1-saturating Sets, Caps and round Sets in Binary Spaces

We show that, for a positive integer r, every minimal 1-saturating set in PG(r − 1, 2) of size at least 11 36 2r + 3 either is a complete cap or can be obtained from a complete cap S by fixing some s ∈ S and replacing every point s ∈ S \ {s} by the third point on the line through s and s. Stated algebraically: if G is an elementary abelian 2-group and a set A ⊆ G \ {0} with |A| > 11 36 |G| + 3 satisfies A ∪ 2A = G and is minimal subject to this condition, then either A is a maximal sum-free set, or there are a maximal sum-free set S ⊆ G and an element s ∈ S such that A = {s} ∪ ( s + (S \ {s}) ) . Since, conversely, every set obtained in this way is a minimal 1-saturating set, and the structure of large sum-free sets in an elementary abelian 2-group is known, this provides a complete description of large minimal 1-saturating sets. Our approach is based on characterizing those large sets A in elementary abelian 2-groups such that, for every proper subset B of A, the sumset 2B is a proper subset of 2A. 1. Saturating Sets and Caps: The Main Result. Let r ≥ 1 be an integer, q a prime power, and A ⊆ PG(r− 1, q) a set in the (r− 1)dimensional projective space over the q-element field. Given an integer ρ ≥ 1, one says that A is ρ-saturating if every point of PG(r−1, q) is contained in a subspace generated by ρ + 1 points from A. Furthermore, A is said to be a cap if no three points of A are collinear; a cap is complete if it is not properly contained in another cap. Since the property of being ρ-saturating is inherited by supersets and that of being a cap is inherited by subsets, of particular interest are minimal ρ-saturating sets and complete caps. In this paper, we are concerned with the case ρ = 1 and the space PG(r−1, 2) whose points are, essentially, non-zero elements of the elementary abelian 2-group of rank r, and whose lines are triples of points adding up to 0. A large random set in PG(r−1, 2) is 1-saturating with very high probability, but the probability that it is a minimal 1saturating set is extremely low; thus, one can expect that large minimal 1-saturating sets are well-structured and can be explicitly described. A similar heuristic applies to large complete caps, and indeed, a classical result of Davydov and Tombak [DT89] establishes the structure of complete caps of size larger than 2+1. Classifying large 2000 Mathematics Subject Classification. 51E20, 11B75, 11P70. The first author was supported by the Austrian Science Fund FWF (Project Number M1014-N13). 1 2 DAVID J. GRYNKIEWICZ AND VSEVOLOD F. LEV 1-saturating sets seems to be considerably more subtle, which is quite natural bearing in mind that complete caps in PG(r− 1, 2) can be characterized as those 1-saturating sets possessing the extra property of having no internal lines (as will be explained shortly). With the exception of the next section where our result is discussed from the projective geometric viewpoint, we mostly use the language of abelian groups. Accordingly, denoting by Fr2 the elementary abelian 2-group of rank r ≥ 1 and writing 2A := {a1 + a2 : a1, a2 ∈ A} for a subset A ⊆ Fr2, we interpret 1-saturating sets in PG(r − 1, 2) as those subsets A ⊆ Fr2 \ {0} satisfying A∪ 2A = F r 2. Similarly, caps in PG(r− 1, 2) are understood as sets A ⊆ Fr2 \ {0} with A ∩ 2A = ∅; such sets are customarily referred to as sum-free. Complete caps are thus identified with maximal (by inclusion) sum-free sets. It is well known and easy to see that a sum-free set A ⊆ Fr2 is maximal if and only if the sets A and 2A partition Fr2; that is, in addition to being disjoint, they satisfy A ∪ 2A = Fr2. Consequently, any maximal sum-free set is a minimal 1-saturating set without internal lines. Beyond this simple observation, the