TITLE: Ordering subsets of the cyclic group to give distinct partial sums
QUESTION [27 upvotes]: Suppose that you are given a set $S$ of $k$ nonzero elements from $\mathbb{Z}_n$. Is it always possible to order the elements of $S$, say $a_1,a_2,\dots,a_k$ in such a way that the partial sums $a_1,a_1+a_2,a_1+a_2+a_3,\dots,a_1+a_2+\cdots+a_k$ are all distinct. Equivalently, you want to avoid runs $a_i+a_{i+1}+\cdots+a_j$ that add to zero (where $1<i<j\le k$).
In the case when $k=n-1$ this is known -- it is covered by the theory of sequenceable and $R$-sequenceable groups. Has anyone looked at the above generalisation before? Even though $k=n-1$ feels like it should be the hardest case, it doesn't seem to imply the other cases. I have some hope of a proof for the case when $n$ is prime, but before I go to the effort of sorting out the details I want to be sure I'm not reinventing the wheel.
It is tempting to think the problem will be easy when $k$ is very much smaller than $n$. However, it is possible that $S$ might be all the nonzero elements of some subgroup of $\mathbb{Z}_n$, in which case you are just looking for a sequencing of that subgroup. This will exist, but are probably very rare in some sense. 
This problem was told to me by Jeff Dinitz, and I think he might have got it from Dan Archdeacon. Any pointers to relevant literature will be much appreciated.

REPLY [7 votes]: This is not a complete solution, but I believe it can be completed with some further effort for the case where $n$ is prime. (As you have noted, for $n$ composite we can run into troubles with subgroups, and it is not clear whether the assertion holds true in this case.)
Assuming thus that $n$ is prime, let
  $$ P(x_1,\ldots, x_k):= \prod_{1<i<j\le k} (x_i+\dotsb+x_j) \cdot \prod_{1\le i<j\le k} (x_j-x_i) \in {\mathbb F}_n[x_1,\ldots x_k]. $$
This is a polynomial in $k$ variables of degree $(k-1)^2$; it is homogeneous, so that none of its monomials is dominated by any other monomial.
The crucial observation is that if a set $S\subset{\mathbb F}_n^\times$ cannot be ordered the way you want it, then $P$ vanishes on the cartesian product $S^k=S\times\dotsb\times S$. Consequently, by the "generalized Combinatorial Nullstellnsatz" (Theorem 2 of this paper by Michal Lason), $P$ does not contain any monomial $ax_1^{d_1}...x_k^{d_k}$ with $d_1,\dotsc,d_k<k$. In other words, if we can show that $P$ contains a monomial with the individual degree in each variable smaller than $k$, then we are done.
Playing around with small values of $k$ suggests that $P$ actually contains
lots of such monomials; for instance, $x_1^{k-1}\ldots x_{k-1}^{k-1}$ has
coefficient $\pm 1$ in $P$ for $k\le 5$. I would speculate that the coefficients of $P$ can be computed explicitly, and this should give the result.