Sidon sets and C 4 -saturated graphs

Transcription

1 Sidon sets and C 4 -saturated graphs arxiv: v1 [math.co] 11 Oct 018 David F. Daza Carlos A. Trujillo Universidad del Cauca, A.A. 755, Colombia. davidaza@unicauca.edu.co - trujillo@unicauca.edu.co October 15, 018 Abstract The problem of determine the Turán number of C 4, is a well studied problem that date back to a paper of Erdős from It is known that Sidon sets can be used to construct C 4 -free graphs. If A is a Sidon set in the abelian group X, the sum graph with vertices X and edges E = {{x, y} : x y, x + y A} is C 4 -free. Using the sum graph of a Sidon set type Singer we will prove that ex(q + q + 1, C 4 ) 1 q(q + 1), for each prime power q. Further, we will prove a theorem that gives conditions for that the sum graph of a Sidon set be C 4 - saturated. In addiction, we present examples of Sidon sets that satisfy this theorem. Keywords: Cycle of length 4, Sidon set, Turán number, C 4 -Saturated, Polarity graph, Projective plane P G(, F q ). 1 INTRODUCTION Let F be a fixed graph. A graph G is F -free if G does not contain a subgraph that is isomorphic to F. The Turan number of a fixed graph F, denoted by ex(n, F ) is the maximun number of edges in an graph on n vertex that is F -free. Determine Turán numbers for different graphs is one of the most studied problems in extremal graph theory. A case of particular interest is when F = C 4, the cycle on four vertices. A counting argument in [1] gives, ex(n, C 4 ) n 4 (1 + 4n 3). (1.1) In [] and [3] independently constructed graphs that show 1.1 is asymptotically best possible. These graphs are often called Erdős-Rényi orthogonal polarity graphs and are constructed using an orthogonal polarity of the proyective plane P G(, q). The construction is as follows. Let q be a prime power. The Erdős-Rényi graph ER q is the 1

2 graph whose vertices are the points of P G(, q). Two distinct vertices (x 0, x 1, x ) and (y 0, y 1, y ) are adjacent if x 0 y 0 + x 1 y 1 + x y = 0. It is well know that these graphs have q + q + 1 vertices, have 1 q(q + 1) edges and are C 4 -free. Hence for any prime power q, ex(q + q + 1, C 4 ) 1 q(q + 1). In [4] it is shown that if q > 13 is a prime power then ex(q + q + 1, C 4 ) 1 q(q + 1) and that the only C 4 -free graphs with q +q +1 vertices and 1q(q +1) egdes are graphs orthogonal polarity of finite projective planes. Consequently, ex(q + q + 1, C 4 ) = 1 q(q + 1), for all prime power q > 13. Another contribution to the Turán number of C 4 is that, ex(q + q, C 4 ) = 1 q(q + 1) q, provided that q is a power of two. In [5] shows that ex(q + q, C 4 ) 1 q(q + 1) q if q is even and in [6] that ex(q + q, C 4 ) 1 q(q + 1) q if q is a power of two, thus the result follows. These are the only cases in which an exact formula for ex(n, C4) is known. For more on the problem of the Turán number of C 4, see [7]. The study of Sidon sets has its origins in [8]. According to [9], Simon Sidon introduced Sidon sets to Erdos in 193 o A subset A of abelian group X is a Sidon set in X if and only if, a + b = c + d {a, b} = {c, d}, for all a, b, c and d in A. Sidon sets are studied objects in combinatorial number theory. For more on Sidon sets, see [10]. The following constructions of Sidon sets and their respective lemma can be consulted in [11]. We present the proof of the lemmas for completeness. Theorem 1.1. (Construction Bose-Chowla[1]). Let (F q, +, ) the finite field with q elements and θ a primitive element of F q h. The set, B = log θ (θ + F q ) := {log θ (θ + a) : a F q }, is a Sidon set in (Z q h 1, +), with q elements. Remark 1.1. If θ is a primitive element of F q 3, the previous theorem implies that, is a Sidon set in Z q 3 1, with q elements. A := {log θ (θ + a) : a F q },

