On Block Monoid Atomic Structure 1 Bryson W. Finklea a ,Terri Moore b ,Vadim Ponomarenko c,∗ , Zachary J. Turner d a St. b Department
John’s College
of Mathematics, University of Washington
c Department
of Mathematics, Trinity University, One Trinity Place, San Antonio, Texas 78212-7200 d Department
of Mathematics, University of Houston
Abstract Several properties of atoms of block monoids are developed. We count the number of atoms of small cardinality. We also develop various properties of atoms of maximal cardinality (the Davenport constant). We use these properties to find new bounds on the Davenport constant both in general and for several important classes of groups. We also determine the Davenport constant (with computer assistance) for a variety of specific groups. Key words: block monoid, Davenport constant, minimal zero sequence
1
Introduction
Let G be a finite abelian group, with operation +. Consider B(G), the collection of all multisets of elements of G whose sum is 0, the identity in G. This forms the so-called block monoid with operation multiset union and identity ∅. This has been the subject of considerable study; for a survey, see Refs. [3,12]. For B ∈ B(G), we write B = g1a1 . . . gkak , and define the cardinality of B as |B| = a1 + · · · + ak . We write g ∈ B to mean that g appears in B at least once. ∗ Corresponding author. Email address:
[email protected] (Vadim Ponomarenko). 1 Research supported in part by NSF grant 0097366.
Preprint submitted to Elsevier Science
10 May 2005
Let A(G) ⊆ B(G) be the atoms of the block monoid; that is, the irreducible nonempty elements. Set Ac (G) ⊆ A(G) to be the collection of all atoms of cardinality c. We will calculate |Ac (G)| for certain cases. Consider the maximal c so that Ac (G) is nonempty. This is D(G), the Davenport constant of G. D(G) has a long history and has appeared in a variety of contexts in number theory, group theory, Krull domains, graph theory, and Ramsey theory; see Refs. [4–6,9,10,16]. For convenience, we set D(G) = AD(G) (G). We will determine various properties of D(G) and D(G). These properties greatly speed up computer search, allowing calculation of D(G) for groups where it was previously unfeasible. Writing G canonically as G = Zd1 ⊕ Zd2 ⊕ · · · ⊕ Zdk with d1 |d2 | · · · |dk , set D? (G) = d1 + d2 + · · · + dk + (1 − k). For many groups, D(G) = D? (G); in fact, it had been conjectured by Baayen in Ref. [18] that this was true for all groups. Though false in general, D(G) = D? (G) holds for many classes of groups such as all groups of rank 1 or 2, all p-groups, and various other examples given in Ref. [9]. Many classes of groups, however, have unknown D(G); much work has been devoted to establishing bounds, for examples see Refs. [6,11,13,14,17]. An important class of groups with D(G) unknown is of the form G = Zkp ⊕ Zq , where k ≥ 3 and p, q are prime; for these groups, D? (G) = (k−1)(p−1)+pq. A famous example introduced in Ref. [1] and expanded in Ref. [11] has p = 2, q = 3. In this case, D? (G) = k + 5. It is known that D(G) = D? (G) for 1 ≤ k ≤ 4, but D(G) ≥ D? (G) + 1 for 5 ≤ k. Our results allow us to determine D(G) exactly (with computer assistance) for 5 ≤ k ≤ 8. We also give new lower bounds for D(G) for p = 2, q ∈ {3, 5, 7}, and a wide range of k. It is known that D(G)−D? (G) can be arbitrarily large. In Ref. [17], an infinite sequence of groups is constructed with D(G) − D? (G) ≥ log96 |G| ≈ 0.2 ln |G|. We give another such sequence with D(G) − D? (G) ≥ 2 log768 |G| ≈ 0.3 ln |G|. There are also a variety of known upper bounds for D(G), and we produce a new one that is better in certain circumstances.
