Index of Sequences
Weidong Gao
Let $Z_n$ denote the cyclic group of $n$ elements. For every $x\in Z_n$, we denote by $|x|_n$ the smallest positive integer in the residual class $x$. Let $S=(a_1, \ldots, a_k)$ be a sequence of elements in $Z_n$. We say that $S$ has index $n$ if there is an integer $m$ co-prime to $n$ such that $\sum_{i=1}^k|ma_i|_n=n$. In this paper, we prove that a sequence of $n$ elements in $Z_n$ with repetition at least $\frac{n}{2}$ contains a subsequence with index $n$, and if $n=p$ is a prime the restriction of repetition can be relaxed to $\frac{p-2}{10}$. On the other hand, for every $n=4k+2\geq 22$, we give a sequence of length $n+\lfloor \frac{n}{4}\rfloor-5$ and with repetition $\frac{n}{2}-1$ which contains no subsequence with index $n$. This disproves a conjecture given twenty years ago by Lemke and Kleitman.
0909.2461.tex
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\begin{document}

\title{Index of Sequences}

\author{Weidong Gao}
\address{Center for Combinatorics\\
 Nankai University\\
 Tianjin 300071, P.R. China}
 \email{wdgao1963@yahoo.com.cn}
\author{Yuanlin Li}
\address{Department of Mathematics\\
Brock University \\
St.~Catharines, Ontario \\
Canada L2S 3A1} \email{yli@brocku.ca}
\author{Jiangtao Peng}
\address{Center for Combinatorics\\
 Nankai University\\
 Tianjin 300071, P.R. China}
 \email{jtpeng1982@yahoo.com.cn}
 \author{Chris Plyley}
\address{Department of Mathematics\\
Brock University \\
St.~Catharines, Ontario \\
Canada L2S 3A1} \email{cp07rp@brocku.ca}
\author{Guoqing Wang}
\address{Center for Combinatorics\\
 Nankai University\\
 Tianjin 300071, P.R. China}
 \email{gqwang1979@yahoo.com.cn}
%\date{}
\maketitle

\begin{abstract}
 Let $Z_n$ denote the cyclic group of $n$ elements. For every $x\in Z_n$, we denote by $|x|_n$ the smallest positive integer in the
residual class $x$. Let $S=(a_1, \ldots, a_k)$ be a sequence of
elements in $Z_n$. We say that $S$ has index $n$ if there is an
integer $m$ co-prime to $n$ such that $\sum_{i=1}^k|ma_i|_n=n$.  In
this paper, we prove that a sequence of $n$ elements in $Z_n$ with
repetition at least $\frac{n}{2}$ contains a subsequence with index
$n$, and if $n=p$ is a prime the restriction of repetition can be
relaxed to $\frac{p-2}{10}$. On the other hand, for every
$n=4k+2\geq 22$, we give a sequence of length $n+\lfloor
\frac{n}{4}\rfloor-5$ and with repetition $\frac{n}{2}-1$ which
contains no subsequence with index $n$. This disproves a conjecture
given twenty years ago by Lemke and Kleitman.
\end{abstract}

{\sl Key Words}: Index of sequences; Zero-sum; Minimal zero-sum
sequence

\section{Introduction and Notations}



Let $Z_n$ be the cyclic group of $n$ elements. We can regard $Z_n$
as the residual classes group of $Z$ modulo $n$. For every $x\in
Z_n$, let $|x|_n$ denote the smallest positive integer in the
residual class $x$. Let $S=(a_1, \ldots, a_k)$ be a sequence of
elements in $Z_n$. Let $\sigma(S)=\sum_{i=1}^ka_i$. We denote by
$|S|_n$ the sequence $(|a_1|_n, \ldots, |a_k|_n)$ of positive
integers. For a subset $A$ of $Z_n$, we denote $S\cap
A=\prod\limits_{a_i\in A} a_i$ to be the restriction of $S$ to $A$.
By $S\subseteq A$, we mean that $S\cap A=S$. For any integer $m$,
let $mS=(ma_1, \ldots, ma_k)$. Define
$$
Index(S)=\min_{(m,n)=1}\{\sigma(|mS|_n)\}.
$$
Let
$$
\sum_{Index}(S)=\{\sigma(|T|_n): T|S, 1\leq \sigma(|T|_n) \leq n \}.
$$
Let $h=h(S)$ be the maximal repetition value of $S$.

\medskip
The concept of {\it Index} was introduced by Chapman, Freeze and Smith
\cite{CFS} in 1999, and research on {\it Index} can go back to Lemke and
Kleitman \cite{LK} in 1989 when they made the following conjecture:

\begin{conj} \cite{LK} Let $d$, $n$ be two positive integers with $d|n$, and
let $S$ be a sequence of $n$ elements in $Z_n$. Then, there is a
nonempty subsequence $T$ of $S$ and an integer $m$ co-prime to $n$
such that $$d\big |\sigma(|mT|_n) \big |n.$$
\end{conj}
When $d=n$ Conjecture 1.1 becomes the following:

\begin{conj}
If $S$ is a sequence of $n$ elements (repetition allowed) in $Z_n$
then $S$ contains a subsequence $T$ with $Index(T)=n$.
\end{conj}
In this paper we demonstrate that Conjecture 1.2 and therefore
Conjecture 1.1 is not true in general (Section 2), we prove that the
conclusion of Conjecture 1.2 is true if $S$ contains some element at
least $n/2$ times (Section 3), and we also prove that if $n=p$ is a
prime then the Conjecture is true if $S$ contains some element at
least $\frac{p-2}{10}$ times (Section 4). In fact, we still believe
that Conjecture 1.2 is true when $n$ is prime.


 We call $S$ a {\sl zero-sum sequence} if
$\sigma(S)=0$, we call $S$ a {\sl minimal zero-sum sequence} if $S$
is zero-sum and any proper subsequence of $S$ is not zero-sum, and
we call $S$ a {\sl zero-sum free sequence} if $S$ contains no
nonempty zero-sum subsequence.

Recently, the study of {\it Index} has attracted many researchers. Let $S$
be a minimal zero-sum sequence in $Z_n$. It has been proven that if
$|S|\leq 3$ then $Index(S)=n$ (see \cite{CFS}) and if $|S|\geq
\frac{n}{2}+1$ then $Index(S)=n$ (\cite{SC}, \cite{Yuan}). It has
also been proven that for $5\leq k\leq \frac{n}{2}$ there is a
minimal zero-sum subsequence $S$ in $Z_n$ such that $|S|=k$ and
$Index(S)\geq 2n$, and for $k=4$ and $(n,6)\neq 1$, there is a
minimal zero-sum subsequence $S$ in $Z_n$ such that $|S|=4$ and
$Index(S)\geq 2n$. The following was conjectured.

\begin{conj} \cite{Pon} Let $n$ be a positive integer with
$(n,6)=1$, and let $S$ be a minimal zero-sum sequence in $Z_n$ with
$|S|=4$. Then, $Index(S)=n$.
\end{conj}

 Li and Plyley \cite{CL} proved that the above conjecture is true for
$n=p$ a prime. We shall provide a new proof of this result in
Section 5.



