } , B ′ = e B/H and C ′ = e C/H where e B = B \ H and e C = C \ H . Note that, since the original coloring { A, B, C } of G israinbow–free, then so it is { A ′ , B ′ , C ′ } in G/H .If some of B ′ or C ′ is an empty set, then part (iii) of Theorem 3 is clearly satisfied.If both B ′ and C ′ are nonempty sets, then { A ′ , B ′ , C ′ } is a rainbow–free coloring of G/H withnonempty color classes and | A ′ | = 1. By Lemma 3, it follows that 2 · B ′ = B ′ = − B ′ and2 · C ′ = C ′ = − C ′ . Thus part (iii) is also satisfied for the coloring { A, B, C } . ✷ We next handle the cases when there is a class with two elements or there are two classes withthree elements.
Lemma 5. If | A | = 2 then the coloring is H –regular for some H < G .Proof.
By Lemma 4 we only have to show that one color is contained in a single coset of aproper subgroup
H < G .We may assume that A = { , a } . Let us show that a generates a proper subgroup H of G .Suppose on the contrary that the cyclic group generated by a is the whole group G = h a i ∼ = Z /n Z .Since {− a, a, a } can not be rainbow, we have c ( − a ) = c (3 a ). Since {− a, , a } can not berainbow we have c ( − a ) = c (3 a ). By iterating this argument, we have c ( − a ) = c (3 a ) = c ( − a ) = c (5 a ) = c ( − a ) = ... = c (( n − a ) = c ( − ( n − a ) , so that the color class of − a has n − A ⊂ h a i = H < G and, by Lemma 4, parts (i), (ii) and (iii) of Theorem 3 are satisfied. ✷ Lemma 6. If | A | = | B | = 3 and | A + B | = 6 , then the coloring is H –regular for some H < G .Proof.
Let A = { , a, a ′ } . It can be shown (see e.g. [8]) that only two possibilities occur if | A + B | = | A | + | B | = 6: either B is a transllate of A or one of the sets, say A , is an arithmeticprogression, and the second one, B , is an arithmetic progression of length four with the samedifference and with one element removed. AMANDA MONTEJANO AND ORIOL SERRA
Suppose that B = A + x = { x, x + a, x + a ′ } for some x ∈ G . Since {− x, , x } can not berainbow we have − x ∈ A ∪ B . If − x ∈ A then 0 ∈ A ∩ B , a contradiction. Thus − x ∈ B and a = − x . Since { , x, x } can not be rainbow we have 2 x ∈ A ∪ B . If 2 x ∈ B then x ∈ A ∩ B (since A = B − x ). Thus 2 x = a which implies B = {− x, x, x } and A = {− x, , x } and bothsets are arithmetic progressions with the same difference contradicting | A + B | = | A | + | B | .Suppose now that A is an arithmetic progression with difference d . If A generates a propersubgroup H of G then the result follows from Lemma 4. Hence we may assume that A = { , , } and G is the cyclic group of order n . Moreover B = { x, x + 2 , x + 3 } for some x ∈ Z /n Z \ { , , , n − , n − , n − } . If { , x, − x } is not rainbow, since − x ∈ { , } can not holdand n is odd, we must have − x = x + 3. But then x = ( n − / { , ( n − / , ( n + 1) / } is rainbow. This contradiction completes the proof. ✷ We next consider the case when two color classes are almost progressions. An almost –progressionis an arithmetic progression with one point removed. Observe that, with this definition, the classof almost progressions contains the class of all arithmetic progressions except the ones whoselength equals the order of the cyclic group generated by the difference.
Lemma 7.