2
Counting Atoms
We now compute |Ac (G)| for arbitrary G and small c. We have A1 (G) = {01 }, and hence |A1 | = 1. Let n = |G| denote the order of G, and let mi denote the number of elements of G of order i; we see that m1 + m2 + · · · + mexp(G) = n. Theorem 1 |A2 (G)| = 12 (n − m1 + m2 ). 2
PROOF. An atom of cardinality 2 must be either of the form g 2 or of the form gh (implicitly g 6= h). There are m2 elements g satisfying g + g = 0, g 6= 0. There are n − m1 − m2 elements g satisfying g 6= 0, g 6= −g. Once g is chosen, h = −g. There are therefore 21 (n − m1 − m2 ) atoms of the form gh. We can easily compute |A2 | for some classes of groups. Corollary 2 |A2 (Zk2 )| = 2k − 1, |A2 (Zk3 )| = (3k − 1)/2, |A2 (Zn )| = bn/2c, |A2 (Zk3 ⊕ Z2 )| = 3k , |A2 (Zk2 ⊕ Zp )| = (p + 1)2k−1 − 1 (p > 2 prime) This method can be continued at least one step further. Despite the tantalizing pattern, we can find no similar formula for A4 . Theorem 3 |A3 (G)| = 16 (n2 − m1 − 3m2 + 2m3 ). PROOF. Atoms of cardinality three fall into three classes: g 3 , g 2 h, ghf (where different letters imply distinct group elements). There are m3 atoms of the first class. In the second class, g cannot have order one or two since otherwise the atom would be reducible. g cannot have order three because then g would equal h and the atom would in fact be of the first class. Any other order for g is possible, and then h = −x − x is determined. Hence there are n − m1 − m2 − m3 atoms of the second class. In the third class, we select the elements sequentially. Once we have done so, we will have counted all atoms six times – once for each ordering of ghf . First, g can be any element not of order one. Then, we select a valid h. Once g and h have been chosen, f is uniquely determined as −g − h. There are five (possibly redundant) restrictions on the choice of h. Since each atom is irreducible, we have h 6= 0, h 6= −g. Because this is the third class, we also insist on g, h, f distinct. These three restrictions correspond to h 6= g, h 6= −2g, −2h 6= g. Each of the first four restrictions eliminates one possible h; although sometimes these eliminated h’s coincide, making some of the restrictions redundant. The last restriction, −2h 6= g, may eliminate multiple h. If g admits at least one h with −2h = g, we call g type I, otherwise type II. We now uniquely write G = G0 ⊕G00 ⊕G000 , where G0 is the maximal elementary 2-group, then G00 is the maximal remaining 2-group (in G/G0 ). Let a be the 3
rank of G00 , and let b be the rank of G0 ⊕ G00 . Here is a table counting atoms of type ghf : Conditions on g
Number of g
Number of h
Comments
|g| = 2, Type I
2a − 1
n − 2 − 2b
0 = −2g, g = −g
|g| = 2, Type II
2b − 2a
n−2
0 = −2g, g = −g
|g| = 3, Type I
m3
n − 2 − 2b
g = −2g
|g| > 3, Type I
n2−b − 2a − m3
n − 4 − 2b
|g| > 3, Type II
n − n2−b − 2b + 2a
n−4
We now combine the above data to get (2a − 1)(n − 2 − 2b ) + (2b − 2a )(n − 2) + m3 (n − 2 − 2b ) + (n2−b − 2a − m3 )(n − 4 − 2b ) + (n − n2−b − 2b + 2a )(n − 4). After some simplification, this is seen to equal n2 − 6n + 2m3 + 3 · 2b + 2. Therefore there are 61 (n2 − 6n + 2m3 + 3 · 2b + 2) atoms of length 3 with distinct elements. Adding this to the previously calculated values for the other two classes of atoms gives 16 (n2 + 2m3 + 3 · 2b + 2 − 6m1 − 6m2 ). Using m1 = 1, m2 = 2b − 1, we can simplify and the result follows.
We can now compute |A3 | for several classes of groups. Corollary 4 |A3 (Zk2 )| = 16 (2k − 1)(2k − 2), |A3 (Zk3 )| = 61 (3k + 3)(3k − 1), |A3 (Zk2 ⊕ Z3 )| = 3 · 22k−1 − 2k−1 + 1, |A3 (Zk3 ⊕ Z2 )| = 2 · 32k−1 + 3k−1 − 1, |A3 (Zk2 ⊕ Zp )| = 13 (p2 22k−1 − 3 · 2k−1 + 1) (p > 3 prime) We can also compute |A3 | for cyclic groups. Corollary 5 |A3 (Zn )| = 16 (n2 − α), where α is given by: (n mod 6) ≡
0 1 2
3
4
5
α=
0 1 4
-3
4
1
PROOF. 1 − m2 = (n mod 2). If 3|n then m3 = 2; else m3 = 0. 4
This formula stands in contrast to the following result, from Ref. [15], counting large atoms: Theorem 6 Let n ≥ 10, and c > 2n/3. Then |Ac (Zn )| = φ(n)pc (n), where φ is Euler’s totient and pc (n) denotes the number of partitions of n into c parts.
