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Volume 38, Issue 2
Some New Results on Purely Singular Splittings

Pingzhi Yuan

DOI: 10.4208/cmr.2020-0048

Commun. Math. Res., 38 (2022), pp. 136-156.

Published online: 2022-02

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  • Abstract

Let $G$ be a finite abelian group, $M$ a set of integers and $S$ a subset of $G$. We say that $M$ and $S$ form a splitting of $G$ if every nonzero element $g$ of $G$ has a unique representation of the form $g=ms$ with $m\in M$ and $s\in S$, while 0 has no such representation. The splitting is called purely singular if for each prime divisor $p$ of $|G|$, there is at least one element of $M$ is divisible by $p$. In this paper, we continue the study of purely singular splittings of cyclic groups. We prove that if $k\geq 2$ is a positive integer such that $[−2k+1, 2k+2]^∗$ splits a cyclic group $\mathbb{Z}_m$, then $m=4k+2$. We prove also that if $M=[−k_1, k_2]^∗$ splits $\mathbb{Z}_m$ purely singularly, and $15 \leq k_1+k_2 \leq 30$, then $m = 1$, or $m = k_1+k_2+1$, or $k_1 = 0$ and $m=2k_2+1$.

  • Keywords

Splitter sets, perfect codes, factorizations of cyclic groups.

  • AMS Subject Headings

20D60, 20K01, 94A17

  • Copyright

COPYRIGHT: © Global Science Press

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@Article{CMR-38-136, author = {Yuan , Pingzhi}, title = {Some New Results on Purely Singular Splittings}, journal = {Communications in Mathematical Research }, year = {2022}, volume = {38}, number = {2}, pages = {136--156}, abstract = {

Let $G$ be a finite abelian group, $M$ a set of integers and $S$ a subset of $G$. We say that $M$ and $S$ form a splitting of $G$ if every nonzero element $g$ of $G$ has a unique representation of the form $g=ms$ with $m\in M$ and $s\in S$, while 0 has no such representation. The splitting is called purely singular if for each prime divisor $p$ of $|G|$, there is at least one element of $M$ is divisible by $p$. In this paper, we continue the study of purely singular splittings of cyclic groups. We prove that if $k\geq 2$ is a positive integer such that $[−2k+1, 2k+2]^∗$ splits a cyclic group $\mathbb{Z}_m$, then $m=4k+2$. We prove also that if $M=[−k_1, k_2]^∗$ splits $\mathbb{Z}_m$ purely singularly, and $15 \leq k_1+k_2 \leq 30$, then $m = 1$, or $m = k_1+k_2+1$, or $k_1 = 0$ and $m=2k_2+1$.

}, issn = {2707-8523}, doi = {https://doi.org/10.4208/cmr.2020-0048}, url = {http://global-sci.org/intro/article_detail/cmr/20268.html} }
TY - JOUR T1 - Some New Results on Purely Singular Splittings AU - Yuan , Pingzhi JO - Communications in Mathematical Research VL - 2 SP - 136 EP - 156 PY - 2022 DA - 2022/02 SN - 38 DO - http://doi.org/10.4208/cmr.2020-0048 UR - https://global-sci.org/intro/article_detail/cmr/20268.html KW - Splitter sets, perfect codes, factorizations of cyclic groups. AB -

Let $G$ be a finite abelian group, $M$ a set of integers and $S$ a subset of $G$. We say that $M$ and $S$ form a splitting of $G$ if every nonzero element $g$ of $G$ has a unique representation of the form $g=ms$ with $m\in M$ and $s\in S$, while 0 has no such representation. The splitting is called purely singular if for each prime divisor $p$ of $|G|$, there is at least one element of $M$ is divisible by $p$. In this paper, we continue the study of purely singular splittings of cyclic groups. We prove that if $k\geq 2$ is a positive integer such that $[−2k+1, 2k+2]^∗$ splits a cyclic group $\mathbb{Z}_m$, then $m=4k+2$. We prove also that if $M=[−k_1, k_2]^∗$ splits $\mathbb{Z}_m$ purely singularly, and $15 \leq k_1+k_2 \leq 30$, then $m = 1$, or $m = k_1+k_2+1$, or $k_1 = 0$ and $m=2k_2+1$.

Yuan , Pingzhi. (2022). Some New Results on Purely Singular Splittings. Communications in Mathematical Research . 38 (2). 136-156. doi:10.4208/cmr.2020-0048
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