only general result which seems to be known about minimal 1-saturating sets in Fr2 is established in [DMP03]; it asserts that the largest possible size of such a set is 2, examples being furnished by the following two constructions: (i) if H < Fr2 is an index-2 subgroup and g ∈ F r 2 \ H , then g + H is a minimal 1-saturating set; (ii) with H and g as in (i), the union {g}∪ (H \ {0}) is a minimal 1-saturating set. An extension of construction (i) has just been mentioned: any maximal sum-free set is a minimal 1-saturating set. Construction (ii) can be extended by observing that if S is a maximal sum-free set and s ∈ S, then A := {s} ∪ ( (S + s) \ {0} ) is a minimal 1-saturating set: for in this case, A ∪ 2A = 2(A ∪ {0}) = 2 ( s+ (S ∪ {0}) ) = 2(S ∪ {0}) = S ∪ 2S = Fr2, and this computation also shows that, for any proper subset B ⊂ A, we have B∪2B 6= F r 2. Indeed, a common description can be given to these two extensions: namely, if S ⊆ F2 is a maximal sum-free set and s ∈ S ∪ {0}, then A := ( s + (S ∪ {0}) ) \ {0} is a minimal 1-saturating set. In this paper, we classify completely minimal 1-saturating sets in Fr2 of size at least 11 36 2 + 3, showing that they all are of this form. Theorem 1. Let r ≥ 1 be an integer. A set A ⊆ F2 \ {0} with |A| > 11 36 2 + 3 is a minimal 1-saturating set if and only if there are a maximal sum-free set S ⊆ Fr2 and an element s ∈ S ∪ {0} such that A = ( s+ (S ∪ {0}) ) \ {0}. 1-SATURATING SETS, CAPS AND ROUND SETS 3 We notice that Theorem 1 provides a comprehensive characterization of large minimal 1-saturating sets, as the structure of large maximal sum-free sets is known due to the result of Davydov and Tombak mentioned at the beginning of this section. We record the following easy corollary of their result. Fact 2 ([DT89]). Let r ≥ 1 be an integer. Every maximal sum-free set in Fr2 of size larger than 9 · 2 either is the non-zero coset of an index-2 subgroup, or has the form B +H, where H < Fr2 is a subgroup of index 16 and B ⊂ F r 2 is a five-element set with F r 2 = 〈B〉 ⊕H such that the elements of B add up to 0. In the statement of Fact 2 and below in the paper, for a set B of group elements, we use 〈B〉 to denote the subgroup generated by B. Furthermore, given yet another subset C of the same group, we write B + C := {b+ c : b ∈ B, c ∈ C}. The set B +C is commonly referred to as the sumset of B and C. Notice that B +B = 2B. We conjecture that the density assumption of Theorem 1 can actually be relaxed to |A| ≥ 2 + 3, provided that r ≥ 6. (The group F2 contains an 11-element minimal 1-saturating set, but no 11-element maximal sum-free sets; see [DMP06].) If true, this is best possible. Example 3. Given an integer r ≥ 4, fix elements e1, e2 ∈ F r 2 and an index-4 subgroup H < F2 with F r 2 = 〈e1, e2〉 ⊕ H , and let A := (〈e1, e2〉 ∪ H) \ {0}. Straightforward verification shows that A is a minimal 1-saturating set. Now, if A = ( s+(S∪{0}) ) \{0} for a subset S ⊆ Fr2 \{0} and an element s ∈ S ∪{0}, then S ∪{0} = s+(〈e1, e2〉∪H). Since this set contains 0, we have s ∈ 〈e1, e2〉 ∪ H . If s ∈ H , then S contains all non-zero elements of H , whence 2S = H in view of |H| ≥ 4, and therefore S is not sum-free. If s = e1, then S = {e1, e2, e1 + e2} ∪ (e1 +H) is evidently not sum-free, and similarly it is not sum-free if s = e2 or s = e1 + e2. Thus A cannot be represented as in Theorem 1. More generally, if F and H are subgroups with Fr2 = F ⊕ H and |F |, |H| ≥ 4, then (F ∪ H) \ {0} is a minimal 1-saturating set which cannot be represented as in Theorem 1. 