A problem on partial sums in abelian groups

TL;DR

This paper introduces a conjecture about partial sums in finite subsets of abelian groups, explores its connections to open problems, and presents initial results on its validity.

Contribution

It proposes a new conjecture related to partial sums in abelian groups and investigates its connections and preliminary results.

Findings

Connection established with open problems in the field

Initial results support the conjecture's validity

Framework for future research on partial sums in abelian groups

Abstract

In this paper we propose a conjecture concerning partial sums of an arbitrary finite subset of an abelian group, that naturally arises investigating simple Heffter systems. Then, we show its connection with related open problems and we present some results about the validity of these conjectures.

Equations

$$\Delta C=\{\pm (c_{h+1}-c_h)|1\le h\le k\},$$

$$\ell (x,y)=min(|x-y|,v-|x-y|).$$

$$\omega=\left\{\begin{array}[]{ll}(a_{1},a_{2},a_{4},a_{3},a_{6},a_{5},a_{7})&\textrm{ if }a_{1}+a_{5}+a_{7}\neq 0,\\ (a_{1},a_{4},a_{2},a_{3},a_{5},a_{6},a_{7})&\textrm{ if }a_{1}+a_{5}+a_{7}=0\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{1},a_{2},a_{3},a_{6},a_{4},a_{5},a_{7},a_{8})&\textrm{ if }a_{2}+a_{3}+a_{6}\neq 0,\\ (a_{1},a_{2},a_{3},a_{7},a_{4},a_{5},a_{6},a_{8})&\textrm{ if }a_{2}+a_{3}+a_{6}=0\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{1},a_{2},a_{3},a_{5},a_{4},a_{6},a_{7},a_{8})&\textrm{ if }a_{3}+a_{4}+a_{5}+a_{6}\neq 0,\\ (a_{1},a_{2},a_{3},a_{5},a_{4},a_{7},a_{6},a_{8})&\textrm{ if }a_{3}+a_{4}+a_{5}+a_{6}=0\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{1},a_{4},a_{2},a_{3},a_{5},a_{6},a_{7},a_{8})&\textrm{ if }a_{1}+a_{4}+a_{8}\neq 0,\\ (a_{1},a_{4},a_{2},a_{3},a_{5},a_{6},a_{8},a_{7})&\textrm{ if }a_{1}+a_{4}+a_{8}=0\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{1},a_{2},a_{4},a_{3},a_{5},a_{7},a_{6},a_{8},a_{9})&\textrm{ if }a_{3}+a_{5}+a_{7}\neq 0,\\ (a_{1},a_{2},a_{4},a_{3},a_{6},a_{7},a_{5},a_{8},a_{9})&\textrm{ if }a_{3}+a_{5}+a_{7}=0\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{1},a_{2},a_{4},a_{3},a_{6},a_{5},a_{7},a_{8},a_{9})&\textrm{ if }a_{1}+a_{8}+a_{9}\neq 0,\\ (a_{1},a_{2},a_{4},a_{3},a_{6},a_{5},a_{8},a_{7},a_{9})&\textrm{ if }a_{1}+a_{8}+a_{9}=0\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{1},a_{2},a_{3},a_{5},a_{4},a_{7},a_{6},a_{8},a_{9})&\textrm{ if }a_{2}+a_{3}+a_{5}\neq 0,\\ (a_{1},a_{2},a_{3},a_{6},a_{4},a_{7},a_{5},a_{8},a_{9})&\textrm{ if }a_{2}+a_{3}+a_{5}=0\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{1},a_{5},a_{3},a_{2},a_{4},a_{6},a_{7},a_{8},a_{9})&\textrm{ if }a_{2}+a_{4}+a_{6}\neq 0,\\ (a_{1},a_{5},a_{3},a_{2},a_{4},a_{7},a_{6},a_{8},a_{9})&\textrm{ if }a_{2}+a_{4}+a_{6}=0\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{lll}(a_{1},a_{3},a_{5},a_{4},a_{7},a_{6},a_{8},a_{9},a_{2})&\textrm{ if }\sum T_{3}\neq 0\textrm{ and }\sum Q_{4}\neq 0,\\ (a_{1},a_{3},a_{5},a_{4},a_{7},a_{6},a_{9},a_{8},a_{2})&\textrm{ if }\sum T_{3}\neq 0\textrm{ and }\sum Q_{4}=0,\\ (a_{1},a_{2},a_{3},a_{5},a_{4},a_{7},a_{6},a_{8},a_{9})&\textrm{ if }\sum T_{3}=0;\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{lll}(a_{1},a_{3},a_{7},a_{4},a_{5},a_{6},a_{8},a_{9},a_{2})&\textrm{ if }a_{1}+a_{3}+a_{7}\neq 0,\\ (a_{1},a_{4},a_{7},a_{3},a_{5},a_{6},a_{8},a_{9},a_{2})&\textrm{ if }a_{1}+a_{3}+a_{7}=0.\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{lll}(a_{1},a_{2},a_{4},a_{7},a_{3},a_{5},a_{6},a_{8},a_{9})&\textrm{ if }\sum T_{4}\neq 0,\\ (a_{1},a_{2},a_{7},a_{4},a_{3},a_{5},a_{8},a_{6},a_{9})&\textrm{ if }\sum T_{4}=0\textrm{ and }\sum T_{5}\neq 0,\\ (a_{1},a_{2},a_{7},a_{4},a_{3},a_{5},a_{9},a_{6},a_{8})&\textrm{ if }\sum T_{4}=\sum T_{5}=0.\\ \end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{2},a_{1},a_{4},a_{5},a_{3},a_{7},a_{9},a_{6},a_{8})&\textrm{ if }\sum T_{4}=0,\\ (a_{2},a_{1},a_{3},a_{5},a_{4},a_{6},a_{9},a_{7},a_{8})&\textrm{ if }\sum T_{4}\neq 0\textrm{ and }\sum T_{6}=0,\\ (a_{1},a_{2},a_{3},a_{5},a_{4},a_{6},a_{9},a_{7},a_{8})&\textrm{ if }\sum T_{6}\neq 0.\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{1},a_{3},a_{2},a_{5},a_{4},a_{6},a_{8},a_{7},a_{9})&\textrm{ if }\sum T_{5}\neq 0,\\ (a_{2},a_{3},a_{1},a_{5},a_{4},a_{6},a_{8},a_{7},a_{9})&\textrm{ if }\sum T_{5}=0\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{1},a_{2},a_{7},a_{3},a_{5},a_{4},a_{6},a_{8},a_{9})&\textrm{ if }\sum T_{3}=\sum T_{7}=0,\\ (a_{2},a_{1},a_{3},a_{5},a_{4},a_{6},a_{8},a_{7},a_{9})&\textrm{ if either }\sum T_{3}=0\textrm{ and }\sum T_{7}\neq 0\\ &\textrm{ or }\sum T_{5}=0\textrm{ and }\sum T_{4}\neq 0,\\ (a_{3},a_{2},a_{1},a_{5},a_{4},a_{6},a_{8},a_{7},a_{9})&\textrm{ if }\sum T_{5}=\sum T_{4}=0,\\ (a_{1},a_{2},a_{3},a_{5},a_{4},a_{6},a_{8},a_{7},a_{9})&\textrm{ if }\sum T_{3},\sum T_{5}\neq 0.\\[2.0pt] \end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{3},a_{1},a_{6},a_{5},a_{4},a_{2},a_{9},a_{7},a_{8})&\textrm{ if }a_{3}+a_{7}+a_{8}\neq 0,\\ (a_{2},a_{1},a_{3},a_{6},a_{4},a_{5},a_{9},a_{7},a_{8})&\textrm{ if }a_{3}+a_{7}+a_{8}=0.\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{1},a_{2},a_{3},a_{4})&\textrm{ if }a_{2}+a_{3}+a_{4}\neq 0,\\ (a_{2},a_{1},a_{3},a_{4})&\textrm{ if }a_{2}+a_{3}+a_{4}=0\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{2},a_{1},a_{3},a_{5},a_{4})&\textrm{ if }a_{3}+a_{5}+a_{4}\neq 0,\\ (a_{5},a_{3},a_{2},a_{4},a_{1})&\textrm{ if }a_{3}+a_{5}+a_{4}=0\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(a_{3},\varepsilon\cdot a_{1},a_{5},a_{4},-\varepsilon\cdot a_{1})&\textrm{ if }a_{5}+a_{4}\neq\varepsilon\cdot a_{1},\\ (\varepsilon\cdot a_{1},a_{5},a_{3},a_{4},-\varepsilon\cdot a_{1})&\textrm{ if }a_{5}+a_{4}=\varepsilon\cdot a_{1}\end{array}\right.$$

