# Proof of the Brown-Erdős-Sós conjecture in groups

###### Abstract.

The conjecture of Brown, Erdős and Sós from 1973 states that, for any , if a -uniform hypergraph with vertices does not contain a set of vertices spanning at least edges then it has edges. The case of this conjecture is the celebrated -theorem of Ruzsa and Szemerédi which implies Roth’s theorem on -term arithmetic progressions in dense sets of integers. Solymosi observed that, in order to prove the conjecture, one can assume that consists of triples of some finite quasigroup . Since this problem remains open for all , he further proposed to study triple systems coming from finite groups. In this case he proved that the conjecture holds also for . Here we completely resolve the Brown-Erdős-Sós conjecture for all finite groups and values of . Moreover, we prove that the hypergraphs coming from groups contain sets of size which span edges. This is best possible and goes far beyond the conjecture.

## 1. Introduction

One of the main research directions in discrete mathematics concerns emergences of certain local sub-structures in objects of high density. Many classical results, such as Szemerédi’s theorem on arithmetic progressions in subsets of integers of constant density or Turán’s theorem on the existence of complete graphs in very dense graphs, belong to this category of problems. In the study of hypergraphs, one of the most important open questions in this direction is the Brown-Erdős-Sós conjecture from 1973.

###### Conjecture 1.1 (Brown-Erdős-Sós [3]).

For any and any integer there exists , such that every -uniform hypergraph with vertices and at least edges contains a subset of vertices which span at least edges.

Already the simplest case of this conjecture, which is usually called the -problem, had many interesting consequences. In particular, in the course of proving it Ruzsa and Szemerédi [16] used Szemerédi’s regularity lemma to obtain an auxiliary result which is now known as the triangle-removal lemma. This lemma and its extensions have many striking application in combinatorics, number theory and theoretical computer science. For example, it implies Roth’s theorem [15] on -term arithmetic progressions in dense sets of integers and its stronger corner version by Ajtai and Szemerédi [1] (see [18]). A removal lemma for larger complete graphs was later obtained by Erdős, Frankl and Rödl [5] in the course of extending the -theorem of Ruzsa and Szemerédi to higher uniformities. Deriving a hypergraph removal lemma was one of the driving forces behind development of the hypergraph regularity method (see, e.g., [14]), with one of the main applications in mind being a simpler proof of Szemerédi’s theorem [21] which generalises Roth’s theorem to arithmetic progressions of arbitrary length.

Despite a lot of research in the last 40 years, the Brown-Erdős-Sós conjecture remains open for all values . The best upper bound on the number of vertices which are known to span edges is , obtained by Sárközy and Selkow [17].

It is not difficult to see that we may assume is linear, that is no two edges share more than one vertex. Indeed, if a pair of vertices in is shared by edges, then this already gives vertices spanning at least edges. Otherwise a simple greedy argument produces a linear subgraph of size at least . Furthermore, by partitioning vertices of at random into three parts we obtain a tripartite hypergraph with edges. These hyperedges can be seen as entries of a partial Latin square. Using a result of Evans [7] which states that every partial Latin square can be embedded into a Latin square, Solymosi [19] observed that, by the previous, the Brown-Erdős-Sós conjecture can be phrased in terms of quasigroups^{1}^{1}1Recall that a finite set forms a quasigroup under a binary operation if it satisfies all group axioms except associativity, or, combinatorially, if its multiplication table forms a Latin square..

###### Conjecture 1.2.

For every integer and , there exists such that if is a finite quasigroup with , then for every set of triples of the form with there exists a subset of elements which spans at least triples from , that is, at least triples from belong to .

###### Remark.

Without loss of generality, here and in the rest of the paper, we assume that every triple (as a set) appears in only once. Moreover for every such triple we fix some ordering in which the third element is the product of the first two.

As a step towards understanding this conjecture, Solymosi [19] suggested to consider the case where is a group. In particular, he showed that the Brown-Erdős-Sós conjecture for groups holds also when . In this paper we completely resolve this problem for all values of . Unlike Conjecture 1.2 which, if true, would be optimal, we show that in the case of groups there are already sets of size spaning triples.

