aa r X i v : . [ m a t h . C O ] J u l ON NON-MINIMAL COMPLEMENTS
ARINDAM BISWAS AND JYOTI PRAKASH SAHA
Abstract.
The notion of minimal complements was introduced by Nathanson in 2011.Since then, the existence or the inexistence of minimal complements of sets have beenextensively studied. Recently, the study of inverse problems, i.e., which sets can or cannotoccur as minimal complements has gained traction. For example, the works of Kwon, Alon–Kravitz–Larson, Burcroff–Luntzlara and also that of the authors, shed light on some of thequestions in this direction. These works have focussed mainly on the group of integers, oron abelian groups. In this work, our motivation is two-fold:(1) to show some new results on the inverse problem,(2) to concentrate on the inverse problem in not necessarily abelian groups.As a by-product, we obtain new results on non-minimal complements in the group of integersand more generally, in any finitely generated abelian group of positive rank and in any freeabelian group of positive rank. Moreover, we show the existence of uncountably manysubsets in such groups which are “robust” non-minimal complements. Introduction
Motivation.
Given two nonempty subsets
A, B of a group G , the set A is said to be aleft (resp. right) complement to B if A · B = G (resp. B · A = G ). If A is a left (resp. right)complement to B and no subset of A other than A is a left (resp. right) complement to B ,then A is said to be a minimal left (resp. right) complement to B . The study of minimalcomplements began with Nathanson in [Nat11], who introduced the notion in the contextof additive number theory as a natural arithmetic analogue of the metric concept of nets.Since then, most of the literature about minimal complements have focussed on the directproblem about which sets admit minimal complements, see the works of Chen–Yang [CY12],Kiss–S´andor–Yang [KSY19], of the authors [BS], [BS19a] etc. Recently, the study of inverseproblems, i.e., which sets occur as minimal complements, has become popular. The worksof Kwon [Kwo19], Alon–Kravitz–Larson [AKL20], Burcroff–Luntzlara [BL20] and also of theauthors [BS19b, BS20a, BS20b] have investigated this direction of research. However, mostof the literature till date, has focussed on abelian groups. In this work, our motivation istwo-fold:(1) To show some new results on the inverse problem.(2) To concentrate on the inverse problem in not necessarily abelian or finite groups.In [BS19b, Theorem C], it has been proved that the “large” subsets of a group cannot bea minimal complement to any subset. In [AKL20], Alon–Kravitz–Larson have establishedseveral interesting results which includes the above statement in the context of finite abeliangroups. For any group G , [BS19b, Theorem C] states that a subset C of G , other than G , is Mathematics Subject Classification.
Key words and phrases.
Additive complements, minimal complements, sumsets, representation of integers,additive number theory. not a minimal complement in G if C is “large” in the sense that(1) | C || G \ C | > . In [BS19b, Theorem C], the set G \ C was assumed to be finite. A refined version of this resultin the context of finite abelian groups is established in [AKL20, Proposition 17], which statesthat a subset C of a finite abelian group G , contained in a subgroup H , is not a minimalcomplement in G if C is “large” in the sense that2 | G || H || H | + 2 | G | < | C | < | H | . Note that the above inequality can be restated as(2) | C || H \ C | > G : H ]together with C ( H (as explained in the proof of Proposition 2.17).We consider the subsets of G which are contained in the subgroups of G and establish anecessary condition (similar to Equations (1), (2)) for them to be non-minimal complementsin G . For a subset C of G , strictly contained in a subgroup H , define the relative quotientof C with respect to H to be λ H ( C ) = | C || H \ C | . Note that [AKL20, Proposition 17] (in the context of finite abelian groups G ), [BS19b,Theorem C] (for any group G with G = H ) can be restated as follows: a subset C of a group G , properly contained in a subgroup H of G , is not a minimal complement in G if its relativequotient with respect to H is greater than the double of the index of H in G , i.e., λ H ( C ) > G : H ] . The aim of this article is to establish that such a statement holds in more general contexts.1.2.
Results obtained.
By suitably adapting the proof of [BS19b, Theorem C], we provethat a subset C of a group G , properly contained in a subgroup H , is not a minimal com-plement in G if the inequality λ H ( C ) > G : H ]holds (when the above inequality is interpreted in an appropriate manner). In fact, ourresults are more general. Under suitable hypothesis, we prove that not only such sets C ,but also the sets of the form ( C \ E ) ∪ F are non-minimal complements for subsets C of H satisfying the above inequality, finite subsets E ⊆ C and subsets F ⊆ H \ C . We referto Theorems 2.2, 2.4, 2.7, 2.10, 2.19, 2.23 and Propositions 2.13, 2.15, 2.17 for the precisestatements. These results are more general than [AKL20, Proposition 17], [BS19b, TheoremC]. Using them, we obtain subsets of groups which are not minimal complements to anysubset. Though the above-mentioned results apply to any group, to motivate the discussion,we provide the examples in the context of the integers. Example 1.1. (1) It follows from Theorem 2.2 that the set ( { , , · · · , , } + 32 Z ) ∪ { p | p ≡ ± , p is a prime } is not a minimal complement in Z . N NON-MINIMAL COMPLEMENTS 3 (2) It follows from Theorem 2.4 that the set ( { , , , , · · · , } + 48 Z ) ∪ { p | p ≡ , , , p is a prime } is not a minimal complement in Z .(3) It follows from Theorem 2.7 that the set ( { , , , , } + 12 Z ) ∪ { p | p ≡ , p is a prime } is not a minimal complement in Z .(4) It follows from Theorem 2.10 that the set ( { , , , , , , } + 9 Z ) ∪ { p | p ≡ ± , p is a prime } is not a minimal complement in Z .(5) It follows from Proposition 2.13 that { , , , , } +12 Z is not a minimal complementin Z . Moreover, it also follows that the set of irrational numbers is not a minimalcomplement in R , and the set of transcendental numbers is not a minimal complementin C .(6) It follows from Proposition 2.15 that for any positive integer k and for any nonemptyfinite subset F of k Z , the set k Z \ F is not a minimal complement in Z .(7) It follows from Theorem 2.19 that the set of real numbers having absolute value greaterthan one is not a minimal complement in R .(8) It follows from Theorem 2.23 that the set of irrational numbers, with a countable num-ber of points removed, is not a minimal complement in R , the set of transcendentalnumbers, with a countable number of points removed, is not a minimal complementin C . There are several immediate questions about the minimal complements in a finite group,for instance, given a group G of order n , what are the sizes of the minimal complements,what are the integers k between 1 and n such that any subset (or some subset) of G of size k is a minimal complement [BS19b, Question 1]. Further, one can study these questions in thecontext of cyclic groups, or abelian groups, or finite groups. Some of these questions wereanswered by Alon, Kravitz and Larson in the context of abelian groups [AKL20, Theorem 1,Proposition 17]. The results obtained in Section 2 apply to groups, which are not assumedto be abelian, and thus they further improve our understanding about [BS19b, Question 1].Following [BL20, Definition 5], one can consider the notion of robust MAC and robustnon-MAC in any abelian group G . A subset of an abelian group G is said to be a robustnon-MAC if it remains a non-minimal complement after the removal or the inclusion offinitely many points (see Definition 3.1). We obtain uncountably many examples of robustnon-MACs in finitely generated abelian groups of positive rank and in any free abelian groupof positive rank (see Theorem 3.4 for a more general statement). Further, one can considerthe analogous notion in non-abelian groups and obtain several examples by applying theresults from Section 2. In particular, we show that for any number field K of degree ≥