$$\omega=\left\{\begin{array}[]{ll}(-a_{3},a_{1},a_{3},-a_{1},a_{5})&\textrm{ if }a_{3}-a_{1}+a_{5}\neq 0,\\ (-a_{3},-a_{1},a_{3},a_{1},a_{5})&\textrm{ if }a_{3}-a_{1}+a_{5}=0\end{array}\right.$$

Full text

The results of this paper will be presented at HyGraDe 2017

1. Introduction

It is well known that difference methods have a primary role in the construction of combinatorial designs of various kinds, see [1, 8]. The continuous search for more efficient ways to use these methods often leads to intriguing problems which are very difficult despite their easy statements. Some examples are the conjectures proposed by Alspach [9] and by Archdeacon et al. [7].

In order to describe these conjectures we introduce the concept of partial sums. Let $A$ be a finite list of elements of a group $(G,+)$. Let $(a_{1},a_{2},\ldots,a_{k})$ be an ordering of the elements in $A$ and define the partial sums $s_{1},s_{2},\ldots,s_{k}$ by the formula $s_{j}=\sum_{i=1}^{j}a_{i}$ $(1\leq j\leq k)$. An ordering of $A$ is said to be simple if all the partial sums are distinct.

Several years ago Alspach made the following conjecture, whose validity would shorten some cases of known proofs about the existence of cycle decompositions.

Conjecture 1**.**

Let $A\subseteq\mathbb{Z}_{v}\setminus\{0\}$ such that $\sum_{a\in A}a\neq 0$. Then there exists an ordering of the elements of $A$ such that the partial sums are all distinct and nonzero.

The only published paper on this problem is by Bode and Harborth [9]. There, the authors proved that Conjecture 1 is valid if $|A|=v-1$ or $|A|=v-2$ or if $v\leq 16$ (the latter was obtained by computer verification). Also they stated, without proof, that the conjecture is true if $|A|\leq 5$.

In [7] Archdeacon et al. proposed a related conjecture.

Conjecture 2**.**

Let $A\subseteq\mathbb{Z}_{v}\setminus\{0\}$. Then there exists an ordering of the elements of $A$ such that the partial sums are all distinct.

In [7] the authors proved that Conjecture 1 implies Conjecture 2. Then they proved that their conjecture is valid if $|A|\leq 6$ and made also a computer verification for $v\leq 25$.

Here, we propose the following conjecture:

Conjecture 3**.**

Let $(G,+)$ be an abelian group. Let $A$ be a finite subset of $G\setminus\{0\}$ such that no $2$-subset $\{x,-x\}$ is contained in $A$ and with the property that $\sum_{a\in A}a=0$. Then there exists an ordering of the elements of $A$ such that the partial sums are all distinct.

Clearly if $G=\mathbb{Z}_{v}$, Conjecture 3 immediately follows from Conjecture 2. Indeed we believe that Conjecture 2 can be stated considering a subset of an abelian group and not necessarily of a cyclic one. About that, we proved by computer that Conjecture 2 is valid for any abelian group of order $\leq 23$. Also, we have to point out that in [7], proving their Theorems 3.1 and 3.2, the authors do not use the hypothesis that $A$ is a subset of a cyclic group. In fact their proofs for $|A|\leq 6$ work, more in general, in an abelian group.

Another motivation for extending Conjectures 1 and 2 from cyclic groups to any (abelian) group is because of their natural connection with the concept of sequenceable (or R-sequenceable) group, see [18]. In particular, Alspach et al. recently proved that any finite abelian group is either sequenceable or R-sequenceable, confirming the Friedlander-Gordon-Miller conjecture [3].