3
Atoms In Certain Elementary Groups
We now turn to |Ac (G)| for elementary 2- and 3-groups. The former can be completely determined. Theorem 7 For c ≥ 3, |Ac (Zk2 )| =
Y 1 c−2 (2k − 2i ) c! i=0
PROOF. Any collection of i group elements strictly contained in an atom must be linearly independent in Zk2 (as a vector space). Further, any atom of cardinality at least 3 must contain distinct elements. Therefore, we select c − 1 elements sequentially. At the time of the ith selection, the previous i − 1 elements span a subspace of cardinality 2i−1 . We must avoid this subspace to avoid forming a smaller atom than desired; hence we have 2k − 2i−1 choices available. The last element is determined uniquely as the sum of the first c − 1 elements. We must, in the end, divide by c! to account for the possible orderings of the c (distinct) elements. We can compute |Ac | for elementary 3-groups, for slightly larger c. Our calculations suggest that this tantalizing pattern, unfortunately, does not continue. Theorem 8 |A2 (Zk3 )| = (1/2!)(3k − 1) |A3 (Zk3 )| = (1/3!)(3k − 1)(3k + 3) |A4 (Zk3 )| = (1/4!)(3k − 1)(3k + 3)(3k − 3) |A5 (Zk3 )| = (1/5!)(3k − 1)(3k + 3)(3k − 3)(3k − 4) PROOF. The first two formulas are from Corollaries 2 and 4. Atoms of cardinality four (in Zk3 ) fall into three classes: f 2 g 2 , f 2 gh, f ghe. However, 2f +2g = 0 implies f + g = 0, so there are no atoms of the first class. In the second class, there are 3k − 1 choices for f , 3k − 3 choices for g, and h is determined perforce. We must divide by 2 to account for permuting g, h, leaving 5
(1/2)(3k −1)(3k −3) atoms of this form. In the third case, the atoms must span a 3-dimensional subspace – otherwise it is 2-dimensional and each of h, e must be one of f + g, f + 2g, 2f + g, but any choice of two from this list leads to a contradiction as the atom would be reducible. There are (3k −1)(3k −3)(3k −9) ways to choose three independent elements (the fourth is determined perforce), but this must be divided by 4! to account for permuting the elements. Adding the two expressions gives the desired formula. We now consider atoms of length 5. They fall into three classes: f 2 g 2 h, f 2 ghe, f ghed. In the first class, any choice of linearly independent f, g leads to a viable h = f + g. Therefore there are (1/2)(3k − 1)(3k − 3) atoms of this form, where we must divide by 2 to account for permuting f, g. In the second class, the atoms must span a 3-dimensional subspace, by the argument given in f ghe class above. Hence, there are (1/3!)(3k − 1)(3k − 3)(3k − 9) atoms of this form, where we must divide by 3! to account for permuting g, h, e. In the last class, the atoms can span either a 4-dimensional subspace or a 3-dimensional subspace. If the former, there are (1/5!)(3k − 1)(3k − 3)(3k − 9)(3k − 27) such atoms. The most interesting atoms of class f ghed span a 3-dimensional subspace. We claim that, by relabeling the atoms if necessary, the sets {f, g, h} and {f, e, d} are each linearly dependent. Therefore, we can choose f in 3k − 1 ways, then choose g in 3k − 3 ways, then choose e in 3k − 9 ways. d, h are chosen perforce. However, we must divide by 8, to swap g, h or e, d or the pair {g, h} with {e, d}. Consequently, there are (1/8)(3k − 1)(3k − 3)(3k − 9) atoms of this form. Adding the four expressions gives the desired formula. To prove the claim, take three linearly independent elements, without loss of generality d, e, f . Express g, h in this basis. The only possibilities are (1, 1, 0), (1, 1, 2), (1, 1, 1), (2, 1, 0) (up to permutation of d, e, f ). We must have g + h = (2, 2, 2), so neither may be (1, 1, 1) since we are assuming five distinct elements. Therefore, either g, h are (1, 1, 0), (1, 1, 2) or (2, 1, 0), (0, 1, 2) (up to permutation of d, e, f ). In the first case, {g, d, e} and {g, h, f } are linearly dependent. In the second case, {e, g, d} and {e, f, h} are linearly dependent. This proves the claim and thus the theorem.
