A set M of nonzero integers is said to split a finite abelian group G if there is a subset S of G for which
M
⋅
S
=
G
∖
{
0
}
M \cdot S = G\backslash \{ 0\}
. If, moreover, each prime divisor of
|
G
|
|G|
divides an element of M, we call the splitting purely singular. It is conjectured that the only finite abelian groups which can be split by
{
1
,
…
,
k
}
\{ 1, \ldots ,k\}
in a purely singular manner are the cyclic groups of order
1
,
k
+
1
1,k + 1
and
2
k
+
1
2k + 1
. We show that a proof of this conjecture can be reduced to a verification of the case
gcd
(
|
G
|
,
6
)
=
1
\gcd (|G|,6) = 1
.