We came across the problem of Conjecture 3 studying Heffter systems, see [19], as explained in Section 2. In Section 3 we show how some known conjectures about graphs with prescribed edge-lengths can be stated in terms of partial sums and we propose a related open problem. In Section 4 we prove the validity of Conjecture 3 for subsets $A$ of size less than $10$; we remark that we have also checked by computer the validity of our conjecture for abelian groups of order not exceeding $27$. Finally, some observations about the above conjectures in the nonabelian case are presented in Section 5.

2. Heffter systems and cyclic cycle systems

Given an odd positive integer $v$, an half-set $A$ of $\mathbb{Z}_{v}$ is a subset of $\mathbb{Z}_{v}\setminus\{0\}$ of size $(v-1)/2$ such that no $2$-subset $\{x,-x\}$ is contained in $A$. A Heffter system $D(v,k)$ is a partition of an half-set of $\mathbb{Z}_{v}$ into parts of size $k$ such that the elements in each part sum to [math]. Heffter himself introduced these systems to construct Steiner triple systems, see [19]. In [4], Archdeacon presented the related concept of a Heffter array, which has various applications, see [5, 6, 14, 16, 17]. A Heffter system is said to be simple if each part admits a simple ordering. Archdeacon in [4] proved that a simple Heffter system $D(v,k)$ gives rise to a cyclic $k$-cycle system of order $v$, namely a decomposition of the complete graph $K_{v}$ of order $v$ into cycles of length $k$ admitting $\mathbb{Z}_{v}$ as an automorphism group acting sharply transitively on the vertices. We recall the following result.

Proposition 2.1**.**

A $k$-cycle system $\mathcal{C}$ of order $v$ is sharply vertex-transitive under $\mathbb{Z}_{v}$ if and only if, up to isomorphisms, the following conditions hold:

• •

the set of vertices of $K_{v}$ is $\mathbb{Z}_{v}$;

• •

for all $C=(c_{1},c_{2},\ldots,c_{k})\in\mathcal{C}$, also $C+1:=(c_{1}+1,c_{2}+1,\ldots,c_{k}+1)\in\mathcal{C}$.

Clearly, to describe a cyclic $k$-cycle system of order $v$ it is sufficient to show a complete system $\mathcal{B}$ of representatives for the orbits of $\mathcal{C}$ under the action of $\mathbb{Z}_{v}$. The elements of $\mathcal{B}$ are called base cycles of $\mathcal{C}$.

The existence of (cyclic) cycle systems has been widely investigated, see [10, 12]. In order to explain how to construct the cycles starting from a Heffter system we have to introduce the concept of list of differences from a cycle and to show its usefulness in constructing cyclic cycle systems, see [11].

Definition 2.2**.**

Let $C=(c_{1},c_{2},\ldots,c_{k})$ be a $k$-cycle with vertices in an abelian group $G$. The multiset

[TABLE]

where the subscripts are taken modulo $k$, is called the list of differences from $C$.

More generally, given a set $\mathcal{B}$ of $k$-cycles with vertices in $G$, by $\Delta\mathcal{B}$ one means the union (counting multiplicities) of all multisets $\Delta C$, where $C\in\mathcal{B}$.

Theorem 2.3**.**

Let $\mathcal{B}$ be a set of $k$-cycles with vertices in $\mathbb{Z}_{v}$. If $\Delta\mathcal{B}=\mathbb{Z}_{v}\setminus\{0\}$ then $\mathcal{B}$ is set of base cycles of a cyclic $k$-cycle system of order $v$.

Suppose now to have a simple Heffter system $D(v,k)$. For $1\leq i\leq\frac{v-1}{2k}$, let $\omega_{i}=(a_{i,1},a_{i,2},$ $\ldots,a_{i,k})$ be a simple ordering of the $i$-th part of $D(v,k)$. Let $C_{i}=(a_{i,1},a_{i,1}+a_{i,2},\ldots,\sum_{j=1}^{k-1}a_{i,j},\sum_{j=1}^{k}a_{i,j}=0)$. Since the ordering $\omega_{i}$ is simple, then $C_{i}$ is a $k$-cycle of $K_{v}$; also it results $\Delta C_{i}=\pm\omega_{i}$. Let $\mathcal{B}=\left\{C_{1},\ldots,C_{\frac{v-1}{2k}}\right\}$, since $D(v,k)$ is a Heffter system then $\Delta\mathcal{B}=\mathbb{Z}_{v}\setminus\{0\}$. Hence, by Theorem 2.3, $\mathcal{B}$ is a set of base cycles of a cyclic $k$-cycle system of order $v$.

Example 2.4**.**

Consider the Heffter system $D(25,6)=\{\{3,1,4,-5,10,12\},\{2,7,$ $-9,6,8,11\}\}$. Since the orderings $\omega_{1}=(1,3,4,-5,10,12)$ and $\omega_{2}=(2,6,7,8,-9,$ $11)$ are simple, there exists a cyclic $6$-cycle system of order $25$. The cycles associated to these orderings are $C_{1}=(1,4,8,3,13,0)$ and $C_{2}=(2,8,15,23,14,0)$. Note that $\Delta C_{1}\cup\Delta C_{2}=\mathbb{Z}_{25}\setminus\{0\}$, so $\{C_{1},C_{2}\}$ is a set of base cycles of a cyclic $6$-cycle system of order $25$, that is $\mathcal{C}=\{C_{1}+i,C_{2}+i:i\in\mathbb{Z}_{25}\}$.

Note that if, in the previous example, we replace the ordering $\omega_{1}$ with $\omega^{\prime}_{1}=(1,4,-5,3,10,12)$, we do not obtain a cycle, but the union of the cycles $(0,1,5)$ and $(0,3,13)$. Hence, in order to obtain a system with cycles of the same length starting from a Heffter system $D$, it is necessary to require for the simplicity of $D$. The validity of Conjecture 3 would imply that any part of a Heffter system admits a simple ordering, namely that any Heffter system is simple.