Assume that ≤ | A | ≤ | B | ≤ | C | . If A and B are almost–progressions with thesame difference d , then the coloring is H –regular for some H < G .Proof. If d generates a proper subgroup H of G then A is contained in a single coset of H andthe result follows by Lemma 4.Thus we may assume that d generates the full group, so that G is the cyclic group Z /n Z andwe may assume d = 1 (since the property of being rainbow–free is invariant by dilations). Wewill show that in this case c contains a rainbow 3–term arithmetic progression.Let b be the minimum circular distance from elements in A to elements in B .If b = 1 we may assume that n − ∈ A and 0 ∈ B . Since max {| A | , | B |} ≤ ( n − / n − / ∈ C giving the rainbow { , ( n − / , n − } .Suppose now that b >
1. Since | A | ≥ { n − , } ⊂ A and { , , . . . , b ′ } ⊆ C and b ′ + 1 ∈ B for some b ′ ≥ b . If b ′ is odd then { , ( b ′ + 1) / , b ′ + 1 } is rainbow and if b ′ is eventhen { n − , ( b ′ + 2) / , b ′ + 1 } is rainbow. ✷ Periodic color classes
In this section we analyze the structure of the color classes when they are close to be periodic.The consideration of these cases arise from the discussion on the size of sumsets of the colorclasses in a ranbow–free 3–coloring and the KST and Grynkiewicz theorems.Throughout the section we keep the notation of the previous one. Thus G denotes an abeliangroup of odd order n and c is a rainbow–free 3–coloring of G with non–empty color classes { A, B, C } . The coloring is H –regular if it satisfies conditions (i), (ii) and (iii) of Theorem 3 forsome subgroup H . AINBOW–FREE 3–COLORINGS OF ABELIAN GROUPS 9
Recall that, for a subset X ⊂ G and a subgroup H < G , we denote by
X/H the image of X bythe natural projection π : G → G/H .The proof of Theorem 3 (in Section 6) is by induction on the number of (not necessarily distinct)primes dividing n = | G | , the initial step being proved in Section 3. Thus, in the remainder ofthis section we will assume that Theorem 3 holds for any group of order a proper divisor of n = | G | .We start with the simplest case. Lemma 8.
If the three color classes
A, B, and C are K –periodic for some subgroup K < G ,then the coloring is H –regular for some H < G .Proof.
Consider the coloring A ′ = A/K , B ′ = B/K and C ′ = C/K of G/K . Note that, since { A, B, C } is a rainbow–free 3–coloring with nonempty color classes, then so it is { A ′ , B ′ , C ′ } .Since G/K is a group of order a proper divisor of n = | G | , the coloring { A ′ , B ′ , C ′ } is H ′ –regularfor some H ′ < G/K . In particular, there is a color class, say A ′ , such that A ′ ⊆ H ′ < G/K .Thus A is contained in a single coset of the proper subgroup H ′ + K in G , and the statementfollows from Lemma 4. ✷ We next consider the case where two of the color classes are quasiperiodic. Recall that a set S ⊂ G is H – quasiperiodic if it admits a decomposition S = S ∪ S , where each of S and S can be empty, S is a maximal H –periodic subset of S and S is (properly) contained in a singlecoset of H . Lemma 9. If A = A ∪ A and B = B ∪ B are K –quasiperiodic decompositions of A and B with K a nontrivial proper subgroup of G , then the coloring is H –regular for some H < G .Proof.
By Lemma 4 we may assume that none of the color classes is contained in a single cosetof a proper subgroup of G . If two of the color classes are periodic then so is the third one andthe result follows from Lemma 8. Therefore, up to renaming the color classes we may assumethat each of the sets A , B , A and B are nonempty, and that | C/K | >
1. We also assumethat 0 ∈ A .Let us show that A /K = B /K . Suppose the contrary and let Z be a K –coset such that X = K, Y = B + K and Z are in arithmetic progression (such a coset always exists since n is odd). Note that X intersects A and C , Y intersects B and C and that Z is monochromatic.This contradicts the 3–cosets Lemma ( Lemma 2). Hence A /K = B /K .Consider the 3–coloring c K of G/K with color classes { A ′ , B ′ , C ′ } where A ′ = A/K , B ′ = B /K and C ′ = G/K \ ( A ′ ∪ B ′ ). Note that C ′ = ( C \ K ) /K . Observe that c K is a 3–coloring of G/K with non empty color classes. Moreover, c K is rainbow–free, otherwise we have three K –cosetsin G in arithmetic progression where at least two of them are monochromatic (since both B and C \ K are K –periodic) contradicting the 3–cosets Lemma (Lemma 2).Since we assume that Theorem 3 holds in G/K , there is a proper subgroup
