On Snevily s Conjecture and Related Topics

Jiangsu University (Nov. 24, 2017) and Shandong University (Dec. 1, 2017) and Hunan University (Dec. 10, 2017) On Snevily s Conjecture and Related Topics Zhi-Wei Sun Nanjing University Nanjing 210093, P. R. China zwsun@nju.edu.cn http://math.nju.edu.cn/ zwsun December 10, 2017

Abstract Let G be a finite abelian group of odd order. In 1999 H. S. Snevily conjectured that for any two subsets A and B of G with A = B = n there is a numbering {a i } n i=1 of the elements of A and a numbering {b i } n i=1 of the elements of B such that a 1 + b 1,..., a n + b n are pairwise distinct. In this talk we first review the initial progress on this conjecture via Alon s Combinatorial Nullstellensatz, and also the Feng-Sun-Xiang work on the related Dasgupta-Karolyi-Serra-Szegedy conjecture via characters of abelian groups. Finally we talk about B. Arsovski s elegant solution of the Snevily conjecture and pose a novel conjecture on finite abelian groups. 2 / 35

Hall s theorem In 1952 M. Hall [Proc. Amer. Math. Soc.] obtained an extension of Cramer s conjecture. M. Hall s theorem Let G = {b 1,..., b n } be an additive abelian group, and let a 1,..., a n be elements of G with a 1 +... + a n = 0. Then there exists a permutation σ S n such that {a σ(1) + b 1,..., a σ(n) + b n } = G. Remark. Hall used induction argument and his method is very technique. Up to now there are no other proofs of this theorem. Observation. If a 1,..., a n Z are incongruent modulo n with a 1 + + a n 0 (mod n), then n divides and hence n is odd. 0 + 1 + + (n 1) = n(n 1) 2 3 / 35

Snevily s Conjecture Snevily s Conjecture for Abelian Groups [Amer. Math. Monthly, 1999]. Let G be an additive abelian group of odd order. Then for any two subsets A = {a 1,..., a k } and B = {b 1,..., b k } of G with A = B = k, there is a permutation σ S k such that a σ(1) + b 1,..., a σ(k) + b k are (pairwise) distinct. Remark. The result does not hold for any group G of even order. In fact, there is an element g G of order 2, and A = B = {0, g} gives a counterexample. Another Form of Snevily s Conjecture. Let G = {a 1,..., a N } be an additive abelian group with G = N odd, and let M be the Latin square (a i + a j ) 1 i,j N formed by the Cayley addition table. Then any n n submatrix of M contains a Latin transversal. Snevily s conjecture looks simple, beautiful and difficult! 4 / 35

Jäger-Alon-Tarsi Conjecture In 1982, motivated by his study of graph theory, F. Jäger posed the following conjecture in the case F = 5 Jäger-Alon-Tarsi Conjecture. Let F be a finite field with at least 4 elements, and let A be an invertible n n matrix with entries in F. There there exists a vector x F n such that both x and A x have no zero component. In 1989 N. Alon and M. Tarsi [Combinarorica, 9(1989)] confirmed the conjecture in the case when F is not a prime. Moreover their method resulted in the initial form of the Combinatorial Nullstellensatz which was refined by Alon in 1999. 5 / 35

Alon s Combinatorial Nullstellensatz Combinatorial Nullstellensatz [Combin. Probab. Comput. 8(1999)]. Let A 1,..., A n be finite nonempty subsets of a field F and let f (x 1,..., x n ) F [x 1,..., x n ]. Suppose that 0 k i < A i for i = 1,..., n, k 1 + + k n = deg f and [x k 1 1 x kn n ]f (x 1,..., x n ) (the coefficient of x k 1 1 x kn n in f ) does not vanish. Then there are a 1 A 1,..., a n A n such that f (a 1,..., a n ) 0. Advantage: This advanced algebraic tool enables us to establish existence via computation. It has many applications. 6 / 35

Alon s contribution for cyclic groups of prime orders Alon s Result [Israel J. Math. 2000]. Let p be an odd prime and let A = {a 1,..., a k } be a subset of Z p = Z/pZ with cardinality k < p. Given (not necessarily distinct) b 1,..., b k Z p there is a permutation σ S k such that a σ(1) + b 1,..., a σ(k) + b k are distinct. Remark. This result is slightly stronger than Snevily s conjecture for cyclic groups of prime order. Proof. Let A 1 = = A k = {a 1,..., a k }. We need to show that there exist x 1 A 1,..., x k A k such that 1 i<j k (x j x i )(x j + b j (x i + b i )) 0. By the Combinatorial Nullstellensatz, it suffices to prove c := [x k 1 1 x k 1 k ] 1 i<j k (x j x i )(x j + b j (x i + b i )) 0. 7 / 35