4
D(G) and D(G)
We now give several structure results on D(G). If x ∈ B ∈ A(G), then the multiplicity of x must be strictly less than its order in G (unless B contains no other element). The following theorem shows that, in some cases, meeting this bound will determine D(G).
6
Theorem 9 Suppose that G = Zn ⊕ G0 , and there is some B ∈ A(G) with g n−1 ⊆ B. Suppose further that g = (a, 0). Then |B| ≤ D(G0 ) + n − 1. If in fact B ∈ D(G), then D(G) = D(G0 ) + n − 1. PROOF. Because B is an atom, we must have gcd(a, n) = 1, so the subsets of g n−1 sum to {(x, 0)|x ∈ Zn }. Suppose, by way of contradiction, that |B|−(n− 1) > D(G0 ). Then the projection of B \ g n−1 onto G0 cannot be an atom of G0 . Hence we can write B \ g n−1 = B1 ∪ B2 , where B1 and B2 are each nonempty and sum to (b1 , 0), (b2 , 0), respectively. Now, choose k with 0 ≤ k ≤ n − 1 and g k = (−b1 , 0). But now g k ∪ B1 violates the irreducibility of B in G. Hence |B| ≤ D(G0 ) + n − 1. If B ∈ D(G), then in fact D(G) ≤ D(G0 ) + n − 1. Finally, we show that D(G) ≥ D(G0 ) + n − 1. Choose B 0 ∈ D(G0 ). Choose one element, b, of B 0 . Consider (1, 0)n−1 (1, b)(0, B 0 \ b), where (0, B 0 \ b) denotes the image of B 0 \ b under the natural injection of G0 into G. It is easily seen that this is an atom of G, and has cardinality D(G0 ) + n − 1. This theorem is very helpful in determining D(G) through computer search, when D(G0 ) is known for all relevant subgroups G0 . One may restrict consideration to atoms not containing such an element to such a multiplicity. For example, for G = Zk2 ⊕ Z3 , assuming that D(G0 ) is known for all relevant subgroups G0 , it is not necessary to consider any elements of order 2 (approximately one-third of the elements) in the search. Further, the two elements of order 3 may each appear only with multiplicity one (of course, they cannot both appear). Finally, the remaining elements, all of order 6, may each appear with multiplicity at most four. The following theorem also substantially shortens the search for D(G) in many cases. Theorem 10 Suppose that G = Zkp ⊕ G0 , for some prime p and k > 0. Let B ∈ D(G). Then there is an automorphism φ of G where φ(B) contains f1 , f2 , . . . , fk , where the projection of fi onto Zkp is the ith standard unit vector for each i. PROOF. It suffices to prove that the projection of B to Zkp has dimension k. Suppose otherwise; then let g ∈ Zkp be outside the span of the projection of B. Set g 0 = (g, 0) ∈ G. Choose any x ∈ B, and set x0 = x − g. Then B \ {x} ∪ {x0 , g 0 } is an atom of G of larger size than B, which is violative of B ∈ D(G). Corollary 11 Suppose that G = Zkp ⊕ G0 , for some prime p and k > 0. Then D(G) ≥ D(G0 ) + k. 7
Using these structure theorems, and a computer algorithm (available upon request), we have determined the following, previously unknown, values of D(G). These particular groups were selected because they are all known to satisfy D(G) > D? (G). G
Calculated Value Previously Best Bound
Z52 ⊕ Z3
D(G) = 11
D(G) > D? (G) = 10
Z62 ⊕ Z3
D(G) = 12
D(G) > D? (G) = 11
Z72 ⊕ Z3
D(G) = 13
D(G) > D? (G) = 12
Z82 ⊕ Z3
D(G) = 15
D(G) > D? (G) = 13
Z52 ⊕ Z5
D(G) = 15
D(G) > D? (G) = 14
Z43 ⊕ Z2
D(G) = 13
D(G) > D? (G) = 12
In Ref. [17], it was proved that D(Gn ) − D? (Gn ) ≥ n(D(G) − D? (G)). It was there applied to G = Z52 ⊕ Z3 . By setting G0 = Gn , D(G0 ) − D? (G0 ) ≥ n. Because |G| = 96, |G0 | = 96n , this gives a bound of D(G0 ) − D? (G0 ) ≥ log96 |G0 | ≈ 0.2 ln |G0 |. This is the best bound previously known. Observe in the above table that for G = Z82 ⊕ Z3 , D(G) = 15 = 2 + D? (G). We can set G to this group, and G0 = Gn . We get D(G0 ) − D? (G0 ) ≥ 2n. Because |G| = 768, |G0 | = 768n , this gives a bound of D(G0 ) − D? (G0 ) ≥ 2 log768 |G0 | ≈ 0.3 ln |G0 |. Hence, this demonstrates faster growth in the gap D(G) − D? (G).