3. Conjectures on graphs with prescribed edge-lengths

In this section we will see how the conjectures on partial sums of a given set presented in the Introduction are closely related to some conjectures on graphs with prescribed edge-lengths. We recall that the length $\ell(x,y)$ of an edge $[x,y]$ of $K_{v}$ is so defined:

[TABLE]

If $\Gamma$ is any subgraph of $K_{v}$, then the list of edge-lengths of $\Gamma$ is the multiset $\ell(\Gamma)$ of the lengths (taken with their respective multiplicities) of all the edges of $\Gamma$. For our convenience, if a list $L$ consists of $m_{1}$ $a_{1}$’s, $m_{2}$ $a_{2}$’s, …, $m_{t}$ $a_{t}$’s, we will write $L=\left\{a_{1}^{m_{1}},a_{2}^{m_{2}},\ldots,a_{t}^{m_{t}}\right\}$, whose underlying set is the set $\{a_{1},a_{2},\ldots,a_{t}\}$. Moreover, with an abuse of notation, by $\sum L$ we will mean $\sum_{a\in L}a$.

A famous conjecture about edge-lengths of a Hamiltonian path of the complete graph has been proposed by Buratti, Horak and Rosa, see [20].

Conjecture 4**.**

Let $L$ be a list of $v-1$ positive integers not exceeding $\left\lfloor\frac{v}{2}\right\rfloor$. Then there exists a Hamiltonian path $H$ of $K_{v}$ such that $\ell(H)=L$ if and only if for any divisor $d$ of $v$, the number of multiples of $d$ appearing in $L$ does not exceed $v-d$.

It is not hard to see that, even if the statement in terms of edge-lengths of a Hamiltonian path is more elegant, this conjecture can be formulated also in terms of partial sums of a given list. In fact Conjecture 4 can be also stated as:

Let $v$ be a positive integer and let $L$ be a list of $v-1$ nonzero elements of $\mathbb{Z}_{v}$. Then, there exists a suitable sequence $(\varepsilon_{1},\ldots,\varepsilon_{v-1})$, where each $\varepsilon_{i}=\pm 1$, and a suitable ordering $(a_{1},\ldots,a_{v-1})$ of $L$ such that the partial sums of the sequence $(\varepsilon_{1}a_{1},\ldots,\varepsilon_{v-1}a_{v-1})$ are exactly the elements of $\mathbb{Z}_{v}\setminus\{0\}$ if and only if for any divisor $d$ of $v$ the number of multiples of $d$ appearing in $L$ does not exceed $v-d$.

For example, given the list $L=\{1^{2},4^{2},5\}$ in $\mathbb{Z}_{6}$, we can consider the sequence $(1,4,-1,4,-5)$ whose partial sums are $1,5,4,2,3$.

Meszka, Pasotti and Pellegrini proposed another related conjecture, see [23].

Conjecture 5**.**

Let $v=2n+1$ be an odd integer and let $L$ be a list of $n$ positive integers not exceeding $n$. Then there exists a near $1$-factor $F$ of $K_{v}$ such that $\ell(F)=L$ if and only if for any divisor $d$ of $v$ the number of multiples of $d$ appearing in $L$ does not exceed $\frac{v-d}{2}$.

Results about Conjectures 4 and 5 can be found in [13, 15, 20, 21, 22] and in [23, 24], respectively.

Looking at Conjecture 3 and at these conjectures it is quite natural to ask which properties have to satisfy a list $L$ of $k$ elements of $\left\{1,2,\ldots,\left\lfloor\frac{v}{2}\right\rfloor\right\}$ in order to have a $k$-cycle $C$ of $K_{v}$ such that $\ell(C)=L$.

Remark 3.1**.**

Let $v$ be a positive integer and let $L=\{a_{1},\ldots,a_{k}\}$ be a list of $k$ elements not exceeding $\left\lfloor\frac{v}{2}\right\rfloor$. A trivial necessary condition for the existence of a cycle $C$ of $K_{v}$ such that $\ell(C)=L$ is the existence of a list $(\varepsilon_{1},\ldots,\varepsilon_{k})$, where each $\varepsilon_{i}=\pm 1$, such that $\sum_{i=1}^{k}\varepsilon_{i}a_{i}\equiv 0\pmod{v}$.

For example, given the list $L=\{1^{2},2,3,5^{2}\}$ we have $1-1-2+3+5+5\equiv 0\pmod{11}$ and $C=(0,5,10,9,1,2)$ is a cycle of $K_{11}$ such that $\ell(C)=L$. It is easy to see that this condition it is not sufficient: in Proposition 3.3 we will show another necessary condition. Firstly, we state the following.

Lemma 3.2**.**

Let $L=\{a_{1}^{m_{1}},a_{2}^{m_{2}},\ldots,a_{t}^{m_{t}}\}$ be a list of $k$ elements of $\{1,2,\ldots,\left\lfloor\frac{v}{2}\right\rfloor\}$ such that $d=\gcd(v,a_{1},a_{2},\ldots,a_{t})>1$. There exists a cycle $C$ of $K_{v}$ such that $\ell(C)=L$ if and only if there exists a cycle $C^{\prime}$ of $K_{\frac{v}{d}}$ such that $\ell(C^{\prime})=\left\{\left(\frac{a_{1}}{d}\right)^{m_{1}},\left(\frac{a_{2}}{d}\right)^{m_{2}},\ldots,\left(\frac{a_{t}}{d}\right)^{m_{t}}\right\}$.

Proof.

Let $C=(0,c_{2},\ldots,c_{k})$ be a $k$-cycle of $K_{v}$ such that $\ell(C)=L$. Note that since $d$ divides each element of $L$, $d$ divides $c_{i}$ for any $i$. Moreover, since $C$ is a cycle, $c_{i}\not\equiv c_{j}\pmod{v}$ for any $i\neq j$, so $k$ cannot exceed $\frac{v}{d}$, since every class of residues modulo $d$ intersects $\{0,\ldots,v-1\}$ in a set of that size. From $c_{i}\not\equiv c_{j}\pmod{v}$ for any $i\neq j$, it immediately follows also that $\frac{c_{i}}{d}\not\equiv\frac{c_{j}}{d}\pmod{\frac{v}{d}}$ for any $i\neq j$, so $C^{\prime}=\left(0,\frac{c_{2}}{d},\ldots,\frac{c_{k}}{d}\right)$ is a $k$-cycle of $K_{\frac{v}{d}}$. Obviously, $\ell(C^{\prime})$ is the list of the elements of $L$ divided by $d$. The converse can be done in a similar way. ∎