2. The Projective Geometry Viewpoint We remark that Theorem 1 can be reformulated in purely geometrical terms, as in the abstract. Theorem 1a. For an integer r ≥ 1, every minimal 1-saturating set in PG(r− 1, 2) of size at least 11 36 2 + 3 either is a complete cap, or can be obtained from a complete cap S by fixing some s ∈ S and replacing every point s ∈ S \ {s} by the third point on the line through s and s. 4 DAVID J. GRYNKIEWICZ AND VSEVOLOD F. LEV Another reformulation, kindly pointed out by Simeon Ball, involves blocking sets. Recall that a set of points in a projective geometry is called a blocking set if it has a non-empty intersection with every line; consequently, a set in PG(r−1, 2) is a (minimal) blocking set if and only if its complement is a (complete) cap. It is easy to derive that Theorem 1 is equivalent to the following assertion. Theorem 1b. For an integer r ≥ 1, every minimal 1-saturating set A in PG(r− 1, 2) of size at least 11 36 2 + 3 either is the complement of a minimal blocking set, or can be obtained from a minimal blocking set B by fixing a point s / ∈ B and letting A consist of s along with all points b ∈ B for which the line through s and b is tangent to B (i.e., passes through precisely one point of B). The equivalence between Theorems 1 and 1b is a simple exercise, left to the reader. Yet another consequence of Theorem 1 concerns the spectrum of possible sizes of minimal 1-saturating sets. As indicated in Section 1, the largest size of a minimal 1-saturating set in PG(r − 1, 2) is 2. The second largest size can be determined as an immediate corollary of Theorem 1 and Fact 2. Corollary 4. If r ≥ 9 is an integer, then the second largest size of a minimal 1saturating set in PG(r − 1, 2) is 5 · 2, and the third largest size is smaller than 11 36 2 + 3. It is observed in [DMP06] that, with a single exception for r = 5, the spectrum of sizes of all known large minimal 1-saturating sets in PG(r − 1, 2) is contained in the spectrum of sizes of sum-free sets in PG(r− 1, 2). Theorem 1 and its above-mentioned conjectured strengthening provide, of course, an explanation to this phenomenon. Finally, we note that Theorem 1 allows one to find all classes of projectively equivalent minimal 1-saturating sets in PG(r − 1, 2). For, it is not difficult to derive from Fact 2 that if r ≥ 6 is an integer, S1 and S2 are (potentially identical) complete caps in PG(r − 1, 2) with |S1| = |S2| > 9 · 2 , and, for i ∈ {1, 2}, the sets S ′ i are obtained from Si as described in Theorem 1a, then S1 and S2 are projectively equivalent, as are S ′ 1 and S ′ 2, while S1 is not equivalent to S ′ 2—regardless of the specific choice of the elements fixed in S1 and S2 to get S ′ 1 and S ′ 2 (for the non-equivalence, one only needs to note that S ′ 2 is not a cap for r ≥ 6.) This leads to the following corollary. Corollary 5. For a positive integer r ≥ 9, there are four projectively non-isomorphic minimal 1-saturating sets in PG(r−1, 2) of size larger than 11 36 2+3: two are complete caps of sizes 2 and 5 ·2, and two more are obtained from them as in Theorem 1a. 3. Round Sets and the Unique Representation Graph. In a paradoxical way, for a minimal 1-saturating set, minimality seems to be more important than saturation. This idea is captured in the notion of a round set, introduced 1-SATURATING SETS, CAPS AND ROUND SETS 5 in the present section. We also bring into consideration unique representation graphs, which are of fundamental importance for our argument, and establish some basic properties of round sets and unique representation graphs. Finally, we state a structure theorem for round sets (Theorem 7 below) and show that it implies Theorem 1. The remainder of the paper is structured as follows. Important auxiliary results are gathered in Section 4. In Section 5, we prove a “light version” of Theorem 1, with the assumption on the size of A strengthened to |A| > 1 3 2+2; besides supplying a proof of Theorem 1 for small dimensions (r ≤ 5), it serves as a simplified model of our