This is a question I have been asking myself some 5 years ago. I later got bored by lack of progress, but maybe some additive combinatorialists here know further. I'm not claiming it is conceptual or objectively interesting, but I wouldn't be surprised to see it studied by the likes of Erdös either.

**Question:** Let $G$ be a finite abelian group such that the sum of all elements of $G$ is zero. Let $n\neq 1$ be an integer such that $n\mid \left|G\right|-1$. Can we partition the set $G\setminus \left\lbrace 0\right\rbrace$ into disjoint $n$-element zero-sum subsets? (A subset of $G$ is said to be *zero-sum* if the sum of its elements is zero.)

*[Question corrected due to a remark by quid.]*

**Remarks:**

**1.** This has a definitely positive answer for $G = \left(\mathbb Z / \left(p\right)\right)^k$ with $p$ a prime and $k$ a positive integer. (In fact, the abelian group $\left(\mathbb Z / \left(p\right)\right)^k$ is isomorphic to the additive group of the finite field with $p^k$ elements; now you can take a primitive root $\zeta$ in this field, set $m=\dfrac{p^k-1}{n}$, and partition $G\setminus \left\lbrace 0\right\rbrace$ into the zero-sum subsets

$\left\lbrace \zeta^0, \zeta^{0+m}, ..., \zeta^{0+\left(n-1\right)m}\right\rbrace$,

$\left\lbrace \zeta^1, \zeta^{1+m}, ..., \zeta^{1+\left(n-1\right)m}\right\rbrace$,

...,

$\left\lbrace \zeta^{m-1}, \zeta^{m-1+m}, ..., \zeta^{m-1+\left(n-1\right)m}\right\rbrace$.)

**2.** In the general case, we can WLOG assume that $n$ is prime, but this doesn't seem to help (me).

`$d_{r-1}$`

is odd and`$d_r$`

even. – quid Feb 20 '12 at 10:40