Zero-sum partitions of Abelian groups of order $2^n$

Sylwia Cichacz, Karol Suchan

Published 2021-11-09, updated 2022-03-15Version 3

The following problem has been known since the 80's. Let $\Gamma$ be an Abelian group of order $m$ (denoted $|\Gamma|=m$), and let $t$ and $m_i$, $1 \leq i \leq t$, be positive integers such that $\sum_{i=1}^t m_i=m-1$. Determine when $\Gamma^*=\Gamma\setminus\{0\}$, the set of non-zero elements of $\Gamma$, can be partitioned into disjoint subsets $S_i$, $1 \leq i \leq t$, such that $|S_i|=m_i$ and $\sum_{s\in S_i}s=0$ for every $i$, $1 \leq i \leq t$. It is easy to check that $m_i\geq 2$ (for every $i$, $1 \leq i \leq t$) and $|I(\Gamma)|\neq 1$ are necessary conditions for the existence of such partitions, where $I(\Gamma)$ is the set of involutions of $\Gamma$. It was proved that the condition $m_i\geq 2$ is sufficient if and only if $|I(\Gamma)|\in\{0,3\}$. For other groups (i.e., for which $|I(\Gamma)|\neq 3$ and $|I(\Gamma)|>1$), only the case of any group $\Gamma$ with $\Gamma\cong(Z_2)^n$ for some positive integer $n$ has been analyzed completely so far, and it was shown independently by several authors that $m_i\geq 3$ is sufficient in this case. Moreover, recently Cichacz and Tuza proved that, if $|\Gamma|$ is large enough and $|I(\Gamma)|>1$, then $m_i\geq 4$ is sufficient. In this paper we generalize this result for every Abelian group of order $2^n$. Namely, we show that the condition $m_i\geq 3$ is sufficient for $\Gamma$ such that $|I(\Gamma)|>1$ and $|\Gamma|=2^n$, for every positive integer $n$. We also present some applications of this result to graph magic- and anti-magic-type labelings.