Hence, without loss of generality, we can consider lists $\left\{a_{1}^{m_{1}},a_{2}^{m_{2}},\ldots,a_{t}^{m_{t}}\right\}$ of elements not exceeding $\left\lfloor\frac{v}{2}\right\rfloor$ such that $\gcd(v,a_{1},a_{2},\ldots,a_{t})=1$. For example, instead of $v=20$ and $L=\{6^{6},8^{2}\}$, we can consider $v^{\prime}=10$ and $L^{\prime}=\{3^{6},4^{2}\}$. Note that $C^{\prime}=(0,4,8,5,2,9,6,3)$ is a cycle of $K_{10}$ such that $\ell(C^{\prime})=L^{\prime}$, so $C=(0,8,16,10,4,18,12,6)$ is a cycle of $K_{20}$ such that $\ell(C)=L$.

Proposition 3.3**.**

Let $L=\left\{a_{1}^{m_{1}},a_{2}^{m_{2}},\ldots,a_{t}^{m_{t}}\right\}$ be a list of $k\leq v$ elements of $\{1,2,\ldots,\left\lfloor\frac{v}{2}\right\rfloor\}$ with $\gcd(v,a_{1},a_{2},\ldots,a_{t})=1$. If there exists a $k$-cycle $C$ of $K_{v}$ such that $\ell(C)=L$ then for any divisor $d>1$ of $v$, the number of multiples of $d$ appearing in $L$ does not exceed $\frac{k}{v}(v-d)$.

Proof.

Let $C$ be a $k$-cycle of $K_{v}$ with $\ell(C)=L$ and let $d>1$ be a divisor of $v$. Denote by $N$ the number of non-multiples of $d$ appearing in $L$. By the hypothesis, $N\geq 1$. Consider the graph $\Gamma$ obtainable from $C$ by deleting all $N$ edges whose length is not divisible by $d$. Then $\Gamma$ has exactly $N$ connected components some of which may be just isolated vertices. It is also clear that all vertices of every connected component $K$ of $\Gamma$ are in the same residue class modulo $d$ so that $K$ has at most $\frac{v}{d}$ vertices. It follows that $k=|V(\Gamma)|\leq N\frac{v}{d}$, i.e. $N\geq\frac{k\cdot d}{v}$. Hence the number of multiples of $d$, namely $k-N$, is at most $k-\frac{k\cdot d}{v}$, so we have the thesis. ∎

We have to point out that the necessary conditions of Remark 3.1 and Proposition 3.3 are not sufficient. In fact if we take $v=8$ and we consider the list $L=\{3^{4},4^{4}\}$, one can check that a cycle $C$ of $K_{8}$ such that $\ell(C)=L$ does not exist. Note that if $v$ is a prime, Proposition 3.3 gives no necessary condition. Although, for example, there exists no cycle $C$ of $K_{7}$ such that $\ell(C)=\{1,2,3^{5}\}$. A special case is considered in the following remark.

Remark 3.4**.**

Let $L=\{a^{k}\}$ with $1\leq a\leq\left\lfloor\frac{v}{2}\right\rfloor$. Then $(0,a,2a,\ldots,(k-1)a)$ is a cycle of $K_{v}$ if and only if the order of $a$ in $\mathbb{Z}_{v}$ is $k$, namely if $k=\frac{v}{\gcd(v,a)}$.

So we propose the following.

Open Problem**.**

Let $v$ be a positive integer and let $L$ be a list of positive integers not exceeding $\left\lfloor\frac{v}{2}\right\rfloor$. Find the necessary and sufficient conditions for $L$ in order to have a cycle $C$ of $K_{v}$ such that $\ell(C)=L$.

4. Proof of main results

In this section we prove that Conjecture 3 holds for sets of small size. Clearly, given an ordering $(a_{1},a_{2},\ldots,a_{k})$ of a finite set $A\subseteq G\setminus\{0\}$, for $1\leq i<j\leq k$, we have $s_{i}=s_{j}$ if and only if $\sum_{h=i+1}^{j}a_{h}=0$. In the proof of the following theorems, we will show that an ordering $\omega$ of $A$ is simple, by checking that there is no subsequence of consecutive elements of $\omega$, viewed as a $k$-cycle, which sums to [math]. We recall that by $\sum A$ we mean $\sum_{a\in A}a$. Clearly, if there exists $B\subsetneq A$ such that $\sum B=0$, then we may assume that $|B|\geq 3$, as we are requiring that $A$ does not contain $2$-subsets of shape $\{x,-x\}$. Furthermore, if $\sum B=0$ then $\sum(A\setminus B)=0$ (as $\sum A=0$) and so we may also assume $|B|\leq\frac{|A|}{2}$.

Theorem 4.1**.**

Conjecture 3 is true for $|A|\leq 8$.

Proof.

Clearly, if there are no proper subsets $B$ of $A$ such that $\sum B=0$ we have the thesis. So, suppose that there exists a subset $B\subsetneq A$ such that $\sum B=0$. As previously remarked, we may assume $3\leq|B|\leq\frac{|A|}{2}$.

If $|A|\leq 5$ the thesis immediately follows by above considerations.

Assume that $A=\{a_{1},a_{2},\ldots,a_{6}\}$ has order $6$ and, without loss of generality, we may suppose that $B=\{a_{1},a_{2},a_{3}\}$. It is now clear that the ordering $(a_{1},a_{2},a_{4},a_{3},a_{5},a_{6})$ is simple, since $A\setminus B$ is the only other proper subset of $A$ that sums to [math].

Now, assume that $A=\{a_{1},a_{2},\ldots,a_{7}\}$ is a set of size $7$ and let $T_{1}\subsetneq A$ such that $\sum T_{1}=0$. We can assume, without loss of generality, $T_{1}=\{a_{1},a_{2},a_{3}\}$. If $T_{1}$ is the unique subset of $A$ of size $3$ such that $\sum T_{1}=0$, then the ordering $(a_{1},a_{2},a_{4},a_{3},a_{5},a_{6},a_{7})$ is simple. Suppose that there is another subset $T_{2}\subsetneq A$ of size $3$ such that $\sum T_{2}=0$. It is easy to see that $|T_{1}\cap T_{2}|=1$. Without loss of generality, we may assume $T_{2}=\{a_{3},a_{4},a_{5}\}$. The ordering

[TABLE]

is simple.