method, exhibiting many of the core ideas. Sections 6–8 are devoted to the proof of Theorem 7: in Section 6, the problem is reduced to the case where the unique representation graph is known to have at least two isolated edges, Sections 7 and 8 present a treatment of this case. Let r ≥ 1 be an integer. We say that a set A ⊆ Fr2 is round if, for every proper subset B ⊂ A, we have 2B 6= 2A; that is, for every a ∈ A, there exists a ∈ A such that a+ a has a unique (up to the order of summands) representation as a sum of two elements of A. It is immediate from the definition that A ⊆ Fr2 \ {0} is a minimal 1-saturating set if and only if it satisfies 2(A∪{0}) = Fr2 and is minimal subject to this condition. The following simple lemma takes this observation a little further. Lemma 6. Let r ≥ 1 be an integer. If A ⊆ F2 \ {0} is a minimal 1-saturating set, then either A or A ∪ {0} is round. Remark. It is easy to derive from Theorem 1 and the observation following the proof below that if A ⊆ Fr2 \ {0} is a large minimal 1-saturating set, then, indeed, A∪ {0} is round. Proof of Lemma 6. Suppose that A ⊆ Fr2 is a minimal 1-saturating set. If A∪{0} is not round, then there exists a0 ∈ A∪{0} such that 2 ( (A∪{0}) \ {a0} ) = 2(A∪{0}) = Fr2. Since a0 ∈ A would contradict the minimality of A, we actually have a0 = 0, whence 2A = Fr2. Now if also A is not round, then there exists a ∈ A with 2(A\{a}) = 2A = F r 2. This yields 2 ( (A\{a})∪{0} ) = F2, which, again, contradicts the minimality of A. Lemma 6 allows us to concentrate on studying large round sets instead of large 1saturating sets; indeed, we will hardly refer to 1-saturating sets from now on, except for the deduction of Theorem 1 from Theorem 7 at the end of this section. We observe that if S ⊆ Fr2 is sum-free, then 0 / ∈ S and, for each g ∈ F r 2, the set g+(S∪{0}) is round. To verify this, we can assume g = 0 (as roundness is translation invariant) and notice that, fixing arbitrarily s0 ∈ S and letting S0 := S \ {s0}, we have s0 / ∈ 2S and s0 / ∈ 2(S0 ∪ {0}), whereas s0 ∈ 2(S ∪ {0}). The heart of our paper is 6 DAVID J. GRYNKIEWICZ AND VSEVOLOD F. LEV the following theorem, showing that, in fact, any large round set has the structure just described. Theorem 7. Let r ≥ 1 be an integer and suppose that A ⊆ F2 is round. If |A| > 11 36 2 + 3, then there is a sum-free set S ⊆ F2 and an element g ∈ F r 2 such that A = g + (S ∪ {0}). We now turn to the notion of a unique representation graph. Given an integer r ≥ 1 and a set A ⊆ Fr2, we define D(A) to be the set of all those elements of F r 2 with a unique, up to the order of summands, representation as a sum of two elements of A. By Γ(A) we denote the graph on the vertex set A in which two vertices a1, a2 ∈ A are adjacent whenever a1+a2 ∈ D(A); if |A| > 1, then Γ(A) is a simple, loopless graph (as all graphs below are tacitly assumed to be). We call Γ(A) the unique representation graph of A. Notice that the number of edges of Γ(A) is |D(A)| and that, for any g ∈ Fr2, we have D(A + g) = D(A), while Γ(g + A) is obtained from Γ(A) by re-labeling the vertices. Evidently, a set A ⊆ Fr2 with |A| ≥ 2 is round if and only if Γ(A) has no isolated vertices. Another indication of the importance of unique representation graphs is given by the following lemma. Lemma 8. Let r ≥ 1 be an integer, let g ∈ Fr2, and suppose that A ⊆ F r 2 satisfies |A| ≥ 2. For Γ(A) to have a spanning star with the center at g, it is necessary and sufficient that A = g + (S ∪ {0}), where S ⊆ Fr2 is sum-free. Proof. If g / ∈ A, then g is not a vertex of Γ(A) and A 6= g+(S∪{0}); thus, the assertion is immediate in this case. If g ∈ A, set S := (A + g) \ {0}, so that A = g + (S ∪ {0}). The graph Γ(A) has a spanning star with the center at g if and only if, for every s ∈ S, we have