Some results on near factorizations of boolean groups

A group in which every element is its own inverse is called a boolean group. We show that if G is an infinite boolean group, then there exist subsets A,B of G such that |A| = |B| = |G| and G \ {0} is the direct sum of A and B, i.e., for each (a, b) ∈ A × B, a + b = 0 and for each nonzero element g ∈ G, there exists a unique pair (a, b) ∈ A×B such that g = a + b. A variant of this result also is derived: any infinite boolean group G has subsets A,B such that min{|A|, |B|} = א0 and G \ {0} is the direct sum of A and B. In [S. Grüter et al., Australas. J. Combin. 49 (2011), 245–254] and [P.N. Balister et al., European J. Combin. 32 (2011), 533–537], by using different methods, it has been shown that if A,B are subsets of a finite nontrivial boolean group G such that G \ {0} is the direct sum of A and B, then |A| or |B| is 1. In this note we present a simple proof for this result and extend this result to infinite boolean groups. Let G be an additive group; let A,B be two subsets of G; the set {a + b : a ∈ A and b ∈ B} is called the sum of A and B and denoted by A+B; if for each g in this set, there is only one pair (a, b) ∈ A× B such that g = a+ b, then this set is called the direct sum of A and B and denoted by A⊕B. For any g ∈ G, in place of {g}+A, we simply write g+A. If for every g ∈ G, g+g = 0, then G is called a boolean group. A well known easily provable fact is that any boolean group is abelian. If A,B are subsets of a boolean group, it is easy to see that 0 / ∈ A+B ⇐⇒ A ∩ B = ∅. Remark 1. Let G be a group; let A,B be subsets of G such that A+ B = A⊕ B. Then it can be verified easily that for any P ⊂ A, P +B = P ⊕B and for any α ∈ G, (α +A) +B = (α +A)⊕ B. If {Bi : i ∈ I} is a collection of subsets of G such that for each i ∈ I, A+Bi = A⊕Bi and for all distinct i, j ∈ I, (A+Bi)∩ (A+Bj) = ∅, then it can be verified that ∪i∈I(A+ Bi) = A⊕ (∪i∈IBi). 136 G.R. VIJAYAKUMAR This note concerns the question of finding subsets A,B of a nontrivial boolean group G such that G \ {0} = A ⊕ B. Choosing a subset of G which is a singleton and its complement in G is an obvious solution. In [4], it has been conjectured that if G is finite, then there is no other solution. (The terminology used in [4] for stating this conjecture is quite different.) In [1] and [3], by using different methods, this conjecture has been settled. An objective of this note is to present a simple short proof for the result of [1, 3] obtained in this regard, and generalize this result to infinite boolean groups. Let N denote the set of all natural numbers. The main objective of this article is to show in an infinite boolean group G, the existence of subsets A,B, P and Q such that |A| = |B| = |G|, min{|P |, |Q|} = |N| and G \ {0} = A⊕ B = P ⊕Q. Proposition 2. Let G be a finite nontrivial boolean group and A,B be subsets of G such that G \ {0} = A⊕B. Then |A| or |B| is 1. Proof. First note that |G| − 1 = |A||B|. Suppose that |A| = 1 = |B|; then (|A| − 1)(|B| − 1) > 0; therefore |A| + |B| < 1 + |A||B| = |G| whence we can find some α ∈ G \ (A ∪ B). Now, let P = α + A and Q = α + B. It is easy to verify that G \ {0} = P ⊕Q. Note that 0 / ∈ P ∪ Q. Now, let p ∈ P . Let us show that there is exactly one element v ∈ P such that p + v ∈ Q. Since p = 0, for some p′ ∈ P and q ∈ Q, p = p′ + q whence p + p′ = q. Suppose that p′′ ∈ P and r ∈ Q such that p + p′′ = r. Then p′ + q = p = p′′ + r whence p′ = p′′. Therefore for each p ∈ P , there is a unique p ∈ P such that p+ p ∈ Q. Since for each p ∈ P , p = p (because 0 / ∈ Q) and (p ) = p, we find that {{p, p } : p ∈ P} is a partition of P into subsets of order 2. Therefore |P | is even. Since |G| = 1+ |P ||Q|, it follows that |G| is odd—a contradiction. Theorem 3. Suppose