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\title{\bf A Generalization of Graham's Conjecture}

% input author, affilliation, address and support information as follows;
% the address should include the country, and does not have to include
% the street address

\author{Huanhuan Guan\thanks{Supported by  NSFC of
     China (Nos.11326053, 11301556, 11271142), the Science
and Technology Foundation of Guizhou Provice (No. [2013]2087) 
and the Guangdong Provincial Natural Science Foundation (Nos. S2012010009942, S2013040013796).
}\\
\small School of Mathematic and Statistics\\[-0.8ex]
\small Guizhou University of Finance and Economics\\[-0.8ex]
\small Guiyang 550025, P.R.China\\
\small\tt guan1110h@163.com\\
\and
Pingzhi Yuan\\
\small School of Mathematics\\[-0.8ex]
\small South China Normal University\\[-0.8ex]
\small Guangzhou 510631, P.R. China\\
\small\tt yuanpz@scnu.edu.cn\\
\and
Xiangneng Zeng\\
\small Sino-French Institute of Nuclear Engineering and Technology\\[-0.8ex]
\small Sun Yat-Sen University\\[-0.8ex]
\small Guangzhou 510275, P.R. China\\
\small\tt junevab@163.com
}

% \date{\dateline{submission date}{acceptance date}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{Apr 15, 2015}{Sep 29, 2015}{Oct 16, 2015}\\
\small Mathematics Subject Classifications: 11B50, 11B75}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
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\begin{abstract}
    Let $G$ be a finite Abelian group of order
    $|G|=n$, and let $S=g_1\cdot\ldots\cdot g_{n-1}$ be a sequence over $G$
    such that  all nonempty zero-sum subsequences
    of $S$ have the same length. In this paper, we completely determine
    the structure of these sequences.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} Graham's
  conjecture; zero-sum subsequence; support of sequence
\end{abstract}

\section{Introduction}
Let $G$ be a finite Abelian group (written additively), and let $\mathcal {F}(G)$ denote
the free Abelian monoid with basis $G$, the elements of which are called \emph{sequences} (over $G$).
A sequence of not necessarily distinct elements from $G$ will be written in the
form $$S=g_1\cdot\, \ldots \,\cdot g_l=\prod_{i=1}^l g_i=\prod_{g\in
G}g^{\mathsf v_g(S)}\in\mathcal{F}(G),$$ where $g_i\in G$ are the terms of $S$ and $\mathsf v_g(S) \geq
0$ is called the $multiplicity$ of $g$ in $S$. Denote by $|S|= l$
the number of terms in $S$ (called the $length$ of $S$) and let
$\supp(S) = \{g\in G: \, \mathsf v_g(S)>0\}$ be the $support$ of
$S$.

We say that $S$ contains some $g\in G$ if $\mathsf v_g(S)\geq 1$, and
a sequence $T\in\mathcal{F}(G)$ is a $subsequence$ of $S$ if
$\mathsf v_g(T) \leq \mathsf v_g(S)$ for every $g\in  G$, denoted
$T|S$. If $T|S$, then let $ST^{-1}$ denote the sequence obtained by
deleting the terms of $T$ from $S$. Furthermore, we denote by $\sigma(S)$ the sum
of $S$, i.e., $\sigma(S)=\sum_{i=1}^l{g_i}=\sum_{g\in G}{\mathsf{v}_{g}(G)g}\in G$. We define
$$\sum(S)=\{\sigma(T): T\ \mbox{is a nonempty subsequence of $S$}\},$$  the set  of subsums
of $S$,  $$\sum_{k}(S)=\left\{\sum_{i\in I}g_i\mid I\subseteq [1, |S|]\ \mbox{with}\ |I|=k\right\},$$ the
 set of $k-$term subsums of $S$, and for all $k\in \mathbb{N}$, $$\sum_{\le k}(S)=
 \bigcup_{j\in [1, k]}\sum_j(S)\  \mbox{and}\ \sum_{\ge k}(S)=\bigcup_{j\ge k}\sum_j(S).$$

Let $S$ be a sequence in $G$. We define $S$ a zero-sum sequence if
$\sigma(S) =
 0$, a zero-sum free sequence if  $0\not\in \sum(S)$,  and a minimal zero-sum sequence if $\sigma(S)=0$
 and $\sigma(T)\neq 0$ for any proper and nontrivial subsequence $T$ of $S$.