g+(g+s) ∈ D(A); that is, g+(g+s) 6= (g+s1)+(g+s2) whenever s1, s2 ∈ S. This is equivalent to S being sum-free. By Lemma 8, to prove Theorem 7, it suffices to show that if A ⊆ Fr2 is a large round set, then Γ(A) contains a spanning star. The following basic result shows that, for the unique representation graph of a large set, containing a spanning star is equivalent to being a star. Proposition 9. Let r ≥ 1 be an integer and suppose that A ⊆ F2. If |A| ≥ 2 r−2 + 3, then Γ(A) is triangle-free. Moreover, if |A| > 2 +3, then, indeed, D(A) is sum-free. Remark. Observe that if a1, a2, a3 ∈ A induce a triangle in Γ(A), then D(A) is not sum-free in view of (a1 + a2) + (a2 + a3) = a1 + a3; thus, “D(A) is sum-free” is a stronger conclusion than “Γ(A) is triangle-free”. We also notice that the bound 2 + 3 is sharp. To see this, suppose that e1, e2, H , and A are as in Example 3, and 1-SATURATING SETS, CAPS AND ROUND SETS 7 set A0 := 〈e1, e2〉 ∪H . Then |A| = 2 r−2 +2 and the vertices e1, e2, and e1 + e2 of Γ(A) induce a triangle, whereas |A0| = 2 r−2 + 3 and D(A0) is not sum-free: for if h1 and h2 are distinct non-zero elements of H , then e1 + h1, e2 + h2 and e1 + e2 + h1 + h2 belong to D(A0). Proof of Proposition 9. Fix two distinct elements d1, d2 ∈ D(A) and consider the subgroup H := 〈d1, d2〉 generated by d1 and d2. Suppose, to begin with, that the edges of Γ(A) corresponding to d1 and d2 are incident; that is, there are a, b1, b2 ∈ A such that d1 = a+ b1 and d2 = a+ b2. It is easy to see that the coset a+H contains exactly three elements of A (namely b1, b2 and a), while every other coset of H contains at most two elements of A—both conclusions in view of d1, d2 ∈ D(A). Thus, the assumption |A| ≥ 2 +3 implies that there is a coset containing exactly two elements of A. These two elements cannot differ by d1 or d2 (again, since d1, d2 ∈ D(A)); therefore they differ by d1 + d2, yielding a representation of d1 + d2 as a sum of two elements of A. Another representation is d1 + d2 = b1 + b2, and the existence of two representations shows that d1+d2 / ∈ D(A). The first assertion follows since if Γ(A) were containing a triangle with two legs corresponding to d1 and d2, then the third leg would correspond to d1 + d2. Assuming now that the edges of Γ(A) corresponding to d1 and d2 are not incident, find a1, a2, b1, b2 ∈ A such that d1 = a1 + b1 and d2 = a2 + b2. (Note that a1, a2, b1 and b2 are all distinct.) Then there are two cosets of H intersecting the set {a1, a2, b1, b2}. Each of these cosets contains exactly two elements of A, while every other coset of H contains at most two elements of A. If |A| > 2 + 3, then there are at least two cosets disjoint with {a1, a2, b1, b2} and containing two elements of A. This yields two distinct representations of d1 + d2, leading, as above, to the conclusion d1 + d2 / ∈ D(A) and proving the second assertion. Given a set A ⊆ Fr2, for each a ∈ A, we use deg(a) to denote the degree of the vertex a in Γ(A). Yet another fundamental property of the unique representation graph is established by the following result. Proposition 10. Let r ≥ 1 be an integer and suppose that A ⊆ Fr2 satisfies |A| > 2 + 3. If (a1, a2) is an edge in Γ(A), then deg(a1) + deg(a2) ≥ |A|+ |D(A)| − 2 . We present two different proofs. First proof of Proposition 10. Let A denote the set of those elements of A neighboring neither a1 nor a2 in Γ(A); thus, |A | = |A| − deg(a1)− deg(a2) by Proposition 9. Then the sets a1 + A , a2 + A , D(A), and a1 + a2 +D(A) 8 DAVID J. GRYNKIEWICZ AND VSEVOLOD F. LEV are easily seen to be pairwise disjoint, with the fact that the last two are disjoint following from a1 + a2 ∈ D(A) and Proposition 9, the fact that the first two are disjoint following from a1 + a2 ∈ D(A), and the rest following from the definition