L < G containing K such that, up to translation, one of the three chromatic classes of c K is contained in L/K andthe remaining two are (
L/K )–periodic outside
L/K . Suppose that A ′ ⊂ L/K . Then A ⊂ L and the statement follows by Lemma 4.Let us show now that C ′ can not be contained in a single coset of L/K in G/K . Suppose on thecontrary that C \ K is contained in a single L –coset X of G . Let Z be a L –coset in arithmeticprogression with X and Y = A + L . Since Y intersects the two colors, A and B , and Z isnecessarily monochromatic with color A or B , we have | Z | + | Y A | , | Z | + | Y B | > | L | contradictingLemma 2.Suppose now that B ′ is contained in a single coset of L/K in G/K . We may assume that B iscontained in a single L –coset X = L in G , otherwise we are done by Lemma 4.Consider the coloring c L of G/L with color classes { A ′′ , B ′′ , C ′′ } where A ′′ = ( A \ X ) /H, B ′′ = B/L, and C ′′ = G/L \ ( A ′′ ∪ B ′′ ). Note that | B ′′ | = 2 and | C ′′ | >
1, otherwise C ⊂ X and C ′ can not be contained in a single coset of L/K as shown in the paragraph above.If A ′′ = ∅ , consider Z an L –coset in arithmetic progression with L and X (exist since the orderof G is odd). Note that L intersects A , X intersects B and Z is monochromatic of color C , acontradiction by Lemma 2.If | A ′′ | >
1, note that c L is a 3–coloring of G/L with non empty color classes. Moreover, c L israinbow–free since otherwise we have three L –cosets in G in arithmetic progression where one ofthem is monochromatic (since C ′′ is L –periodic) a contradiction by Lemma 2. Since | B ′′ | = 2, itfollows from Lemma 5 that B ′′ is contained in a single coset of a proper subgroup H/L < G/L .Thus B ⊆ H and the statement follows from Lemma 4. ✷ We next consider the case where A + B is K –periodic for some subgroup K of G . Observe that,since c is rainbow–free, we also have K = G . Lemma 10. If A + B is K –periodic for some proper nontrivial subgroup K of G , then thecoloring is H –regular for some subgroup H < G .Proof.
We show that, under the assumption of the Lemma, both sets A and B admit a K –quasiperiodic decomposition and thus the statement follows from Lemma 9.By the Theorem of Kneser we have(2) | A/K + B/K | ≥ |
A/K | + | B/K | − . Since A + B ∩ · C = ∅ we have(3) ( A + B ) / ⊂ ( G \ C ) = A ∪ B, where X/ X ⊂ G by the inverse of the automorphism of G defined as x x . This automorphism leaves all subgroups invariant so that ( A + B ) / K –periodic.Let D = (( A ∪ B ) + K ) \ ( A + B ) / . Note that the aperiodic parts of A and of B are contained in D ∪ ( A ∩ B ). By (2) we have | A/K | + | B/K |−|
A/K ∩ B/K | = | ( A ∪ B ) /K | = | ( A + B ) /K | + | D/K | ≥ |
A/K | + | B/K |− | D/K | , which implies | D/K | + | A/K ∩ B/K | ≤ . AINBOW–FREE 3–COLORINGS OF ABELIAN GROUPS 11
Hence each of A and B admits a K –quasiperiodic decomposition. ✷ Now we consider the case where two of the color classes are almost quasiperiodic. A set X ⊂ G is almost H –quasiperiodic (resp. almost H –periodic) if there is x ∈ G such that X ∪ { x } is H –quasiperiodic (resp. H –periodic). Lemma 11. If A and B are almost H –quasiperiodic for some proper nontrivial subgroup H < G ,then the coloring is H ′ –regular for some proper subgroup H ′ < G .Proof. We say that a coset X of a subgroup H < G is punctured if all but one of its elementsare in the same color class U ∈ { A, B, C } . We then say that X is a punctured coset of color U .Since A and B are almost H –quasiperiodic, they admit decompositions A = A ∪ A and B = B ∪ B where each of A and B are subsets of some H –coset and each of A and B arealmost periodic so that each of them contains at most one punctured coset.We may assume that at least one of A or B contains a punctured coset and that 0 < | A | , | B | < | H | since otherwise A and B are quasiperiodic and the result follows from Lemma 9. Wemay also assume that none of A, B and C are periodic since otherwise at least one of A + B or A + C is periodic and the result is implied by Lemma 10. Finally we may assume thatmin {| A/H | , | B/H | , | C/H |} > Case 1 : A + H = B + H . Let Z be a coset in arithmetic progression with X = A + H and Y = B + H .We may assume that one of X, Y , say X , intersects C , since otherwise X is the punctured coset of B and Y is the punctured coset of A , which implies that C is periodic. In particular X ∩ B = ∅ .Moreover, whatever the colors present in Z , the conditions of Lemma 2 are satisfied and Z cannot be a full coset. Since all H –cosets different from X and Y are either monochromatic orpunctured, Z is a punctured coset. Moreover it can not be of color C since Z ∩ A = Z ∩ B = ∅ .Suppose that | Z A | = | Z ∩ A | = | H | −