Our main results are the following:

\begin{theorem} \label{integer} Let $S$ be a sequence of $n$ elements in $Z_n$. If
$h(S)<4$ or $h(S)\geq n/2$ then $S$ contains a subsequence $T$ with
$Index(T)=n$ and $|T|\leq h(S)$.
\end{theorem}


\begin{theorem} \label{prime2} Let $p>24318$ be a prime, and let $S$ be a sequence of $p$ elements in $Z_p\setminus\{0\}$.
If $h(S)\geq \frac{p-2}{10}$ then $S$ contains a subsequence with
index $p$.
\end{theorem}

\begin{theorem}\label{counterexample}
Let $k\in \mathbb{N},\ n=4k+2,\ G=\mathbb{Z}_n$. If $k\geq 5$, then
$$S=1^{\frac{n}{2}-3}\left(\frac{n}{2}\right)
\left(\frac{n}{2}+1\right)^{\frac{n}{2}-1}\left(\frac{n}{2}+2\right)^{\lfloor
\frac{n}{4}\rfloor-2}$$ contains no subsequence with index $n$.
\end{theorem}

\section{Proof of Theorem \ref{counterexample}}







{\sl Proof of Theorem \ref{counterexample}}. Suppose the conclusion
is false, i.e., there exists a zero sum subsequence $T$ of $S$ and
$j\in [1,n-1]$ with $(j,n)=1$ such that
\begin{equation}\label{sum jT=n}
\sigma(|jT|_n)=n.
\end{equation}


Let $$T=1^x\left(\frac{n}{2}\right)^y\left(\frac{n}{2}+1\right)^z
\left(\frac{n}{2}+2\right)^w,$$ where $x\leq \frac{n}{2}-3$, $y\leq
1$, $z\leq \frac{n}{2}-1$ and $w\leq \lfloor \frac{n}{4}\rfloor-2$.

Then
\begin{equation}\label{sum T equvalent 0 mod n}
\sigma(T)=(x+z+2w)+\frac{n}{2}(y+z+w)\equiv 0\pmod n.
\end{equation}

{\bf Case 1.} $j< \frac{n}{4}$.

Then $$|jT|_n=j^x\left(\frac{n}{2}\right)^y
\left(\frac{n}{2}+j\right)^z\left(\frac{n}{2}+2j\right)^w.$$

By \eqref{sum jT=n}, we have $y+z+w\leq 1$. It follows that
$\sigma(|T|_n)\leq x+(\frac{n}{2}+2)\leq
\frac{n}{2}-3+\frac{n}{2}+2<n$, a contradiction.

{\bf Case 2.} $\frac{n}{4}<j< \frac{n}{2}$.

Then $$|jT|_n=j^x\left(\frac{n}{2}\right)^y
\left(\frac{n}{2}+j\right)^z\left(2j-\frac{n}{2}\right)^w.$$

By \eqref{sum jT=n}, we have $x\leq 3$ and $z\leq 1$. Then,
$x+z+2w\leq 3+1+2\times(\lfloor \frac{n}{4}\rfloor-2)<\frac{n}{2}$.
Clearly, $x+z+2w>0$. From \eqref{sum T equvalent 0 mod n} we derive
that $x+z+2w \equiv 0 \pmod {\frac{n}{2}}$, a contradiction.

{\bf Case 3.} $\frac{n}{2}<j< \frac{3n}{4}$.

Then $$|jT|_n=j^x\left(\frac{n}{2}\right)^y
\left(j-\frac{n}{2}\right)^z\left(2j-\frac{n}{2}\right)^w.$$

By \eqref{sum jT=n}, we have $x+y+w\leq 1$. We claim
\begin{equation}\label{x+y+w=1}x+y+w=1.\end{equation} Otherwise, $x=y=w=0$ and $\sigma(T)=z+\frac{n}{2}z \not \equiv 0 \pmod
{\frac{n}{2}}$,  a contradiction with $\sigma(T)\equiv 0\pmod n$.

Note that $0<x+z+2w<n$. By \eqref{sum T equvalent 0 mod n}, we have
that
\begin{equation}\label{x+z+2w=n/2}
x+z+2w=\frac{n}{2}
\end{equation}
and
\begin{equation}\label{y+z+w equiv 1 mod 2}
y+z+w\equiv 1\pmod 2.
\end{equation}
By \eqref{x+y+w=1} and \eqref{x+z+2w=n/2}, we have $y+z+w\equiv
z+w-y=\frac{n}{2}-1\equiv 0\pmod 2$, a contradiction to \eqref{y+z+w
equiv 1 mod 2}.

{\bf Case 4.} $\frac{3n}{4}<j<n$.

Then $$|jT|_n=j^x\left(\frac{n}{2}\right)^y
\left(j-\frac{n}{2}\right)^z\left(2j-\frac{3n}{2}\right)^w.$$

By \eqref{sum jT=n}, we have $x\leq 1$ and $z\leq 3$. Then,
$x+z+2w\leq1+3+2\times(\lfloor \frac{n}{4}\rfloor-2)<\frac{n}{2}$.
Clearly, $x+z+2w>0$. From \eqref{sum T equvalent 0 mod n}, we derive
a contradiction. \qfd



\section{Proof of Theorem \ref{integer}}

\begin{lemma}\label{Lemma of Alon}\cite{Alon} Let $S$ be a sequence of $n$ elements in $Z_n$. Then S
contains a zero-sum subsequence $T$ with length $|T|\in [1,h(S)]$.
\end{lemma}

\begin{lemma}\label{Lemma of Pon}\cite{Pon} Let $S$ be a minimal zero-sum sequence of elements in $Z_n$ with $|S|\in \{1,2,3\}$.
Then $Index(S)=n$.
\end{lemma}



{\sl Proof of Theorem \ref{integer}}. Let $h=h(S)$. The case for
$h<4$ follows from Lemma \ref{Lemma of Alon} and Lemma \ref{Lemma of
Pon}. Thus, we need only to consider the case for $h\geq n/2$.
Assume to the contrary that $S$ contains no subsequence with index
$n$ and with length not exceeding $h$.

Write $S$ in the form $S=a^hb_1\cdot\ldots \cdot b_{n-h}$. If
$o(a)<n$ then $o(a)\leq n/2\leq h$. It follows that $a^{o(a)}$ is a
subsequence of $S$ with index $n$. So, we may assume that $o(a)=n$
and that (up to isomorphism)
$$
S=1^hb_1\cdot\ldots \cdot b_{n-h}.
$$

We show that for every $t\in [1,n-h]$ and any subset $\{i_1, \ldots,
i_t\}\subset [1,n-h]$ the following holds
\begin{equation}\label{basic inequality for h(s) geq n/2}
|b_{i_1}|_n+\cdots+|b_{i_t}|_n\leq n-h+t-1.
\end{equation}

We proceed by induction on $t$. For $t=1$, assume to the contrary
that $|b_i|_n\geq n-h+1$. Then $1^{n-|b_i|_n}b_i$ is a subsequence
of $S$  with index $n$ and with length $n-|b_i|_n+1\leq h$, a
contradiction.

Now assume that \eqref{basic inequality for h(s) geq n/2} is true
for $t=k \in [1,n-h-1]$ and we want to show that \eqref{basic
inequality for h(s) geq n/2} is also true for $t=k+1$. Let
$\tau=|b_{i_1}|_n+\cdots+|b_{i_{k+1}}|_n$. By the induction
hypothesis, $\tau-|b_{i_j}|_n\leq n-h+k-1$ holds for every $j\in
[1,k+1]$. Therefore,

$\tau=\frac{1}{k}(k\tau)=\frac{1}{k}(\sum_{j=1}^{k+1}(\tau-|b_{i_j}|_n)\leq
\frac{(k+1)(n-h+k-1)}{k} \leq n$ (since $k\in [1,n-h-1]$). If $\tau
\geq n-h+k+1$, then $1^{n-\tau} b_{i_1}\cdot\ldots \cdot
b_{i_{k+1}}$ is a subsequence of $S$ with index $n$ and with length
$n-\tau+k+1\leq h$, a contradiction. This proves \eqref{basic
inequality for h(s) geq n/2}. Taking $t=n-h$ in \eqref{basic
inequality for h(s) geq n/2} we have that
$|b_1|_n+\cdots+|b_t|_n\leq n-h+t-1=2t-1$. This forces that $b_i=1$
for some $i\in [1,n-h]$, a contradiction to $h=h(S)$. This proves
the theorem. \qfd

\medskip
Note that the sequence given in Section 2 has maximal repetition
$\frac{n}{2}-1$. So, the restriction that $h(S)\geq \frac{n}{2}$ in
Theorem \ref{integer}  is necessary for $n\equiv 2 \pmod 4$.



