slug: bhanja-2025-restricted-sumset-question
bibkey: bhanja2025distinctterms
doi: null
triage: theorem
motivation_gids:
  - D5/S3/FiniteGroups/BhanjaRestrictedSumset/Result.result
url: https://arxiv.org/abs/2307.03977v2

Unequal minima for sumsets with at most two distinct terms

Problem

J. Bhanja, Sumsets with a minimum number of distinct terms, Integers 25 (2025), A12, arXiv:2307.03977v2, §4 “Further Remarks”, printed p. 11:

Another sumset that may come immediately to the reader’s mind is the sumset h(≤r) A, which contains those elements of the group G that are the sum of h elements of A with at most r distinct elements. The sumset hA is a special case of h(≤r) A for r = h.

Question 1. Is it always the case that the minimum size of both the sumsets hA and h(≤r) A are the same in any abelian group G, where the minimum runs over the subsets of G of size (say) k?

Formal reading

Names have prefix D5.S3.FiniteGroups.BhanjaRestrictedSumset.Result. hSumset h A is Mathlib’s pointwise $h\bullet A$, with exactly $h$ summands. restrictedSumset h r A is the image of functions $f:\mathrm{Fin}h\to G$ taking values in $A$ whose image has at most $r$ elements, under $f\mapsto {\sum }_{i}f(i)$. minSize G k h and minRestrictedSize G k h r are infima in the natural numbers of the corresponding cardinalities over $k$-element finsets. The claim requires a $k$-element subset to exist, so these infima are actual minima. It quantifies over all abelian groups, $h\ge 1$, $r\ge 2$, and all attainable cardinalities $k$. Allowing $k=0$ adds only a case in which both sumsets are empty for $h\ge 1$.

Motivation

The settling declaration is D5/S3/FiniteGroups/BhanjaRestrictedSumset/Result.result. The source note is D5/L/FiniteGroups/bhanja2025distinctterms. Restricting the number of distinct summands can preserve the integer minimum while changing minima in finite groups.

Gap

For $m\ge 2$, odd $h\ge 3$ and $2^{m-1}<k<2^m$, every $k$-element subset of $G=(\mathbb{Z}/2\mathbb{Z}{)}^m$ has $h^{(\le 2)}A=A$ and $hA=G$. Therefore ${\mu }_{\le 2}(G,k,h)=k<2^m=\mu (G,k,h)$. The smallest example has $m=2$, $k=h=3$ and $r=2$.

Route

Every element of $G$ has twice itself equal to zero. A sum with odd length using at most two values is one of those values: exactly one multiplicity is odd. Conversely, repeating any member an odd number of times recovers it. This proves the restricted identity.

Since $2|A|>|G|$, the sets $A$ and $g-A$ intersect for each $g$ by the finite-set pigeonhole principle, so $A+A=G$. For $h\ge 2$, choose $a_0\in A$, represent $g-(h-2)a_0$ as two members of $A$, and append $h-2$ copies of $a_0$. Thus $hA=G$. The identities for every $k$-subset, together with existence of such a subset, determine both minima.

Falsifier

A mismatch between the image-cardinality restriction and “at most r distinct elements”, or between the minima and optimization over all $k$-element subsets, would invalidate the conclusion. A single convenient subset alone would not establish the ordinary minimum.

Evidence

PROVED. elementary_two_group_gap proves both sumset identities for all $m\ge 2$, odd $h\ge 3$ and subsets with cardinality strictly between $2^{m-1}$ and $2^m$. elementary_two_group_minima proves both exact minima. result states exactly $\neg \mathrm{claim}$.

The immutable experiment directory contains bhanja-q1.py, SHA-256 4c041a3988bd92a78819b081d13dfded6cdf3dda78b8988221311271e576220a. Command: python3 bhanja-q1.py; exit code 0. It tests every nonempty proper-subset cardinality $1\le k<|G|$ for the groups in the following survey at $h=3$, $r=2$, and checks the elementary 2-group identities for $m=2,3$, $h=3,5$ in the strict half-to-full range.

The supplied literature search covers the current arXiv v2 and journal version, title and identifier searches, and searches combining Bhanja with Question 1, counterexample and at-most-distinct sumset terminology. No settlement was found in that supplied scope. Goswami–Mistri, arXiv:2603.04572, concerns all-distinct summands, a different restriction.

Triage

What the settlement shows

• PROVED (Lean). The entire elementary 2-group family and its exact minima are elementary_two_group_gap and elementary_two_group_minima. The universal equality is refuted by result.
• PROVED (prose consequence). At $h=3$ and $k=2^{m-1}+1$, the gap is $2^m-k=2^{m-1}-1$. For $m\ge 2$ this cardinality lies in the strict range. Since powers of two are unbounded, so are the gaps.
• COMPUTED. At $h=3,r=2$ the minima differ in $(\mathbb{Z}/2{)}^2$, $(\mathbb{Z}/2{)}^3$, $\mathbb{Z}/2\times \mathbb{Z}/4$, $\mathbb{Z}/2\times \mathbb{Z}/6$, and the cyclic groups $\mathbb{Z}/7$ ($k=3$: $6$ versus $7$), $\mathbb{Z}/8$ ($k=5$: $7$ versus $8$), $\mathbb{Z}/9$ ($k=4$: $8$ versus $9$), and $\mathbb{Z}/12$ ($k=7$: $10$ versus $12$). The first number is the restricted minimum. Evidence is the exhaustive experiment above, with its pinned program, command, exit code and hash.
• COMPUTED. In the same survey the minima agree for every nonempty proper-subset cardinality $1\le k<|G|$ in $\mathbb{Z}/3$, $\mathbb{Z}/4$, $\mathbb{Z}/5$, $\mathbb{Z}/6$, $\mathbb{Z}/10$ and $\mathbb{Z}/3\times \mathbb{Z}/3$. This is finite computational evidence, with the same experiment entry.
• PROVED (prose argument). For the endpoint $k=|G|$ in a finite abelian group, the only subset is $A=G$. At $h=3,r=2$, every $g\in G$ is $g+0+0$, using at most two distinct values from $A$. Thus both sumsets equal $G$, and both minima equal $|G|$. This endpoint is a prose argument, separate from the computed survey.
• PROVED (mechanism). Parity cancellation in exponent two is the mechanism of elementary_two_group_gap. The cyclic survey shows that failure is not confined to exponent two. For $\mathbb{Z}/7$, Cauchy–Davenport gives $|2A|\ge 5$ and $|3A|\ge 7$ for $|A|=3$; the computed restricted minimum is $6$. Thus cyclic wrap-around can separate the restricted minimum from the ordinary bound. The Cauchy–Davenport explanation is a prose application, not a Lean theorem in this module.
• PROVED (literature). Over the integers, the paper’s increasing chain of $hk-h+1$ sums uses at most two values (§4, printed p. 11). Hence both minima are at least $hk-h+1$ for $k,h\ge 1,r\ge 2$. An arithmetic progression attains that size for $hA$, and its restricted sumset is contained in $hA$, so both minima equal $hk-h+1$.
• OPEN. Classifying the abelian groups giving equality for all attainable $k$, all $h\ge 1$ and $r\ge 2$ remains unresolved here. No positive conclusion for other open questions follows from this counterexample; the source’s integer results retain their hypotheses.

ASSUMED-UNVERIFIED

Exhaustive literature coverage and priority are unverified. Google Scholar’s complete cited-by list was unavailable in the supplied search (403/429). No new network search is asserted. The cyclic and mixed-group survey is finite computation, not a classification theorem.