3 Let S =: A mod (q + q + 1) and S 0 := S {0}. The set S is a Sidon set in Z q +q+1 with q elements. Theorem 1.. (Construction Singer [13]). With the previous notation, S 0 is a Sidon set in Z q +q+1 with q + 1 elements. Lemma 1.1. S 0 S 0 := {a b : a, b S, a b} = Z q +q+1\{0}. Proof. Since S S = q q, S = S = q, the sets S S, S, S are pairwise disjoint and S 0 S 0 = (S S) S ( S) then, por lo tanto, S 0 S 0 = Z q +q+1\{0}. S 0 S 0 = (S S) S ( S), = (S S) + S + ( S), = (q q) + q + q, = q + q, = Z q +q+1\{0}, Theorem 1.3. (Construction Ruzsa [14]). If θ is a primitive element of finite field Z p then, R =: {x ip θ i (p 1)(modp p) : 1 i p 1}, is a Sidon set in Z p p with p 1 elements. Let M p = {x Z p p : x 0(mod p)} and M p 1 = {y Z p p : y 0(modp 1)}. Lemma 1.. R R := {a b : a, b A, a b} = Z p p\{m p M p 1 }. Proof. Note that M p = p 1, M p 1 = p and M p M p 1 = {p(p 1)} because M p and M p 1 are subsets of Z p p. The above implies that, M p M p 1 = M p + M p 1 M p M p 1 = (p 1). On the other hand, since (R R) M p = and (R R) M p 1 = then, (R R) (M p M p 1 ) =. Now as R R = ( p 1 ) = (p 1)(p ) = p 3p + = p p (p 1) and R R Z p p then, R R = Z p p\{m p M p 1 }. Let (F p, +) the finite field with p elements and (F p, ) its multiplicative group. 3

4 Theorem 1.4. If p is a add prime, then C = {(a, a ) : a F p } is a Sidon set in (F p, +) (F p, +) with p elements. Lemma 1.3. C C = F p F p \{(0, z) : z F p }. Proof. Indeed, C C = {(x, x ) (y, y ) : x, y F p, (x, x ) (y, y )}, = {(x y, x y ) : x, y F p, (x, x ) (y, y )}. The elements (0, z), with z F p does not belong to this set. Indeed, if (0, z) = (x y, x y ), for some x, y and z in F p, then (x, x ) = (y, y ). To see that in the set {(0, z) : z F p } are all the elements that does not belong to C C. We do a count. Since F p F p = p, C = p, {(0, z) : z F p } = p y C is a Sidon set then, ( ) p C C = = p p. This completes the proof. Theorem 1.5. If p is a add prime, then I = {(a, a) : a F p} is a Sidon set in (F p, +) (F p, ) with p 1 elements. We denote the operation of the group (F p, +) (F p, ) by. Let I I := {(a, a) (b, b) 1 : a, b F p, (a, a) (b, b)}. Lemma 1.4. If A 1 = {(0, z) : z F p} and A = {(z, 1) : z F p} then, Proof. Indeed, I I = F p F p\(a 1 A ). I I := {(a, a) (b, b) 1 : a, b F p, (a, a) (b, b)} = {(a b, ab 1 ) : a, b F p, (a, a) (b, b)}. As (a, a) (b, b) then the elements of the difference set never have 0 in its first component, so (a b, ab 1 ) (0, z), for all z F p. On the other hand, since a b and b F p then ab 1 1 and hence the elements of the form (z, 1), z F p does not belong to the set difference. 4