1. Then, again by Lemma 2, | Z A | + | X C | = | H | , whichimplies | X C | = | X ∩ C | = 1 and | Y B | = | Y ∩ B | = 1. Thus both X and Z are punctured cosetsof color A . Since A can not contain more than two partially filled cosets, Y is a punctured cosetof color C . Finally, the other color in Z must also be C since Z is not the coset containing B .Since | B/H | > Y ′
6∈ {
X, Y, Z } which intersects B . Moreover, Y ′ is either a fullcoset or a punctured coset of B . Let Z ′ be a third coset in arithmetic progression with X and Y ′ . Whatever the colors present in Z ′ , the conditions of Lemma 2 are satisfied, so that both Y ′ and Z ′ must be punctured cosets. Thus Z ′ must intersect C (there are no punctured cosetswith colors A and B ) and | X A | + | Y B | > | H | , contradicting Lemma 2.Suppose now that | Z B | = | Z ∩ B | = | H | −
1. If Y ∩ A = ∅ then Y is a punctured coset of color A and | Y A | + | Z B | > | H | contradicting Lemma 2. Otherwise Y intersects C and application ofLemma 2 implies | Y C | = | X A | = 1. Thus both Y and Z are punctured cosets of B with secondcolor C and X is a punctured coset of C with second color A , the same structure as in the caseabove with colors A and B exchanged, and we again obtain a contradiction with Lemma 2. Case 2 : A + H = B + H . We may assume that at least one of A or B contains a puncturedcoset which is not X , otherwise A and B are quasiperiodic and the results follows from Lemma 9.So let Y be a punctured coset of color A (observe that Y B = ∅ since B is contained in X , thus | Y C | = 1). Let Z be a coset in arithmetic progression with X and Y .We fist prove that Z is not monochromatic. If Z is monochromatic of color B (resp. C or A )then | Z B | + | Y C | > | H | (resp. | Z C | + | Y A | > | H | or | Z A | + | Y C | > | H | ) and we get a contradictionby Lemma 2 since X A (resp. X B ) is not empty.Thus Z must be a punctured coset of color B with | Z C | = 1. Since | Y A | + | Z B | > | H | then X C = ∅ . Since | Y A | + | Z C | = | H | then | B | = 1, but also | Y C | + | Z B | = | H | implies | A | = 1which is a contradiction. This completes the proof. ✷ Proof of Theorem 3
The next proposition proves the ‘if’ part of Theorem 3.
Proposition 2.
Let { A, B, C } be a coloring of an abelian group G of odd order. If there is aproper subgroup H of G and a color class, say A , such that the three following conditions hold: (i) A ⊆ H , and the –coloring induced in H is rainbow–free, (ii) both e B = B \ H and e C = C \ H are H –periodic sets, and (iii) e B = − e B = 2 e B and e C = − e C = 2 e C .Then the -coloring is rainbow–free.Proof. If C = G \ H then A + B is contained in H , and thus it is disjoint from 2 · C . Moreover,each of 2 · A and 2 · B are contained in H and thus disjoint from A + C = B + C = C .Suppose that C = G \ H . Since, by (i), a rainbow 3–term arithmetic progression in G can notbe contained in H , it gives rise, by conditions (ii) and (iii), to a rainbow 3–term arithmeticprogression in G/H with the coloring { A/H, e B/H, e C/H } . However, since every color class X in this last coloring verifies X = − X = 2 · X any 3–term arithmetic progression of G/H containing
A/H has its remaining members in the same color class. Hence { A/H, e B/H, e C/H } is rainbow–free and so it is c . ✷ It remains to prove that, if c is a rainbow–free coloring of an abelian group G of odd order, thenthe color classes verify conditions (i), (ii) and (iii) of Theorem 3 with some proper subgroup H < G . To prove this we use the results in sections 3, 4 and 5 together with the theorems ofKneser, Kemperman and Grynkiewicz.
