Maximal-elasticity length sets
for families of two-groups of exponent eight
Henry Zweiman
September 9, 2026
Abstract
Let be noncyclic, with . We prove that the minimum distance of the signed support of any minimal zero-sum sequence of length divides three. This distance is one if , and it is also one whenever . Consequently, for either of these families, sets of lengths of maximal elasticity in transfer Krull monoids over are intervals apart from uniformly bounded end pieces. The proof combines relations along the order filtration with a classification of signed supports whose atom lengths are all two modulo three. A group-algebra argument excludes the order-two terms that could otherwise survive that classification. The result settles these families of the maximal-elasticity conjecture; the general exponent-eight case remains open.
1. Introduction and main results
A central problem in factorization theory asks whether the signed support of every longest minimal zero-sum sequence over a noncyclic finite abelian group, other than an elementary two-group, admits consecutive factorization lengths. Bashir, Geroldinger, and Zhong formulate this as Conjecture 1.1 in [1]. Their small-exponent theorem covers exponents from three through eleven except eight. The equivalent maximal-elasticity formulation remains Conjecture 2.3 in the February 2026 version of [4].
We establish two families within the exponent-eight case. All groups are finite and abelian, and are written additively. Sequences over a subset are written multiplicatively, so that means occurrences of . An atom is a nonempty zero-sum sequence with no proper nonempty zero-sum subsequence. The Davenport constant is the greatest atom length. For a subset , let be its monoid of zero-sum sequences and let be the union of the sets of successive differences in its factorization-length sets.
Theorem 1.1.
Let
For every atom of length , put . Then is nonempty and
Moreover, if either
The assertion with includes for every , as well as all groups obtained by adjoining elementary two-group factors. The cyclic group is excluded. The two conclusions are proved together: the filtration argument leaves only distance three, and a signed-residue argument eliminates this possibility when there is one order-eight factor.
For an atomic monoid , let denote the set of numbers of atoms in factorizations of . Its elasticity is the supremum of over nontrivial length sets. For a transfer Krull monoid over , its system of length sets agrees with that of , and its elasticity is .
Corollary 1.2.
Let satisfy either of the two conditions in Theorem 1.1, and let be a transfer Krull monoid over . Then . There is an integer , depending only on , such that every length set of elasticity has the form
Here , , and all intervals contain integers only.
The invariant records the differences of arbitrarily long almost arithmetical progressions among length sets of maximal elasticity. Geroldinger and Zhong [2, Corollary 3.3] show that it equals when every signed support in Theorem 1.1 has minimum distance one. The asserted uniform end bound then follows from [1, Corollary 2.3].
The previously known cases include rank two, homocyclic -groups, the condition , and two-level groups with . Thus there is overlap with earlier theorems, but those statements do not cover the families established here in full. We do not assert a resolution for all groups of exponent eight, nor for arbitrary finite abelian groups.
2. Distance relations and extremal sequences
We first record the elementary tools and the one structural lemma used from the literature.
Lemma 2.1.
Let contain an element of order greater than two. Then . If , every atom over satisfies
In particular for every .
Proof.
Differences of lengths of two factorizations of the same element form a subgroup of : products add such differences, and exchanging the factorizations changes their signs. Its positive generator, when nonzero, is the minimum successive distance, since every length difference is its multiple and the generator itself is realized. For an atom , the sequence has a factorization into two atoms and a factorization into the atoms . This proves (1). Apply it to ; the two lengths are distinct if .
Lemma 2.2 (Geroldinger–Zhong).
Suppose has rank at least two and exponent at least three, and is an atom of length . If independent elements and an element lie in , where , and
then . The same conclusion holds if contains a basis of .
This is [2, Lemma 3.10], restated as [1, Lemma 3.1]. We will use its relation criterion only when , , and two is the least positive multiple placing in the subgroup generated by the .
Lemma 2.3.
