# Zero-sum partition of an abelian group

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).

• You need to take care that p is not 2, or otherwise modify the conditions. Otherwise there are no choices for n. Gerhard "Ask Me About System Design" Paseman, 2012.02.19 Feb 19, 2012 at 20:06
• I should add the words "Mersenne prime" to my previous comment. Gerhard "Ask Me About System Design" Paseman, 2012.02.19 Feb 19, 2012 at 20:09
• If there are no proper nontrivial divisors of |G|-1, the better... in this case the problem holds vacuously. Feb 19, 2012 at 20:10
• About the zero-sum assumption: We know that $G\sim({\mathbb Z}/(d_1))\times\cdots({\mathbb Z}/(d_r))$ with $d_1|d_2$, ..., $d_{r-1}|d_r$. Then the sum of all elements in $G$ equals zero unless one $d_\ell$ is even and all the other ones are odd. Feb 20, 2012 at 9:19
• @Denis Serre: yes, as I said, the two-rank must not be one. But you are right the explict form is helpful. Perhaps even more explcitly: $d_{r-1}$ is odd and $d_r$ even.
– user9072
Feb 20, 2012 at 10:40

No. The sum of all (nonzero) elements of a finite abelian group is not necessarily $0$, which is however necessary for what you want.

A small counterexample would be the cyclic group with ten elements (and $n=3$).

Remark: To avoid this reason the sharp condition is to restrict to groups whose two-rank is not one.

Personal comment: sorry for the 'stupid answer'; not sure what happens with the suggested restriction in place, but seems interesting.

• Thanks for this answer. "Stupid" as you claim it to be, it explains why most of my approaches failed! I have restricted the question now, as you expected. Feb 19, 2012 at 20:50

Answer rewritten in light of comments received: Original answer essentially did what Zack suggested in the case that $|G|$ was odd and $n$ was even, but in an unnecessarily complicated manner. This edit treats the suggested amendment by Will in a slightly different fashion. Suppose that $G$ admits an automorphism $\alpha$ of order prime order $n$ which fixes only $0$ (the additive identity of $G$). Denote $G \backslash \{ 0 \}$ by $G^{\#}.$ Then $G^{\#}$ is a disjoint union of orbits under $\langle \alpha \rangle$, each of which has length $n$ by hypothesis. The sum across each of these orbits is zero because for any $g \in G$, the element $\sum_{k=0}^{n-1} \alpha^{k}(g)$ is fixed by $\alpha$, so must be $0$ by assumption. This covers the case that $G$ has odd order, since in that case inversion is an automorphism which fixes only $0.$ It is not necessary to suppose that $\alpha$ has prime order. More generally, the argument works if each power of $\alpha$ other than the identity fixes only $0$.

Third edit: it may be of interest to note that this argument does not generalize directly to the case that $\alpha$ is an automorphism of prime order $n$ which has non-zero fixed points on $G$. If we could partition $G^{\#}$ into disjoint "zero-sum" sets of size $n,$ then we have $|G| \equiv 1$ (mod $n$), so certainly $n$ does not divide $|G|.$ By standard results on coprime automorphisms, we may wirt $G = H \oplus K$, where $H$ is the fixed-point subgroup of $\alpha$ and $K$ is an $\alpha$-invariant complement (in this special case, this can be proved in the same way as Maschke's theorem). Note that $|H| \equiv |G|$ (mod $n$), so that $H$ contains no element of order $n.$ Then for any non-zero element $h \in H$ and any $k \in K,$ we have $\sum_{i=0}^{n-1} \alpha^{i}(h +k) = n . h \neq 0.$ However, it does reduce the problem to finding a suitable partition of $H$ into zero-sum subsets of size $n.$ For example, if $\sum_{i=1}^{n} s_{i} = 0$ for distinct elements $\{s_{1},s_{2},\ldots ,s_{n} \}$ of $H$, then for each orbit of $\langle \alpha \rangle$ on $K^{\#}$, we can add the $i$-th element of the orbit to $s_i,$ and we get a zero sum subset of $G$ disjoint from $\{s_{1},s_{2},\ldots ,s_{n} \}$. Hence if we can partition $H^{\#}$ into disjoint zero sum sets of size $n,$ we can do the same for $G^{\#}.$ Fourth Edit: It might be worth noting that when $n$ is prime, this reduces the problem in the case that $|G|$ is Abelian and $|G| \equiv 1$ (mod $n$) ( the latter case obviously being necessary if the desired partition is to exist) to the case that ${\rm Aut}(G)$ is group of order prime to $n.$ For if $G$ has an automorphism $\alpha$ of order $n,$ then we can work with the smaller group $C_{G}(\alpha)$ of fixed points of $\alpha$. If that group has a partition of the required form, so does $G.$ If the smaller groups still has an automorphism of order $n,$ then we can replace it by an even smaller group, and so on. If a finite Abelian group $G$ has no automorphism group of order $n,$ then (among other things) when it is decomposed as a direct summand of cyclic groups of prime power order, no $n-1$ summands can be mutually isomorphic.

• For G odd, no element has order 2, hence we can pair each non-identity element with its inverse to get zero sum sets of size 2. For multiples of 2, combine the correct number of these pairs. Feb 20, 2012 at 0:09
• As I said, it is enough to consider the case of $n$ prime - this is so for the very reason Zack gave in his comment. Unfortunately I didn't get any farther than $n=2$ with this approach. Feb 20, 2012 at 0:42
• An obvious generalization is that if $n=p$ and the order of the group has only prime factors that are $1$ mod $p$, then you can set up multiplication by $p$th roots of unity and make this work. Feb 20, 2012 at 2:03

there is paper by Zeng about that : Zeng X., On zero-sum partitions of abelian groups. Integers 15 (2015), Paper No. A44, 16 pp.

• Thanks -- but this doesn't settle the question in its general case, does it? Mar 4, 2018 at 0:41

Alexey Pokrovskiy and I addressed this problem for sufficiently large groups. See Section 1.1.2 and 6.3 in https://arxiv.org/pdf/2204.09666.pdf