For σ S k let ε(σ) be the sign of σ which takes 1 or 1 according as the permutation σ is even or odd. Recall that det(a ij ) 1 i,j k = σ S k ε(σ) Now we have c =[x k 1 1 x k 1 k ] =[x k 1 =[x k 1 k a i,σ(i), per(a ij ) 1 i,j k = σ S k i=1 1 i<j k (x j x i ) 2 ](det(x i 1 j ) 1 i,j k ) 2 1 x k 1 k 1 x k 1 ] k ε(σ) σ S k k j=1 x σ(j) 1 j τ S k ε(τ) k j=1 k a i,σ(i). i=1 x τ(j) 1 j = σ S k ε(σ)ε(σ ) = σ S k ( 1) (k 2) = k!( 1) ( k 2) 0 (in Zp ). where σ (j) = k σ(j) + 1 for j = 1,..., k. 8 / 35

An extension of Alon s result by Hou and Sun Theorem. (Qing-Hu Hou and Z.-W. Sun [Acta Arith. 102(2002)]) Let k n 1 be integers, and left F be a field whose characteristic is zero or greater than max{n, (k n)n}. Let A 1,..., A n be subsets of F with cardinality k, and let b 1,..., b n F. Then the sumset {a 1 +... + a n : a i A i, a i a j and a i + b i a j + b j if i j} have more than (k n)n elements. Actually, Hou and Sun proved a much more general result including the above theorem as a special case. 9 / 35

Snevily s Conjecture for cyclic groups For odd composite number n, Z n = Z/nZ is not a field. How to prove Snevily s conjecture for the cyclic group Z n? Dasgupta, Károlyi, Serra and Szegedy [Israel J. Math., 2001] Snevily s conjecture holds for any cyclic group of odd order. Their key observation is that a cyclic group of odd order n can be viewed as a subgroup of the multiplicative group of the finite field F 2 ϕ(n). Thus, it suffices to show that c := [x k 1 1 x k 1 k ] 1 i<j k (x j x i )(b j x j b i x i ) 0. Now c depends on b 1,..., b k so that the condition 1 i<j k (b j b i ) 0 might be helpful. 10 / 35

Comuting c For σ S k let ε(σ) be the sign of σ. Then 1 i<j k =( 1) (k 2) x k j i =( 1) (k 2) σ S k ε(σ) Therefore ( 1) (k 2) c = (x j x i )(b j x j b i x i ) σ S k ε(σ) 2 = σ S k ε(σ) 1 i,j k b j 1 i k i=1 k i=1 k i=1 x k σ(i) i b σ(i) 1 i = b i 1 j 1 i,j k = x j 1 i 1 i,j k τ S k ε(τ) k i=1 b τ(i) 1 i x τ(i) 1 i. = per((b j 1 i ) 1 i,j k ) b σ(i) 1 i (because ch(f ) = 2) 1 i<j k (b j b i ) 0 (Vandermonde). 11 / 35

My related work Theorem [Z. W. Sun, J. Combin. Theory Ser A 103(2003)] Let G be an additive abelian group whose finite subgroups are all cyclic. Let b 1,..., b n be pairwise distinct elements of G, and let A 1,..., A n be finite subsets of G with cardinality k m(n 1) + 1 where m is a positive integer. (i) There are at least (k 1)n m ( n 2) + 1 multisets {a1,..., a n } such that a i A i for i = 1,..., n and all the ma i + b i are pairwise distinct. (ii) If b 1,..., b n are of odd order, then the sets {{a 1,..., a n }: a i A i, a i a j and ma i + b i ma j + b j if i j} and {{a 1,..., a n }: a i A i, ma i ma j and a i + b i a j + b j if i j} have more than (k 1)n (m + 1) ( n 2) (m 1) ( n 2) elements. 12 / 35