5
An Upper Bound on D(G)
A variety of upper bounds have been found for D(G), as summarized below. Theorem 12 From Refs. [14,2,17,7], respectively. Let e = exp(G), and r = rank(G). (1) (2) (3) (4)
If G = H ⊕ K and |H| divides |K|, then D(G) ≤ |H| + |K| − 1. D(G) ≤ |G|(r + 1)/2r . D(G) ≤ e(1 + ln |G| ). e D(G) < D? (G)(500r ln r)r .
We now present a simple new upper bound that beats all of these for certain groups. Theorem 13 Let G = H ⊕ K. Then D(G) ≤ D(H)D(K). 8
PROOF. Let B ∈ D(G). Partition B = B1 ∪B2 ∪· · · Bk , where each Bi is the preimage of an element of A(H) under the natural projection from G to H. Hence |Bi | ≤ D(H) for all i. Now, set bi to be the sum of all the elements of Bi . Its projection onto H is 0 (for any i), by construction. Let us denote by ai the projection of bi onto K. If some collection of these ai sums to 0 in K, then the corresponding collection of Bi sums to 0 in G. Hence, since B ∈ D(G) ⊆ A(G), the entire collection of ai is in A(K), and therefore k ≤ D(K). Putting these together, we have |B| = |B1 | + |B2 | + · · · + |Bk | ≤ k max |Bi | ≤ D(K)D(H). 1≤i≤k
Example 14 Consider G = Zkp ⊕ Z2 , for p > 2 prime. By taking H = Z2 , Theorem 13 gives D(G) ≤ 2((p − 1)k + 1) = (2p − 2)k + 2. Theorem 12(1) gives D(G) ≤ pm + 2 · pk−m − 1, for any 0 ≤ m ≤ k/2, which is worse for k ≥ 5. Theorem 12(2) gives D(G) ≤ (2k + 2)( p2 )k , which is worse for k ≥ 2. Theorem 12(3) give D(G) ≤ (2p ln p)k + 2p(1 − ln p), which is worse for all k. Theorem 12(4) gives D(G) < ((p − 1)k + p + 1)(500k ln k)k , which is worse for all k ≥ 2. Example 15 Consider G = Zk3 ⊕ Z2 ⊕ Z2 . By taking H = Z2 ⊕ Z2 , Theorem 13 give D(G) ≤ 3(2k + 1) = 6k + 3. Theorem 12(1) gives a variety of possible bounds, all worse for k ≥ 5. Theorem 12(2) gives D(G) ≤ (4k + 4)( 23 )k , which is worse for k ≥ 2. Theorem 12(3) gives D(G) ≤ (6 ln 3)k + 6(1 + ln 2), which is worse for all k. Theorem 12(4) gives D(G) < (2k + 7)(500k ln k)k , which is worse for all k ≥ 2.
6
D(G) for a Special Class of Groups
In this section, we restrict attention to groups of the class Gk = Zk2 ⊕ Z3 . For convenience, set dk = D(Gk ) − D? (Gk ), the gap. Because G? (Gk ) = k + 5, we have equivalently D(Gk ) = dk + k + 5. It has been shown in Ref. [11] that dk = 0 for 1 ≤ k ≤ 4, and dk ≥ 1 otherwise. In Section 4 we showed that dk = 1 for 5 ≤ k ≤ 7, and dk = 2 for k = 8. This last fact actually means that dk ≥ 2 for k ≥ 8, as shown by the following. Theorem 16 Let dk = D(Gk ) − D? (Gk ). Then dk ≥ dk−1 , for k ≥ 2. PROOF. We have dk − dk−1 = D(Gk ) − D(Gk−1 ) − 1. By Corollary 11, D(Gk ) ≥ D(Gk−1 ) + 1.