Proposition 3.
Let { A, B, C } be a rainbow–free coloring of an abelian group G of odd order.There is a proper subgroup H of G and a color class, say A , such that the three followingconditions hold: (i) A ⊆ H , and the –coloring induced in H is rainbow–free, (ii) both e B = B \ H and e C = C \ H are H –periodic sets, and AINBOW–FREE 3–COLORINGS OF ABELIAN GROUPS 13 (iii) e B = − e B = 2 e B and e C = − e C = 2 e C .Proof. The proof is by induction on the number of (not necessarily distinct) primes dividing n = | G | . If n is prime, the statement holds by Proposition 1. We assume that Theorem 3 holdsfor any group of order a proper divisor of n = | G | , so that we can use the results in Section 5.We first note the following remark. Remark 1.
For any pair of distinct color classes
X, Y ∈ {
A, B, C } we have | X + Y | ≤ | X | + | Y | . Proof.
Suppose on the contrary that | X + Y | ≥ | X | + | Y | + 1 for some distinct color classes X, Y ∈ {
A, B, C } . Since n = | G | is odd, we have | · Z | = | Z | (where Z is the remaining colorclass). It follows from the condition | X + Y | ≥ | X | + | Y | + 1 that ( X + Y ) ∩ (2 · Z ) = ∅ , whichimplies that there is a rainbow 3–term arithmetic progression. ✷ It follows from the above remark that(4) | A + B | ≤ | A | + | B | . By Lemma 10 we can assume that A + B is aperiodic. Then it follows from Kneser’s theoremthat | A + B | ≥ | A | + | B | −
1. According to (4) we have to consider two cases.
Case 1: | A + B | = | A | + | B | −
1. It follows from the simplified version of the KST, Theorem 5,that one of the following holds:(i) min {| A | , | B |} = 1. In this case the result follows by Lemma 3.(ii) Both A and B are arithmetic progressions with the same common difference d . Theresult follows by Lemma 7.(iii) Both A and B are H –quasiperiodic for some nontrivial proper subgroup H < G . Theresult follows by Lemma 9.
Case 2: | A + B | = | A | + | B | . Then the simplified version of the Theorem by Grynkiewicz,Theorem 6, implies that one of the following holds:(i) min {| A | , | B |} = 2 or | A | = | B | = 3. In this case the result follows by Lemmas 5 and 6respectively.(ii) Both A and B are H –quasiperiodic for some nontrivial proper subgroup H < G . Theresult follows by Lemma 11.(iii) There are a, b ∈ G such that | A ′ + B ′ | = | A ′ | + | B ′ | − A ′ = A ∪ { a } and B ′ = B ∪ { b } . According to Kemperman’s Theorem 5, either A ′ + B ′ is periodic, inwhich case A + B is also periodic and the result follows by Lemma 10, or A ′ , B ′ are bothquasiperiodic, in which case A and B are almost periodic and we can apply Lemma 11,or A ′ , B ′ are both arithmetic progressions and then A and B are almost arithmeticprogressions, a case handled in Lemma 7.This completes the proof of the Proposition and of Theorem 3. ✷ The description of rainbow–free 3–colorings of abelian groups of odd order can be used to proveConjecture 1. Actually Conjecture 1 holds for general abelian groups of odd order as shown inthe next Corollary.
Corollary 1.