\section{ Proof of Theorem \ref{prime2}}
 Throughout this section, let $p>24138$ be a prime and let $S$ be a
sequence of $p$ elements in $Z_p\setminus \{0\}$. Up to isomorphism,
we may assume that $S$ contains 1 (note  that 1 does not necessarily
have the maximal repetition value).

Let $M=M(S)$ be the maximal integer $t$ such that
$\sum_{Index}(T)=[1,t]$ holds for some subsequence $T|S$. Let
$$
m(S)=\max_{(r,p)=1}\{M(|rS|_p)\}
$$
where $r$ runs over all positive integers in $[1,p-1]$. For
convenience, in the rest of this section, we shall always assume
that
$$m(S)=M(S).$$

\begin{lemma}\label{Lemma of x geq M+2} Let $p$ be a prime, and let $S$ be a sequence of $p$ elements in
$Z_p\setminus \{0\}$ with $1\in S$. Put $M=M(S)$. Let $T$ be a
subsequence of $S$ such that $\sum_{Index}(T)=[1,M]$. Then, $|T|\leq
M$ and every term $x$ of $ST^{-1}$ satisfies $|x|_p\geq M+2.$
Furthermore, if $M= p$ or there is one  term $y$ of $ST^{-1}$ with
$|y|_p\geq p-M$ then $S$ contains a subsequence with index $p$.
\end{lemma}

\proof  $|T|\leq |T|_p=M.$ If there is some $x$ of $ST^{-1}$
satisfies $|x|_p\leq M+1$, clearly, $\sum_{Index}(xT)=[1,M+|x|_p]$,
a contradiction on the maximality of $M$. The second part of this
lemma is clear. \qfd


 Let $k\geq 2$ be a positive integer, and let
$F[\frac{1}{k}, \frac{k-1}{k}]$ be all irreducible fractions between
$\frac{1}{k}$ and  $\frac{k-1}{k}$ and with denominators in $[2,k]$,
i.e.
$$
F\left [\frac{1}{k}, \frac{k-1}{k} \right ]=\left \{\frac{a}{b}:
(a,b)=1, \frac{1}{k}\leq \frac{a}{b}\leq \frac{k-1}{k}, 2\leq b \leq
k \right \}.
$$


\begin{lemma}\label{Lemma for two adjacet fractions} Let $\frac{a}{b}$ and $\frac{c}{d}$ be two adjacent fractions in
$F[\frac{1}{k}, \frac{k-1}{k}]$. If $\frac{a}{b}<\frac{c}{d}$ then,
(i) $b+d\geq k+1$ and (ii) $bc-ad=1$.
\end{lemma}

\proof  (i) Note that $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$.
Since $\frac{a}{b}$ and $\frac{c}{d}$ are adjacent it follows that
the irreducible fraction with value $\frac{a+c}{b+d}$ is not in
$F[\frac{1}{k}, \frac{k-1}{k}]$. This forces that $b+d\geq k+1$.

(ii) Since $(a,b)=1$, there are two integers $u$ and $v$ such that
$bu+av=1$. Note that $b(u+ma)+a(v-mb)=1$ holds for any integer $m$.
Let $x=u+ma$ and $y=mb-v$. Then, $bx-ay=1$. By choosing $m$ suitably
we may assume that $y\leq k$ and $y+b\geq k+1$. It follows that
$y\geq k+1-b>0$ and $x>0$. From $bx-ay=1$ we get
$$
\frac{x}{y}-\frac{a}{b}=\frac{1}{by}.
$$
If $y>1$ then $\frac{x}{y}$ is a fraction in $F[\frac{1}{k},
\frac{k-1}{k}]$. So, either $\frac{c}{d}=\frac{x}{y}$ and we are
done, or $\frac{c}{d}<\frac{x}{y}$. For the latter case we have

$\frac{1}{by}=\frac{x}{y}-\frac{a}{b}=(\frac{x}{y}-\frac{c}{d})+(\frac{c}{d}-\frac{a}{b})=
\frac{b(dx-cy)+y(cb-ad)}{byd}\geq \frac{b+y}{byd}$. This implies
that $d\geq b+y\geq k+1$, a contradiction.

Now assume that $y=1$ and we must have $b=k$. It follows from
$bx-ay=1$ that $a=kx-1$. Therefore, $x=1$ and $a=k-1$. So,
$\frac{a}{b}=\frac{k-1}{k}$ is the biggest fraction in
$F[\frac{1}{k}, \frac{k-1}{k}]$, a contradiction. This completes the
proof. \qfd

\begin{lemma}\label{Lemma for S_i} Let $S$ be a sequence $p$ elements in
$Z_p\setminus\{0\}$ containing no subsequence with index $p$, and
let $k=\lfloor\frac{p}{M}\rfloor$. Let us arrange all fractions in
$F[\frac{1}{k}, \frac{k-1}{k}]$ increasingly. Let $\frac{a_i}{b_i}$
be the i-th fraction in $F[\frac{1}{k}, \frac{k-1}{k}]$ and let
$f=|F[\frac{1}{k}, \frac{k-1}{k}]|$. Let $S_1=S\cap [1,M]$,
$S_2=[M+2, \frac{p-1}{b_1}]$. For every $i\in [1,f]$, let
$S_{2i+1}=S\cap [\frac{a_ip+1}{b_i}, \frac{a_ip+M}{b_i}]$ and let
$S_{2i+2}=S\cap [\frac{a_ip+M+1}{b_i}, \frac{a_{i+1}p-1}{b_{i+1}}]$.
Then, $S=\prod_{j=1}^{2f+1}S_j$.
\end{lemma}

\begin{lemma}\label{Lemma the length of S} Let $S$ be a sequence $p$ elements in
$Z_p\setminus\{0\}$ containing no subsequence with index $p$, and
let $f$, $S_i$ be defined as in Lemma \ref{Lemma for S_i}. Suppose
$4\leq M\leq \frac{p-3}{2}$ and
$\max\{\frac{p-M-2}{M},\frac{p-M}{M+1}\}\leq k\leq \frac{p+1}{M}$.
Then $|S_{2i+2}|\leq b_{i+1}-1$ for every $i\in [0,f-1]$.
Furthermore, for every $i\in [2,k]$, define $R_i=\{x|S, 1\leq
|ix|_p\leq M, (|ix|_p,i)=1\}$. Then we have,

$p=|S| \leq M+\sum_{i=2}^k\sum_{(j,i)=1, 1\leq j\leq
i-1}(i-1)+\sum_{i=2}^k|R_i|.$
\end{lemma}