Finally, we suppose that $A=\{a_{1},a_{2},\ldots,a_{8}\}$ is a set of size $8$. We split the proof into two cases.

Case 1. Assume that there exists a subset $Q\subsetneq A$ such that $|Q|=4$ and $\sum Q=0$.

First suppose that there exists a subset $T\subsetneq A$ such that $|T|=3$ and $\sum T=0$. Clearly, $|Q\cap T|=1,2$ and replacing $Q$ with $A\setminus Q$ if $|Q\cap T|=1$, we may assume that $Q=\{a_{1},a_{2},a_{3},a_{4}\}$ and $T=\{a_{3},a_{4},a_{5}\}$. In this case the ordering

[TABLE]

is simple. Hence, we may assume that there are no subsets $T\subsetneq A$ of size $3$ such that $\sum T=0$. In this case the ordering

[TABLE]

is simple.

Case 2. Assume that there are no subsets $Q\subsetneq A$ such that $|Q|=4$ and $\sum Q=0$.

Hence, there is a subset $T_{1}\subsetneq A$ such that $|T_{1}|=3$ and $\sum T_{1}=0$: we may assume $T_{1}=\{a_{1},a_{2},a_{3}\}$. If there exists another subset $T_{2}\subsetneq A$ with $\sum T_{2}=0$, by our assumptions it must be that $|T_{2}|=3$. As in case $|A|=7$ we have $|T_{1}\cap T_{2}|=1$ and so we may assume $T_{2}=\{a_{3},a_{4},a_{5}\}$. In this case the ordering

[TABLE]

is simple. Finally, suppose that $T_{1}$ is the unique subset of $A$ that sums to [math]. In this case, the ordering $(a_{1},a_{2},a_{4},a_{3},a_{5},a_{6},a_{7},a_{8})$ is obviously simple. ∎

Theorem 4.2**.**

Conjecture 3 is true for $|A|=9$.

Proof.

Clearly, if there are no proper subsets $B$ of $A$ such that $\sum B=0$ we have the thesis. Hence in the following we will assume that there exists $B\subsetneq A$ such that $\sum B=0$. As already remarked, we may assume $3\leq|B|\leq 4$.

In the proof by $Q_{i}$ and $T_{i}$ we always mean a subset of $A$ of size $4$ and $3$, respectively. Also by $Q_{i}$, $Q_{j}$ with $i\neq j$, we will mean $Q_{i}\neq Q_{j}$; analogously for $T_{i}$, $T_{j}$. Let $Q_{i},Q_{j}$ be such that $\sum Q_{i}=\sum Q_{j}=0$. Note that $|Q_{i}\cap Q_{j}|\neq 0$ otherwise $0\in A$, also $|Q_{i}\cap Q_{j}|\neq 3$ otherwise $Q_{i}=Q_{j}$. Hence $\sum Q_{i}=\sum Q_{j}=0$ implies $|Q_{i}\cap Q_{j}|=1,2$. Let $Q_{i},T_{j}$ be such that $\sum Q_{i}=\sum T_{j}=0$. Note that $|Q_{i}\cap T_{j}|\neq 0$, otherwise $A$ contains a $2$-subset $\{x,-x\}$; also $|Q_{i}\cap T_{j}|\neq 3$ otherwise $0\in A$. Hence $\sum Q_{i}=\sum T_{j}=0$ implies $|Q_{i}\cap T_{j}|=1,2$. Let $T_{i},T_{j}$ be such that $\sum T_{i}=\sum T_{j}=0$. Note that $|T_{i}\cap T_{j}|\neq 2$, otherwise $T_{i}=T_{j}$, hence $|T_{i}\cap T_{j}|=0,1$. Also if $\sum T_{i}=\sum T_{j}=0$ and $T_{i}\cap T_{j}=\emptyset$, then the triple $A\setminus(T_{i}\cup T_{j})$ sums to [math].

We set $A=\left\{a_{1},a_{2},\ldots,a_{9}\right\}$. Firstly, we suppose that there is no $Q_{i}$ which sums to [math]. Let $T_{1}=\left\{a_{1},a_{2},a_{3}\right\}$ such that $\sum T_{1}=0$. If there is no other triple in $A$ which sum to [math], the ordering $(a_{1},a_{2},a_{4},a_{3},a_{5},a_{6},a_{7},a_{8},a_{9})$ is simple.

Case a. Assume there exists $T_{2}$ with $\sum T_{2}=0$ and $|T_{1}\cap T_{2}|=0$. We can suppose $T_{2}=\left\{a_{4},a_{5},a_{6}\right\}$: the ordering

[TABLE]

is simple.

Case b. We now assume that for all $T_{i}\neq T_{1}$ which sums to [math], $|T_{1}\cap T_{i}|=1$. Let $T_{2}=\left\{a_{3},a_{4},a_{5}\right\}$ with $\sum T_{2}=0$. In this case the ordering

[TABLE]

is simple.

Suppose now that there exists $Q_{i}$ which sums to [math]. We split the proof into 3 cases.

There exists only one $Q_{1}$ which sums to [math].

• 2)

There exist $Q_{1}$ and $Q_{2}$ which sum to [math] such that $|Q_{1}\cap Q_{2}|=2$,

• 3)

There exist at least two quadruples in $A$ which sum to [math]. For all such quadruples $Q_{i}\neq Q_{j}$, we have $|Q_{i}\cap Q_{j}|=1$.

Case 1. We split this case into 3 subcases.

• 1.1)

There is no $T_{1}$ which sums to [math].

• 1.2)

There exists $T_{1}$ which sum to [math] such that $|Q_{1}\cap T_{1}|=1$,

• 1.3)

Otherwise.

Case 1.1. We can suppose $Q_{1}=\{a_{1},a_{2},a_{3},a_{4}\}$. Then $(a_{1},a_{2},a_{3},a_{5},a_{4},a_{6},a_{7},a_{8},$ $a_{9})$ is a simple ordering of $A$.