5 To conclude the proof, note that F p F p = p(p 1), I = p 1, A 1 = A = p 1 and A 1 A = (p 1) since A 1 A =. Therefore, ( ) p 1 I I = = p 3p + = p(p 1) (p 1). Theorem 1.6. If p is a add prime, then I α = {(a α, a) : a, α F p, a α} is a Sidon set in (F p, ) (F p, ) with p elements. Let I α I α := {((a α)(b α) 1, ab 1 ) : a, b, α F p, a α b, (a α, a) (b α, b)}. Lemma 1.5. I α I α = F p F p\(a 1 A A 3 ), where A 1 = {(1, z) : z F p}, A = {(z, 1) : z F p} and A 3 = {(z, z) : z F p}. Proof. Note that (a α, a) (b α, b) implies that a α b α or a b. If a α b α then (a α)(b α) 1 1 since b α 0. Hence, the elements of the form (1, z) with z F p does not belong to I α I α. On the other hand, if a b then ab 1 1 because b 0. Thus, the elements (z, 1), z F p does not belong to the difference set. Other type of elements that are not in I α I α, are of the form (z, z) with z F p. To see this, suppose that, for some z F p. Now, ((a α)(b α) 1, ab 1 ) = (z, z), a α = ab 1 (b α), = a ab 1 α. Therefore, (ab 1 1)α = 0, which is not possible. As F p F p = (p 1), I α = p and A B C = 3(p 1) since A 1 = A = A 3 = p 1 and A B = A C = B C = A B C = {(1, 1)} then, ( ) p I α I α =, This completes the proof. = p 5p + 6, = p p + 1 3p + 3 +, = (p 1) (3(p 1) ). 5

6 Sum graph of Sidon sets. Definition.1. Let X an abelian group noticed additively and A a subset of X. The sum graph G X,A has set of vertices X and set of edges, E = {{a, b} X X : a + b A, a b}. Besides, a vertex a X is called an absolute point if and only if a + a A. When the set A is a Sidon set in X, the sum graph G X,A is C 4 -saturated. The next lemma is known [15], we give the proof for completeness. Lemma.1. If A is a Sidon set in the group X, then the sum graph G X,A does not contain a 4-cycle. Proof. If ijkl is a 4-cycle in G X,A, then are elements a, b, c, d A such that, i + j = a, j + k = b, k + l = c and l + i = d. This implies that a + c = b + d, but as A is a Sidon set, {a, c} = {b, d}. If a = b then i + j = j + k thus i = k. If a = d then i + j = l + i so i = l. In either case we have a contradiction, thus G X,A does not contain a 4-cycle. In 1966, Brown, Erdős, Rényi, and Sós independently constructed graphs that have q + q + 1 vertices, have 1 q(q + 1) edges, and are C 4 -free. These graphs and a result of [4] imply the Theorem.1. We will prove this theorem using a result of [4] and the sum graph of a Sidon set type Singer. Theorem.1. For each prime power q > 13, ex(q + q + 1, C 4 ) = 1 q(q + 1). A graph G is F -saturated if G is F -free and adding any edge to G creates a copy of F. In [6] is proved that if B is a Sidon set type Bose-Chowla in the group X = Z q 1, q a prime power, then the sum graph G X,B is C 4 -saturated. We extend this result to other Sidon sets. The next theorem gives conditions for the sum graph G X,A be C 4 -saturated. Theorem.. Let A a Sidon set in the finite group X. If for any x X\A there are four ordered triples (a, b, c) A 3 such that, x = a b + c, with a b and b c. Then the sum graph G X,A is C 4 -saturated. Proof. Let x, y X\A, x y such that x + y / A, by supposition there are four ordered triples (a, b, c) A 3 verifying, x + y = a b + c, 6