3, thegroup GL n ( O K ) contains uncountably many robust non-minimal complements where O K denote the ring of integers of K . We refer to Section 3 for the details.2. Non-minimal complements in groups
The principal results of this Section are Theorems 2.2, 2.4, 2.7, 2.10, 2.19, 2.23. They areaimed at establishing that a subset C of a group G , properly contained in a subgroup H , is ARINDAM BISWAS AND JYOTI PRAKASH SAHA not a minimal complement in G if the inequality λ H ( C ) > G : H ]holds (when the above inequality is interpreted in an appropriate manner). Moreover, theseresults not only deal with such sets C , but also deal with the sets of the form ( C \ E ) ∪ F where C is a subset of H satisfying the above inequality, E is a finite subset of C and F ⊆ H \ C . We refer to Theorems 2.2, 2.4, 2.7, 2.10, 2.19, 2.23 for the precise statements.These results are illustrated by applying them to subsets of certain groups, and therebyobtaining examples of non-minimal complements, see Remarks 2.3, 2.6, 2.8, 2.12, 2.20, 2.24,see also Section 3. Some of their important consequences are stated in Propositions 2.13,2.15, 2.17.We remark that no group is assumed to be abelian or finite unless otherwise stated.In the following, H denotes a finite index subgroup of a group G , K denotes a normalsubgroup of H . If X is a subset of G and X is the union of certain K -right cosets, thendenote the number of K -right cosets contained in X by [ X : K ]. Let C denote a propersubset of H . Suppose C is a union of certain right cosets of K in H and H \ C is the unionof finitely many right cosets of K in H . Henceforth, we assume that the relative quotient of C with respect to H is greater than the double of the index of H in G , i.e., the inequality λ H ( C ) > G : H ]holds in the following sense. Assumption 2.1.
The number of the K -right cosets contained in C is greater than theproduct of G : H ] and the number of K -right cosets contained in H \ C . Let E be a finite subset of C and F be a subset of H \ C . Theorem 2.2.
If(1) the set F does not intersect with some K -right coset in H \ C ,(2) the number of elements of K is greater than G : H ] + 1) | E | ,and Assumption 2.1 holds, then ( C \ E ) ∪ F is not a minimal complement in G .Proof. On the contrary, let us assume that ( C \ E ) ∪ F is a minimal left complement to asubset S of G . Let ℓ denote the index of H in G . Let s , · · · , s ℓ be elements of S such that Hs i ∩ Hs j = ∅ for all i = j. For 1 ≤ i ≤ ℓ , let S i denote the subset of S defined by S i := { s ∈ S | Hs = Hs i } . By the first condition, it follows that ( C \ E ) ∪ F and K · (( C \ E ) ∪ F ) are proper subsetsof H . So, for each 1 ≤ i ≤ ℓ , there exists an element s ′ i in S i such that( K · (( C \ E ) ∪ F )) s ′ i = ( K · (( C \ E ) ∪ F )) s i for all 1 ≤ i ≤ ℓ . Since K is normal in H , it follows that(3) Ks i = Ks ′ i for any i .Note that there exists a subset C of C consisting of certain K -right cosets such that C contains at most ℓ [( H \ C ) : K ] many K -right cosets and ( C ∪ F ) · S contains ( H \ C ) · A subset A of a set B is said to be a proper subset if B \ A is nonempty. N NON-MINIMAL COMPLEMENTS 5 { s , · · · , s ℓ } . Moreover, there exists a subset E of C \ E containing at most | E | elementssuch that (( C \ E ) ∪ E ∪ F ) · S contains ( H \ C ) · { s , · · · , s ℓ } . Further, the set C ∪ (( H \ C ) · { s ′ s − , · · · , s ′ ℓ s − ℓ } )contains at most 2 ℓ [( H \ C ) : K ] many K -right cosets. By Assumption 2.1, it follows thatthe set C contains a K -right coset Kh which is disjoint from the set C ∪ (( H \ C ) · { s ′ s − , · · · , s ′ ℓ s − ℓ } ) . Note that there exists a subset E ′ of C \ E containing at most ℓ | E | elements such that( E ′ ∪ F ) · S contains E · { s ′ , · · · , s ′ ℓ } . Further, note that there exists a subset E ′′ of C \ E containing at most ℓ | E | elements such that ( E ′′ ∪ F ) · S contains E · { s , · · · , s ℓ } .We claim that Hs i ⊆ ((( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ ∪ F ) · S for any 1 ≤ i ≤ ℓ . Since ( C \ E ) ∪ F is contained in ( C \ E ) ∪ F , the set (( C \ E ) ∪ E ∪ F ) · S contains ( H \ C ) · s i and Kh does not intersect with C \ E , it follows that ( H \ C ) · s i iscontained in ((( C \ E ) \ Kh ) ∪ E ∪ F ) · S. Note that ( C \ Kh ) · s i is contained in( E · s i ) ∪ ((( C \ E ) \ Kh ) · S ) , which is contained in (( C \ E ) \ Kh ) ∪ E ′′ ∪ F ) · S. Note that
Khs i does not intersect with ( H \ C ) · { s ′ i } . Since Hs i = Hs ′ i , it follows that Khs i is contained in C · s ′ i . Further, note that Khs i does not intersect with Khs ′ i , otherwise, Khs i = Khs ′ i . Since K is normal in H , it follows that hKs i = hKs ′ i , which yields Ks i = Ks ′ i ,contradicting Ks i = Ks ′ i . So Khs i is contained in ( C \ Kh ) · s ′ i . Since ( E ′ ∪ F ) · S contains E · { s ′ , · · · , s ′ ℓ } , it follows that Khs i is contained in((( C \ E ) \ Kh ) ∪ E ′ ∪ F ) · S. This proves the claim that Hs i ⊆ ((( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ ∪ F ) · S for all 1 ≤ i ≤ ℓ . So (( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ ∪ F is a left complement to S . By thesecond condition, (( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ is a proper subset of C \ E . Hence ( C \ E ) ∪ F is not a minimal left complement to S .If ( C \ E ) ∪ F is a minimal right complement to some subset T of G , then ( C − \ E − ) ∪ F − is a minimal left complement to T − , which is impossible. (cid:3) Remark . Taking G = Z , H = 2 Z , K = 32 Z , C = ( { , , · · · , , } + 32 Z ) −
1, and F = { p | p ≡ ± , p is a prime } −
1, it follows from Theorem 2.2 that the set( { , , · · · , , } + 32 Z ) ∪ { p | p ≡ ± , p is a prime } is not a minimal complement in Z .In the proof of Theorem 2.2, Equation (3) played a crucial role. This equation was obtainedby using the hypothesis that F does not intersect with some K -right coset contained in H \ C .In the following result, we prove that even if F intersects with each K -right coset containedin H \ C , one may obtain a similar result under an alternate hypothesis. ARINDAM BISWAS AND JYOTI PRAKASH SAHA
Theorem 2.4.