Case 1.2. We can suppose $Q_{1}=\{a_{1},a_{2},a_{3},a_{4}\}$ and $T_{1}=\{a_{4},a_{5},a_{6}\}$. By the hypothesis, $\sum Q_{1}=\sum T_{1}=0$. Let $T_{2}=\{a_{1},a_{2},a_{9}\}$ and $T_{3}=\{a_{1},a_{8},a_{9}\}$. If $\sum T_{2}=0$ set $\omega=(a_{1},a_{2},a_{5},a_{4},a_{7},a_{6},a_{8},a_{9},a_{3})$ and if $\sum T_{3}=0$ set $\omega=(a_{1},a_{2},a_{3},a_{5},a_{4},a_{8},a_{6},a_{7},a_{9})$. In both cases, $\omega$ is a simple ordering of $A$. So, assume that $\sum T_{2},\sum T_{3}\neq 0$. In this case,

[TABLE]

is a simple ordering of $A$.

Case 1.3. We can suppose $Q_{1}=\{a_{1},a_{2},a_{3},a_{4}\}$ and $T_{1}=\{a_{3},a_{4},a_{5}\}$. We note that

[TABLE]

is a simple ordering of $A$.

Case 2. We can suppose $Q_{1}=\{a_{1},a_{2},a_{3},a_{4}\}$ and $Q_{2}=\{a_{3},a_{4},a_{5},a_{6}\}$. We recall that, by the hypothesis, $\sum Q_{1}=\sum Q_{2}=0$. Let $T_{1}=\{a_{2},a_{4},a_{7}\}$, $T_{2}=\{a_{3},a_{4},a_{7}\}$ and $Q_{3}=\{a_{1},a_{6},a_{8},a_{9}\}$. If $\sum T_{1}=0$, set $\omega=(a_{1},a_{2},a_{3},a_{7},a_{4},a_{5},a_{6},a_{8},a_{9})$; if $\sum T_{2}=0$ set $T_{3}=\{a_{1},a_{3},a_{5}\}$, $Q_{4}=\{a_{4},a_{6},a_{7},a_{8}\}$ and

[TABLE]

if $\sum Q_{3}=0$, set

[TABLE]

In all cases, $\omega$ is a simple ordering of $A$. So, assume $\sum T_{1},\sum T_{2},\sum Q_{3}\neq 0$. Set $T_{4}=\{a_{3},a_{5},a_{7}\}$, $T_{5}=\{a_{1},a_{6},a_{9}\}$; then

[TABLE]

is a simple ordering of $A$.

Case 3. By the hypothesis, $\sum Q_{1}=\sum Q_{2}=0$ and $|Q_{1}\cap Q_{2}|=1$. Let $Q_{1}=\{a_{1},a_{2},a_{3},a_{4}\}$ and $Q_{2}=\{a_{4},a_{5},a_{6},a_{7}\}$. Suppose firstly that does not exist $Q_{3}$ which sums to [math]. Let $T_{1}=\{a_{4},a_{6},a_{8}\}$, $T_{2}=\{a_{3},a_{4},a_{5}\}$, $T_{3}=\{a_{2},a_{3},a_{5}\}$, $T_{4}=\{a_{1},a_{3},a_{5}\}$ and $T_{5}=\{a_{1},a_{7},a_{9}\}$.

Assume $\sum T_{1}=0$. If either $\sum T_{2}=0$ or $\sum T_{3}=0$ take $\omega=(a_{2},a_{1},a_{4},a_{6},a_{3},a_{5},$ $a_{8},a_{7},a_{9})$; if $\sum T_{2},\sum T_{3}\neq 0$, take $T_{6}=\{a_{1},a_{7},a_{8}\}$ and

[TABLE]

In all these cases, $\omega$ is a simple ordering of $A$.

So, from now on, assume $\sum T_{1}\neq 0$. If $\sum T_{2}=0$, the ordering

[TABLE]

is simple, so we may also suppose $\sum T_{2}\neq 0$. To conclude, it suffices to take $T_{7}=\{a_{2},a_{7},a_{9}\}$ and

[TABLE]

Now, we have to suppose that there exists $Q_{3}$ which sums to [math]. Clearly, this implies that there is no $Q_{4}$ such that $\sum Q_{4}=0$ and $|Q_{4}\cap Q_{i}|=1$ for all $i=1,2,3$. By the assumptions it follows that $|Q_{1}\cap Q_{2}|=|Q_{1}\cap Q_{3}|=|Q_{2}\cap Q_{3}|=1$. So we can suppose $Q_{1}=\{a_{1},a_{2},a_{3},a_{4}\}$, $Q_{2}=\{a_{4},a_{5},a_{6},a_{7}\}$ and $Q_{3}=\{a_{1},a_{7},a_{8},a_{9}\}$. We set $T_{1}=\{a_{1},a_{3},a_{5}\}$, $T_{2}=\{a_{4},a_{6},a_{8}\}$ and $T_{3}=\{a_{2},a_{7},a_{9}\}$. If $\sum T_{1},\sum T_{2},\sum T_{3}\neq 0$ then $(a_{2},a_{1},a_{3},a_{5},a_{4},a_{6},a_{8},a_{7},a_{9})$ works. We can focus our attention on the case $\sum T_{1}=0$, namely the case $\sum T_{2}=0$ ($\sum T_{3}=0$, respectively) can be done in a similar way replacing each $a_{i}$ with $a_{i+3}$ (with $a_{i+6}$, respectively) where the subscripts are taken modulo $9$. So in the following we assume $\sum T_{1}=0$ and set $T_{4}=\{a_{4},a_{5},a_{8}\}$. If $\sum T_{3}=0$, take $\omega=(a_{2},a_{1},a_{4},a_{5},a_{3},a_{7},a_{9},a_{6},a_{8})$ and if $\sum T_{4}=0$ take

[TABLE]

In all cases $\omega$ is a simple ordering of $A$. So, assume $\sum T_{3},\sum T_{4}\neq 0$. In this case, $(a_{2},a_{1},a_{3},a_{6},a_{4},a_{5},a_{8},a_{7},a_{9})$ is a simple ordering of $A$. ∎

5. Further developments

Looking at Conjectures 1 and 2 presented in the Introduction, a natural question is to ask what happens if one considers finite subsets $A\subseteq G\setminus\{0\}$, where $G$ is any group, not necessarily cyclic. For instance we verified, with the help of a computer, the validity of Conjecture 1, for all abelian groups of order $|G|\leq 21$. On the other hand, Conjecture 1 cannot be generalized to nonabelian groups. For instance, consider the symmetric group $G=\mathrm{Sym}(3)$ and its subset $A=G\setminus\{0\}$ (we keep using the additive notation): any ordering of $A$ is such that $\sum A\neq 0$, but there is no ordering of $A$ such that all the partial sums are distinct and nonzero.