7 with a b and b c. If (a 1, b 1, c 1 ) and (a, b, c ) are two such triples, then their first coordinate must be different unless they are the same triple. Indeed, if a 1 = a, then c 1 b 1 = c b. As b 1 c 1 and A is a Sidon set, we must have b 1 = b and c 1 = c. A similar argument applies to the third coordinate of each such triple. Choose a triple (a, b, c) so that a x + x and c y + y. There are at least two of such triples. Let z = a x and t = c y. Then x + z = a, y + t = c and z + t = a x + c y = b, hence x is adjacent to z, y is adjacent to t and z is adjacent to t. In order to conclude that xzty is a path of length three, we must show that x, z, t and y are all distinct. If x = z then a = x + z = x + x which does not occur. We conclude that x z and similarly t y. Note that, z = t can only occur for a unique triple, otherwise a four cycle is formed. In consequence, there is at least one triple (a, b, c) such that xzty is a path of length three between x and y. So adding the edge xy gives a 4-cycle. This completes the proof of theorem. As an application of the Theorem., we will prove that the sum graphs associated to the Sidon sets S, R, C, I and I α are C 4 -saturated. Theorem.3. Si p 11 es primo y R es el conjunto de Sidon en X = Z p p, entonces el grafo suma S X,R es C 4 -saturado. Theorem.4. Si p 7 es un número primo y C es el conjunto de Sidon en X = F p F p, entonces el grafo S X,C es C 4 -saturado. Theorem.5. Si p 11 es un número primo y I es el conjunto de Sidon en X = F p F p, entonces el grafo S X,I es C 4 -saturado. Theorem.6. Si p 11 es primo y I α es el conjunto de Sidon en X = F p F p, entonces el grafo S X,Iα es C 4 -saturado. Proof of Theorem.1. In order to proof this theorem, we need some auxiliary results. Theorem.7. Let X an finite abelian group noticed additively and A a subset of X. If G X,A is the sum graph and P = {a X : a + a A}, then E = X A P. Proof. We are interested in how many different elements there are in, E = {{a, b} X X : a + b A, a b}. 7

8 For purposes of counting, in the following table we illustrated the sum of the group X = {x 1, x,, x k } with itself: + x 1 x x k x 1 x 1 + x 1 x + x 1 x k + x 1 x x 1 + x x + x x k + x x k x 1 + x k x + x k x k + x k Table 1: Sum set X + X. Since X is a group, the elements of each row in the Table 1 are all distinct. As each row has X elements, then the elements of set A appear exactly once in each row. Thus we have to A X pairs (a, b) of X X that verify a + b A. As X is abelian, we must divide by to avoid repetitions. Let S 0 the Sidon set of the Theorem 1. and X = Z q +q+1. In order to calculate the exact number of edges of the sum graph G X,S0, we require the following lemma. Lemma.. If P = {x X : x + x S 0 } then P = q + 1. Proof. If x P, then x a(modq + q + 1) for some a S 0. As q + q + 1 is odd, the previous congruence has a unique solution for each a. Of the above and the fact that S 0 = q + 1 it follows that P = q + 1. Theorem.8. The number of edges of the graph sum G X,S0 is 1 q(q + 1). Proof. By the Theorem.7 and the Lemma. the number of egdes of sum graph G X,S0 is, (q + q + 1)(q + 1) (q + 1) = (q + 1)(q + q), = 1 q(q + 1). Proof of Theorem.1. The Theorem.8 implies that for all prime power q, 1 q(q + 1) ex(q + q + 1, C 4 ). On the other hand, in [4] it is shown that if q > 13 is a prime power, then ex(q + q + 1, C 4 ) 1 q(q + 1). In consecuense, ex(q + q + 1, C 4 ) = 1 q(q + 1) for all prime power q > 13. 8

9 Remark.1. In [6] it is established that ex(q + q, C 4 ) 1 q(q + 1) q if q is a power of two. This bound can be obtained by removing a vertex of degree q of the sum graph of a Sidon set type Singer, if q is a power of two. In [5] it is established that when q is a power of two, the free graphs of C 4 with q + q vertex and 1 q(q + 1) q edges are obtained by eliminating a vertex of degree q from an orthogonal polarity graph. Proof of Theorem.3. In order to prove the theorem we need the following lemma. Lemma.3. If p is an add prime then for any x Z p p\r there is at least p 3 ordered triples (a, b, c) R 3 such that a b + c = x with a b and b c. Proof. We consider two cases. Case 1. Let x M p := {x Z p p : x 0(mod p)} and a R. Note that x a / M p and that there is at most one a R such that x a M p 1 := {y Z p p : y 0(modp 1)}. Suppose that R = {a 1, a,, a p 1 }, without loss of generality it can be assumed that x a i / M p 1 for i p 1. Therefore, x a i / (M p M p 1 ) for i p 1. In consecuense by the Lemma1., x a i R R, so there is a unique pair b i, c i R such that x a i = c i b i. Indeed, if x a i = c i b i then {c i, b i} = {c i, b i } by property Sidon. Rewriting the equation we have that x = a i b i +c i. If a i = b i or b i = c i then x R, which contradicts the hypothesis. Finally note that the triples (a, b, c ),, (a p 1, b p 1, c p 1 ) are all all distinct since a, a 3,, a p 1 are all distinct. In this case there are at least p triples that satisfy the required. Case. Let x / M p and R = {a 1, a,, a p 1 }. Consider the p 1 distinct elements, x a 1, x a,, x a p 1. Note that there is at most one a R such that x a M p and there is at most one a R such that x a M p 1. Besides that a a, otherwise, x a M p M p 1 = {0}. Without loss of generality, it can be assumed that x a 1 M p and x a M p 1. Then x a i / (M p M p 1 ), 3 i p 1. Therefore, by the Lemma 1., x a i R R for 3 i p 1, so there is a unique pair b i, c i R such that x a i = c i b i. Rewriting the equation we have that x = a i b i + c i. If a i = b i and b i = c i then x R, contradicting the hypothesis. Finally note that the triples (a 3, b 3, c 3 ),, (a p 1, b p 1, c p 1 ) are all distinct since a 3,, a p 1 are all distinct. In this case, there are at least p 3 triples that satisfy the required. Proof of Theorem.3. It follows from the Theorem. and the Lemma.3. Proof of Theorem.4. In order to prove the theorem we need the following lemma. 9