If(1) F is a proper subset of H \ C and given G : H ] many elements x , y , · · · , x [ G : H ] , y [ G : H ] of G with x i = y i for any i , there exists a finite index subgroup L of K such that Lx i = Ly i for any i and L is normal in H .(2) for any finite index subgroup L of K , the number of elements of L is greater than G : H ] + 1) | E | ,and Assumption 2.1 holds, then ( C \ E ) ∪ F is not a minimal complement in G .Proof. On the contrary, let us assume that ( C \ E ) ∪ F is a minimal left complement to asubset S of G . Let ℓ denote the index of H in G . Let s , · · · , s ℓ be elements of S such that Hs i ∩ Hs j = ∅ for all i = j. For 1 ≤ i ≤ ℓ , let S i denote the subset of S defined by S i := { s ∈ S | Hs = Hs i } . By the first condition, ( C \ E ) ∪ F is a proper subset of H . It follows that S i contains anelement other than s i . Let 1 ≤ i ≤ ℓ be an integer and s ′ i = s i be an element of S i . Bythe first condition, there exists a finite index subgroup L of K such that L is normal in H and Ls i = Ls ′ i for any i . Replacing K by L (if necessary), we may (and do) assume that Ks i = Ks ′ i for any i . Note that the same condition was obtained in Equation (3) in thecourse of the proof of Theorem 2.2. Proceeding in a similar fashion, we obtain the result. (cid:3) Corollary 2.5.
Suppose C is a subset of Z and it is the union of translates of a nonzerosubgroup K of Z . If λ Z ( C ) > , then ( C \ E ) ∪ F is not a minimal complement in Z for anyfinite subset E of C and for any proper subset F of Z \ C .Remark . Taking G = Z , H = 2 Z , K = 48 Z , C = { , , , , · · · , } + 48 Z and F = { p | p ≡ , , , p is a prime } −
1, it follows from Theorem 2.4 that the set( { , , , , · · · , } + 48 Z ) ∪ { p | p ≡ , , , p is a prime } is not a minimal complement in Z .Note that the proofs of Theorems 2.2, 2.4 crucially relied on the observation that S i contains two elements which lies in two disjoint K -right cosets. However, if we consider aset of the form ( C \ E ) ∪ F and assume that it is a minimal left complement to some set S ,it is not clear whether each S i has this property. We show in the following result that evenin such a situation, one may obtain a similar result under an alternate hypothesis. Theorem 2.7.
If(1) the set F is either empty or it is contained in a single K -right coset,(2) the set F ∪ X does not contain any K -right coset for any subset X of H of size ≤ G : H ] + 1) | E | ,(3) the set E · F − does not contain any K -right coset,and Assumption 2.1 holds, then ( C \ E ) ∪ F is not a minimal complement in G .Proof. On the contrary, let us assume that ( C \ E ) ∪ F is a minimal left complement to asubset S of G . Let ℓ denote the index of H in G . Let s , · · · , s ℓ be elements of S such that Hs i ∩ Hs j = ∅ for all i = j. N NON-MINIMAL COMPLEMENTS 7
For 1 ≤ i ≤ ℓ , let S i denote the subset of S defined by S i := { s ∈ S | Hs = Hs i } . By the first and second condition, ( C \ E ) ∪ F is a proper subset of H . It follows that S i contains an element other than s i . Let P, Q denote the sets defined by P = { i | ≤ i ≤ ℓ, Cs ′ = Cs i for some s ′ ∈ S i } ,Q = { i | ≤ i ≤ ℓ, i / ∈ P } . For i ∈ P , let s ′ i denote an element of S i such that Cs ′ i = Cs i . For i ∈ Q , let s ′ i denote anelement of S i other than s i .Note that there exists a subset C of C consisting of certain K -right cosets such that C contains at most ℓ [( H \ C ) : K ] many K -right cosets and ( C ∪ F ) · S contains ( H \ C ) ·{ s , · · · , s ℓ } . There exists a subset E of C \ E containing at most | E | elements such that(( C \ E ) ∪ E ∪ F ) · S contains ( H \ C ) · { s , · · · , s ℓ } . Further, the set C ∪ (( H \ C ) · { s ′ s − , · · · , s ′ ℓ s − ℓ } )contains at most 2 ℓ [( H \ C ) : K ] many K -right cosets. By Assumption 2.1, it follows thatthe set C contains a K -right coset R which is disjoint from the set C ∪ (( H \ C ) · { s ′ s − , · · · , s ′ ℓ s − ℓ } ) . Note that there exists a subset E ′ of C \ E containing at most ℓ | E | elements such that( E ′ ∪ F ) · S contains E · { s ′ , · · · , s ′ ℓ } . Further, note that there exists a subset E ′′ of C \ E containing at most ℓ | E | elements such that ( E ′′ ∪ F ) · S contains E · { s , · · · , s ℓ } .Assume that F is contained in Kα for some α ∈ H . Let h be an element of R such that h / ∈ ( E · F − ) α (if there were no such h , then ( E · F − ) α would contain R , which wouldimply that E · F − contains a K -right coset, contradicting the third condition.). We claimthat Hs i ⊆ ((( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ ∪ ( hα − F ) ∪ F ) · S for any 1 ≤ i ≤ ℓ . Since ( C \ E ) ∪ F is contained in ( C \ E ) ∪ F , the set (( C \ E ) ∪ E ∪ F ) · S contains ( H \ C ) · s i and Kh does not intersect with C \ E , it follows that ( H \ C ) · s i iscontained in ((( C \ E ) \ Kh ) ∪ E ∪ F ) · S. Note that ( C \ Kh ) · s i is contained in( E · s i ) ∪ (((( C \ E ) \ Kh ) ∪ F ) · S ) , which is contained in (( C \ E ) \ Kh ) ∪ E ′′ ∪ F ) · S. Let i be an element of P . Note that Khs i does not intersect with ( H \ C ) · { s ′ i } . Since Hs i = Hs ′ i , it follows that Khs i is contained in C · s ′ i . Further, note that Khs i does notintersect with Khs ′ i , otherwise, Khs i = Khs ′ i . Since K is normal in H , it follows that hKs i = hKs ′ i , which yields Ks i = Ks ′ i , and consequently K e hs ′ i = K e hs i holds for any e h ∈ H , contradicting i ∈ P . So Khs i is contained in ( C \ Kh ) · s ′ i . Since ( E ′ ∪ F ) · S contains E · { s ′ , · · · , s ′ ℓ } , it follows that Khs i is contained in(( C \ E ) \ Kh ) ∪ E ′ ∪ F ) · S for i ∈ P . ARINDAM BISWAS AND JYOTI PRAKASH SAHA
