Representing Sequence Subsums as Sumsets of Near Equal Sized Sets

TL;DR

This paper demonstrates that the subsum set of a sequence in an abelian group can be partitioned into nearly equal-sized subsets, enhancing the structural understanding of sumsets and subsums, with applications to sumset theorems.

Contribution

It proves that the partitioning of subsum sets into nearly equal-sized parts is always possible, except for a specific structured counterexample, and extends these results for large n.

Findings

Partition sets $A_i$ can be made nearly equal in size.

A structured counterexample exists when $| ext{Sigma}_n(S)|=|S|-n+1$ with $n=2$.

Stronger results are proven for large $n$ relative to $|S|$.

Abstract

For a sequence $S$ of terms from an abelian group $G$ of length $|S|$, let $\Sigma_n(S)$ denote the set of all elements that can be represented as the sum of terms in some $n$-term subsequence of $S$. When the subsum set is very small, $|\Sigma_n(S)|\leq |S|-n+1$, it is known that the terms of $S$ can be partitioned into $n$ nonempty sets $A_1,\ldots,A_n\subseteq G$ such that $\Sigma_n(S)=A_1+\ldots+A_n$. Moreover, if the upper bound is strict, then $|A_i\setminus Z|\leq 1$ for all $i$, where $Z=\bigcap_{i=1}^{n}(A_i+H)$ and $H=\{g\in G:\; g+\Sigma_n(S)=\Sigma_n(S)\}$ is the stabilizer of $\Sigma_n(S)$. This allows structural results for sumsets to be used to study the subsum set $\Sigma_n(S)$ and is one of the two main ways to derive the natural subsum analog of Kneser's Theorem for sumsets. In this paper, we show that such a partitioning can be achieved with sets $A_i$ of as near equal a size as possible, so $\lfloor \frac{|S|}{n}\rfloor \leq |A_i|\leq \lceil\frac{|S|}{n}\rceil$ for all $i$, apart from one highly structured counterexample when $|\Sigma_n(S)|= |S|-n+1$ with $n=2$. The added information of knowing the sets $A_i$ are of near equal size can be of use when applying the aforementioned partitioning result, or when applying sumset results to study $\Sigma_n(S)$. We also give an extension increasing the flexibility of the aforementioned partitioning result and prove some stronger results when $n\geq \frac12|S|$ is very large.