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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} }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 $m=2k_2+1$.

*Communications in Mathematical Research*.

*38*(2). 136-156. doi:10.4208/cmr.2020-0048