9
However, the dk cannot grow too fast either. Theorem 17 Let dk = D(Gk ) − D? (Gk ). Then dk ≤ dk−1 + 2, for k ≥ 2. Further, if dk = dk−1 + 2, then 3|D(Gk ).
PROOF. Let B ∈ D(Gk ). Consider the projection of each element onto Z3 . There must be some collection {x1 , . . . , xi } with 1 ≤ i ≤ 3 where x = x1 + · · · + xi has order 2 (that is, x is in the kernel of the projection onto Z3 ). Set B 0 = B \ {x1 , . . . , xi } ∪ {x}. We have B 0 ∈ A(Gk ). By Theorem 9, |B 0 | ≤ D(Gk−1 ) + 1. We also have D(Gk ) = |B| ≤ |B 0 | + 2. Combining, we have D(Gk ) ≤ D(Gk−1 )+3, and the theorem follows since D(Gk ) = dk +k +5. The only circumstances under which the above {x1 , . . . , xi } cannot be of cardinality 2 is if all elements of B have the same (nonzero) projection onto Z3 . We may assume (by taking an automorphism if necessary) that this projection is 1. However, since B ∈ B(Gk ), 3 must divide |B| and hence D(Gk ).
These results can be combined with the table of Section 4 to bound D(Gk ) as follows. k=
5
6
7
8
D(Gk ) =
11
12 13 15
9
10
k > 10
[16..18] [17..21] [(k+7)..(3k−9)]
Further, if we knew the single fact that D(G9 ) 6= 18, then the upper bound could be improved to D(Gk ) ≤ 3k − 12, for k ≥ 10. Our lower bound of k +7 improves on the current best lower bound for D(Gk ), found in Ref. [13], of D(Gk ) ≥ max(k + 6, k + log2 k − 1). Our bound is better for k ∈ [8, 64). This is further improved in the following section, where it is shown that D(Gk ) ≥ k + 9 for k ≥ 57, and D(Gk ) ≥ k + 11 for k ≥ 166. Our bounds are therefore better for all k ∈ [8, 4096).
7
Large Atom Construction
We now turn to a construction that improves the known lower bounds for various D(Zk2 ⊕ Zp ). Definition 18 Fix p prime, odd s > 0, t > s, and r with 0 ≤ r < p. Set ³´ t ⊕ Zp . We will indicate elements as m = s . Let G = G(m, r, p) = Zm+r 2 ordered (m+r+1)-tuples, in terms of the standard basis ei , 1 ≤ i ≤ (m+r+1). 10
We define B = {f1 , . . . , fm+r , g1 , . . . , gt }, a sequence of elements from G, as follows. Set fi = ei + em+r+1 , for 1 ≤ i ≤ m + r. Fix the set T = {1, 2, . . . , t}, and arbitrarily associate ³ ´the s-subsets of T (that is, those subsets of cardinality s) with the first m = st coordinates. We define gi , for i ∈ T , by setting the j th coordinate of gi to be 1 if element i ∈ T is contained in the j th s-subset of T . We set the last r + 1 coordinates of each gi to be 1. This construction produces a sequence, |B|, of cardinality m + r + t. Hence, if there is no B 0 ( B with B 0 ∈ A(G), then this construction proves that D(G) ≥ m + r + t. Fortunately, this condition can be tested by the following theorem. Theorem 19 Suppose that, if t is even, then p - (t + m) (if t is odd, no restriction). Then, the following conditions are equivalent: (1) No proper subset of B is in A(G). Ã ! P ³n´³t−n´ (2) For all n with 1 ≤ n < t, p - n + r (n mod 2) + . j s−j oddj
If these conditions hold and t is even, then B itself is not in A(G).