Let , where every is a power of two, and put . Then , and the support of every atom of length generates . For a decomposition
the projections onto of the order-two terms of any zero-sum-free sequence of length are linearly independent over .
Proof.
With basis , the group algebra has the presentation
Its greatest nonzero total degree is . For a zero-sum-free sequence , the product
is nonzero: its coefficient at in the group basis is one, since no nonempty subsequence has sum zero. Each factor has positive augmentation degree, so . Conversely, the sequence is zero-sum free and adjoining makes it an atom. This proves the Davenport formula.
To prove generation, suppose an atom of greatest length has support in a proper subgroup . Choose and a term . Replace this term by . A zero-sum subsequence uses both new terms or neither, because their nonzero images in are opposites. Either possibility reduces to a zero-sum subsequence of , so the new sequence is an atom longer than , a contradiction.
Finally, if , all terms of beyond the product of its linear factors have degree greater than and vanish. The linear factor of an order-two term is precisely its projection: the coordinates in the factors of order four and eight are even. If these linear forms were dependent, their product would vanish in the algebra on the variables for , whose squares are zero. Indeed, express one dependent form as a sum of the others and expand; every summand contains a square of a linear form, and that square is zero in characteristic two. Then , a contradiction.
The truncated group-algebra method and the extremal-degree identities are classical; see also [3, Lemma 3.4] for these identities and their application to extremal sequences. The last assertion above is the specific consequence needed here.
3. The order filtration leaves only distance three
Proposition 3.1.
Under the hypotheses of Theorem 1.1, the minimum distance divides three.
Proof.
By Lemma 2.3, generates . Select from this support whose projections to form a basis. This is possible by choosing a basis modulo two and lifting it; the resulting matrix is invertible over . Their span is a direct summand, and
Select from whose projections onto the order-four part of this quotient form a basis. Each has order four. Suppose first that some such has order eight. Then
for a nonempty . Put . Both sequences
are atoms. In fact a zero-sum subsequence must use either zero or four occurrences of (respectively ), by the quotient, and independence forces all remaining terms in the latter case. The identities
give length differences and . Hence .
It remains to consider the case where every selected has order four. Together with the they form an independent family spanning a direct summand
Indeed, independence follows first in the quotient by and then in ; the same basis matrices exhibit a complement of type . Select from with basis images in . If all have order two, these elements together with the basis of form a basis of in , and Lemma 2.2 gives .
Otherwise choose one such element of order four or eight. Two is the least positive multiple placing it in . Write, with respect to the selected independent basis of ,
Here . If , the atoms and have product , with lengths two and three. Thus .
If , all . The atoms
satisfy
Their length differences are and , so again . Minimality follows from the least-multiple property and independence, as above. In the remaining case and , apply Lemma 2.2 with multiplier two. This completes the proof.
Since , Proposition 3.1 immediately proves the first condition in Theorem 1.1. We now remove the remaining possibility when .
4. Signed supports with one order-eight factor
For an integer and a group element , let denote the sequence if , and otherwise.
Lemma 4.1.
Let
Put , , and . Suppose contains and every atom over has length congruent to two modulo three. Then every order-eight element of belongs to .
Proof.
Every word
is zero-sum free. Its coordinates force any zero-sum subsequence to omit each , after which its coordinate forces it to be empty.
First let have order eight and lie outside . Then and is odd. For each , represent modulo four by if its residue is zero, or by either or otherwise. Represent modulo eight by if its residue is zero, or by either or when its residue is . The sequence
is an atom. A zero-sum subsequence uses zero or two copies of , by projection onto . The zero-copy case is excluded by (2); a proper two-copy case leaves a nonempty zero-sum complement in that word, also impossible. Hence
for every representative choice. The residue of is even. If it is two or six, exchanging its two representatives changes the length by four, contradicting (3). Thus the residue is zero or four. If any is nonzero, changing its sign preserves and interchanges these two residues, changing between zero and four. This again contradicts (3). If all , the residue is or , already excluded. Thus no such lies outside .
