Purely singular splittings of cyclic groups

Let $G$ be a finite abelian group. We say that $M$ and $S$ form a \textsl{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 $\mathit{\text{purely}}\mathit{\text{singular}}$ if for each prime divisor $p$ of $\mathrm{\mid }G\mathrm{\mid }$, there is at least one element of $M$ is divisible by $p$. In this paper, we mainly study the purely singular splittings of cyclic groups. We first prove that if $k\ge 3$ is a positive integer such that $[-k+1, ,k]^$splitsacyclicgroup$$\mathbb{Z}$_m$,then$m=2k$. Next, we have the following general result. Suppose $M=[-k_1,k_2{]}^{\ast }$ splits ${\mathbb{Z}}_{n(k_1+k_2)+1}$ with $1\le k_1<k_2$. If $n\ge 2$, then $k_1\leq n-2$and$k_2\leq 2n-5$\mathrm{.}Applyingthisresult,weprovethatif$M=[-k_1, ,k_2]^$splits$$\mathbb{Z}$_m$purelysingularly,andeither$(i)$ $\gcd(s, ,m)=1$forall$s\in S$or$(ii)$ $m=2^{\alpha }{p}^{\beta }$ or $2^{\alpha }{p}_1{p}_2$ with $\alpha \ge 0$, $\beta \ge 1$ and $p$, $p_1$, $p_2$ odd primes, then $m=k_1+k_2+1$ or $k_1=0$ and $m=k_2+1$ or $2k_2+1$.