Counterexamples to Gao’s
zero-sum invariant conjecture
Henry Zweiman
September 9, 2026
Abstract
For every odd integer , we prove that . This disproves Gao’s conjecture that for every nontrivial finite abelian group. The construction is a zero-sum-free sequence of length with exactly six omitted nonzero subsequence sums. These sums form the product of the three nonzero elements of an elementary two-group of rank two with two adjacent cyclic residues. Their differences force every coset containing them to contain zero. We also give explicit extremal extensions and show that the index-two refinement and the associated local invariant attain their larger possible values.
1. The conjecture and the counterexamples
Let be a nontrivial finite abelian group, written additively. A sequence over is a finite multiset of elements, written multiplicatively. Thus denotes occurrences of . For a sequence , put
The sequence is zero-sum free if . Its set of holes is
Write for the greatest length of a zero-sum-free sequence and for the Davenport constant.
The invariant is the least integer such that every zero-sum-free sequence of length at least satisfies
The exclusion of zero from the coset is part of the definition. Merely putting the holes in a proper subgroup does not establish (1).
Gao conjectured that ; equivalently, . The latter formulation appears in Gao and Thangadurai [3, Definition 3 and Conjecture 4]. Geroldinger and Yang [4, Introduction and Section 6] restate it as an open conjecture. Their recent results include the family and certain groups with four elementary two-group factors. We give counterexamples with three such factors.
Theorem 1.1.
Let be odd, and let
Set
Then the sequence
is zero-sum free, has length , and has exactly the six holes
No coset excluding zero contains these holes. Consequently,
The Davenport formula used here is classical. Baayen [1, p. 1] proves that every sequence of length in this group has a nonempty zero-sum subsequence for odd . On the other hand, is zero-sum free. Thus . The same formula is recorded in [4, Proposition 2.1]; it is not a new result of this paper.
2. A coset obstruction and the exact holes
We start with the obstruction that makes the construction effective.
Lemma 2.1.
Let be distinct nonzero elements of order two in an abelian group, let be any group element, and let . No coset with contains
Proof.
Suppose all displayed elements lie in . The difference of and gives . The difference of and gives . Therefore . This element also lies in , forcing , a contradiction.
Put . The four elements of form a circuit in : their sum is zero, and no proper nonempty subset has sum zero. Also,
Any zero-sum subsequence of must therefore use none or all four of the terms . In the former case it is with , and has nonzero sum. In the latter case its sum is with , again nonzero. Thus is zero-sum free, and its length is .
We determine all its holes, including those that will not be needed to disprove the conjecture. Enumerate . The map
is a surjective homomorphism of elementary two-groups with kernel consisting of the zero vector and the all-one vector. Hence each is represented by precisely two subsets of , which are complements. Let their sizes be and .
Selecting either subset and between zero and copies of shows that the cyclic coordinates of above are
Consequently the omitted coordinates above form the intersection
Matching the first element of one pair with the second of the other would require or , which is impossible. The other two matches occur together, precisely when
Since and , this is equivalent to . For odd and , the only possibility is . The sums of pairs from are exactly
each represented by complementary pairs. When , the two omitted coordinates in (6) are
This proves (3). Its six elements are distinct and nonzero.
Apply Lemma 2.1 with , , and . Thus , of length , violates (1), so . For completeness, every zero-sum-free sequence of length has : if , then would be zero-sum free of greater length. The empty set satisfies (1), so . This proves (4).
3. Extremal extensions and the parity of the construction
Geroldinger and Yang define by requiring the subgroup in (1) to have index , for a prime divisor of . They also define a local invariant for a maximal zero-sum-free sequence : it is the least threshold for which all subsequences of beyond that threshold have their holes in a coset excluding zero. Here maximal means that no additional term can be adjoined while retaining zero-sum freeness.
Corollary 3.1.
For the group in Theorem 1.1, we have . Let . Then
is zero-sum free of length and satisfies . Moreover,
is a minimal zero-sum sequence of length .
Proof.
The inequality follows from the definitions, and the reverse inequality follows from the empty-hole property at length . Since is a hole of and equals its negative, is zero-sum free. It has greatest possible length, hence is maximal and has no holes. Its subsequence violates the coset property, so the local threshold is exactly .
The sum of is , by (5). Therefore the sum of is , whose negative is . Adjoining this term gives a zero-sum sequence. A proper zero-sum subsequence either lies in , or has a nonempty zero-sum complement in ; both are impossible. Thus is minimal.
The parity restriction in the theorem is intrinsic to this particular sparse construction, rather than an omitted even case.
Proposition 3.2.
Let , let with , and use the same circuit . For , the sequence
is zero-sum free if and only if . Such a exists if and only if is odd.
Proof.
A zero-sum subsequence must use either none or all four circuit terms. The former possibility is excluded by the number of copies of . In the latter case the available cyclic sums are . This interval modulo excludes zero exactly when is congruent to one or two. The first possibility is impossible by parity, giving the stated criterion. The congruence is soluble exactly when divides two, or equivalently when is odd.
4. The smallest example and remaining questions
For , the counterexample lies in a group of order and has the particularly simple form
Its nonempty subsequences have distinct sums, all nonzero. The six holes are
They lie in the proper subgroup , but in no coset that excludes zero. This distinction explains why containment in a proper subgroup would not suffice in the conjecture.
The proof above is uniform in and does not depend on finite computation. The accompanying verification checks the exact hole set, the difference-generated subgroup obstruction, the extremal extension, and the longest atom for every odd from three through . It also separately enumerates all nonempty subsequences in the smallest example.
The theorem disproves the universal conjecture, but it does not classify all finite abelian groups with . In particular, Proposition 3.2 only rules out this construction for even ; it does not determine in all those cases. A revised classification must distinguish the established positive families from the rank-four family exhibited here. The explicit six-hole configuration provides a local obstruction that such a classification must accommodate.
References
[1] P. C. Baayen, , Report ZW 1969-006, Stichting Mathematisch Centrum, Amsterdam, 1969. Institutional full text.
[2] W. Gao, On Davenport’s constant of finite abelian groups with rank three, Discrete Mathematics 222 (2000), 111–124. doi:10.1016/S0012-365X(00)00010-8.
[3] W. D. Gao and R. Thangadurai, Davenport constant and non-Abelian version of Erdős–Ginzburg–Ziv theorem, in The Riemann Zeta Function and Related Themes, Ramanujan Mathematical Society Lecture Notes Series, vol. 2, 2006, pp. 57–64. Author-hosted version.
[4] A. Geroldinger and W. Yang, On a classical zero-sum invariant, arXiv:2608.19090v1, August 19, 2026.