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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 \textit{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 mainly study the purely singular splittings of cyclic groups. We first prove that if $k\ge3$ is a positive integer such that $[-k+1, \,k]^*$ splits a cyclic group $\mathbb{Z}_m$, then $m=2k$. Next, we have the following general result. Suppose $M=[-k_1, \,k_2]^*$ splits $\mathbb{Z}_{n(k_1+k_2)+1}$ with $1\leq k_1< k_2$. If $n\geq 2$, then $k_1\leq n-2$ and $k_2\leq 2n-5$. Applying this result, we prove that if $M=[-k_1, \,k_2]^*$ splits $\mathbb{Z}_m$ purely singularly, and either $(i)$ $\gcd(s, \,m)=1$ for all $s\in S$ or $(ii)$ $m=2^αp^β$ or $2^αp_1p_2$ with $α\geq 0$, $β\geq 1$ 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 $2k_2+1$.