of A. Hence 2 ≥ 2|A|+ 2|D(A)| = 2 ( |A|+ |D(A)| − deg(a1)− deg(a2) ) . Second proof of Proposition 10. Since a1 + a2 ∈ D(A) and the set D(A) is sum-free by Proposition 9, it contains at most one element from each coset of the two-element subgroup 〈a1 + a2〉. On the other hand, D(A) has exactly deg(a1)+deg(a2)−2 elements in common with the set {a1, a2} + (A \ {a1, a2}), the size of which is 2(|A| − 2), and which is a union of cosets of 〈a1 + a2〉. It follows that |D(A)| ≤ (deg(a1) + deg(a2)− 2) + 1 2 ( 2 − 2(|A| − 2) ) = deg(a1) + deg(a2) + 2 r−1 − |A|. We conclude this section deducing Theorem 1 from Theorem 7. To this end, we first derive from Proposition 9 an interesting property of sum-free sets. Thinking projectively, if A is a large cap in PG(r − 1, 2) and the point p / ∈ A lies on the line determined by a pair of points in A, then in fact there are many pairs of points in A determining a line through p. (We remark that, for a generic subset of PG(r − 1, 2), not assumed to be a cap, the same conclusion requires a much stronger assumption; cf. Lemma 12.) In the definitions of a round set and the set D(A) given above in this section, we consider unordered representations of elements of F2, that is, representations which differ by the order of summands are considered identical. This convention is extended onto the following corollary. Corollary 11. Let r, κ ≥ 2 be integers and suppose that S ⊆ F2 is a sum-free set with |S| > 2 + κ. Then every element of the sumset 2S has at least κ representations (distinct under permutation of summands) as a sum of two elements of S. Proof. Assuming that an element c ∈ 2S has fewer than κ representations as a sum of two elements from S, we find a subset S0 ⊆ S with |S0| ≥ |S| − (κ− 2) such that c has exactly one representation as a sum of two elements of S0. Let A := S0 ∪ {0}. Since |S| > 2 r−2 + κ, we have |A| > 2 + 3, so in view of S0 ⊆ D(A) and Proposition 9, we get (2S0) ∩D(A) = ∅. Thus, every element of 2S0 has at least two representations as a sum of two elements from A, and therefore at 1-SATURATING SETS, CAPS AND ROUND SETS 9 least two representations as a sum of two elements from S0 (since (2S0) ∩ S0 = ∅), contradicting the choice of S0. Deduction of Theorem 1 from Theorem 7. As we have already observed, if S ⊆ F2 is a maximal sum-free set and s ∈ S ∪ {0}, then ( s + (S ∪ {0}) ) \ {0} is a minimal 1-saturating set. Suppose now that r ≥ 1 is an integer and A ⊆ Fr2 \ {0} is a minimal 1-saturating set with |A| > 11 36 2 + 3. By Lemma 6, either A ∪ {0} or A is round. We show that, in the former case, A is of the form required, while the latter case cannot occur. If A ∪ {0} is round, then by Theorem 7 there exist a sum-free set S ⊆ F2 and an element g ∈ Fr2 such that A ∪ {0} = g + (S ∪ {0}). From 0 ∈ g + (S ∪ {0}), it follows that g ∈ S ∪ {0}, and Fr2 = 2(A ∪ {0}) = 2(S ∪ {0}) = S ∪ 2S implies that S is a maximal sum-free set (as remarked in Section 1), proving the assertion in this case. Suppose now that A is round, so that by Theorem 7 there exist a sum-free set S ⊆ Fr2 and an element g ∈ Fr2 with A = g + (S ∪ {0}). In view of the previous paragraph, we may assume that A∪{0} = g+(S∪{0, g}) is not round, whence S∪{g} is not sum-free (see the comment just above Theorem 7); that is, g ∈ 2S, and we write g = s1 + s2 with s1, s2 ∈ S. Notice that 0 / ∈ A yields g 6= 0 and thus s1 6= s2, and that 2A = S∪2S and A ∪ 2A = S ∪ (g + S) ∪ 2S. Let S1 := S \ {s1} and A1 := g + (S1 ∪ {0}). Since |S| = |A| − 1 > 2 + 2, it follows from Corollary 11 (applied with κ = 2) that 2S1 = 2S. Consequently, A1 ∪ 2A1 = S1 ∪ (g + S1) ∪ 2S. On the other hand, as g = s1+s2 with s1 6= s2, we have s1, s2 ∈ S1∪ (g+S1), implying S1∪ (g+S1) = S ∪ (g+S); therefore, A1 ∪2A1 = A∪2A, contradicting the minimality of A. 