My related work In the proof I used Alon s Combinatorial Nullstellensatz and Dirichlet s unit theorem in Algebraic Number Theory. The theorem follows from my stronger results on sumsets with polynomial restrictions for which we need the following auxiliary result. Lemma (Sun). Let R be a commutative ring with identity. Let A = (a ij ) 1 i,j n be a matrix over R, and let k, m 1,..., m n N. (i) If m 1 m n k, then we have [x k 1 x k n ] a ij x m i j ( k ) kn n i=1 1 i,j n x m i s s=1 (ii) If m 1 < < m n k then [x k 1 x k n ] a ij x m i j 1 i,j n 1 i<j n = (kn n i=1 m i)! n i=1 (k m det(a). i)! ( n ) kn ( n 2) n i=1 m i (x j x i ) x s (kn ( n = ( 1) (n 2) 2) n i=1 m i)! n i=1 m i <j k (j m i ) per(a), j m i+1,...,mn 13 / 35 s=1

My result of dimension three Theorem [Z. W. Sun, Math. Res. Lett., 15(2008)] Let G be any additive abelian group with cyclic torsion subgroup, and let A, B, C be subsets of G with cardinality n Z +. Then we can write A = {a 1,..., a n }, B = {b 1,..., b n } and C = {c 1,..., c n } such that a 1 + b 1 + c 1, a 2 + b 2 + c 2,..., a n + b n + c n are pairwise distinct. Corollary Let N be any positive integer. For the N N N Latin cube over Z/NZ formed by the Cayley addition table, each n n n subcube with n N contains a Latin transversal. Conjecture (Z. W. Sun [Math. Res. Lett. 15(2008)]) Every n n n Latin cube contains a Latin transversal. 14 / 35

A general result on restricted sumsets (joint with L. Zhao) Theorem (Z. W. Sun & L. Zhao [J. Combin. Theory Ser A 119(2012)] Let P(x 1,..., x n ) be a polynomial over a field F. Suppose that k 1,..., k n are nonnegative integers with k 1 + + k n = deg P and [x k 1 1 x n kn ]P(x 1,..., x n ) 0. Let A 1,..., A n be finite subsets of F with A i > k i for i = 1,..., n. Then, for the restricted sumset C = {x 1 + +x n : x 1 A 1,..., x n A n, and P(x 1,..., x n ) 0}, we have C min{p(f ) deg P, A 1 + + A n n 2 deg P + 1}, where p(f ) = if F is of characteristic zero, and p(f ) = p if F is of prime characteristic p. Remark. In the proof we apply the Combinatorial Nullstellensatz twice. 15 / 35

The DKSS Conjecture The DKSS Conjecture (Dasgupta, Károlyi, Serra and Szegedy [Israel J. Math., 2001]). Let G be a finite abelian group with G > 1, and let p(g) be the smallest prime divisor of G. Let k < p(g) be a positive integer. Assume that A = {a 1, a 2,..., a k } is a k-subset of G and b 1, b 2,..., b k are (not necessarily distinct) elements of G. Then there is a permutation π S k such that a 1 b π(1),..., a k b π(k) are distinct. Remark. When G = Z p = Z/pZ, the DKSS conjecture reduces to Alon s result. DKSS proved their conjecture for Z p n and Z n p via the Combinatorial Nullstellensatz. W. D. Gao and D. J. Wang [Israel J. Math. 2004]: The DKSS conjecture holds when k < p(g), or G is an abelian p-group and k < 2p. 16 / 35

A Recent Result of Feng, Sun and Xiang Theorem [T. Feng, Z. W. Sun & Q. Xiang, Israel J. Math., 182(2011)]. Let G be a finite abelian group with G > 1. Let A = {a 1,..., a k } be a k-subset of G and let b 1,..., b k G, where k < p = p(g). Then there is a permutation π S k such that a 1 b π(1),..., a k b π(k) are distinct, provided either of (i)-(iii). (i) A or B is contained in a p-subgroup of G. (ii) Any prime divisor of G other than p is greater than k!. (iii) There is an a G such that a i = a i for all i = 1,..., k. Remark. By this result, the DKSS conjecture holds for any abelian p-group! Tools: Characters of abelian groups, exterior algebras. Below I ll introduce the work of Feng-Sun-Xiang by avoiding exterior algebra. 17 / 35