that the set of all nonzero elements of a nontrivial boolean group G is the direct sum of some subsets A,B of G. Then either one of the sets A, B is a singleton or both are infinite. Proof. We can assume that |A| < ∞ and |B| 2. Let b1, b2 be distinct elements of B. Let H be the subgroup generated by A ∪ {b1, b2}. [For any nonempty finite subset X of G, let σ(X) denote the sum of all elements in X; let σ(∅) = 0; note that for any finite subsets X, Y of G, σ(X) + σ(Y ) = σ(XΔY ). Therefore for any S ⊂ G, {σ(X) : X ⊆ S and |X| < ∞} is a subgroup of G, known as the group generated by S.] Since A ∪ {b1, b2} is finite, H also is finite. Let g be a nonzero element of H; then for some a ∈ A and b ∈ B, g = a + b whence b = g + a ∈ H. Therefore H \ {0} = A + (H ∩ B) whence by Remark 1, H \ {0} = A ⊕ (H ∩ B). Since |H ∩ B| 2, by Proposition 2, it follows that |A| = 1. Let G be a nontrivial group and g be an element of G. If A,B are subsets of G such that G \ {g} is the direct sum of A and B, then the pair (A,B) is called a near factorization of G. (For some basic results on this notion, see [2]; in [6, 7] by using this notion, an important class of graphs known as ‘partitionable graphs’ has been studied.) A near factorization (X, Y ) is called trivial if min{|X|, |Y |} = 1; in NEAR FACTORIZATIONS OF BOOLEAN GROUPS 137 this case, note that for some a, b ∈ G, {X, Y } = {{a}, G \ {b}}. If G is boolean and (A,B) is a near factorization of G, then G \ {0} = (α + A)⊕ B where α is the element which does not belong to A + B; from this observation and Proposition 2, we have the following. Corollary 4. Let G be a finite nontrivial boolean group; then any near factorization of G is trivial. A natural question in connection with the above result is the following. Is there an infinite boolean group that admits a nontrivial near factorization? This is answered by Theorem 7; to derive this result, we need the following two set theoretic results. (A proof for the first one, a basic result in set theory, is given in [5].) Theorem 5. If A is an infinite set and B is a non-empty set, then |A × B| = |A ∪B| = max{|A|, |B|}. Lemma 6. Any non-empty set X can be endowed with a well ordering such that for all a ∈ X, |{x ∈ X : x ≺ a}| < |X|. (For any x, y ∈ X, when we write x ≺ y we mean that x = y and x y.) Proof. Let be a well ordering of X. For any a ∈ X, let Sa = {x ∈ X : x ≺ a}. If for each a ∈ X, |Sa| < |X|, then has the required property; so assume that {a ∈ X : |Sa| = |X|} is non-empty. Let m be the smallest element of this set. Since |Sm| = |X|, there is a bijection θ : X → Sm. Now define a relation ′ on X as follows: for any x, y ∈ X, x ′ y ⇐⇒ θ(x) θ(y). It is easy to verify that ′ is a well ordering of X with the required property. Theorem 7. Let G be an infinite boolean group. Then there exist subsets A,B of G such that |A| = |B| = |G| and G \ {0} = A⊕ B. Proof. Let H = G \ {0}. Let be a well ordering of H having the property mentioned in the statement of Lemma 6. By using transfinite induction, let us construct two maps from H to itself such that for each g ∈ H the following hold. (For any x ∈ H, its images under these mappings are denoted by x′ and x′′, respectively.) (1) g ∈ {x′ : x g}+ {x′′ : x g}. (2) {x′ : x g} ∩ {x′′ : x g} = ∅. (3) For each a in H such that a ≺ g, a′ = g′ and a′′ = g′′. (4) {x′ : x g}+ {x′′ : x g} = {x′ : x g} ⊕ {x′′ : x g}. Let f be the first element of (H, ). Choose f ′, f ′′ arbitrarily such that f = f ′ + f ′′. Obviously, (1), (2), (3) and (4) hold when g = f ; now, let h be any element in H \ {f}. Assume that {x′, x′′ : x ≺ h} is known and for each g ∈ {x ∈ H : x ≺ h}, (1), (2), (3) and (4) hold. 138 G.R. VIJAYAKUMAR Let X = {x′, x′′ : x ≺ h} ∪ {0, h} and Y = X +X +X +X. Since |{x ∈ H : x ≺ h}| < |H|, by Theorem 