10 Lemma.4. For any (x, y) F p F p \C there are at least p 1 ordered triples, ((a, a ), (b, b ), (c, c )) C 3, such that (x, y) = (a, a ) (b, b ) + (c, c ), a b and b c. Proof. Suppose that C = {(a 0, a 0), (a 1, a 1),, (a p 1, a p 1)}. Let (x, y) F p F p \C. By the Lemma 1.3 (x, y) (a k, a k ) C C, 0 k p 1, unless x a k 0 mod p, that is, if x = a k for some k. Without loss of generality suppose that x = a 0. So, (x, y) (a k, a k ) C C, for 1 k p 1. Therefore, there are unique (b i, b i ), (c i, c i ), 1 i p 1, such that, Indeed, if (x, y) (a k, a k ) = (c k, c k ) (b k, b k ), 1 k p 1. (x, y) (a k, a k ) = (c k, c k ) (b k, b k ) and (x, y) (a k, a k ) = (c k, c k ) (b k, b k ), then, (c k, c k) (b k, b k) = (c k, c k ) (b k, b k ), and therefore, (c k, c k) + (b k, b k ) = (c k, c k ) + (b k, b k), thus (c k, c k ) = (c k, c k ) and (b k, b k ) = (b k, b k ) because C is a Sidon set and (b k, b k ) (c k, c k ), otherwise (x, y) C. Similarly, (a k, a k ) (b k, b k ). So, there are at least p 1 ordered triples that satisfy the required. Proof of Theorem.4. It follows from the Theorem. and the Lemma.4. Proof of Theorem.5. In order to prove the theorem we need the following lemma. Lemma.5. For any (x, y) F p F p\i there are at least p 3 triple ordered ((a, a), (b, b), (c, c)) I 3 such that (x, y) = (a, a) (b, b) 1 (c, c), a b y b c. Proof. We consider two cases. Case 1. Let (x, y) = (0, z), z F p and (a, a) I. By definition (x, y) / I. Note that (x, y) (a, a) 1 = ( a, za 1 ), then there is p 1 elections for a, because a F p. On the other hand, by the Lemma 1.4 ( a, za 1 ) I I, unless, ( a, za 1 ) = (0, t), t F p or ( a, za 1 ) = (s, 1), s F p, the first never occurs and the second occurs only if z = a, therefore, there are at least p elections of a that satisfy that (x, y) (a, a) 1 I I, so there are unique (b, b), (c, c) I such that (x, y) (a, a) 1 = (c, c) (b, b) 1. As (x, y) (a, a) 1 (0, z), then (b, b), (c, c) are unique since I is a Sidon set. If a = b or b = c then (x, y) I which is not possible. In this case there are at least p ordered triples that satisfy the required. Case. Let (x, y) (0, z) F p F p\i and (a, a) I. Note that by Lemma 1.4 (x, y) (a, a) 1 I I unless, 10

