A proof of purely singular splitting conjecture
Fuente:
arXiv
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| Autores principales: | , |
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| Formato: | Preprint |
| Publicado: |
2026
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| _version_ | 1866917479601668096 |
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| author | Leung, Ka Hin Zhang, Tao |
| author_facet | Leung, Ka Hin Zhang, Tao |
| contents | A set $M$ of nonzero integers is said to split a finite abelian group $G$ if there exists a subset $S\subseteq G$ such that $M\cdot S = G\setminus\{0\}$. Such a splitting is called purely singular if every prime divisor of $|G|$ divides some element of $M$. In 1995, Woldar \cite{W1995} conjectured that the finite abelian groups admitting a purely singular splitting by the set $\{1,2,\dots,k\}$ are precisely the cyclic groups of orders $1$, $k+1$, and $2k+1$. In this paper, we prove this conjecture. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2605_09871 |
| institution | arXiv |
| publishDate | 2026 |
| record_format | arxiv |
| spellingShingle | A proof of purely singular splitting conjecture Leung, Ka Hin Zhang, Tao Combinatorics 52C22, 05A18, 11H71, 20K01 A set $M$ of nonzero integers is said to split a finite abelian group $G$ if there exists a subset $S\subseteq G$ such that $M\cdot S = G\setminus\{0\}$. Such a splitting is called purely singular if every prime divisor of $|G|$ divides some element of $M$. In 1995, Woldar \cite{W1995} conjectured that the finite abelian groups admitting a purely singular splitting by the set $\{1,2,\dots,k\}$ are precisely the cyclic groups of orders $1$, $k+1$, and $2k+1$. In this paper, we prove this conjecture. |
| title | A proof of purely singular splitting conjecture |
| topic | Combinatorics 52C22, 05A18, 11H71, 20K01 |
| url | https://arxiv.org/abs/2605.09871 |