For a sequence $S$ over $G$, we define
$$h(S) =\max\{\mathsf v_g(S)\mid g \in  G\}\in [0, |S|],
\mbox{  the\  maximum\ of\ the\ multiplicities\ of\ } S.$$

Let $C_n$ denote the cyclic group of order $n$, where $n\in \mathbb{N}$, and $G$ be a
finite Abelian group (written additively) with $|G|>1$. By the Structure Theorem of
Abelian Groups, we have that $G\cong C_{n_1}\oplus \ldots \oplus C_{n_r}$, where
$1< n_1\mid \ldots \mid n_r \in \mathbb{N}$, $r=r(G)$ is the rank of $G$, and
$n_r=\exp(G)$ is the exponent of $G$. If $n_1=\ldots =n_r$, we denote $G=C_n^r$.

Here and henceforth, $n$ is a fixed integer greater than $1$, and the cyclic group
of order $n$ is identified with the additive group $\mathbb{Z}/n\mathbb{Z}$ of
the integers modulo $n$.


Graham \cite{ES{76}} stated the following conjecture.
\begin{conjecture}\label{c1} Let $ p$ be a prime and $S$ be a sequence
over $\mathbb{Z}/p\mathbb{Z}$ of length $p$.
If all nontrivial zero-sum subsequences of $S$ is of the same
length, then the number of distinct terms in $S$ is at most 2.
\end{conjecture}

In 1976, Erd{\H{o}}s and Szemer{\'{e}}di \cite{ES{76}} verified Conjecture \ref{c1}
for sufficiently large prime $p$. However, the proof was so complicated
 that the details for small primes were never worked out.
Recently Gao {\it et al} \cite{GHW09} proved the following result.


\begin{theorem}(\cite{GHW09})
  \label{Theorem A}
Let $S$ be a sequence over $\mathbb{Z}/n\mathbb{Z}$ of length $n$.
If all nontrivial zero-sum subsequences of $S$ are of the same
length, then the number of distinct terms in $S$ is at most $2$.
\end{theorem}


Our objects of study can be characterized in very simple terms. To
be more specific, let us recall several standard notions, see \cite{SC07}.

If $S=a_1\cdot \ldots \cdot a_k$ is a sequence over $\mathbb{Z}/n\mathbb{Z}$,
let $\overline{a_i}$ be the unique integer in the set $\{0, 1, \ldots, n-1\}$
which belongs to the congruence class $a_i$ modulo $n$, $i=1,\ldots, k$. The number
$\overline{a_i}$ is called the least nonnegative representative of $a_i$ modulo $n$.

Let $S=a_1 \cdot \ldots \cdot a_t$ be a sequence of integers. We
write $m\ast S=(ma_1)\cdot \ldots \cdot (ma_t)$ where $m$ is a
integer.

If $g$ is an integer coprime to $n$, multiplication by $g$ preserves the
zero sums in $\mathbb{Z}/n\mathbb{Z}$ and does not introduce new ones.
Hence a sequence $S=a_1\cdot \ldots \cdot a_k$ is zero-free if and only if the
sequence $g*S=(ga_1)\cdot \ldots \cdot (ga_k)$ is zero-free. This fact motivates the
following definition.