11 (x a, ya 1 ) = (0, t), t F p or (x a, ya 1 ) = (s, 1), s F p, the first occurs only if x = a, so there is p elections for a because a F p, the second occurs if ya 1 = 1, that is, if y = a, hence there is p 3 elections for a which guarantees that (x, y) (a, a) 1 I I, so there are unique (b, b), (c, c) I such that (x, y) (a, a) 1 = (c, c) (b, b) 1. In this case there are at least p 3 ordered triples that satisfy the required. Proof of Theorem.5. It follows from the Theorem. and the Lemma.5. Proof of Theorem.6. In order to prove the theorem we need the following lemma. Lemma.6. For any (x, y) X = F p F p \I α there are at least p 5 ordered triples ((a α, a), (b α, b), (c α, c)) Iα 3 such that (x, y) = (a α, a)(b α, b) 1 (c α, c), with a b and b c. Proof. Let (x, y) X and (a α, a) I α. By the Lemma 1.5 (x, y)(a α, a) 1 I α I α, unless, (x, y)(a α, a) 1 = (1, z), z F p, (x, y)(a α, a) 1 = (s, 1), s F p, (x, y)(a α, a) 1 = (t, t), t F p. o That is, if x (a α) mod p, y a mod p or xy 1 (1 a 1 α) mod p; these congruences have unique solution. From the above and the fact that I α = p it follows that there are at least p 5 elements (a α, a) in I α with (x, y)(a α, a) 1 I α I α. Therefore, there are unique elements (b α, b), (c α, c) I α that satisfy, (x, y)(a α, a) 1 = (c α, c)(b α, b) 1. Uniqueness is because (x, y)(a α, a) 1 (1, 1) and I α is a Sidon set. On the other hand, (a α, a) (b α, b) and (b α, b) (c α, c), since otherwise (x, y) I α. Proof of Theorem.6. It follows from the Theorem. and the Lemma.6. References [1] I. Reiman, Űber ein Problem von K. Zarankiewicz, Acta Math Acad Sci Hungar, 9 (1958), no. 1-4, [] W. G. Brown. On graphs that do not contain a Thomsen graph, Math. Bull, 9 (1966), no. 3, [3] P. Erdős, A. Rényi and V. T. Sós. On a problem of graph theory, Stud. Sci. Math, 1 (1966),

12 [4] Z. Fűredi. On the number of edges of quadrilateral-free graphs, Journal of Combinatorial Theory, Series B, 60, no. 1, 1-6. [5] F. Firke, P. Kosek, E. Nash and J. Williford. Extremal graphs without 4-cycles, Journal of Combinatorial Theory, Series B, 103, no. 3, [6] M. Tait and C. Timmons, Orthogonal Polarity Graphs and Sidon Sets, Journal of Graph Theory, 8 (016), no. 1, [7] Z. Fűredi and M. Simonovits. The history of degenerate (bipartite) extremal graph problems, Bolyai Society Mathematical Studies, 5 (013), [8] P. Erdős and P. Turán, on a problem of Sidon in additive number theory, and on some related problems, J. London Math. Soc, 16 (1941), no. 1, [9] P. Erdős, A survey of problems in combinatorial number theory, Annals of Discrete Mathematics, 6 (1980), no. 1, [10] K. O Bryant, A complete annotated bibliography of work related to Sidon sequences, Electr. J. Combin. DS 11 (004). [11] N. Y. Caicedo and C. A. Trujillo. Conjuntos de Sidon en dos dimensiones. Tesis Doctoral, Universidad del Valle, 016. [1] R. C. Bose, An affine analogue Singer s theorem, J. Ind. Math. Soc. (new series), 6 (194), [13] J. Singer, A Theorem in Finite Projective Geometry and Some Applications to Number Theory, Trans. Amer. Math. Soc, 43 (1938), no. 3, [14] I. Z. Ruzsa, Solving a linear equation in a set of integers I, Acta Arithmetica, 65 (1993), no. 3, [15] J. Cilleruelo, Infinite C 4 -free graphs. Personal page. Preprint. 1