  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 $(ii)$ $m=2^{\alpha}p^{\beta}$ or
$2^{\alpha}p_1p_2$ with $\alpha\geq 0$, $\beta\geq 1$ 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 $2k_2+1$.