We have already remarked in the Introduction what happens for Conjecture 2 if $G$ is an abelian group. Note that if we consider Conjectures 1 and 2 in the case of abelian groups then, again, Conjecture 1 implies Conjecture 2: in fact it suffices to apply the same proof of [7, Proposition 1.1].

For Conjecture 2 it is also natural to investigate the nonabelian case. We made a computer verification of this conjecture for all groups of order $|G|\leq 19$. We have also the following theoretical result.

Theorem 5.1**.**

Let $(G,+)$ be a group and let $A\subseteq G\setminus\{0\}$ with $|A|\leq 5$. Then there exists an ordering of the elements of $A$ such that the partial sums are all distinct.

Proof.

If $|A|\leq 2$ it is obvious. Suppose $|A|>2$ and let $p$ be the number of distinct $2$-subsets $\{x,-x\}$ contained in $A$.

Let $A=\{a_{1},a_{2},a_{3}\}$ be a subset of size $3$. If $p=0$, then the ordering $(a_{1},a_{2},a_{3})$ is simple. If $p=1$ we may assume $a_{2}=-a_{1}$ and take the ordering $(a_{1},a_{3},-a_{1})$.

Let $A=\{a_{1},a_{2},a_{3},a_{4}\}$ be a subset of size $4$. If $p=0$, the ordering

[TABLE]

is simple. If $p=1$, we may assume $a_{2}=-a_{1}$: in this case, $(a_{3},a_{1},a_{4},-a_{1})$ is a simple ordering. If $p=2$, we may assume $a_{2}=-a_{1}$ and $a_{4}=-a_{3}$: it is easy to see that $(a_{1},a_{3},-a_{1},-a_{3})$ is a simple ordering of $A$.

Now, let $A=\{a_{1},a_{2},a_{3},a_{4},a_{5}\}$ be a subset of size $5$. First, we consider the case $p=0$. If $a_{2}+a_{3}+a_{4}+a_{5}\neq 0$, $a_{3}+a_{4}+a_{5}\neq 0$ and $a_{2}+a_{3}+a_{4}\neq 0$, the ordering $(a_{1},a_{2},a_{3},a_{4},a_{5})$ is simple. So, suppose that $a_{2}+a_{3}+a_{4}+a_{5}=0$. If $a_{1}+a_{3}+a_{4}\neq 0$, then the ordering $(a_{2},a_{1},a_{3},a_{4},a_{5})$ is simple. So, assume also that $a_{1}+a_{3}+a_{4}=0$ and observe that

[TABLE]

is a simple ordering of $A$. Next, suppose $a_{3}+a_{4}+a_{5}=0$. The ordering $(a_{5},a_{1},a_{3},a_{4},$ $a_{2})$ is simple, except when $a_{1}+a_{3}+a_{4}+a_{2}=0$. However, if this holds, in order to find a simple ordering of $A$ it suffices to reapply (1) to the set $\{a_{1}^{\prime},a_{2}^{\prime},a_{3}^{\prime},a_{4}^{\prime},a_{5}^{\prime}\}$, where $a_{1}^{\prime}=a_{5}$, $a_{2}^{\prime}=a_{1}$, $a_{3}^{\prime}=a_{3}$, $a_{4}^{\prime}=a_{4}$ and $a_{5}^{\prime}=a_{2}$. Similarly, if $a_{2}+a_{3}+a_{4}=0$, we consider the set $\{a_{1}^{\prime},a_{2}^{\prime},a_{3}^{\prime},a_{4}^{\prime},a_{5}^{\prime}\}$, where $a_{1}^{\prime}=a_{1}$, $a_{2}^{\prime}=a_{5}$, $a_{3}^{\prime}=a_{2}$, $a_{4}^{\prime}=a_{3}$, $a_{5}^{\prime}=a_{4}$, and proceed as done previously.

If $p=1$, we may suppose that $a_{2}=-a_{1}$. If $\pm a_{1}+a_{3}+a_{4}\neq 0$, then $(a_{5},a_{1},a_{3},a_{4},-a_{1})$ is a simple ordering of $A$. Suppose $\varepsilon\cdot a_{1}+a_{3}+a_{4}=0$ for some $\varepsilon=\pm 1$: in this case, the ordering

[TABLE]

is simple.

If $p=2$, then we may suppose that $a_{2}=-a_{1}$ and $a_{4}=-a_{3}$. If $a_{1}+a_{3}-a_{1}-a_{3}\neq 0$, then the ordering $(a_{5},a_{1},a_{3},-a_{1},-a_{3})$ is simple. Assume $a_{1}+a_{3}-a_{1}-a_{3}=0$. Then the ordering

[TABLE]

is simple. In fact, since $a_{1}+a_{3}=a_{3}+a_{1}$, we have $\pm a_{1}+a_{3}\mp a_{1}+a_{5}=a_{3}+a_{5}\neq 0$. ∎

Alspach [2] recently proposed the following definition, closely related to Conjecture 2 in its generalized formulation: a finite group $G$ is said to be strongly sequenceable if every Cayley digraph on $G$ admits either an orthogonal directed path or an orthogonal directed cycle. When this path (respectively, cycle) has length $|G|-1$, we retrieve the concept of sequenceable (respectively, R-sequenceable) group, see [3]. The problem that Kalinowski and he propose is the classification of the strongly sequenceable groups. In this direction, Theorem 5.1 can be viewed as an intermediate step.

Acknowledgements

The authors would like to thank Marco Buratti and Jeff Dinitz for useful discussion on this topic.