For each i ∈ Q , we have F · S i = ( H \ C ) · s i . Note that for any β ∈ H , we obtain βα − F ⊆ βα − Kα = Kβα − α = Kβ and ( βα − F ) · S i = ( βα − ) · ( F · S i ) ⊇ ( βα − ) · Kαs i = Kβα − αs i = Kβs i for i ∈ Q . It follows that Khs i is contained in ( hα − F ) · S for i ∈ Q .This proves the claim that Hs i ⊆ (( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ ∪ ( hα − F ) ∪ F ) · S for all i . So ( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ ∪ ( hα − F ) ∪ F is a left complement to S . Since h / ∈ ( E · F − ) α , using the second condition, (( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ ∪ ( hα − F ) is aproper subset of C \ E . Hence ( C \ E ) ∪ F is not a minimal left complement to S .If ( C \ E ) ∪ F is a minimal right complement to some subset T of G , then ( C − \ E − ) ∪ F − is a minimal left complement to T − , which is impossible. (cid:3) Remark . Taking G = Z , H = 2 Z , K = 12 Z , C = { , , , , } + 12 Z and F = { p | p ≡ , p is a prime } −
1, it follows from Theorem 2.7 that the set( { , , , , } + 12 Z ) ∪ { p | p ≡ , p is a prime } is not a minimal complement in Z .Note that in the proof of Theorem 2.7, the hypothesis that F is either empty or is containedin a single K -right coset, played a crucial role. It would be interesting to consider the subsetsof H of the form ( C \ E ) ∪ F for “large” C and for any proper subset F of H \ C . In Theorem2.10, we prove that even if F intersects with each K -right coset contained in H \ C , one mayobtain a similar result under an alternate hypothesis. Proposition 2.9.
Let
X, Y be two nonempty disjoint subsets of a group G with X ∪ Y = G .Let L be a subgroup of G such that X is the union of certain right cosets of L . Then theinclusion (4) ( X · Y − ) ∪ ( Y · X − ) ⊆ G \ L holds. Moreover, the following conditions are equivalent.(1) The inclusion in Equation (4) is a proper inclusion.(2) For each y ∈ Y , there exists an element y ′ ∈ Y \ ( Ly ) such that y ′ y − · Y = Y. (3) For some y ∈ Y , there exists an element y ′ ∈ Y \ ( Ly ) such that y ′ y − · Y = Y. N NON-MINIMAL COMPLEMENTS 9 (4) The set Y is the union of certain right cosets of some subgroup of G which properlycontains L .(5) The set X is the union of certain right cosets of some subgroup of G which properlycontains L .The set G \ (cid:0) ( X · Y − ) ∪ ( Y · X − ) (cid:1) is a subgroup of G and it is the maximal subgroup of G such that Y is a union of its rightcosets. If Y is finite, then the inclusion (5) ( X · Y − ) ∪ ( Y · X − ) ⊆ G \ { e } is an equality under any one of the following conditions.(a) The order of y ′ y − is greater than the size of Y for any y, y ′ ∈ Y with y = y ′ .(b) The size of Y is not divisible by the size of any nontrivial finite subgroup of G .Proof. Since
X, Y are disjoint and each of them can be expressed as the union of certain L -right cosets, it follows that the inclusion in Equation (4) holds.Note that (( X · g ) · ( Y · g ) − ) ∪ (( Y · g ) · ( X · g ) − ) = ( X · Y − ) ∪ ( Y · X − )for any g ∈ G . Thus the inclusion( X · Y − ) ∪ ( Y · X − ) ⊆ G \ L is an equality if and only if the inclusion(( X · g ) · ( Y · g ) − ) ∪ (( Y · g ) · ( X · g ) − ) ⊆ G \ L is an equality.Note that for y, y ′ ∈ Y , the element y ′ y − does not belong to the set (( X · y − ) · ( Y · y − ) − ) ∪ (( Y · y − ) · ( X · y − ) − ) if and only if y ′ y − · ( Y · y − ) ⊆ Y · y − , and y ′ y − · ( X · y − ) ⊆ X · y − , which holds if and only if y ′ y − · Y = Y. Assume that the first condition holds. Choose an element y ∈ Y . Note that the set(( X · y − ) · ( Y · y − ) − ) ∪ (( Y · y − ) · ( X · y − ) − ) contains X · y − . So this set does not contain y ′ y − for some y ′ ∈ Y \ ( Ly ). We obtain y ′ y − · Y = Y. So the first condition implies the second condition. Note that the second condition impliesthe third condition. Now, assume that the third condition holds, i.e., y ′ y − · Y = Y holds with y, y ′ ∈ Y , Ly = Ly ′ . Let L ′ denote the subgroup of G generated by L and y ′ y − .Since x · Y = Y for any x ∈ L , and y ′ y − · Y = Y , it follows that x · Y = Y for any x ∈ L ′ .So, Y is union of certain right cosets of L ′ . Since Ly = Ly ′ , it follows that L is properlycontained in L ′ . Thus the fourth condition follows. Assume that the fourth condition holds,i.e., Y is the union of certain translates of some subgroup L of G which properly contains L . Then the set ( X · Y − ) ∪ ( Y · X − ) does not contain L . Since L properly contains L , the first condition follows. Since X, Y are disjoint and X ∪ Y = G , the fourth and the fifthconditions are equivalent. This proves the equivalence of the five conditions.Consider the subgroups L ′ of G such that L ′ contains L and Y can be expressed as theunion of right cosets of L ′ . Let L denote the subgroup of G generated by such subgroups.Note that Y can be expressed as the union of the right cosets of L . It follows that( X · Y − ) ∪ ( Y · X − ) ⊆ G \ L . By the construction of L , it follows that the above inclusion is an equality, and it alsofollows that L is the maximal subgroup of G such that Y is a union of its right cosets.Suppose Y is finite and the order of y ′ y − is greater than the size of Y for any y, y ′ ∈ Y with y = y ′ . Assume that the inclusion in Equation (5) is not an equality. So, there existtwo distinct elements y , y ∈ Y such that y y − · Y = Y. Let y be an element of Y . Denote the order of y y − by r . Then the set Y contains the r distinct elements yy , y y , y y , · · · , y r y where y = y y − , which is impossible, since r isgreater than the size of Y . Hence, the inclusion in Equation (5) is an equality. Moreover, if Y is finite and the size of Y is not divisible by the size of any nontrivial finite subgroup of G ,then Y cannot be expressed as the union of certain right cosets of some nontrivial subgroupof G . Hence, the inclusion in Equation (5) is an equality. (cid:3) Theorem 2.10.