Now let have order eight, and suppose . Represent modulo four by zero, or by either or for . Represent modulo eight by in the same fashion as above. The sequence
is an atom: a proper zero-sum subsequence either lies in the zero-sum-free word or leaves a nonempty zero-sum complement in that word. Consequently
for every choice.
If the residue of is , exchanging its representatives changes length by . Thus the only permitted residues of are . Changing the representative of any nonzero changes that residue by four. If some is nonzero, both residues must therefore be zero or four. Choose a configuration with . Changing just one nonzero changes to four and changes the total length by . Divisibility by three forces . If two are nonzero, changing both returns the residue of to zero and increases the sum of their absolute values by four, contradicting (4). There is exactly one nonzero , and then gives , contrary to its choice. If all , then ; after excluding , we have or , and the corresponding atom or has length four. This is the final contradiction.
By Proposition 3.1, only needs to be excluded. Assume this value. Lemma 2.1 says that every atom over has length two modulo three. In particular, contains no element of order four.
Choose an order-eight element , and split off . Select whose projections onto the order-four part of form a basis. They have order eight, since order four is excluded. As there is only one order-eight factor, . Therefore has order four, and
To see the direct-summand assertion explicitly, the order-four projection matrix of the is invertible. Their lower and higher coordinates give a graph over that order-four summand; adjoining leaves the elementary two-summand as a complement.
Lemma 4.1 implies that every order-eight term of belongs to , where , and hence lies in . Remove one order-eight term from to obtain a zero-sum-free sequence of length . By Lemma 2.3, the projections of all order-two terms of are independent. But their sum is zero, by projecting the sum of onto , since its order-eight terms lie in and there are no order-four terms. Thus has no order-two terms at all, and it is an atom over .
Every atom over has length at most eight. To verify this, an atom containing opposite terms is their pair. Otherwise it uses at most one sign of each generator. A multiplicity at least eight contains a pure atom. All remaining nonzero multiplicities must be four, by the coordinates. Two such blocks already sum to zero, since ; one such block requires exactly four occurrences of one sign of to have sum zero. With no such block, the only possibility is or . Hence . This contradicts
since . The theorem follows.
5. Scope and further questions
The proof isolates a precise residual obstruction for arbitrary exponent-eight groups: only minimum distance three can occur. The groups left outside the two sufficient conditions have and , after removing separately known cases. When several order-eight basis elements are present, a relation can have without forcing distance one. The signed-support classification above uses essentially that there is one such coordinate. Extending that classification, or bounding the longest atom supported on these residual configurations, is a concrete remaining problem.
The accompanying computation checks every constructed witness in Lemma 4.1 for , both inside and outside it in . It independently verifies zero sum and minimality for 4,756 sequences and finds a forbidden atom length for every element excluded by the lemma. These finite diagnostics are not used as a substitute for any universal step of the proof.
References
[1] A. Bashir, A. Geroldinger, and Q. Zhong, On a zero-sum problem arising from factorization theory, in Combinatorial and Additive Number Theory IV, Springer, 2021, pp. 11–24. doi:10.1007/978-3-030-67996-5_2. Preprint: arXiv:2007.10094.
[2] A. Geroldinger and Q. Zhong, Long sets of lengths with maximal elasticity, Canadian Journal of Mathematics 70 (2018), 1284–1318. doi:10.4153/CJM-2017-043-4. Preprint: arXiv:1706.06907.
[3] A. Geroldinger and W. Yang, On a classical zero-sum invariant, arXiv:2608.19090v1, August 2026.
[4] D. Yazdonov, On the structure of length sets with maximal elasticity, Communications in Algebra 54 (2026), 2145–2158. doi:10.1080/00927872.2025.2578208. Revised preprint: arXiv:2508.21383v2, February 25, 2026.