###### Theorem 1.3.

For every integer and , there exists such that if is a finite group with , then for every set of triples of the form with there exists a subset of of size at most

which spans at least triples from .

Note that, since our hypergraphs are linear, the bound of is tight up to a constant factor. Interestingly, as does not depend on we have that triple systems coming from groups are much denser locally than globally.

## 2. Proof of Theorem 1.3

In the proof of Theorem 1.3 we utilise two classical theorems in additive combinatorics: the density version of the Gallai-Witt theorem [8, 10, 14] (also known as the multidimensional Szemerédi’s theorem) and the multidimensional density Hales-Jewett theorem [4, 9, 12]. Let us recall them here, starting with the former.

###### Theorem 2.1.

Let be a positive integer, be a finite subset of , and . If is sufficiently large, then every subset of size contains a homothetic copy of , that is there exist and an integer such that .

Let and be integers, with . A -dimensional combinatorial subspace of a cube is defined as follows: partition the ground set into sets such that are non-empty; the subspace consists of all sequences such that whenever and is constant on each set , that is if then . There is an obvious isomorphism between and any -dimensional combinatorial subspace: the sequence is sent to the sequence such that whenever and whenever . With this notion at hand, we are ready to state the multidimensional density Hales-Jewett theorem.

###### Theorem 2.2.

For every and every pair of integers and there exists a positive integer such that, for ), every subset of size contains a -dimensional combinatorial subspace of .

We now prove Theorem 1.3.

###### Proof of Theorem 1.3.

Let be a finite group with , for some sufficiently large . Let be an arbitrary subgroup of . Recall that the sets of both left and right cosets of partition the elements of the group . Therefore direct product of such cosets partitions into sets of size . Thus, by averaging there exist such that the set

is of size at least . Let

and note that . Crucially, for any sets and such that

(2.1) |

we have

Therefore, to prove the theorem it suffices to find such sets and in with .

By an observation of Erdős and Straus [6], contains an abelian subgroup with . A better (and tight) estimate on the size of a largest abelian subgroup was obtained by Pyber [13], however this results relies on the classification of finite simple groups and for our purposes a much more elementary result of Erdős and Straus suffices. In fact, any estimate which allows us to assume that is sufficiently large, provided is large, would do as well.

Let and be sufficiently large constants ( will depend on ) which we choose later. From the fundamental theorem of finite abelian groups we have that is isomorphic to a direct sum of the form

where all ’s are distinct. Therefore, by choosing to be large enough we can assume that there exists some such that either or . For brevity let us call and . By the above discussion, in the first case we can reduce problem to and in the second to . In both cases, for the rest of the proof we switch to additive notation.

### Case 1:

Let be a subset consisting of all such that , and note that . Choose sufficiently large, so that we can apply Theorem 2.1 to find and some positive integer such that , where .

Let and , for . Note that for every and we have and thus . Therefore, by the choice of , the sets and satisfy (2.1). As , we have . Thus , with room to spare.

We apply a similar approach to find sets with the sum of sizes at most . For that choose , this time with , and . If is even set , and otherwise . A routine check shows that in both cases the obtained sets satisfy (2.1) and .

### Case 2: .

Choose to be sufficiently large so that we can apply density Hales-Jewett theorem (Theorem 2.2) with , and . Our aim is to show there exists a -dimensional vector space and such that for every and we have . Before we prove that such and exist, let us first show how it implies the existence of desired sets and .

Let be an arbitrary basis of . Let be the largest integer such that , and then let be the smallest integer such that . In particular, we have . Set

For and we have for every and , thus the choice of and implies that the sets and satisfy (2.1). As is of size at most , we have . If then , thus . Otherwise, for we have

which implies . In either case, we have , as desired.