If(1) (6) ( C − ( H \ C )) ∪ (( H \ C ) − C ) = H \ K (2) the set F ∪ X does not contain a K -right coset for any subset X of H of size ≤ G : H ] + 1) | E | ,(3) the set E · F − does not contain any K -right coset,and Assumption 2.1 holds, then ( C \ E ) ∪ F is not a minimal complement in G . In particular, ( C \ E ) ∪ F is not a minimal complement in G if(1) any subgroup K ′ of H such that K is contained in K ′ and H \ C can be expressed asa union of right cosets of K ′ , is normal in H ,(2) the set F ∪ X does not contain a K -right coset for any subset X of H of size ≤ G : H ] + 1) | E | ,(3) the set E · F − does not contain any K -right coset,and Assumption 2.1 holds.Proof. On the contrary, let us assume that ( C \ E ) ∪ F is a minimal left complement to asubset S of G . Let ℓ denote the index of H in G . Let s , · · · , s ℓ be elements of S such that Hs i ∩ Hs j = ∅ for all i = j. For 1 ≤ i ≤ ℓ , let S i denote the subset of S defined by S i := { s ∈ S | Hs = Hs i } . N NON-MINIMAL COMPLEMENTS 11
By the second condition, ( C \ E ) ∪ F is a proper subset of H . It follows that S i contains anelement other than s i . Let P, Q denote the sets defined by P = { i | ≤ i ≤ ℓ, Cs ′ = Cs i for some s ′ ∈ S i } ,Q = { i | ≤ i ≤ ℓ, i / ∈ P } . For i ∈ P , let s ′ i denote an element of S i such that Cs ′ i = Cs i . For i ∈ Q , let s ′ i denote anelement of S i other than s i .Note that there exists a subset C of C consisting of certain K -right cosets such that C contains at most ℓ [( H \ C ) : K ] many K -right cosets and ( C ∪ F ) · S contains ( H \ C ) ·{ s , · · · , s ℓ } . Moreover, there exists a subset E of C \ E containing at most | E | elementssuch that (( C \ E ) ∪ E ∪ F ) · S contains ( H \ C ) · { s , · · · , s ℓ } . Further, the set C ∪ (( H \ C ) · { s ′ s − , · · · , s ′ ℓ s − ℓ } )contains at most 2 ℓ [( H \ C ) : K ] many K -right cosets. By Assumption 2.1, it follows thatthe set C contains a K -right coset class R which is disjoint from the set C ∪ (( H \ C ) · { s ′ s − , · · · , s ′ ℓ s − ℓ } ) . Note that there exists a subset E ′ of C \ E containing at most ℓ | E | elements such that( E ′ ∪ F ) · S contains E · { s ′ , · · · , s ′ ℓ } . Further, note that there exists a subset E ′′ of C \ E containing at most ℓ | E | elements such that ( E ′′ ∪ F ) · S contains E · { s , · · · , s ℓ } .Assume that F ∩ Kα is properly contained in Kα for some α ∈ H . Let h be an elementof R such that h / ∈ ( E · F − ) α (if there were no such h , then ( E · F − ) α would contain R ,which would imply that E · F − contains a K -right coset, contradicting the third condition.).We claim that Hs i ⊆ ((( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ ∪ ( hα − F ∩ Kh ) ∪ F ) · S for any 1 ≤ i ≤ ℓ . Since ( C \ E ) ∪ F is contained in ( C \ E ) ∪ F , the set (( C \ E ) ∪ E ∪ F ) · S contains ( H \ C ) · s i and Kh does not intersect with C \ E , it follows that ( H \ C ) · s i iscontained in (( C \ E ) \ Kh ) ∪ E ∪ F ) · S. Note that ( C \ Kh ) · s i is contained in( E · s i ) ∪ (((( C \ E ) \ Kh ) ∪ F ) · S ) , which is contained in (( C \ E ) \ Kh ) ∪ E ′′ ∪ F ) · S. Let i be an element of P . Note that Khs i does not intersect with ( H \ C ) · { s ′ i } . Since Hs i = Hs ′ i , it follows that Khs i is contained in C · s ′ i . Further, note that Khs i does notintersect with Khs ′ i , otherwise, Khs i = Khs ′ i . Since K is normal in H , it follows that hKs i = hKs ′ i , which yields Ks i = Ks ′ i , and consequently K e hs ′ i = K e hs i holds for any e h ∈ H , contradicting i ∈ P . So Khs i is contained in ( C \ Kh ) · s ′ i . Since ( E ′ ∪ F ) · S contains E · { s ′ , · · · , s ′ ℓ } , it follows that Khs i is contained in(( C \ E ) \ Kh ) ∪ E ′ ∪ F ) · S for i ∈ P .We choose an element i ∈ Q . Note that the set S i is contained in Ks i . Otherwise, thereexists an element s ′ i ∈ S i such that Ks ′ i = Ks i . Let β denote the element s ′ i s − i of H .Note that β does not lie in K . By the first condition (i.e. Equation (6)), β ± = γ − δ with γ ∈ C, δ ∈ H \ C . If β = γ − δ , then Cβ intersects with H \ C , and hence Cβs i intersects with( H \ C ) s i , i.e., Cs ′ i intersects with ( H \ C ) s i , which implies that Cs i = Cs ′ i . If β − = γ − δ ,then Cβ − intersects with H \ C , and hence Cβ − s ′ i intersects with ( H \ C ) s ′ i , i.e., Cs i intersects with ( H \ C ) s ′ i , which implies that Cs ′ i = Cs i . This shows that i ∈ P , which is acontradiction. So S i is contained in Ks i . Since Kαs i is contained in ( C ∪ F ) · S i , it followsthat Kαs i is contained in ( F ∩ Kα ) · S i , and hence( hα − F ∩ Kh ) · S i = ( hα − ( F ∩ ( αh − Kh ))) · S i = ( hα − ( F ∩ ( Kαh − h ))) · S i = ( hα − ( F ∩ Kα )) · S i = ( hα − )(( F ∩ Kα ) · S i ) ⊇ ( hα − ) Kαs i = K ( hα − ) αs i = Khs i . It follows that