\proof  If $i=0$ then  $S_2=S\cap [M+2, \frac{p-1}{b_1}]$ and
$b_1=k$. If $|S_2|\geq b_1=k$ then we can take a $k$-term
subsequence $U$ of $S_2$. Note that $p-1\geq |U|_p\geq k(M+2)\geq
p-M$ and one can find a subsequence $V$ of $S_1$ such that $UV$ has
index $p$, a contradiction. Now assume that $i\geq 1$. Now
$S_{2i+2}=S\cap [\frac{a_ip+M+1}{b_i}, \frac{a_{i+1}p-1}{b_{i+1}}]$.
If $|S_{2i+2}|\geq b_{i+1}$, take an arbitrary $b_{i+1}$-term
subsequence $X$ of $S_{2i+2}$. Consider $|b_iS|_p$. It follows from
Lemma \ref{Lemma for two adjacet fractions} that
$a_{i+1}b_i-a_ib_{i+1}=1$, and so
$b_i(\frac{a_{i+1}p-1}{b_{i+1}})-a_ip=\frac{p-b_i}{b_{i+1}}$.
Therefore, every term $x$ of $|b_iS_{2i+2}|_p$ is in $[M+1,
\frac{p-b_i}{b_{i+1}}]$ and $x\equiv -a_ip \pmod {b_i}$. Therefore,
$p-b_i\geq |b_iX|_p\geq b_{i+1}(M+1)\geq p-b_iM$ (since by Lemma
\ref{Lemma for two adjacet fractions}, $b_i+b_{i+1}\geq k+1$) and
$|b_iX|_p\equiv -b_{i+1}a_ip=(1-a_{i+1}b_i)p\equiv p \pmod {b_i}$.
Therefore, there is a subsequence $Y$ of $S_1$ such that
$|b_i(XY)|_p=p$, a contradiction. This proves the first part of this
lemma. For every $\ell \in [2,k]$, clearly,
$R_{\ell}=\prod_{b_i=\ell}S_{2i+1}$. Therefore,
$S=S_1\prod_{i=0}^{f-1}S_{2i+2}\prod_{\ell=2}^kR_{\ell}$ is a
disjoint decomposition of $S$ into subsequences. Now the second part
follows from the first part. \qfd

\begin{lemma}\label{Lemma n intergers addition} Let $n\in \mathbb{N}_{\geq 2}$, and let
$a_1,a_2,\ldots,a_n$ (not necessarily distinct) be integers coprime
to $n$. For any integer $m$, there exists a subset $\emptyset\neq
I\subseteq [1,n]$ such that $$\sum\limits_{i\in I} a_i\equiv m\pmod
n.$$ Moreover, if $m\not\equiv 0\pmod n$, we can choose such $I$
with $\emptyset\neq I\subseteq [1,n-1]$.
\end{lemma}

\proof We consider $a_1,\ldots,a_n$ as the elements in group $Z_n$.
It suffices to prove that $\{a_1,0\}+\cdots+\{a_{n-1},0\}=Z_n$. Let
$H=St(\{a_1,0\}+\cdots+\{a_{n-1},0\})$. If $|H|=n$, then
$|\{a_1,0\}+\cdots+\{a_{n-1},0\}|=n$ and the lemma follows. Now
assume $|H|<n$. By Kenser's Theorem, we have that
$|\{a_1,0\}+\cdots+\{a_{n-1},0\}|\geq
\sum\limits_{i=1}^{n-1}|\{a_i,0\}+H|-(n-2)|H|=(n-1)\times
2|H|-(n-2)|H|\geq n$, a contradiction. \qfd



\begin{lemma}\label{Lemma of R_t and R_l} Let $p, S, M,k, R_i$ be defined as in Lemma \ref{Lemma for S_i}, and let $2\leq
t<\ell< k$ with $d=\gcd(t,\ell)<t$. Let $u\in[2, M]$ be an integer.
If $\frac{(t-d)p-\ell}{t\ell}\leq M\leq\frac{dp}{\ell}-t(u-1)$, then
either $|R_t|=0$ or $|R_\ell|\leq \frac{p-\ell M-2\ell
+1}{u}+2\ell-1$.
\end{lemma}

\proof Suppose $|R_t|>0$. Take an arbitrary term $x$ of $R_t$. By
the definition of $R_t$ we get

$$\ell| x|_p\in \bigcup\limits_{\gcd(i,t)=1, 0<i<t}[\frac{\ell
ip+\ell}{t},\frac{\ell ip+\ell M}{t}],$$ and thus,
$$\begin{array}{llll}|\ell x|_p\in\bigcup\limits_{d|i, 0<i<t}[\frac{ ip+\ell}{t},\frac{ ip+\ell M}{t}]\\
\ \ \ \ \ \ \ \ \subseteq [\frac{ dp+\ell}{t},\frac{ (t-d)p+\ell
M}{t}]\subseteq [p-\ell M, p-\ell(u-1)].
\end{array}$$

If $||\ell R_{\ell}|_p\cap [1,u-1]|\geq \ell$, by Lemma \ref{Lemma n
intergers addition} and the definition of $R_t$, we may choose a
subsequence $W$ of $R_{\ell}$ of length at most $\ell$ such that
$|\ell W|_p\subseteq [1,u-1]$ and $|\ell x|_p+\sigma(|\ell
W|_p)\equiv p\pmod \ell$. Since $\sigma(|\ell W|_p)\leq \ell(u-1)$,
we have $|\ell x|_p+\sigma(|\ell W|_p)\in [p-\ell M,p]$. Thus, we
can construct a subsequence of $xWS_1$ of index $p$, a
contradiction. Therefore,
\begin{equation}\label{Lemma 2.6 for u-1}
||\ell R_{\ell}|_p\cap [1,u-1]|\leq \ell-1.
\end{equation}

If $|R_{\ell}|<\ell$ then we are done. Otherwise, by Lemma
\ref{Lemma n intergers addition}, we get a subsequence $R_0$ of
$R_{\ell}$ with $\sigma(|\ell R_0|_p)\equiv p\pmod \ell$ and
\begin{equation}\label{Lemma 2.6 relation} |R_0|\geq |R_{\ell}|-\ell.
\end{equation}

 We show that
\begin{equation}\label{Lemma 2.6 for R_0}
\sigma(|\ell R_0|_p)\leq p-\ell M-\ell.
\end{equation} Otherwise,
$\sigma(|\ell R_0|_p)\geq p-\ell M$, choose $T$ to be the minimal
subsequence of $R_0$ such that $\sigma(|\ell T|_p)\geq p-\ell M$ and
$\sigma(|\ell T|_p)\equiv p\pmod \ell$. If $\sigma(|\ell T|_p)\leq
p$, then we can construct a subsequence of $TS_1$ with index $p$, a
contradiction. If $\sigma(|\ell T|_p)>p$, note that every term $y$
of $R_{\ell}$ satisfies $1\leq |\ell y|_p\leq M$ and $\gcd\{|\ell
y|_p,\ell\}=1$. By Lemma \ref{Lemma n intergers addition}, by
dropping at most $\ell$ terms from $T$, we get a proper subsequence
$\tilde{T}$ such that $\sigma(|\ell \tilde{T}|_p)\geq p-\ell M$ and
$\sigma(|\ell \tilde{T}|_p)\equiv p\pmod \ell$, a contradiction with
the minimality of $T$. Therefore, $\sigma(|\ell R_0|_p)\leq p-\ell
M-\ell.$ By \eqref{Lemma 2.6 for u-1}, we have that $\sigma(|\ell
R_0|_p)\geq 1\times (\ell-1)+u\times (|R_0|-\ell+1)$. By
\eqref{Lemma 2.6 for R_0}, then $|R_0|\leq \frac{p-\ell M-2\ell
+1}{u}+\ell-1$. Now the lemma follows from \eqref{Lemma 2.6
relation}. \qfd

\begin{lemma}\label{Lemma the estimate by the intergers sequence} Let $p, S, M, k, R_i$ be defined as
Lemma \ref{Lemma for S_i}. For any $t\in [2,k]$, let
$1=a_1<a_2<a_3<\cdots$ be all positive integers coprime to $t$. If
$M\leq \frac{p-2t+w a_{u+1}+2}{t+\sum\limits_{i=2}^u a_i}$ for some
$w,u\in \mathbb{N}_0$, then $|R_t|\leq \frac{p-(t+
\sum\limits_{i=2}^u a_i)M-2t+2}{a_{u+1}}+ \delta_u(u-1)M+2t+w$,
where $\delta_u=0$ for $u=0$ and $\delta_u=1$ for $u\geq 1$.
\end{lemma}