Let G be an abelian group of odd order n . Let p denote the smallest prime factorof n in P and let q be the smallest prime factor of n in P . If { A, B, C } is a rainbow–free –coloring of G then (5) min {| A | , | B | , | C |} ≤ (cid:22) n min { p, q } (cid:23) . Moreover, there are rainbow–free –colorings of G for which equality holds.Proof. We first observe that (5) is equivalent to:(6) min {| A | , | B | , | C |} ≤ max (cid:26)(cid:22) n p (cid:23) , nq (cid:27) . Note that, since the smallest prime factor of n is either p or q , then the largest proper subgroupof G has size either np or nq .By Theorem 3 (i), there is a proper subgroup H < G and one color class, say A , contained in H . Hence, | A | ≤ | H | .If | H | ≤ nq then (6) is satisfied.Suppose that | H | > nq . Then necessarily p < q and the size of the largest proper subgroup of G is np . Hence | H | ≤ np .If | H | < np , since n is an odd number, then | H | = na where a > p . Thus | H | < ⌊ n p ⌋ and (6) issatisfied.Suppose that | H | = np . Then G/H is a cyclic group of prime order p ∈ P . By Theorem 3 (ii),each of the two sets e B = B \ H and e C = C \ H is a (possibly empty) union of H –cosets.Consider the 3–coloring of G/H with color classes A ′ = { } , B ′ = e B/H and C ′ = e C/H . Notethat, if the original coloring
A, B, C of G is rainbow–free, then the induced coloring A ′ , B ′ , C ′ of G/H ≃ Z /p Z is also rainbow–free. By Proposition 1, since p ∈ P , one of the color classes B ′ or C ′ must be empty. Hence either B/H or C/H is an empty set which implies that G \ H ismonochromatic and thus H contains two colors. It follows that min {| A | , | B | , | C |} ≤ ⌊ n p ⌋ . Thiscompletes the first part of the Corollary.Let us show now that there are rainbow–free 3–colorings of G for which equality holds in (5).If 2 p ≤ q then choose a subgroup H < G with cardinality np . Consider a partition A ∪ B = H where | A | = ⌊ n p ⌋ , and let C = G \ H . This coloring clearly satisfies parts (i), (ii) and (iii) ofTheorem 3 and therefore, by Proposition 2, it is rainbow–free.If q < p then choose a subgroup H < G with cardinality nq . We define a coloring A, B, C of G as follows. Since q ∈ P , by Theorem 1, there is a rainbow–free 3–coloring of Z /q Z withnonempty color classes A ′ , B ′ and C ′ . Let π : G → G/H ≃ Z /q Z denote the natural projection AINBOW–FREE 3–COLORINGS OF ABELIAN GROUPS 15 and define A = π − ( A ′ ), B = π − ( B ′ ) and C = π − ( C ′ ). By Proposition 1 this coloring satisfiesparts (i), (ii) and (iii) of Theorem 3 and therefore, by Proposition 2, it is rainbow–free. ✷ The even case
In this Section we shall prove Conjecture 1 for cyclic groups Z /n Z of even order. We recallthat, by Theorem 1, if n = 2 m then there are no rainbow–free 3–colorings of Z /n Z . Therefore,throughout this Section, we assume that n = 2 m l for some m ≥ l > Lemma 12.
Let { A, B, C } be a rainbow–free –coloring of a cyclic group G . Suppose that thereis a subgroup H such that one of the colors, say A , is contained in H , and each of e B = B \ H and e C = C \ H are H -periodic.There is a proper subgroup K of G containing H such that B + C ⊇ G \ K and each of B \ K and C \ K is K –periodic.Proof. We consider two cases.
Case 1 : H = { } . We have A = { } and | B | + | C | = | G | − | B + C | = | B | + | C | = | G | − K = { } .Suppose that | B + C | = | B | + | C | − { , x } = G \ ( B + C ). Since the coloring israinbow–free we have − B = B and − C = C . Since x B + C we have x − B ∩ C = ∅ and x − C ∩ B = ∅ . Hence { , x } + B = { , x } − B ⊂ G \ C. It follows that(7) |{ , x } + B | ≤ | B | + 1 . Let K = h x i be the cyclic subgroup generated by x and let B = B ∪ · · · ∪ B t be a decompositionof B into maximal arithmetic progressions with difference x . By the maximality of each B i wehave B i + { , x } = min {| K | , | B i | + 1 } . By (7), we see that all but at most one of the B i ’s are cosets of K . Moreover, if there is one ofthe B i ’s, say B , which is not a K –coset, then B is an arithmetic progression of