Khs i is contained in ( hα − F ∩ Kh ) · S for any i ∈ Q .This proves the claim that Hs i ⊆ (( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ ∪ ( hα − F ∩ Kh ) ∪ F ) · S for all 1 ≤ i ≤ ℓ . So ( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ ∪ ( hα − F ∩ Kh ) ∪ F is a left complement to S . By the second condition, (( C \ E ) \ Kh ) ∪ E ∪ E ′ ∪ E ′′ ∪ ( hα − F ∩ Kh ) is a proper subsetof C \ E . Hence ( C \ E ) ∪ F is not a minimal left complement to S .If ( C \ E ) ∪ F is a minimal right complement to some subset T of G , then ( C − \ E − ) ∪ F − is a minimal left complement to T − , which is impossible.Now we establish the second part. Since H \ C is the union of finitely many right cosetsof K , it follows from Proposition 2.9 that( C − ( H \ C )) ∪ (( H \ C ) − C ) = H \ K ′ and H \ C is the union of certain left cosets of K ′ for some subgroup K ′ of H containing K as a finite index subgroup. Since K ′ contains K , it follows from the hypothesis that the set F ∪ X does not contain a K ′ -right coset for any subset X of H of size ≤ G : H ] + 1) | E | ,and the set E · F − does not contain any K ′ -right coset. Hence from the first part, the resultfollows. (cid:3) Remark . By Proposition 2.9, the first condition in Theorem 2.10 is equivalent to requir-ing that H \ C cannot be expressed (or equivalently, C cannot be expressed) as the union ofcertain left cosets of some subgroup L of H satisfying L ) K . Remark . Taking G = Z , H = 2 Z , K = 2 n Z , it follows from Theorem 2.10 that for anyinteger n ≥
11, for any 1 ≤ a < b ≤ n with(7) 2( a − b ) n ) , the set (( { , , , · · · , n } \ { a, b } ) + 2 n Z ) ∪ F N NON-MINIMAL COMPLEMENTS 13 is not a minimal complement in Z for any proper subset of F of { a, b } + 2 n Z since(( { , , , · · · , n } \ { a, b } ) + 2 n Z ) + ( {− a, − b } + 2 n Z )= (( { , , , · · · , n } \ { , b − a ) } ) + 2 n Z ) ∪ (( { , , , · · · , n } \ { a − b ) , } ) + 2 n Z )= { , , , · · · , n } \ { } + 2 n Z . Note that Equation (7) holds when n is odd. One can obtain a more general example thanthe above. Taking G = Z , H = 2 Z , K = 2 n Z , it follows from Proposition 2.9 and Theorem2.10 that for any integer k ≥ n ≥ k + 1 such that n is not divisible by any integer1 < i ≤ k and for any 1 ≤ a < a < · · · < a k ≤ n , the set(( { , , , · · · , n } \ { a , a , · · · , a k } ) + 2 n Z ) ∪ F is not a minimal complement in Z for any proper subset of F of { a , a , · · · , a k } + 2 n Z .When F is the empty set, Theorems 2.2, 2.7 are equivalent. One obtains the followingconsequences. Proposition 2.13.
If(1) the number of elements of K is greater than G : H ] + 1) | E | ,and Assumption 2.1 holds, then C \ E is not a minimal complement in G .Remark . It follows from Proposition 2.13 that { , , , , } + 12 Z is not a minimalcomplement in Z . Taking G = H = R , K = Q , it follows from Proposition 2.13 that theset of irrational numbers is not a minimal complement in R . Taking G = H = C , K = Q ,it follows from Proposition 2.13 that the set of transcendental numbers is not a minimalcomplement in C .When K is the trivial subgroup and E is the emptyset in Proposition 2.13, one obtainsthe following results. Proposition 2.15. If H \ C is finite and C contains more than G : H ] | H \ C | elements,i.e., the relative quotient of C with respect to H satisfies λ H ( C ) > G : H ] , then C is not a minimal complement to any subset of G . In particular, if D is a propersubset of G such that G \ D is finite and D contains more than | G \ D | elements, then D is not a minimal complement to any subset of G .Proof. The first part follows from Proposition 2.13. The second past follows from the firstpart as a consequence. (cid:3)
Remark . It follows from Proposition 2.15 that for any positive integer k and for anynonempty finite subset F of k Z , the set k Z \ F is not a minimal complement in Z .In the context of finite groups, one has the following consequence of Proposition 2.15. Seealso [AKL20, Proposition 17]. Proposition 2.17. If G is finite and the relative quotient of C with respect to H satisfies λ H ( C ) > G : H ] , i.e., C is a subset of H satisfying | H | > | C | > G : H ] | H \ C | , then C is not a minimal complement to any subset of G . Equivalently, no subset C of asubgroup H of a finite group G satisfying | H | G : H ]1 + 2[ G : H ] = 2 | G || H || H | + 2 | G | < | C | < | H | is a minimal complement to some subset of G . In particular, if C is a proper subset of afinite group G containing more than | G \ C | elements, then C is not a minimal complementin G .Proof. The first statement and the third statement follow from Proposition 2.15.To obtain the second statement, note that for a subset C of H , the inequality | C | > G : H ] | H \ C | is equivalent to(2[ G : H ] + 1) | C | > G : H ] | C | + 2[ G : H ] | H \ C | = 2[ G : H ] | H | = 2 | G | , which is equivalent to | C | > | G || H || H | + 2 | G | . Then the second part follows from the first part. (cid:3)
Remark . It follows from Proposition 2.17 that the set { , , , , } is not a minimalcomplement in Z / Z . Theorem 2.19.