\proof Suppose $|R_t|>\frac{p-(t+ \sum\limits_{i=2}^u
a_i)M-2t+2}{a_{u+1}}+\delta_u(u-1)M+2t+w$. It follows from $M\leq
\frac{p-2t+w a_{u+1}+2}{t+\sum\limits_{i=2}^u a_i}$ that $|R_t|\geq
2t+1$. By Lemma \ref{Lemma n intergers addition}, there exists a
nonempty subsequence $R_0$ of $R_t$ with
$$\sigma(|tR_0|_p)\equiv p\pmod t$$ and
\begin{equation}\label{Lemma 2.7 relation}
|R_0|\geq |R_t|-t. \end{equation}

Similarly to Lemma \ref{Lemma of R_t and R_l}, we can prove that
\begin{equation}\label{Lemma 2.7 for R_0} \sigma(|tR_0|_p)\leq p-tM-t.
\end{equation}

Note that $tR_0$ contains $a_1=1$ at most $t-2$ times, otherwise,
$m(S)\geq M(tS)\geq tM+t-1>M$, a contradiction with $m(S)=M(S)$.
Since $v_{a_i}(S)\leq h(S)\leq M$ for all $i\geq 2$, it follows that
$\sigma(|tR_0|_p)\geq 1\times (t-2)+  \sum\limits_{i=2}^u a_i\times
M +a_{u+1}\times(|R_0|-(u-1)M-(t-2))$. By \eqref{Lemma 2.7 for R_0},
we have $|R_0|\leq \frac{p-(t+
\sum\limits_{i=2}^u)M-2t+2}{a_{u+1}}+\delta(u-1)M+t-2$. By
\eqref{Lemma 2.7 relation}, we derive a contradiction. \qfd

\medskip
Now we are in a position to prove Theorem \ref{prime2}.

%In fact, we shall prove the following slightly stronger result than
%Theorem \ref{prime2}.

{\sl Proof of Theorem \ref{prime2}}. Assume to the contrary that $S$
contains no subsequence with index $p$. Clearly, $M\leq p-1$. By
Lemma \ref{Lemma of x geq M+2}, every term $x$ of $ST^{-1}$
satisfies
$$M+2\leq |x|_p\leq p-M-1.$$
This gives
$$
M\leq \frac{p-3}{2}.
$$
We distinguish several cases.

{\bf Case 1.} $\frac{p-2}{3}\leq M\leq \frac{p-3}{2}$.

With $k=2$ in Lemma \ref{Lemma the length of S}, we have
$$
p\leq M+1+|R_2|.
$$

Applying Lemma \ref{Lemma the estimate by the intergers sequence}
with $u=0$ and $w=6$, we have

$$|R_2|\leq p-2M+8.$$

It follows that $p\leq M+1+|R_2|=M+1+p-2M+8<p,$ a contradiction.

{\bf Case 2.} $\frac{p+3}{4}\leq M\leq \frac{p-4}{3}$.


With $k=3$ in Lemma \ref{Lemma the length of S}, we have

$p\leq M+1+2+2+|R_2|+|R_3|$.

Applying Lemma \ref{Lemma the estimate by the intergers sequence}
with $u=1$ and $w=6$, we have

$$|R_2|\leq \frac{p-2M+28}{3},$$
$$|R_3|\leq \frac{p-3M+20}{2}.$$
It follows that $p\leq
M+5+\sum_{i=2}^3|R_i|=M+5+\frac{p-2M+28}{3}+\frac{p-3M+20}{2}<p,$ a
contradiction.

{\bf Case 3.}  $\frac{p-2}{5}\leq M\leq \frac{p+1}{4}$.

With $k=4$ in Lemma \ref{Lemma the length of S}, we have

$p\leq M+1+2\times 2+3\times 2+|R_2|+|R_3|+|R_4|.$

Applying Lemma \ref{Lemma the estimate by the intergers sequence}
with $u=1$ and $w=6$, we have


$$|R_2|\leq \frac{p-2M+28}{3},$$
$$|R_3|\leq \frac{p-3M+20}{2},$$
$$|R_4|\leq \frac{p-4M+36}{3}.$$ It follows that
$p\leq M+11+\frac{p-2M+28}{3}+\frac{p-3M+20}{2}+\frac{p-4M+36}{3}<
p$, a contradiction.

\medskip
{\bf Case 4.} $\frac{p-1}{6}\leq M\leq \frac{p-3}{5}$.

With $k=5$ in Lemma \ref{Lemma the length of S}, we have

$p\leq M+27+\sum_{i=2}^5|R_i|$.

Applying Lemma \ref{Lemma the estimate by the intergers sequence}
with $u=1$ and $w=6$, we have

$$|R_2|\leq \frac{p-2M+28}{3},$$
$$|R_3|\leq \frac{p-3M+20}{2},$$
$$|R_4|\leq \frac{p-4M+36}{3},$$
$$|R_5|\leq \frac{p-5M+24}{2}.$$

Applying Lemma \ref{Lemma of R_t and R_l} with $t=2,\ell=3$ and
$u=12$, we have that either $|R_2|=0$, or $|R_3|\leq
\frac{p-3M+55}{12}$. It follows that $|R_2|+|R_3|\leq
\max\{\frac{p-2M+28}{3}+\frac{p-3M+55}{12},\frac{p-3M+20}{2}\}=\frac{5p-11M+167}{12}$.

It follows that $p\leq
M+27+\sum_{i=2}^5|R_i|=M+27+(|R_2|+|R_3|)+|R_4|+|R_5|\leq
\frac{5p-11M+167}{12}+\frac{p-4M+36}{3}+\frac{p-5M+24}{2}+27<p,$ a
contradiction.

\medskip
{\bf Case 5.} $\frac{p-5}{7}\leq M\leq \frac{p-5}{6}$.

With $k=6$ in Lemma \ref{Lemma the length of S}, we have

$p\leq M+37+\sum_{i=2}^6|R_i|$.

Applying Lemma \ref{Lemma the estimate by the intergers sequence}
with $u=2$ and $w=0$, we have
$$|R_2|\leq \frac{p+18}{5},$$
$$|R_3|\leq\frac{p-M+20}{4}.$$

Applying Lemma \ref{Lemma the estimate by the intergers sequence}
with $u=1$ and $w=6$, we have

$$|R_4|\leq \frac{p-4M+36}{3},$$
$$|R_5|\leq \frac{p-5M+24}{2},$$
$$|R_6|\leq \frac{p-6M+80}{5}.$$

It follows that $p\leq
M+37+\sum_{i=2}^6|R_i|=M+37+\frac{p+18}{5}+\frac{p-M+20}{4}+\frac{p-4M+36}{3}+\frac{p-5M+24}{2}+\frac{p-6M+80}{5}<p$,
a contradiction.

\medskip
{\bf Case 6.} $\frac{p-2}{8}\leq M\leq \frac{p-3}{7}$.

With $k=7$ in Lemma \ref{Lemma the length of S}, we have

$p\leq M+73+\sum_{i=2}^7|R_i|$.

Applying Lemma \ref{Lemma the estimate by the intergers sequence}
with $u=2$ and $w=0$, we have
$$|R_2|\leq \frac{p+18}{5},$$
$$|R_3|\leq\frac{p-M+20}{4}.$$

Applying Lemma \ref{Lemma the estimate by the intergers sequence}
with $u=1$ and $w=6$, we have

$$|R_4|\leq \frac{p-4M+36}{3},$$
$$|R_5|\leq \frac{p-5M+24}{2},$$
$$|R_6|\leq \frac{p-6M+80}{5},$$
$$|R_7|\leq \frac{p-7M+28}{2}.$$

By Lemma \ref{Lemma of R_t and R_l}, take $t=2$, $\ell=5$ and
$u=10$, we have $|R_2|+|R_5|
\leq\max\{\frac{p-5M+4}{2},\frac{p+18}{5}+\frac{p-5M-9}{10}+9\}
=\frac{3p-5M+117}{10}$.