As in Case 1, a similar approach is also used to find a subsets with the sum of sizes . Let , and set

and

Note that and . Again, for and we have . This time we do not take , which would be too large, but only a subset of it, namely

Moreover, if is not even then . Note that when is even and otherwise. Thus, it is easy to verify that . Let us briefly check that (2.1) is also satisfied. We do this only in the case is even. The other case is done analogously. First, for every and every

we have . Overall this amounts to triples in . Similarly, for every

we again have , which contributes additional triples (notice that we have already counted in the previous step). Using we conclude that this amounts to triples in total.

It remains to show, using Theorem 2.2, that a desired -dimensional vector subspace and elements exist. Consider a subset , where , which contains an element

if and only if for and . As and is sufficiently large, by Theorem 2.2 the set contains a -dimensional combinatorial subspace. Let and be the partition of corresponding to this subspace. Define by setting for every and, similarly, for every . For all set . Furthermore, let be vectors defined as for and otherwise, for . It is clear that they are independent in and therefore span a -dimensional vector subspace, which we denote by .

Let us briefly check that the obtained and have the desired property. Consider some and . Then belongs to a -dimensional combinatorial subspace of given by the partition and . As this combinatorial subspace lies in , from the definition of we conclude . ∎

## 3. Concluding remarks

Theorem 1.3 shows that triples coming from groups contain much denser subsets than conjectured. Determining the best possible constant in the -term of Theorem 1.3 remains an interesting problem. We were able to do it for cyclic groups , where we obtain . We believe that the proof, which is presented in the Appendix, is interesting in its own right as it establishes a correspondence between the Brown-Erdős-Sós conjecture for and the following discrete isoperimetric problem. Recall that the edge-boundary of a vertex set in a graph is defined as , that is the number of edges leaving . The edge-isoperimetric problem for a graph (that may be infinite) and an integer asks to find the minimum edge-boundary of vertex sets of size in .

Here we are particularly concerned with being the two-dimensional triangular lattice , where the above question was answered by Harper [11, Theorem 7.2].

###### Theorem 3.1 ([11]).

There exists a nested family of vertex sets in , , with , such that each minimizes the edge-boundary over all sets of size . The family contains all balls in the lattice metric of , i.e. regular hexagons.

The following figure visualizes : the set consists of all vertices labeled through .

The proof in [11] is not elementary, as it makes use of some powerful abstract tools that can be applied to various other isoperimetric problems. It might be therefore of independent interest that as a by-product of determining the correct constant in Theorem 1.3 for , we, somewhat unexpectedly, obtain a short elementary proof of Harper’s theorem. Recently Angel, Benjamini and Horesh [2, Theorem 2.4] proved an edge-isoperimetric inequality for planar triangulations which generalises Harper’s theorem. Our result can be viewed as an extension of Theorem 3.1 in a different direction, as we determine the minimum number of axis-parallel lines occupied by any set of points in .

Finally, it would be interesting to determine the correct constant also in the -case as this would in turn yield the optimal constant in Theorem 1.3.

### Note added in proof.

While writing this paper we learned that Solymosi and Wong [20], using very different methods, independently obtained the following result on the Brown-Erdős-Sós conjecture in groups. They proved that for any group and every set of quadratically many triples there are infinitely many values of , which depend on the group and the set of triples, such that there is a set of size spanning at least triples.

## Acknowledgement

We would like to thank Asaf Shapira for arranging the third author’s visit to ETH Zurich where the present research was conducted.

## References

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## Appendix

Here we shall determine the sharp constant in Theorem 1.3 for cyclic groups. For a set of points define

That is, measures the total number of rows, columns and lines of form (‘diagonals’ for short – note that we completely ignore the diagonals of the type ) occupied by points in . Let

The following lemma shows that precisely determines the size of a smallest subset which is guaranteed to span at least edges, in the case where .

###### Lemma .1.

For every integer and , there exists such that if then for every set of triples of the form with , there exists a subset of of size which spans at least triples from . Moreover, there exists a set of triples in which every subset of size spans less than edges.

###### Proof.

Let be a given set of triples, and let be the set of points containing all such that . In other words, each point in corresponds to a triple from . Next, let be a set of points such that , where , clearly, depends only on . By Theorem 2.1, contains a homothetic copy of , that is there exist some and an integer such that . We claim that the edges corresponding to points in span at most elements of . This easily follows from the observation that and the number of different elements where is at most the size of the set

where the addition is done in instead of (hence, cam be strictly smaller than ).