Let H be a finite index subgroup of a topological group G . Let C be a propersubset of H . Suppose H \ C is compact and closed in H . If H is not a union of finitelymany translates of H \ C , then C is not a minimal complement in G . In particular, if C is aproper subset of a topological group of G such that G \ C is closed and compact, and
G \ C is“small” in the sense that G is not a union of finitely many translates of G \ C , then C is notminimal complement in G .Proof. On the contrary, let us assume that C is a minimal left complement to some subset T of G . Let S be a subset of T such that H · S = G , and H s ∩ H s = ∅ for any twodistinct elements s , s ∈ S . Since C is a proper subset of H , for each s ∈ S , there exists anelement t s ∈ T such that t s = s and H s = H t s . Since C is compact and S is finite, there isa nonempty finite subset T ′ of T such that {C · s } s ∈ T ′ is an open cover of C · S . From thehypothesis, it follows that the subgroup H strictly contains( H ∩ ( ∪ x ∈ S · T ′− C · x )) ∪ ( ∪ s ∈ S C · t s s − ) ∪ C , and hence there is an element h ∈ H lying outside this union. We claim that C \ { h } is a leftcomplement to T . Note that C \ { h } is nonempty. It suffices to show that H s is containedin ( C \ { h } ) · T for each s ∈ S . Let k be an element of C . Then ks is equal to ct ′ for some c ∈ C , t ′ ∈ T ′ . So c lies in the above union and hence h = c . Thus ks lies in ( C \ { h } ) · T . So C · s is contained in ( C \ { h } ) · T . Note that hs lies in H t s and does not lie in C t s . Thus hs lies in C · t s . Since s = t s , it follows that hs lies in ( C \ { h } ) · t s . Clearly, ( C \ { h } ) s iscontained in ( C \ { h } ) · T . So H · s is contained in ( C \ { h } ) · T . Thus C \ { h } is a minimalleft complement to T . Note that h lies in H and does not lie in C . So C \ { h } is a propersubset of C . Hence C is not a minimal left complement to T . Similarly, assuming C to be a Note that the compact subsets need not be closed unless the ambient topological space is assumed to beHausdorff.
N NON-MINIMAL COMPLEMENTS 15 right minimal complement to some subset of G will lead to a contradiction. Hence C is nota minimal complement in G .The second statement follows from the first statement. (cid:3) Remark . From Theorem 2.19, it follows that the set of real numbers having absolutevalue greater than one is not a minimal complement in R . Corollary 2.21.
Let H be a finite index subgroup of an infinite group and C be a propersubset of H such that H \ C is finite. Then C is not a minimal complement in G .Proof. If G is endowed with the discrete topology, then this corollary follows from Theorem2.19.It can also be seen as an immediate consequence of Proposition 2.15 (and also of Propo-sition 2.13). (cid:3) Corollary 2.22.
For any positive integer k and for any nonempty finite subset F of k Z , theset k Z \ F is not a minimal complement in Z . It turns out that the set of irrational numbers is not a minimal complement in R (seeRemark 2.14). It is natural to ask whether the set of irrational numbers, with a countablenumber of points removed, is a minimal complement in R . Theorem 2.19 does not seem toshed any light on this question since the set of irrational numbers, with a countable numberof points removed, does not form a closed or a compact set under the Euclidean topology. Theorem 2.23.
Let H be a subgroup of a group G . Let C be a proper subset of H . SupposeAssumption 2.1 holds in the sense that no map from { , }× ( H\C ) × ( G / H ) to C is surjective.Then C is not a minimal left complement in G .Proof. On the contrary, let us assume that C is a minimal left complement to a subset S of G . Let { s i } i ∈ Λ be elements of S such that G = ∪ i ∈ Λ H s i and H s i ∩ H s j = ∅ for all i = j. Since C is a proper subset of H , it follows that for each i ∈ Λ, there exists an element s ′ i suchthat H s ′ i = H s i and C s ′ i = C s i . For each ( a, i ) ∈ ( H \ C ) × Λ, choose an element ( c ( a,i ) , s ( a,i ) )in C × S such that c ( a,i ) s ( a,i ) = as i . Consider the map( H \ C ) × Λ → C defined by ( a, i ) c ( a,i ) . Denote the image of this map by C . Note that C · S contains ( H \ C ) · { s i | i ∈ Λ } . Considerthe map { , } × ( H \ C ) × Λ → C defined by ( ∗ , a, i ) ( c ( a,i ) if ∗ = 0 ,as ′ i s − i if ∗ = 1 . By the hypothesis, C contains an element h which lies outside the image of this map, i.e., h avoids the set C ∪ (( H \ C ) · { s ′ i s − i | i ∈ Λ } ) . We claim that H s i ⊆ ( C \ { h } ) · S for any i ∈ Λ. Since C is contained in C , the set C · S contains ( H \ C ) · s i and h does notlie in C , it follows that ( H \ C ) · s i is contained in ( C \ { h } ) · S . Note that hs i does notlie in ( H \ C ) s ′ i . Since H s i = H s ′ i , it follows that hs i belongs to C · s ′ i . Hence hs i lies in( C \ { h } ) · s ′ i . Moreover, ( C \ { h } ) · s i is contained in ( C \ { h } ) · S . This proves the claim that H s i is contained in ( C \ { h } ) · S for any i ∈ Λ. Hence C is not a minimal left complement to S .If C is a minimal right complement to some subset T of G , then C − is a minimal leftcomplement to T − , which is impossible. (cid:3) Remark . It follows from Theorem 2.23 that given any uncountable group G , no propersubset C of G having countable set-theoretic complement in G is a minimal complement in G . In particular,(1) the set of irrational numbers, with a countable number of points removed, is not aminimal complement in R ,(2) the set of transcendental numbers, with a countable number of points removed, isnot a minimal complement in C .3. On robust non-minimal complements
In an abelian group, a minimal complement is often called a minimal additive complement,abbreviated as MAC [BL20]. Following [BL20, Definition 5], one can consider the notion ofrobust MAC and robust non-MAC in any abelian group G . Definition 3.1.
Let G be an abelian group. A subset C of G is said to be a robust MAC ifany non-empty subset D of G having finite symmetric difference with C is a MAC in G . Asubset C of G is said to be a robust non-MAC if any non-empty subset D of G having finitesymmetric difference with C is a non-MAC in G . Kwon proved that the finite subsets of the integers are robust MACs [Kwo19, Theorem 9].In [BS19b], the authors showed that the finite subsets of any free abelian group of rank ≥ Z which are robust MACs and severalexamples of infinite subsets of Z which are robust non-MACs [BL20, Theorems 3, 5]. As acorollary of Theorem 2.4, one also obtains examples of robust non-MACs. Corollary 3.2.
Suppose C is a subset of Z and it is the union of translates of a nonzerosubgroup K of Z . If λ Z ( C ) > , then ( C \ E ) ∪ F is a robust non-MAC in Z for any finitesubset E of C and for any subset F of Z \ C such that ( Z \ C ) \ F is infinite. Moreover, there are infinite sets which are neither bounded below nor above, and do notsatisfy the hypothesis of [BL20, Theorem 5]. Indeed, consider the set( { , , } + 4 Z ) ∪ F ∪ F for any subset F of F where the sets F , F are defined by F i = 4 Z \ { (4 n )! + 4 k | n ≥ , n ≡ i (mod 2) , ≤ k ≤ n } for i = 0 ,
1. Note that ( { , , } + 4 Z ) ∪ F ∪ F does not satisfy the hypothesis of [BL20,Theorem 5]. From Theorem 2.7, it follows that it is a robust non-MAC. Since F is an infiniteset, it has uncountably many subsets. Thus Theorem 2.7 yields examples of uncountably N NON-MINIMAL COMPLEMENTS 17 many robust non-MACs in Z , none of them satisfying the hypothesis of [BL20, Theorem5]. This provides a partial answer to [BL20, Question 2]. More generally, we establish thefollowing Theorem 3.4, which shows in particular that any finitely generated abelian group ofpositive rank and any free abelian group of positive rank contain uncountably many robustnon-MACs. Moreover, it follows from Theorem 2.23 that given any uncountable abeliangroup G , any proper subset C of G having countable set-theoretic complement in G is arobust non-MAC.In the context of groups which are not assumed to be abelian, the minimal complementsare more precisely called minimal multiplicative complements to indicate the underlyingstructure of the ambient group. The minimal multiplicative complements are abbreviatedas MMCs. In the spirit of robust MACs and non-MACs, one can also define the notion ofrobust MMCs and robust non-MMCs. Definition 3.3.