It follows that $p\leq
M+73+\sum_{i=2}^7|R_i|=M+73+(|R_2|+|R_5|)+|R_3|+|R_4|+|R_6|+|R_7|\leq
M+73+\frac{3p-5M+117}{10}+\frac{p-M+20}{4}+\frac{p-4M+36}{3}+\frac{p-6M+80}{5}+\frac{p-7M+28}{2}<p$,
a contradiction.



\medskip
{\bf Case 7.} $\frac{p-2}{9}\leq M\leq \frac{p-3}{8}$.

With $k=8$ in Lemma \ref{Lemma the length of S}, we have

$p\leq M+111+\sum_{i=2}^8|R_i|$.

Applying Lemma \ref{Lemma the estimate by the intergers sequence}
with $u=2$ and $w=0$, we have
$$|R_2|\leq \frac{p+18}{5},$$
$$|R_3|\leq\frac{p-M+20}{4},$$
$$|R_4|\leq\frac{p-2M+34}{5},$$
$$|R_5|\leq \frac{p-4M+22}{3}.$$

Applying Lemma \ref{Lemma the estimate by the intergers sequence}
with $u=1$ and $w=6$, we have
$$|R_6|\leq \frac{p-6M+80}{5},$$
$$|R_7|\leq \frac{p-7M+28}{2},$$
$$|R_8|\leq\frac{p-8M+52}{3}.$$

Applying Lemma \ref{Lemma of R_t and R_l} with $t=2$, $\ell=5,7$ and
$u=20$, we can prove that either $|R_2|=0$, or $|R_i|\leq
\frac{p-iM-2i+1}{20}+2i-1$ for both $i=5,7$. It follows that
$|R_2|+|R_5|+|R_7| \leq\max\{
\frac{p-4M+22}{3}+\frac{p-7M+28}{2},\frac{p-M+20}{4}+\frac{p-5M-9}{20}+9+\frac{p-7M-13}{20}+13\}
=\frac{5p-29M+128}{6}$.

Applying Lemma \ref{Lemma of R_t and R_l} with $t=4,\ell=6$ and
$u=10$, we obtain that either $|R_4|=0$, or $|R_6|\leq \frac{p-6
M-11 }{10}+11$. It follows that $|R_4|+|R_6|\leq
\max\{\frac{p-2M+34}{5}+\frac{p-6 M-11
}{10}+11,\frac{p-6M+80}{5}\}=\frac{3p-10M+167}{10}$.


It follows that $p\leq
M+111+\sum_{i=2}^8|R_i|=M+111+(|R_2|+|R_5|+|R_7|)+(|R_4|+|R_6|)+|R_3|+|R_8|\leq
M+111+\frac{5p-29M+128}{6}+\frac{3p-10M+167}{10}+\frac{p-M+20}{4}+\frac{p-8M+52}{3}<p$,
a contradiction.

\medskip
{\bf Case 8.} $\frac{p-2}{10}\leq M\leq \frac{p-4}{9}$.

With $k=9$ in Lemma \ref{Lemma the length of S}, we have

$p\leq M+159+\sum_{i=2}^9|R_i|$.

Applying Lemma \ref{Lemma the estimate by the intergers sequence}
with $u=2$ and $w=0$, we have
$$|R_2|\leq \frac{p+18}{5},$$
$$|R_3|\leq\frac{p-M+20}{4},$$
$$|R_4|\leq\frac{p-2M+34}{5},$$
$$|R_5|\leq \frac{p-4M+22}{3}.$$

Applying \ref{Lemma the estimate by the intergers sequence} with
$u=1$ and $w=6$, we have
$$|R_6|\leq \frac{p-6M+80}{5},$$
$$|R_7|\leq \frac{p-7M+28}{2},$$
$$|R_8|\leq\frac{p-8M+52}{3},$$
$$|R_9|\leq \frac{p-9M+32}{2}.$$

Applying Lemma \ref{Lemma of R_t and R_l} with $t=2,\ell=5,7$ and
$u=10$, we obtain that either $|R_2|=0$, or $|R_i|\leq \frac{p-i
M-2i +1}{10}+2i-1$ for both $i=5,7$. It follows that
$|R_2|+|R_5|+|R_7|\leq
\max\{\frac{p-4M+22}{3}+\frac{p-7M+28}{2},\frac{p+18}{5}+\frac{p-5
M-9}{10}+9+\frac{p-7 M-13}{10}+13\}=\frac{5p-29M+128}{6}$.

Applying Lemma \ref{Lemma of R_t and R_l} with $t=3,\ell=8$ and
$u=5$, we obtain that either $|R_3|=0$, or $|R_8|\leq \frac{p-8 M-15
}{8}+15$. It follows that $|R_3|+|R_8|\leq
\max\{\frac{p-M+20}{4}+\frac{p-8 M-15
}{8}+15,\frac{p-8M+52}{3}\}=\frac{3p-10M}{8}+20$.

It follows that $p\leq
M+159+\sum_{i=2}^9|R_i|=M+159+(|R_2|+|R_5|+|R_7|)+(|R_3|+|R_8|)+|R_4|+|R_6|+|R_9|\leq
M+159+\frac{5p-29M+128}{6}+(\frac{3p-10M}{8}+20)+\frac{p-2M+34}{5}+\frac{p-6M+80}{5}+\frac{p-9M+32}{2}<p$,
a contradiction. \qfd

\section{The sum of four elements in $Z_p$}

In this section, we shall give a new proof of the following result
due to Li and Plyley \cite{CL}.

\begin{theorem} \label{foursum}
Let $p$ be a prime, and let $S$ be a minimal zero-sum sequence of 4
elements in $Z_p$. Then $Index(S)=p$.
\end{theorem}

For any real numbers $a\leq b$, let $[a,b]$ denote the set of
integers between $a$ and $b$, $|[a,b]|$ denote the number of
integers in $[a,b]$.

Define
$$S_{(p,j)}=\left\{i:|ij|_p<\frac{p}{2},
i\in \left[1, \frac{p}{2}\right]\right\}$$ for every $j\in[1,p-1]$,
where $p$ is a prime.

It is easy to show
\begin{center}
$S_{(p,j)}\cap S_{(p,p-j)}=\emptyset$ and $S_{(p,j)}\cup
S_{(p,p-j)}=\left[1, \frac{p}{2}\right]$
\end{center}
for every $j\in [1,p-1]$.

\begin{observation}\label{observation: S(p,j) and S(p,p-j)}
$|S_{(p,j)}|+|S_{(p,p-j)}|=\frac{p-1}{2}$.
\end{observation}

\begin{lemma}\label{lemma:>p-1/6}
Let $p\geq 19$ be a prime and $j\in [2,p-2]$. Then $|S(p,j)|\geq
\frac{p-1}{6}$ and we have strict inequality  if $j \not\in
\{p-3,\frac{p-1}{3}\}$.
\end{lemma}

\proof Since $S_{(p,2)}=\left[1,\frac{p}{4}\right]$, we have
$|S_{(p,2)}| =\left\lfloor\frac{p}{4}\right\rfloor\geq\frac{p-3}{4}>
\frac{p-1}{6}$ and $|S_{(p,2)}|
=\left\lfloor\frac{p}{4}\right\rfloor\leq\frac{p-1}{4}<\frac{p-1}{3}$,
by Observation \ref{observation: S(p,j) and S(p,p-j)},
$|S_{(p,p-2)}|>\frac{p-1}{2}-\frac{p-1}{3}=\frac{p-1}{6}$.


Since $S_{(p,3)}=\left[1,\frac{p}{6}\right]\cup
\left[\frac{p}{3},\frac{p}{2}\right]$, we have
$|S_{(p,3)}|=\left\lfloor\frac{p}{6}\right\rfloor+\lfloor\frac{p}{2}\rfloor
-\left\lceil\frac{p}{3}\right\rceil+1$. It is easy to verify that
$\frac{p-1}{6}< |S_{(p,3)}|\leq \frac{p-1}{3}$. By Observation
\ref{observation: S(p,j) and S(p,p-j)},
$|S_{(p,p-3)}|=\frac{p-1}{2}-|S_{(p,3)}|\geq \frac{p-1}{6}$.