Let $S$ and $T$ be sequences over $\mathbb{Z}/n\mathbb{Z}$. We say that $S$ is equivalent
to $T$ and write $S\cong T$ if $T$ can be obtained from $S$ through multiplication by
an integer coprime to $n$ and rearrangement of terms. Clearly $\cong$ is an equivalence
relation. Otherwise, we say that $S$ is not equivalent to $T$ and write $S \not\cong
T$.

More recently, Grynkiewicz \cite{DG09} gave an exhaustive list
detailing the precise structure of $S$ and showed that the result
holds in an arbitrary finite Abelian group. He proved the following result.


\begin{theorem}(\cite{DG09})
  \label{Theorem B}
Let $G$ be an Abelian group of
order $n$ and let  $S\in\mathcal{F}(G)$. Suppose there
is a unique $r\in[1, n]$ such that $0\in\sum_r(S)$. Then
$|\supp(S)|\le2$.

If $G$ is non-cyclic, then $G=\langle h\rangle\oplus\langle g\rangle
\cong C_2\oplus C_{2m}, r=\frac{n}{2}=2m$ and $$S=g^{n-1}g' \quad
\mbox{or} \quad S=g^{n/2+x}(h+g)^{n/2-x} \quad \mbox{or} \quad
S=g^{n/2+x}\left( h+\frac{n+4}{4}g\right)^{n/2-x},$$ where $g \in G,
h, g'\in G\setminus\langle g\rangle, ord(g)=\frac{n}{2}, ord(h)=2$
and $x\in[1, \frac{n}{2}-1]$ is odd.

If $G\cong \mathbb{Z}/n\mathbb{Z}$, then $S$ is one of the following:
\begin{enumerate}[(i)]
\item $S\cong 1^{n-1}g$, where $g\in \mathbb{Z}/n\mathbb{Z}$.
\item $S\cong 2^{n-1}g$, where $n$ is even and $g\in \mathbb{Z}/n\mathbb{Z}$ is odd.
\item $S\cong 1^{n-2}(\frac{n+2}{2})^2$, where $n$ is odd and $r=\frac{n+1}{2}$.
\item $S\cong 2^{n/2+x}(\frac{n+4}{2})^{n/2-x}$, where $n\equiv2\pmod{4}$ and
$x\in [0, \frac{n}{2}-1]$ is even, and $r=\frac{n}{2}$.
\item $S\cong 1^{n/2+x}(\frac{n+2}{2})^{n/2-x}$, where $n$ is even and
$x\in [0, \frac{n}{2}-1]$ with $\frac{n}{2}-x$ odd, and $r=\frac{n}{2}$.
\end{enumerate}
\end{theorem}


For the proof in the case where $G$ is a general Abelian group, Grynkiewicz
\cite{DG09} used the  result of the Devos-Goddyn-Mohar
Theorem \cite{DGM{09}}. The main purpose of the present paper is to
give an exhaustive list detailing the precise structure of the sequences for a
slight generalization of Graham's Conjecture without using the
Devos-Goddyn-Mohar Theorem.
The following are our main results.

\begin{theorem}\label{1} Let $G$ be a finite Abelian  group of order $n$
 with $r(G)\geq 2$, and let $S$ be a sequence of length $n-1$ over $G$.
 If all nonempty zero-sum subsequences have
the same length $r\in [1, n-1]$, then $G\cong C_2\oplus C_{2m}$. Moreover, either $m=1,
r=3$ and $S=\prod_{g\in G\setminus \{0\}}g$, or $S=0g_1g_2$, where $r=m=1$ and $g_1, g_2\in G\setminus \{0\}$
are distinct, or $r=\frac{n}{2}=2m$ and $S$ is one
of the following:
\begin{enumerate}[(i)]
\item $S=g^{n-1}$,
\item $S=g^{n-2}g'$,
\item $S=g^{n-3}g'(2g-g')$,
\item $S=g^{\frac{n}{2}+x}(h+g)^{\frac{n}{2}-1-x}$,
\end{enumerate}
where $g \in G$, $h, g'\in G\setminus\langle g\rangle$, $\ord(g)=\frac{n}{2}$, $\ord(h)=2$ and $x\in[0, \frac{n}{2}-4]$.