Let us now exhibit a set of triples which shows the optimality of . Consider two intervals in : and . Then and, in particular, , and are disjoint. Let the set of all triples where and , and note that . Identify each triple with a point . Then for a set of triples in , corresponding to a set of points in , the involved vertices in , and correspond to rows, columns and diagonals occupied by , respectively, owing to disjointness of these sets. Hence, any set of edges necessarily span at least elements. ∎

The following theorem determines the growth rate of and, by Lemma .1, tight bounds on the size of a smallest set which spans edges in additive triples coming from .

###### Theorem .2.

.

We prove Theorem .2 by considering a dual problem: given an integer , what is the size of a largest set of points such that ? Note that and are inverse functions, in the sense that if and then .

To determine the growth rate of , consider some fixed sets , with , and let be the largest number of points in occupying at most diagonals. The following lemma, which is the heart of the proof of Theorem .2, shows that we can assume and to be intervals.

###### Lemma .2.

.

###### Proof.

We apply induction on ; for there is nothing to prove. Suppose that the statement holds for , for some . Let be a set of diagonals (recall that we only consider diagonals of the form , for some integer ) and suppose, towards a contradiction, that satisfies .

Let and such that and . For each point consider the point , and let

be the set of all such points. Let be a set of points which is a certificate for , and recall that is a union of diagonals intersecting . In particular, if then for every and such that . Note that

thus there exists some . Let and . In other words, consists of all the points of which lie on the same diagonal as , and are all the points which remain after removing this diagonal.

###### Claim .3.

###### Proof.

Let

Note that and define . For all with we have at most one such for every and vice versa, and if then . Since there are at most values and at most values , we get

Analogously, for we obtain

and the claim follows by addition. ∎

By the induction hypothesis we have

and since , by an earlier observation, we obtain

thus a contradiction. This proves the induction step, and the statement follows. ∎

The previous lemma reduces the problem of estimating to finding integers , such that , which maximize .

###### Proof of Theorem .2.

To prove the theorem it suffices to determine the growth rate of . By Lemma .2, for any integer we have that

For brevity, we write .

###### Claim .4.

is realised by , where .

###### Proof.

Let be the size of the -th largest intersection of a diagonal with . Then, for , we have

A term by term comparison of and shows that to achieve we must have . Similarly, by comparing with and , and with and (alternatively, it is not difficult to see that always holds) we obtain that is realised when are within of each other. We omit the straightforward calculations. ∎

By the above discussion

Therefore

Inverting the function yields , completing the proof of Theorem .2. ∎

Note that as a corollary of Theorem .2 we immediately obtain an asymptotic version of Harper’s theorem (Theorem 3.1). Too see this, observe that the triangular lattice is isomorphic to the square lattice with all the diagonals ‘drawn in’ (formally: the Cayley graph on generated by , and ), where in the latter the edge-boundary of a set satisfies

(.1) |

Thus, by Theorem .2 we obtain , which asymptotically matches the edge-boundary of the regular hexagons.

To derive Theorem 3.1 in full from here, note that in the course of the proof we determine via precisely. Since the extremal sets claimed in Theorem 3.1 are also extremal sets for (as the corresponding unions of rows, columns and diagonals are extremal for ), Theorem 3.1 follows.

Finally, note that Harper’s theorem does not claim a complete classification of extremal sets for the edge-isoperimetric problem on . In fact, for most values for it is easy to see that even up to isometry there is more than one extremal example. That said, the extremal examples are unique for values of that are volumes of balls in . This can be deduced from our argument as follows. If is extremal, by (.1) it has to have no ‘gaps’ (the intersection with each of the three axes has to be an interval), and be extremal for . However, the regular hexagon of radius in corresponds in to the union of the longest diagonals in , which, by a uniqueness analysis in Claim .4, is the unique up to dilation extremal set for and therefore the unique gap-free extremal set for .