It remains to consider $|S_{(p,j)}|$ and $|S_{(p,p-j)}|$ for $j\in
[4,\frac{p}{2}]$.

{\bf Case 1.} $4\leq j=2k< \frac{p}{2}$.

It is easy to show
\begin{equation}\label{interval for S(p,2k)}
S_{(p,2k)}=\left[1,\frac{p}{4k}\right]\cup
\left[\frac{2p}{4k},\frac{3p}{4k}\right]\cup\cdots \cup
\left[\frac{(2k-2)p}{4k},\frac{(2k-1)p}{4k}\right]
\end{equation}
\begin{equation}\label{interval for S(p,p-2k)}
S_{(p,p-2k)}=\left[\frac{p}{4k},\frac{2p}{4k}\right]\cup
\left[\frac{3p}{4k},\frac{4p}{4k}\right]\cup\cdots \cup
\left[\frac{(2k-1)p}{4k},\frac{(2k)p}{4k}\right].
\end{equation}

{\bf Subcase 1.1} $k\geq\frac{p-1}{6}$.

Noting that $\frac{p}{4k}>1$, we have each interval contains at
least one integer in \eqref{interval for S(p,2k)} and
\eqref{interval for S(p,p-2k)}. It follows that $
|S_{(p,2k)}|,|S_{(p,p-2k)}|\geq k\geq \frac{p-1}{6}$, and moreover,
equality implies $j=2\frac{p-1}{6}=\frac{p-1}{3}$.

{\bf Subcase 1.2} $\frac{p-1}{6}>k>\frac{p}{8}$.

Then $\frac{p}{4k}<2$. It follows that there are at most two
integers in each interval in \eqref{interval for S(p,2k)} and
\eqref{interval for S(p,p-2k)}. It follows that
$|S_{(p,2k)}|,|S_{(p,p-2k)}|\leq 2k<\frac{p-1}{3}$. By Observation
\ref{observation: S(p,j) and S(p,p-j)}, we have
$|S_{(p,2k)}|,|S_{(p,p-2k)}|>\frac{p-1}{6}$.

{\bf Subcase 1.3} $\frac{p}{8}>k>\frac{p}{12}$.

Since $\frac{p}{4k}>2$, we have that each interval in
\eqref{interval for S(p,2k)} and \eqref{interval for S(p,p-2k)}
contains at least two integers. Then
$|S_{(p,2k)}|,|S_{(p,p-2k)}|\geq 2k> \frac{p-1}{6}$.

{\bf Subcase 1.4} $k<\frac{p}{12}$.

We see that each interval in \eqref{interval for S(p,2k)} and
\eqref{interval for S(p,p-2k)} contains at most
$\lfloor\frac{(i+1)p}{4k}\rfloor-\lceil\frac{ip}{4k}\rceil+1\leq
\frac{(i+1)p-1}{4k}-\frac{ip+1}{4k}+1=\frac{p-2}{4k}+1$ integers,
where $i=0,1,\ldots,2k-1$. It follows that
$|S_{(p,2k)}|,|S_{(p,p-2k)}|<k\times
(\frac{p-2}{4k}+1)<\frac{p-1}{3}$. By Observation \ref{observation:
S(p,j) and S(p,p-j)}, we have
$|S_{(p,2k)}|,|S_{(p,p-2k)}|>\frac{p-1}{6}$.


{\bf Case 2.} $5\leq j=2k+1<\frac{p}{2}$.

Observe that
\begin{equation}\label{interval for S(p,2k+1)}
S_{(p,2k+1)}=\left[1,\frac{p}{2(2k+1)}\right]\cup
\left[\frac{2p}{2(2k+1)},\frac{3p}{2(2k+1)}\right]\cup\cdots \cup
\left[\frac{(2k)p}{2(2k+1)},\frac{(2k+1)p}{2(2k+1)}\right]
\end{equation}
\begin{equation}\label{interval for S(p,p-2k-1)}
S_{(p,p-2k-1)}=\left[\frac{p}{2(2k+1)},\frac{2p}{2(2k+1)}\right]\cup
\left[\frac{3p}{2(2k+1)},\frac{4p}{2(2k+1)}\right]\cup\cdots \cup
\left[\frac{(2k-1)p}{2(2k+1)},\frac{(2k)p}{2(2k+1)}\right].
\end{equation}

{\bf Subcase 2.1} $k\geq\frac{p-1}{6}$.

Noting that $\frac{p}{2(2k+1)}>1$, we have that at least one integer
belongs to each interval in \eqref{interval for S(p,2k+1)} and
\eqref{interval for S(p,p-2k-1)}.

If $k>\frac{p-1}{6}$, then $|S_{(p,2k+1)}|,|S_{(p,p-2k-1)}|\geq
k>\frac{p-1}{6}$. Hence, we may assume $k=\frac{p-1}{6}$.

Note that
\begin{align*}
\left[\frac{(2k-1)p}{2(2k+1)},\frac{(2k)p}{2(2k+1)}\right]&=
\left[\frac{(2k-1)(6k+1)}{2(2k+1)},\frac{(2k)(6k+1)}{2(2k+1)}\right]\\
&=\left[3k-2-\frac{2k-3}{4k+2},3k-\frac{4k}{4k+2}\right],
\end{align*}

Since $k=\frac{p-1}{6}>1$, we have that $3k-2,3k-1\in
\left[\frac{(2k-1)p}{2(2k+1)},\frac{(2k)p}{2(2k+1)}\right]$. It
follows that $|S_{(p,2k+1)}|,|S_{(p,p-2k-1)}|\geq
k+1>\frac{p-1}{6}$.

{\bf Subcase 2.2} $\frac{p-1}{6}>k>\frac{p-4}{8}$.

Since $p$ is a prime, we have $k\leq \frac{p-5}{6}$.

Since $\frac{p}{2(2k+1)}<2$, we have that at most two integers
belong to each interval in \eqref{interval for S(p,2k+1)} and
\eqref{interval for S(p,p-2k-1)}. Moreover, there are just one
integer in $\left[1,\frac{p}{2(2k+1)}\right]$. It follows that
$|S_{(p,2k+1)}|,|S_{(p,p-2k-1)}|\leq 2k+1<\frac{p-1}{3}$.

{\bf Subcase 2.3} $\frac{p-4}{8}>k>\frac{p-1}{12}$.

Since $\frac{p}{2(2k+1)}>2$, we have that at least two integers
belong to each interval in \eqref{interval for S(p,2k+1)} and
\eqref{interval for S(p,p-2k-1)}. Then
$|S_{(p,2k+1)}|,|S_{(p,p-2k-1)}|\geq 2k>\frac{p-1}{6}$.

{\bf Subcase 2.4} $\frac{p-1}{12}\geq k>\frac{p-1}{18}$.


If $k\in \{\frac{p-1}{12},\frac{p-5}{12}\}$, then
$2<\frac{p}{2(2k+1)}<3$, which implies that each interval in
\eqref{interval for S(p,2k+1)} and \eqref{interval for S(p,p-2k-1)}
contains at least two and at most three integers. It follows that
$|S(p,2k+1)|\geq (k+1)\times 2>\frac{p-1}{6}$. Moreover, since
$\left[1,\frac{p}{2(2k+1)}\right]$ contains just two integers, it
follows that $|S(p,2k+1)|\leq 2+3k<\frac{p-1}{3}$. By Observation
\ref{observation: S(p,j) and S(p,p-j)}, we have
$|S(p,p-2k-1)|>\frac{p-1}{6}$.