5, |X| < |H| whence we have |Y | |X ×X ×X ×X| < |H| also, by Theorem 5. Therefore we can find some h′ ∈ (H \ Y ). If h / ∈ {x′ + y′′ : x ≺ h and y ≺ h}, taking h′′ = h + h′, it can be verified that when g = h, (1), (2), (3) and (4) hold; so, suppose that for some p, q ∈ {x ∈ H : x ≺ h}, p′ + q′′ = h. Now let P = {x′, x′′ : x ≺ h} ∪ {h′} and Q = P + P + P . Using Theorem 5, it is easy to verify that |Q| < |H|; so, let h′′ be any element in H \ Q. It is easy to verify that when g = h, (1), (2), (3) and (4) hold. Thus by transfinite induction, we obtain two subsets A := {g′ : g ∈ H} and B := {g′′ : g ∈ H} of H such that for each g ∈ H, (1), (2), (3) and (4) hold. Since for every g ∈ H, (3) holds, the maps obtained are injective; therefore |A| = |B| = |G|. Since (2) holds for all g ∈ H, A ∩ B = ∅; therefore 0 / ∈ A + B. Since (1) holds for each g ∈ H, H ⊆ A+B. Now from the fact that (4) holds for each g ∈ H, it follows that H = A⊕ B. A result [8] on Z, the set of all integers, which Theorems 7 and 8 are somewhat reminiscent of, is the following: If A,B are finite subsets of Z such that 0 ∈ A ∩ B and A + B = A ⊕ B, then there exist infinite subsets P,Q of Z such that A ⊂ P , B ⊂ Q and Z = P ⊕Q. Let A,B be subsets of a boolean group G such that |A| |B| > 1 and G \ {0} = A⊕B. By Theorem 3, |B| |N| whence it is natural to ask whether this lower bound for |B| can be attained. The following result answers this question affirmatively. Theorem 8. Any infinite boolean group G has subsets A,B such that min{|A|, |B|} = |N| and G \ {0} = A⊕B. Proof. Let H = {0, h1, h2, . . .} be a countably infinite subgroup of G. (Let S be a countably infinite subset of G and F be the collection of all finite subsets of S; it is a well known fact that such a collection is countable; it is easy to show that there exists a surjective map from F to the subgroup generated by S; therefore this subgroup also is countable; we can take H to be this subgroup.) By using induction, let us construct three sequences (An) ∞ n=1, (Xn) ∞ n=1 and (Yn) ∞ n=1 whose terms are finite subsets of H such that for each k ∈ N, Ak ⊂ Ak+1, Xk ⊂ Xk+1 and Yk ⊂ Yk+1 and (1) and (2) given below hold. Let A1 = {0}, X1 = {0, h1} and Y1 = {h1, h2}. It is easy to verify that (1) and (2) given below hold when k = 1. Let us suppose that for some n ∈ N, chains A1 ⊂ A2 ⊂ · · · ⊂ An, X1 ⊂ X2 ⊂ · · · ⊂ Xn and Y1 ⊂ Y2 ⊂ · · · ⊂ Yn whose terms are finite subsets of H are known such that for each k ∈ {1, 2, . . . , n}, the following hold. (1) hk ∈ Ak +Xk = Ak ⊕Xk. (2) 0 / ∈ Ak + Yk and hk ∈ Ak + Yk = Ak ⊕ Yk. Let us construct An+1, Xn+1 and Yn+1 as follows. Let P = Q + Q + Q + Q where Q = An ∪Xn ∪ Yn ∪ {hn+1}. Let a be any element in H \ P ; let An+1 = An ∪ {a}. If hn+1 ∈ An + Xn, let Xn+1 = Xn; otherwise let Xn+1 = Xn ∪ {a + hn+1}. If hn+1 ∈ An + Yn, let Yn+1 = Yn; otherwise let Yn+1 = Yn ∪ {a + hn+1}. It can be verified that (1) and (2) hold when k = n+ 1. NEAR FACTORIZATIONS OF BOOLEAN GROUPS 139 Therefore by induction, we get the desired three sequences. Now let A = ∪i=1Ai, X = ∪i=1Xi and Y = ∪i=1Yi. Since 0 ∈ A1 + X1 and (1) holds for each k ∈ N, it follows thatH = A⊕X; since (2) holds for each k ∈ N, it follows thatH\{0} = A⊕Y . Now let J be a subset of G such that {H + α : α ∈ J} is the collection of all cosets of H, other than H itself. Let B = Y ∪ [∪α∈J(X + α)]. Since for each α ∈ J , H + α = A ⊕ (X + α) and (H \ {0}) ∩ (H + α) = ∅ and for any two distinct α, β ∈ J , (H + α) ∩ (H + β) = ∅, by Remark 1 (taking {Y } ∪ {X + α : α ∈ J} in place of {Bi : i ∈ I}) we get (A + Y ) ∪ [∪α∈J(A + (X + α))] = A ⊕ B; i.e., (H \ {0}) ∪ [∪α(H + α)] = A ⊕ B; i.e., G \ {0} = A ⊕ B. By construction itself, A is infinite whereas the fact that |Y1| = 2 and Theorem 3 imply that Y is infinite whence min{|A|, |B|} = |N|.