difference x properly contained in one K –coset.Likewise { , x } + C ⊂ G \ B implies the analogous structure for C . Since none of B and C contains the whole subgroup K , the only coset where the proper arithmetic progressions cansit in is K itself. If both colors meet K then we have a rainbow–free 3–coloring of this cyclicgroup with all three colors arithmetic progressions. But we can not partition K \ { } into twoarithmetic progressions B ′ , C ′ with B ′ = − B and C ′ = − C . Therefore only one of the twocolors meets K . This shows that K is a proper subgroup of G . Moreover, B \ K and C \ K are K –periodic. By the definition of x , we also have B + C ⊃ G \ K . Finally suppose that | B + C | < | B | + | C | −
1. By Kneser’s theorem there is a proper subgroup
K < G such that B + C is K –periodic and | B + C | = | B + K | + | C + K | − | K | . Since 0 B + C we have B + C ⊂ G \ K . Hence, | G | − | B | + | C | ≤ | B + K | + | C + K | ≤ | G | . The last inequalities imply that one of the sets, say B , satisfies B = B + K and the second onesatisfies | C | = | C + K | −
1. Thus B is K –periodic, C + K = C ∪ { } and C \ K is K –periodicor empty, and B + C = G \ K . Case 2 : H = { } .Suppose first that e C = G \ H , so that e B = ∅ and B ⊂ H . Then B + C = C and the statementholds with K = H .Suppose now that both e B and e C are nonempty. Then { A ′ = A/H, B ′ = e B/H, C ′ = e C/H } is arainbow–free 3–coloring of G/H . It follows from Case 1 that there is a proper subgroup K of G containing H such that B ′ \ ( K/H ) and C ′ \ ( K/H ) are K –periodic and A ′ + B ′ ⊇ ( G/H ) \ ( K/H ).It follows that each of B \ K and C \ K are K –periodic (one of the two may be empty) and B + C ⊇ G \ K . ✷ Let n = 2 m l with l > m ≥
1. Then the cyclic group G = Z /n Z can be written as G = L ⊕ Z / m Z where L has odd order l and m ≥ { A, B, C } denotes a rainbow–free 3–coloring of G . Let P = 2 · G . Since the even factorof G is cyclic we have P ∼ = L ⊕ Z / m − Z . For each color X ∈ { A, B, C } we write X = X ∩ P and X = X ∩ P where P is the secondcoset of P in G . Lemma 13.
With the assumptions above, none of the two cosets of P is monochromatic.Proof. Suppose the contrary and choose the minimal m for which there is a counterexample tothe statement. We may assume that P = C . Since A + B ⊂ P \ · C = P \ · P we have m ≥ · P = P ) and A + B is contained in the proper subgroup 2 · P of P . Thus A ∪ B is contained in one coset of 2 · P and the second coset of this subgroup in P must becolored only with C contradicting the minimality of m . ✷ We next give the structural result analogous to Theorem 3 for cyclic groups of even order.
Theorem 7.
Let G = L ⊕ Z / m Z where L has odd order and m ≥ . There is a proper subgroup H ′ of L such that one of the colors, say A , is contained in one coset of H = H ′ ⊕ Z / m Z andeach of B \ H and C \ H is H –periodic.Proof. By Lemma 13 we may assume that none of the two cosets of P is monochromatic. Theproof is by induction on m . By Theorem 3 the result follows for m = 0. Assume m ≥
1. Weconsider three cases.
AINBOW–FREE 3–COLORINGS OF ABELIAN GROUPS 17
Case 1 : One of the two cosets of P is bichromatic.We may assume that P = B ∪ C and 0 ∈ A = A . Thus | B | + | C | = | P | and B + C ⊂ P . Since c is rainbow–free we have 2 · A ⊂ P \ ( B + C ). In particular, B + C = P . It followsfrom Lemma 1 (ii) that there is a proper subgroup H of P such that 2 · A ⊆ H , and that both B and C are H –periodic. Let H = H ′ ⊕ Z / m Z , where H ′ < L has odd order and Z / m Z denotes the cyclic subgroup of Z / m Z of order 2 m forsome m ≤ m . We next consider two cases according to P being bichromatic or trichromatic. Case 1.1 : P is bichromatic. We may assume that P = A ∪ B . Since | A | + | B | = | P | and2 · C ⊂ P \ ( A + B ), it follows from Lemma 1 (ii) that there is a proper subgroup H of P such that 2 · C is contained in a single coset of H , and that both A and B are H –periodic.Let H = H ′ ⊕ Z / m Z , with H ′ < L of odd order.Now C being H –periodic and 2 · C contained