Characterize the distinction of elements a 1,..., a k (in a field) are distinct 1 i<j k (a j a i ) 0. Let a 1,..., a k be elements of a finite abelian group G. How to characterize that a 1,..., a k are distinct? We need the character group Ĝ = {χ : G K \ {0} : χ(ab) = χ(a)χ(b) for any a, b G}, where K is a field having an element of multiplicative order G. It is well known that Ĝ = G. Lemma 1 (Feng-Sun-Xiang) a 1,..., a k G are distinct if and only if there are χ 1,..., χ k Ĝ such that det(χ i (a j )) 1 i,j k 0. Proof. If a s = a t for some 1 s < t k, then for any χ 1,..., χ k Ĝ the determinant det(χ i(a j )) 1 i,j k vanishes since the sth column and tth column of the matrix (χ i (a j )) 1 i,j k are identical. 18 / 35

Continue the proof Now suppose that a 1,..., a k are distinct. If the characteristic of K is a prime p dividing G, then (x G /p 1) p = x G 1 for all x K, which contradicts the assumption that K contains an element of multiplicative order G. So we have G 1 0, where 1 is the identity of the field K. It is well known that { 0, if a G \ {e}, χ(a) = G 1, if a = e. χ Ĝ To show that there are χ 1,..., χ k Ĝ such that det(χ i (a j )) 1 i,j k 0, we make the following observation. 19 / 35

Continue the proof χ 1,...,χ k Ĝ = χ 1,...,χ k Ĝ χ 1 (a1 1 ) χ k(a 1 ) det(χ i(a j )) 1 i,j k k χ 1 (a1 1 ) χ k(a 1 ) k ε(π) χ i (a π(i) ) π S k = k ε(π) χ 1,...,χ k Ĝ π S k i=1 = k ε(π) π S k =ε(i ) i=1 χ i Ĝ k χ i (a π(i) a 1 i ) χ i (a π(i) a 1 i ) k ( G 1) = ( G 1) k 0, i=1 where I is the identity permutation in S k. i=1 20 / 35

Another Lemma Lemma 2 (Feng-Sun-Xiang). Let a 1,..., a k, b 1,..., b k G and χ 1,..., χ k Ĝ. If det(χ i (a j )) 1 i,j k ) and per(χ i (b j )) 1 i,j k ) are nonzero, then for some π S k the products a 1 b π(1),..., a k b π(k) are distinct. Proof. By Lemma 1 it suffice to show that det(χ i (a j b π(j) )) 1 i,j k 0 for some π S k. Note that det(χ i (a j b π(j) )) 1 i,j k π S k = k ε(σ) χ i (a σ(i) b π(σ(i)) ) π S k σ S k i=1 = k ε(σ) χ i (a σ(i) ) k χ i (b πσ(i) ) σ S k i=1 π S k i=1 = det(χ i (a j )) 1 i,j k per(χ i (b j )) 1 i,j k 0. 21 / 35

One more lemma Lemma 3 (Z. W. Sun, Trans. AMS 1996; Combinatorica, 2003) Let λ 1,..., λ k be complex nth roots of unity. Suppose that c 1 λ 1 + + c k λ k = 0, where c 1,..., c k are nonnegative integers. Then c 1 + + c k can be written in the form p n px p, where the sum is over all prime divisors of n and the x p are nonnegative integers. Tools for the proof. Galois group of cyclotomic extension, Newton s identity for symmetric functions. Corollary. Let p be a prime and let a Z +. If λ 1,... λ k are p a th roots of unity with λ 1 + + λ k = 0, then k 0 (mod p). 22 / 35

Proof of the DKSS conjecture for abelian p-groups Let G be an abelian p-group with G = p a > 1 and let a 1,..., a k be distinct elements of G with k < p. Let b 1,..., b k be a sequence of (not necessarily distinct) elements of G. Let Ĝ be the group of all complex-valued characters of G. As a 1,..., a k are distinct, by Lemma 1 there are χ 1,..., χ k Ĝ such that det(χ i (a j )) 1 i,j k 0. Since all those χ i (b j ) are p a th roots of unity and S k = k! 0 (mod p), by Lemma 3 we have per(χ i (b j )) 1 i,j k = π S k k χ i (b π(i) ) 0. Applying Lemma 2 we see that for some π S k the products a 1 b π(1),..., a k b π(k) are distinct! i=1 23 / 35

Open Problem How to prove the DKSS conjecture for general finite abelian groups? In particular, how to prove the DKSS conjecture for the cyclic group Z/nZ? 24 / 35