Therefore, we assume $\frac{p-7}{12}\geq k>\frac{p-1}{18}$. Then
$\frac{p}{2(2k+1)}>3$. It follows that each interval in
\eqref{interval for S(p,2k+1)} and \eqref{interval for S(p,p-2k-1)}
contains at least three integers, which implies $|S_{(p,2k+1)}|,
|S(p,p-2k-1)| \geq 3k>\frac{p-1}{6}$.

{\bf Subcase 2.5} $\frac{p-1}{18}\geq k>1$.





Note that each interval in \eqref{interval for S(p,2k+1)} and
\eqref{interval for S(p,p-2k-1)} contains
$\left\lfloor\frac{(i+1)p}{2(2k+1)}\right\rfloor-\left\lceil\frac{ip}{2(2k+1)}\right\rceil+1\geq
\frac{(i+1)p-2(2k+1)+1}{2(2k+1)}-\frac{ip+2(2k+1)-1}{2(2k+1)}+1=\frac{p-2(2k+1)+2}{2(2k+1)}$
integers, where $i=1,2,\ldots,2k+1$. It follows that
\begin{equation}\label{>p-1/6} |S_{(p,2k+1)}|, |S(p,p-2k-1)| \geq
k\times\frac{p-2(2k+1)+2}{2(2k+1)}.
\end{equation}

For $k=2$, since $k\leq \frac{p-1}{18}$, then $p\geq 37$. If $p>43$,
by \eqref{>p-1/6}, we have $|S_{(p,2k+1)}|, |S(p,p-2k-1)|
>\frac{p-1}{6}$. If $p\in \{37,41,43\}$, it is easy to verify the conclusion.

Therefore, we may assume that $k\geq 3$. It follows from
\eqref{>p-1/6} that \begin{align*} |S_{(p,2k+1)}|, |S(p,p-2k-1)|
>\frac{p-1}{6}.
\end{align*}
This completes the proof. \qfd

\bigskip



{\sl Proof of Theorem \ref{foursum}}. For $p<19$, it is easy to
check the theorem directly. So, we may assume $p\geq 19$. Let
$S=a_1\cdot a_2\cdot a_3\cdot a_4$, where $1=a_1\leq a_2\leq a_3\leq
a_4<p$.

If $h(S)\geq 2$, say $a_1=a_2=1$, then
$a_1+a_2+a_3+a_4<1+1+(p-2)+(p-2)<2p$, which implies that
$a_1+a_2+a_3+a_4=p$. Hence, we assume $$h(S)=1.$$

If $S_{(p,a_2)}\cap S_{(p,a_3)}\cap S_{(p,a_4)}\neq \emptyset$,
i.e., there exists $r\in S_{(p,a_2)}\cap S_{(p,a_3)}\cap
S_{(p,a_4)}$, then $\sigma(|rS|_p)=|r a_1|_p+|r a_2|_p+|r a_3|_p+|r
a_4|_p<4\times \frac{p}{2}=2p$, which implies $\sigma(|rS|_p)=p$.
Hence, we assume
\begin{equation}\label{cap=emptyset}
S_{(p,a_2)}\cap S_{(p,a_3)}\cap S_{(p,a_4)}=\emptyset.
\end{equation}

Since $h(S)=1$, we have that at least one element of
$\{a_2,a_3,a_4\}$ doesn't belong to $\{p-3,\frac{p-1}{3}\}$. It
follows from Lemma \ref{lemma:>p-1/6} that
\begin{equation}\label{three sets cardinality}
|S_{(p,a_2)}|+|S_{(p,a_3)}|+
|S_{(p,a_4)}|>3\times\frac{p-1}{6}=\frac{p-1}{2}.
\end{equation}
Since $\frac{p-1}{2} \geq |S_{(p,a_2)}\cup S_{(p,a_3)}\cup
S_{(p,a_4)}| =|S_{(p,a_2)}|+|S_{(p,a_3)}|+ |S_{(p,a_4)}|
-(|S_{(p,a_2)}\cap S_{(p,a_3)}|+|S_{(p,a_3)}\cap S_{(p,a_4)}|+
|S_{(p,a_2)}\cap S_{(p,a_4)}|)+|S_{(p,a_2)}\cap S_{(p,a_3)}\cap
S_{(p,a_4)}|$, it follows from \eqref{cap=emptyset} and \eqref{three
sets cardinality} that $|S_{(p,a_2)}\cap
S_{(p,a_3)}|+|S_{(p,a_3)}\cap S_{(p,a_4)}|+ |S_{(p,a_2)}\cap
S_{(p,a_4)}|>0$, say $r\in S_{(p,a_2)}\cap S_{(p,a_3)}$. By
\eqref{cap=emptyset}, $r\notin S_{(p,a_4)}$.

For $i=1,2,3$, since $|ra_i|_p<\frac{p}{2}$, it follows that
$|2ra_i|_p=2|ra_i|_p$. Since $|ra_4|_p>\frac{p}{2}$, it follows that
$|2ra_4|_p=2|ra_4|_p-p$. Then
$\sigma(|2rS|_p)=|2ra_1|_p+|2ra_2|_p+|2ra_3|_p+|2ra_4|_p=2|ra_1|_p+2|ra_2|_p+2|ra_3|_p+2|ra_4|_p-p$
is odd. It follows that $\sigma(|2rS|_p)\in \{p,3p\}$. If
$\sigma(|2rS|_p)=p$, we are done. If $\sigma(|2rS|_p)=3p$, then
$\sigma(|(p-2r)S|_p)=(p-|2ra_1|_p)+(p-|2ra_2|_p)+(p-|2ra_3|_p)+(p-|2ra_4|_p)=p$.
This completes the proof of the theorem. \qfd



\section{Remarks and Open Problems}
It seems plausible to make the following generalization of
Conjecture 1.2 when $n=p$ is a prime.

\begin{conj} If $S$ is a sequence of $p$ elements in $Z_p$ then $S$
contains a subsequence with index $p$ and with  length not exceeding
$h(S)$.
\end{conj}

Let $t(n)$ be the smallest integer $t$ such that every sequence of
$t$ elements in $Z_n$ contains a subsequence of index $n$. From
Theorem \ref{counterexample} we know that $t(n)\geq n+\lfloor
\frac{n}{4} \rfloor -4$ for $n=4k+2\geq 22$.

{\bf Open Problem 1} {\sl To determine $t(n)$ for all positive
integers $n$.}



Let $T(n)$ be the smallest integer $t$ such that every subset of $t$
distinct elements in $Z_n$ contains a subset of index $n$.

{\bf Open Problem 2} {\sl To determine $T(n)$ for all positive
integers $n$.}

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\end{document}

0909.2461.extracted_bib
\begin{thebibliography}{99}

\bibitem{Alon} N. Alon, \emph{Subset sums}, J. Number Theory,
27(1987) 196-205.

\bibitem{CFS} S.T. Chapman, M. Freeze and W.W. Smith, \emph{Minimal zero-sequences and the strong Davenport constant},
Discrete Math., 203(1999) 271-277.

\bibitem{Gao1} W.D. Gao, \emph{Zero sums in finite cyclic groups},
Integers, 0(2000) A12.

\bibitem{LK} P. Lemke and D.J. Kleitman, \emph{An addition theorem on the integers modulo
n}, J. Number Theory, 31(1989) 335-345.

\bibitem{CL}  Y.L. Li and C. Plyley, \emph{four sum}, Preprint, 2008.

\bibitem{Pon} V. Ponomarenko, \emph{Minimal zero sequences of finite
cyclic groups}, Integers, 4(2004) A24.

\bibitem{SC} S. Savchev and F. Chen, \emph{Long zero-free sequences in finite cyclic groups}, Discrete
Mathematics, 307(2007) 2671-2679.

\bibitem{Yuan} P.Z. Yuan, \emph{On the index of minimal zero-sum sequences over finite cyclic
groups}, J. Combinatorial Theory, A, 114(2007) 1545-1551.
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