Snevily's conjecture

Importance: High ✭✭✭
Author(s): Snevily, Hunter S.
Recomm. for undergrads: yes
Posted by: mdevos
on: October 13th, 2007
Conjecture   Let $ G $ be an abelian group of odd order and let $ A,B \subseteq G $ satisfy $ |A| = |B| = k $. Then the elements of $ A $ and $ B $ may be ordered $ A = \{a_1,\ldots,a_k\} $ and $ B = \{b_1,\ldots,b_k\} $ so that the sums $ a_1+b_1, a_2+b_2 \ldots, a_k + b_k $ are pairwise distinct.

The motivation for this question comes from the study of latin squares. The addition table of every (additive) group forms a latin square, and this gives us a rich source of interesting squares. To explain further, we require a couple of definitions. A transversal of a $ k \times k $ matrix is a collection of $ k $ cells, no two of which are in the same row or column, and we say that a transversal is latin if no two of its cells contain the same element. Latin transversals are nice structures to find in latin squares. In particular, note that the cells of a $ k \times k $ latin square $ L $ may be partitioned into $ k $ latin transversals if and only if there is a latin square orthogonal to $ L $ (see this for a definition of orthogonal latin squares). The above conjecture is perhaps most naturally phrased in terms of latin transversals as follows.

Conjecture  (Snevily's conjecture - version 2)   Every $ k \times k $ submatrix of the addition table of every abelian group of odd order has a latin transversal.

Snevily's conjecture was proved by Alon [A] for abelian groups of prime order using a fairly standard application of the Alon-Tarsi polynomial technique. Later, Dasgupta, Karolyi, Serra, and Szegedy [DKSSz] used a sneaky application of the same technique to prove the conjecture for cyclic groups of odd order (the key to their approach is the fact that for $ n $ odd, $ {\mathbb Z}_n $ is a subgroup of the multiplicative group of the field of order $ 2^{\phi(n)} $ where $ \phi $ is Euler's totient function). The conjecture is still open for non-cyclic groups.

The full addition table of $ {\mathbb Z}_{2n} $ does not have a latin transversal. To see this, note that the sum of the elements in this group is equal to $ n $ (here we identify $ \{0,1,\ldots,2n-1\} $ with $ {\mathbb Z}_{2n} $ in the usual manner). So, if $ a_1,\ldots,a_{2n} $ and $ b_1,\ldots,b_{2n} $ are two orderings of $ {\mathbb Z}_{2n} $, then $ \sum_{i=1}^{2n} (a_i + b_i) = 0 $, and therefore $ a_1 + b_1,\ldots,a_{2n} + b_{2n} $ cannot be an ordering of $ {\mathbb Z}_{2n} $. This parity problem is the only obstruction known, and the following conjecture asserts that apart from it, the above conjectures holds for cyclic groups of even order.

Conjecture  (Snevily)   Every $ k \times k $ submatrix of the addition table of $ {\mathbb Z}_{2n} $ has a latin transversal, unless it is a translate of a cyclic subgroup of $ {\mathbb Z}_{2n} $ of even order.

In fact, it appears that the above conjecture might hold with $ {\mathbb Z}_{2n} $ replaced by any abelian group.


[A] N. Alon, Additive Latin transversals. Israel J. Math. 117 (2000), 125--130. MathSciNet

[DKSSz] S. Dasgupta, Gy. Károlyi, O. Serra, B. Szegedy, Transversals of additive Latin squares. Israel J. Math. 126 (2001), 17--28. MathSciNet

*[S] H. S. Snevily, Unsolved Problems: The Cayley Addition Table of Z$ \sb n $. Amer. Math. Monthly 106 (1999), no. 6, 584--585. MathSciNet.

* indicates original appearance(s) of problem.

Proof of Snevily's Conjecture

Snevily's Conjecture was in fact proved in 2009. See: Bodan Arsovski, 'A proof of Snevily's Conjecture', Israel Journal of Mathematics, vol. 182 (2011), pp. 505-508. See also Gergely Harcos, Gyula Károlyi and Géza Kós, 'Remarks to Arsovski's proof of Snevily's Conjecture', Annales Univ. Sci. Budapest., vol. 54 (2011), pp. 57-61.

Comment viewing options

Select your preferred way to display the comments and click "Save settings" to activate your changes.