A conjecture implying Snevily s conjecture Conjecture (Feng, Sun and Xiang, 2009) Let G be a finite abelian group, and let A = {a 1,..., a k } and B = {b 1,..., b k } be two k-subsets of G. Let K be any field containing an element of multiplicative order G, and let Ĝ be the character group of all group homomorphisms from G to K = K \ {0}. Then there are χ 1,..., χ k Ĝ such that det(χ i(a j )) 1 i,j k and det(χ i (b j )) 1 i,j k are both nonzero. Remark. When G is cyclic, we may take χ i = χ i for i = 1,..., k, where χ is a generator of Ĝ. Then the two determinants det(χ i (a j )) 1 i,j k and det(χ i (b j )) 1 i,j k in the above conjecture are both nonzero since they are Vandermonde determinants. Therefore the conjecture is true for cyclic groups. 25 / 35

A Result of Feng, Sun and Xiang Theorem (FSX) (i) The conjecture of FSX implies Snevily s conjecture. (ii) Let G, A, B, Ĝ be as in the conjecture of FSX. Assume that there is a π S k such that C = {a 1 b π(1),..., a k b π(k) } is a k-set with {a 1 b τ(1),..., a k b τ(k) } C for all τ S k \ {π}. Then there are χ 1,..., χ k Ĝ such that det(χ i (a j )) 1 i,j k det(χ i (b j )) 1 i,j k 0. Also, there are χ 1,..., χ k Ĝ such that det(χ i (a j )) 1 i,j k per(χ i (b j )) 1 i,j k 0, and there are ψ 1,..., ψ k Ĝ such that per(ψ i (a j )) 1 i,j k per(ψ i (b j )) 1 i,j k 0. 26 / 35

Arsovski solved the Snevily conjecture In 2010 B. Arsovski [Israel J. Math. 182(2011)] proved Snevily s conjecture fully! A key lemma is closely related to the conditions in part (ii) of the above result of FSX. Combinatorial Lemma of Arsovski. Let A = {a 1,..., a k } and B = {b 1,..., b k } be k-subsets of an arbitrary abelian group G. Then, there exists a permutation π S k such that for any permutation σ S k \ {π}, the multisets {a 1 b π(1),..., a k b π(k) } and {a 1 b σ(1),..., a k b σ(k) } are different. Motivated by Arsovski s solution to Snevily s conjecture, G. Harcos, G. Károlyi, G. Kós [arxiv:1004.0253] confirmed the conjecture of FSX. 27 / 35

Proof of Arsovski s Combinatorial Lemma The lemma is trivial for k = 1. For k = 2 we may just let π be the identity of S k = S 2 since the multi-sets {a 1 b 1, a 2 b 2 } and {a 1 b 2, a 2 b 1 } are different. Now let k > 2 and assume that the lemma holds for smaller values of k. Take g = a 1 b 1, and set I = {1 i k : a i b j = g for some 1 j k} and I = {1,..., k} \ I. For each i I, let π(i) be the unique j {1,..., k} with a i b j = g. Clearly, all those π(i) with i I are distinct. If I = {1,..., k}, then a 1 b π(1) =... = a k b π(k) = g and the two multi-sets {a i b σ(i) : i = 1,..., k} and {a i b π(i) : i = 1,..., k} are different for any σ S k \ {π}. 28 / 35

Proof of Arsovski s Combinatorial Lemma (continued) Now let I. By the induction hypothesis to the (k I )-subsets A = {a i : i Ī } and B = {b j : j {1,..., k} \ {π(i) : i I }}, there is a permutation π on I such that for any other permutation σ on I the multi-sets {a i b π (i) : i I } and {a i b σ (i) : i Ī } are different. Let π(i) = π (i) for all i I. Then π S k. Suppose that σ is a permutation in S k with the two multi-sets {a 1 b π(1),..., a k b π(k) } and {a 1 b σ(1),..., a k b σ(k) } equal. As g appears in the first multi-set exactly I times, it also appears in the second multi-set exactly I times. So σ(i) = π(i) for all i I. Since the muti-sets {a i b π (i) : i I } and {a i b σ(i) : i I } are equal, σ(i) = π (i) for all i I. Therefore σ = π. This concludes the proof. 29 / 35

