A permutation problem for finite subsets of an abelian group

Question

Here I ask the following question in additive combinatorics.

QUESTION: Let $A$ be any finite subset of an additive abelian group $G$ with $|A|=n>3$. Can we write $A$ as $\{a_1,\ldots,a_n\}$ so that all the $n$ sums \begin{gather*}a_1+a_2+a_3,\ a_2+a_3+a_4,\ \ldots, \\ a_{n-2}+a_{n-1}+a_n,\ a_{n-1}+a_n+a_1,\ a_n+a_1+a_2\end{gather*} are pairwise distinct?

Note that if $G=\{a_1,\ldots,a_n\}$ is an addtive abelian group with $|G|=n>3$ then $2(a_1+\ldots+a_n)=0$ and hence \begin{align*}&(a_1+a_2+a_3)+\ldots+(a_{n-2}+a_{n-1}+a_n) \\&+(a_{n-1}+a_n+a_1)+(a_n+a_1+a_2) \\=&3(a_1+\ldots+a_n)=a_1+\ldots+a_n.\end{align*} This is the main motivation of my question.

In 2013 I conjectured that the question always has a positive answer, and showed this when $G$ is torsion-free (cf. http://arxiv.org/abs/1309.1679). However, I'm even unable to answer the question for $G=\mathbb Z/p\mathbb Z$ with $p$ prime.

Any comments are welcome!

Answer 1

Let $A$ consist of $6$ elements of $(\mathbb Z_2)^n$ that sum to $0$. Then $a_1 + a_2 + a_3 = a_4 + a_5 + a_6$, no matter how you permute $A$.