Let G be a group. A subset C of G is said to be a robust MMC if anynon-empty subset D of G having finite symmetric difference with C is a MMC in G . Asubset C of G is said to be a robust non-MMC if any non-empty subset D of G having finitesymmetric difference with C is a non-MMC in G . Note that in the context of abelian groups, the MMCs (resp. robust MMCs, robust non-MMCs) coincide with the MACs (resp. robust MACs, robust non-MACs).
Theorem 3.4.
Any group that admits Z as a quotient, contains uncountably many robustnon-MMCs.Proof. Note that there exists a normal subgroup G ′ of G such that G/G ′ is isomorphic to Z .Let p ≥ a be a positive integer satisfying p ≥ a +1 and C denote asubset of { , , · · · , p } of size p − a . Denote the set { , , · · · , p }\ C by C ′ . Let ψ : G/G ′ → Z be a group isomorphism. Let K denote the subgroup ψ − ( p Z ) of G . Note that K is a normalsubgroup of G . For any subset F of C ′ + p N , the subset ψ − (( C + p Z ) ∪ F ) is a robustnon-MAC by Proposition 2.9 and Theorem 2.10. Since the set C ′ + p N contains infinitelymany elements, it has uncountably many subsets. Thus the group G contains uncountablymany robust non-MACs. (cid:3) Corollary 3.5.
For any number field K of degree ≥ , the group GL n ( O K ) contains un-countably many robust non-MMCs where O K denotes the ring of integers of K .Proof. Note that the group GL n ( O K ) admits O × K as a quotient. Since K has degree ≥
3, bythe Dirichlet’s unit theorem, O × K admits Z as a quotient. Hence GL n ( O K ) admits Z as aquotient. By Theorem 3.4, the result follows. (cid:3) Corollary 3.6.
Any finitely generated abelian group of positive rank and any free abeliangroup of positive rank contain uncountably many robust non-MACs.Proof.
It follows from Theorem 3.4. (cid:3)
It seems plausible that any infinite abelian group contains uncountably many robust non-MACs.We conclude this section with the following remarks.
Remark . If a subset A of H (for example, the sets of form C ∪ F considered in Section2) is not a minimal left complement in G , then so are its left translates, i.e., the sets of theform g · A for any g ∈ G . Remark . Note that the subsets which are shown to be non-minimal complements arenot a part of any co-minimal pair . By Theorem 3.4, any finitely generated abelian group ofpositive rank contains uncountably many infinite subsets which are robust non-MACs andin particular, not minimal complements. We contrast this result with [BS20b, Theorem 2.2],which states that any such group also contains uncountably many infinite subsets whichadmit minimal complements. In [BS20b], we considered lacunary sequences in Z d for d ≥ d = 1.It would be interesting to consider the situations when the sets C, F (as in Section 2) aresomewhat modified. For instance,(1) What happens if C is taken to be a “large” set in H and F is taken to be a “small”set lying outside H (or more generally, to be a “small” set in G \ C )?(2) What happens if C intersects with several right cosets of H in G and the intersectionof C with each such (or one such) right coset is “large”?4. Acknowledgements
The first author is supported by the ISF Grant no. 662/15. He wishes to thank theDepartment of Mathematics at the Technion where a part of the work was carried out.The second author would like to acknowledge the Initiation Grant from the Indian Instituteof Science Education and Research Bhopal, and the INSPIRE Faculty Award from theDepartment of Science and Technology, Government of India.
References [AKL20] Noga Alon, Noah Kravitz, and Matt Larson,
Inverse problems for minimal complements and max-imal supplements , Preprint available at https://arxiv.org/abs/2006.00534 , 2020.[BL20] Amanda Burcroff and Noah Luntzlara,
Sets arising as minimal additive complements in the inte-gers , Preprint available at https://arxiv.org/abs/2006.12481 , 2020.[BS] Arindam Biswas and Jyoti Prakash Saha,
On minimal complements in groups , Ramanujan J.,Accepted for publication, Preprint available at https://arxiv.org/abs/1812.10285 .[BS19a] ,
Minimal additive complements in finitely generated abelian groups , Preprint available at https://arxiv.org/abs/1902.01363 , 2019.[BS19b] ,
On additive co-minimal pairs , Preprint available at https://arxiv.org/abs/1906.05837v3 , 2019.[BS20a] ,
Infinite co-minimal pairs in the integers and integral lattices , Preprint available at https://arxiv.org/abs/2005.11095 , 2020.[BS20b] ,
Infinite co-minimal pairs involving lacunary sequences and generalisations to higher di-mensions , Preprint available at https://arxiv.org/abs/2006.02429 , 2020.[CY12] Yong-Gao Chen and Quan-Hui Yang,
On a problem of Nathanson related to minimal additivecomplements , SIAM J. Discrete Math. (2012), no. 4, 1532–1536. MR 3022150[KSY19] S´andor Z. Kiss, Csaba S´andor, and Quan-Hui Yang, On minimal additive complements of integers ,J. Combin. Theory Ser. A (2019), 344–353. MR 3875615 A pair (
A, B ) of nonempty subsets of a group G is called a co-minimal pair if A is a minimal leftcomplement to B and B is a minimal right complement to A [BS19b, Definition 1.1]. A sequence t < t < t < · · · of elements of Z d is said to be lacunary if t > λ ≥ t n > λt n − for any n ≥
1, where “ > ” denotes the lexicographic order on Z d . N NON-MINIMAL COMPLEMENTS 19 [Kwo19] Andrew Kwon,
A note on minimal additive complements of integers , Discrete Math. (2019),no. 7, 1912–1918. MR 3937752[Nat11] Melvyn B. Nathanson,
Problems in additive number theory, IV: Nets in groups and shortest length g -adic representations , Int. J. Number Theory (2011), no. 8, 1999–2017. MR 2873139 Department of Mathematics, Technion - Israel Institute of Technology, Haifa 32000,Israel
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