P
P
PROOF. Given X ⊆ G, we define X = α∈X α. Consider B 0 ⊆ B. Set P X = B 0 ∩ {g1 , g2 , . . . , gt }, n = |X|, x = X, and n0 = |B 0 | − n. We first consider the case n < t. Consider the first m coordinates of x. Each is 1 precisely when an odd number of elements of X³ are in the s-set corresponding P n´³t−n´ to that coordinate. Altogether, there are such, where the sum j s−j is taken over all odd j (j corresponds to the number of elements of X in the s-set). Now consider the next r coordinates of x. They are all equal to (n mod 2). Hence, for B 0 to be a zero-sequence, it must contain precisely those fi corresponding to the coordinates where x is³one. B 0 is a zero-sequence ´³ Therefore, ´ P n t−n if and only if n0 = r (n mod 2) + , if and only if |B 0 | = n + j s−j P ³ ´³
´
n t−n . However, because all elements of B have 1 in the r (n mod 2) + j s−j last coordinate, we also know that B 0 is a zero-sequence if and only if |B 0 | ≡ 0 (mod p).
We now consider the remaining case, those B 0 where n = t. In this case, because s is odd, each of the first m coordinates of x is 1. If t is odd, then so are the next r coordinates, and hence B 0 is a zero sequence if and only if it is all of B. If t is even, then B 0 is a zero sequence if and only if |B 0 | = t + m is a multiple of p, which is prohibited by hypothesis. 11
Finally, if t is even and r > 0, then B ∈ / A(G), because t + 1 elements have a th 1 in the m + 1 coordinate; however t + 1 is odd. If t is even and r = 0, then P B∈ / A(G), because |B| = t + m and hence the last coordinate of B is not 0.
We can apply this to the family considered in the previous section, namely Gk = Zk2 ⊕ Z3 . As mentioned previously, for k < 4096, these bounds are new. Example 20 Let dk = D(Gk ) − D? (Gk ). Then (1) For all k ≥ 57, dk ≥ 4; equivalently D(Gk ) ≥ k + 9, and (2) For all k ≥ 166, dk ≥ 6; equivalently D(Gk ) ≥ k + 11.
PROOF. For (1), we take r = 1, s = 8, t = 3. For (2), we take r = 1, s = 11, t = 3.
Many similar examples can be generated, improving the known lower bounds 0 k for D(G(m, r, p)). For example, ³ consider p = 5. Set´Gk = Z2 ⊕ Z5 . The current best bound is D(G0k ) ≥ max k + 9, k + log2 k − 2 . The following beats this bound for all k ∈ [464, 235 ). Example 21 Let G0k = Zk2 ⊕ Z5 . Then (1) For all k ≥ 464, D(G0k ) ≥ k + 11. (2) For all k ≥ 6188, D(G0k ) ≥ k + 17. (3) For all k ≥ 237337, D(G0k ) ≥ k + 33.
PROOF. For (1), we take r = 2, t = 7, s = 1. For (2), we take r = 0, t = 17, s = 5. For (3), we take r = 1, t = 33, s = 5.
For one more example, ³consider p = 7. Set ´G00k = Zk2 ⊕ Z7 . The current best bound is D(G00k ) ≥ max k + 13, k + log2 k − 3 . The following beats this bound for all k ∈ [6435, 237 ). Example 22 Let G00k = Zk2 ⊕ Z7 . Then (1) For all k ≥ 6435, D(G00k ) ≥ k + 15. (2) For all k ≥ 245159, D(G00k ) ≥ k + 23. (3) For all k ≥ 5379619, D(G00k ) ≥ k + 34. 12
PROOF. For (1), we take r = 0, t = 15, s = 7. For (2), we take r = 2, t = 23, s = 7. For (3), we take r = 3, t = 34, s = 7.
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Open Problems
In our work with block monoids, we have encountered some tantalizing problems that we could not solve. Conjecture 23 Let dk = D(Gk ) − D? (Gk ). Then dk − dk−1 ≤ 2, for all k. The construction of Definition 18 appears to have some limitations; there are approximately k conditions to fulfil, which becomes increasingly difficult as k grows. Conjecture 24 In the construction of Definition 18, for any fixed p, there are only finitely many k where the conditions of Theorem 19 apply. Our numerical computations of A(G) have led us also to the following two unusual conjectures. Conjecture 25 Let G, H be finite abelian groups of the same order, with rank(G) ≤rank(H). Then |A(G)| ≥ |A(H)|. It seems surprising that δ(n), defined below, can ever be less than 0. In fact, in Ref. [8] we calculated δ(23) = −1100. δ(n) seems particularly small when n is prime and n + 1 has many factors, such as in this example. Conjecture 26 Set δ(n) = |A(Zn+1 )|−|A(Zn )|. Then, as n → ∞, lim sup δ(n) = ∞, and lim inf δ(n) = −∞.
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