Arsovski s proof of Snevily s conjecture Let F be a field of characteristic 2 containing an element of multiplicative order m = G and consider its purely transcendental extension K = F (t 1,..., t m ). The character group Ĝ = {χ : G K \ {0} χ(ab) = χ(a)χ(b) for all a, b G} is isomorphic to G, and the characters in Ĝ form a basis of the vector space V := {maps from G to K} over the field K. Write G = {g 1,..., g m } and define f (g i ) = t i for i = 1,..., m. By Arsovski s combinatorial lemma, there is a permutation π S k such that for any σ S k \ {π} the two multi-sets {a i b π(i) : i = 1,..., k} and {a i b σ(i) : i = 1,..., k} are different. Suppose that each g j (1 j m) occurs in the multi-set {a i b π(i) : i = 1,..., k} exactly k j times. Then D =: det(f (a i b j )) 1 i,j n 0 since its expansion contains the monomial t k 1 1... tkm m only once. 30 / 35

Arsovski s proof of Snevily s conjecture (continued) Write Ĝ = {χ 1,..., χ m } and f = c 1 χ 1 +... + c m χ m with c 1,..., c m K. Then D = det(σ m s=1c s χ s (a i b j )) = = = = = 1 s 1,...,s k m distinct 1 s 1 <...<s k m m... s 1 =1 det(c si χ si (a i b j )) 1 s 1 <...<s k m σ S k 1 s 1 <...<s k m m det(c si χ si (a i b j )) s k =1 σ S k det(c sσ(i) χ sσ(i) (a i b j )) τ S k ε(τ) k c sσ(i) χ sσ(i) (a i b τ(i) ) i=1 c s1 c sk ε(τ) τ S k σ S k i=1 k χ sσ(i) (a i b τ(i) ). 31 / 35

Arsovski s proof of Snevily s conjecture (continued) Recall that K is of characteristic 2. Thus D = 1 s 1 <...<s k m c s1 c sk τ S k det(χ sj (a i b τ(i) ). Since D 0, for some 1 s 1 <... < s k m and τ S k we have det(χ sj (a i b τ(i) ) 0 and hence a 1 b τ(1),..., a k b τ(k) are distinct. Comments from a book of D. J. Grynkiewicz: Snevily s conjecture was finally solved by Arsovski, aided by the preparatory work of Feng, Sun and Xiang who had already shown that Snevily s conjecture could be deduced from a weakened version of Theorem 18.2, which remained a conjecture at the time. 32 / 35

Two conjectures implying Snevily s conjecture Conjecture 1 (Z.-W. Sun, 2017). Let G be any additive abelian group, and let b 1,..., b n be distinct elements of G with odd order. If A 1,..., A n are subsets of G with cardinality n, then there are distinct a 1 A 1,..., a n A n such that a 1 + b 1,..., a n + b n are pairwise distinct. Remark. By my work in [JCTA 2003] and [Math. Res. Lett. 2008], this conjecture holds for cyclic groups. Conjecture 2 (Z.-W. Sun, 2017). Let G be any additive abelian group, and let A and B be two subsets of G with A = B = n Z + then we can write A = {a 1,..., a n } and B = {b 1,..., b n } such that either a 1 + 2b 1,..., a n + 2b n are pairwise distinct or 2a 1 + b 1,..., 2a n + b n are pairwise distinct. Remark. By the author s work in arxiv:1309.1679, this conjecture holds if G is torsion-free and A is equal to B. When G is odd, Conjecture 2 is equivalent to Snevily s conjecture which has been confirmed. 33 / 35

A conjecture for general abelian groups Conjecture (Sun, 2013-09-04). Let G be any abelian group, and let A be a finite subset of G with A = n > 3. Then there is a numbering a 1,..., a n of all the n elements of A such that a 1 +a 2 +a 3, a 2 +a 3 +a 4,..., a n 2 +a n 1 +a n, a n 1 +a n +a 1, a n +a 1 +a 2 are pairwise distinct. Remark. For a finite abelian group G = {a 1, a 2,..., a n }, it is easy to see that 2(a 1 + + a n ) = 0. Theorem (Sun, 2013-09-19). The conjecture holds for any torsion-free abelian group G. Remark. (1) I became too tired and ill immediately after I spent the whole day to finish the proof of this theorem. (2) The conjecture is even open for cyclic groups of prime orders. I d like to offer a prize of 500 US dollars for the first complete proof of the conjecture. 34 / 35

Thank you! 35 / 35