in a single coset of H implies H ′ ≤ H ′ . By theanalogous argument on A we get H ′ ≤ H ′ . Thus H ′ = H ′ and, by symmetry, we may assume H ≤ H . Therefore each color class is H –periodic.It follows that H ′ is a proper subgroup of L since otherwise we get the rainbow–free 3–coloring { A/L, B/L, C/L } of the cyclic group G/L of order 2 m , contradicting Theorem 1.Consider the subgroup H = H ′ ⊕ Z / m Z . Observe that H contains A since 2 · A ⊂ H ≤ H and the two subgroups H and H have the same odd factor. Similarly, since 2 · C ⊂ H thecolor C is also contained in H . Hence B does not intersect H , since otherwise, as all colorclasses are H ′ –periodic, we would get the rainbow-free 3–coloring { ( A ∩ H ) /H ′ , ( B ∩ H ) /H ′ , ( C ∩ H ) /H ′ ) } , of the cyclic group G ′ /H ′ of order 2 m contradicting again Theorem 1. Thus B = G \ H and thestatement of the Theorem holds with H ′ = H ′ . Case 1.2 : P is trichromatic. By the induction hypothesis there is a subgroup H = H ′ ⊕ Z / m − Z , such that A ⊆ H and B \ H and C \ H are H –periodic. Choose a minimal H ′ with thisproperty. We may assume that C \ H is nonempty. Since2 · A ⊂ H ′ ⊕ Z / m Z , we have A ⊂ H ′ ⊕ Z / m − Z with H ′ = H ′ ∩ H ′ . By the minimality of H ′ we have H ′ = H ′ ≤ H ′ .Thus, A ⊂ H . Suppose that, for some x ∈ L \ { } , the coset X = H + (2 x,
0) is colored B . Since X ⊂ A + B is disjoint from 2 · C , then the coset H + ( x,
1) is also monochromatic of color B . By switching the roles of B and C we conclude that B and C are also H –periodic. Let H = H ′ ⊕ Z / m Z . If C = G \ H the statement holds with H and we are done. Otherwise the 3–coloring { A/H , B/H , C/H } , is rainbow–free with the color A ′ = A/H consisting only of zero. Since the coloring is rainbow–free, every color X satisfies 2 · X ⊂ X ∪ { } . This implies that each of B \ H and C \ H are notonly H –periodic but in fact H –periodic. This concludes this case. Case 2 : Both cosets of P are trichromatic.By the induction hypothesis there is a subgroup H = H ′ ⊕ Z / m − Z of P such that A is contained in a single coset H and B \ H and C \ H are H –periodic.Choose a minimal H ′ with this property.It follows from Lemma 12 that there is a proper subgroup T = T ′ ⊕ Z / m − Z < P containing H such that B + C ⊃ P \ T and each of B \ T and C \ T is T –periodic.We have 2 · A ⊂ P \ ( B + C ) , so that 2 · A ⊂ T . It follows that A ⊂ H with H = T ′ ⊕ Z / m Z . Thus A ⊂ H. We now use a similar argument to Case 1.2. For each x ∈ L \ T the coset X = T + (2 x, ⊂ P \ T is monochromatic and, since the coloring is rainbow–free, so that X = A + X is disjointfrom T + (2 x, T + ( x,
1) is also monochromatic. Hence each of B \ ( T + A )and C \ ( T + A ) is also T –periodic. Hence, either G \ H is monochromatic and we aredone, or { A/H, B/H, C/H } is a rainbow–free 3–coloring of G/H with the color class A ′ = A/H consisting only of zero. Hence, every color X satisfies 2 · X ⊂ X ∪ { } . This implies that each of B \ H and C \ H are not only T –periodic but in fact H –periodic. This completes the proof. ✷ Theorem 7 provides a proof of Conjecture 1. The proof is completely analogous to the one inCorollary 1 for the case of abelian groups of odd order except that we invoke Theorem 7 insteadof Theorem 3.
Corollary 2.
Let G be cyclic group of order n . Let p denote the smallest odd prime factor of n in P and let q be the smallest odd prime factor of n in P . If { A, B, C } is a rainbow–free –coloring of G then (8) min {| A | , | B | , | C |} ≤ (cid:22) n min { p, q } (cid:23) . AINBOW–FREE 3–COLORINGS OF ABELIAN GROUPS 19
Moreover, there are rainbow–free –colorings of G for which equality holds. Acknowledgements
The authors are very grateful to Yahya O. Hamidoune for fruitful and colorful discussions onthe problem considered in this paper.
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E-mail address : [email protected] Departament de Matem`atica Aplicada IV, Universitat Polit`ecnica de Catalunya.
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