4. Notation and Auxiliary Results. In this section, we deviate slightly from the flow of the proof to introduce some important notation and results, preparing the ground for the rest of the argument. We start with an easy consequence of the pigeonhole principle; see, for instance, [N01, Lemma 2.1] or [GH01, Lemma 5.29]. Lemma 12. Let B and C be non-empty subsets of a finite abelian group G. If |B| + |C| ≥ |G| + κ with an integer κ ≥ 1, then every element of G has at least κ representations as a sum of an element from B and an element from C. 10 DAVID J. GRYNKIEWICZ AND VSEVOLOD F. LEV We remark that, in Lemma 12 and in the vast majority of situations below, we consider representations of elements of Fr2 as sums of elements from two potentially distinct sets; therefore (in contrast with Section 3), representations are considered ordered. Given a subgroup H of an abelian group G, by φH we denote the canonical homomorphism from G onto the quotient group G/H . For a subset B of an abelian group G, the (maximal) period of B will be denoted by π(B); recall that this is the subgroup of G defined by π(B) := {g ∈ G : B + g = B}, and that B is called periodic if π(B) 6= {0} and aperiodic otherwise. Thus, B is a union of π(B)-cosets, and π(B) lies above every subgroup H ≤ G such that B is a union of H-cosets. Observe also that π(B) = G if and only if either B = ∅ or B = G, and that φπ(B)(B) is an aperiodic subset of the group G/π(B). Theorem 13 (Kneser, [K53, K55]; see also [M76, N01, GH01]). Let B and C be finite, non-empty subsets of an abelian group G. If |B + C| ≤ |B|+ |C| − 1, then, letting H := π(B + C), we have |B + C| = |B +H|+ |C +H| − |H|. Corollary 14. Let r ≥ 1 be an integer and suppose that the sets B,C ⊆ Fr2 are disjoint and non-empty. If |B|+ |C| > 2, then B ∪ C is not disjoint with B + C. Remark. If the elements e1, e2 ∈ F r 2 and the subgroup H < F r 2 of index 4 are so chosen that Fr2 = 〈e1, e2〉⊕H , then the sets B := e1 +H and C := e2 +H are disjoint, and so are their union B ∪ C = {e1, e2} +H and sumset B + C = e1 + e2 +H ; at the same time, |B|+ |C| = 2. This shows that the bound 2 in Corollary 14 is sharp. Proof of Corollary 14. We proceed by induction on r. The case r = 1 is immediate, and so we assume r ≥ 2. Assuming, furthermore, that B ∪ C and B + C are disjoint, whereas |B|+ |C| > 2, we derive |B + C| ≤ 2 − |B| − |C| < |B|+ |C| − 1. Set H := π(B + C). By Theorem 13, the subgroup H is non-trivial and |(B +H) \B|+ |(C +H) \ C| = |B + C| − |B| − |C|+ |H| < |H| − 1. The left-hand side can be interpreted as the total number of “H-holes” in B and C, showing that B +H and C +H are disjoint (since B and C are themselves disjoint). By the same reasoning, these two sets are also disjoint with B + C (as B + C is disjoint with both B and C, and π(B +C) = H , so there are no “H-holes” in B +C). 1-SATURATING SETS, CAPS AND ROUND SETS 11 Consequently, φH(B) and φH(C) are disjoint, non-empty subsets of the group F r 2/H , and φH(B) ∪ φH(C) is disjoint with φH(B) + φH(C) = φH(B + C). This contradicts the induction hypothesis in view of |φH(B)|+ |φH(C)| = (|B +H|+ |C +H|)/|H| ≥ (|B|+ |C|)/|H| > 2/|H| = 1 2 |Fr2/H|. For an integer k and subsets B and C of an additively written group, let B k + C denote the set of all those group elements with at least k representations as b+ c with b ∈ B and c ∈ C; thus, for instance, B 1 + C = B + C. We need a corollary of the following theorem, which is (a refinement of) a particular case of the main result of [G]. Theorem 15 (Grynkiewicz, [G, Theorem 1.2]). Let G be an abelian group and suppose that B,C ⊆ G are finite and satisfy min{|B|, |C|} ≥ 2. Then either |B 1 + C|+ |B 2 + C| ≥ 2|B|+ 2|C| − 4, or there exist subsets B ⊆ B and C ′ ⊆ C with l := |B \B|+ |C \ C | ≤ 1, B + C ′ = B 2 + C ′ = B 2 + C,