\end{theorem}




\begin{theorem}\label{2}
Let $S$ be a sequence of length $n-1$ over $G=\mathbb{Z}/n\mathbb{Z}$.
Suppose that all nontrivial zero-sum subsequences of $S$ have the same
length $r\in [1, n-1]$, then $S$ is
one of the following:

$\bullet$ $S\cong 1^{n-2}\cdot g$, where $g\in \mathbb{Z}/n\mathbb{Z}$ and $\bar{g}\neq 1$.

$\bullet$ $S\cong 1^{n-3}\cdot 0\cdot 2$ and $r=1$.

$\bullet$ $S\cong 1^{n-3}\cdot 2\cdot (n-1)$ and $r=2$.

$\bullet$ $S\cong 1^{n-3}\cdot 2^2$ and $r=n-2$.

$\bullet$ $S\cong 1^{n-3}\cdot (\frac{n+1}{2})^2$, where $n$ is odd, and $r=\frac{n+1}{2}$.

$\bullet$ $S\cong 2^{n-1}$, where $n$ is even, and $r=\frac{n}{2}$.

$\bullet$ $S\cong 2^{n-2}\cdot g$, where $n$ is even and $\bar{g}$ is odd,
and $r=\frac{n}{2}$.

$\bullet$ $S\cong 2^{n-3}\cdot g\cdot (4-g)$, where $n$ is even and $\bar{g}$
is odd, and $r=\frac{n}{2}$.

$\bullet$ $S\cong 2^u \cdot (\frac{n}{2}+2)^{n-1-u}$, where $n$ is even, $\frac{n}{2}$ is odd,
 $u\geq \frac{n}{2}-1$ and $r=\frac{n}{2}$.

$\bullet$ $S\cong 1^u \cdot (\frac{n}{2}+1)^v$, where $n$ is even, $u\geq \frac{n}{2}-1$ and
$r=\frac{n}{2}$.

$\bullet$ $n=6$ and $S\cong 1^2\cdot 3^3$, $1\cdot 3^3\cdot 4$ or $2^2 \cdot 3^3$.

$\bullet$ $n=7$ and $S\cong 1^3\cdot 3^3$.

$\bullet$ $n=8$ and $S\cong 1^4\cdot 3^3$.
\end{theorem}


\begin{corollary} Let $G$ be a finite Abelian group of order $n$ and let
$S$ be a  sequence of length $n-1$ over $G$. If there is an integer
$r\ge 0$ such that all nonempty zero-sum subsequences of $S$ have
length $r$, then $|\supp(Sx^{-1})|\leq 2$ for some $x\in \supp(S)$.
\end{corollary}



\section{Preliminaries}

Given two subsets $A$ and $B$ of an Abelian group $G$, their $sumset$ is the set of
all pairwise sums, denoted $A+B=\{a+b: a\in A, b\in B\}$. From the basic properties
of addition, we have that $A+B=B+A$, and that the sumsets of more than two sets, denoted
$\sum_{i=1}^l A_i=\{\sum_{i=1}^l a_i: a_i\in A_i\}$, is well defined. We often use the convention
that $\sum_{i\in \varnothing }A_i=\{0\}$. For sumsets with a single element set, we abbreviate
$\{x\}+A$ to $x+A$. Substraction of sets is defined similarly, for instance, $-A=\{-a: a\in A\}$
and $A-B=\{a-b: a\in A, b\in B\}$.

Arithmetic progressions over an arbitrary Abelian  group $G$ (with length $l$ and difference $d$) are
sets of the form $\{\alpha +id: i=1,2,\cdots,l \}$ with $\alpha, d\in G$ and $l\in \mathbb{Z}$, and
are closely related to the prototypical cases that arise when studying sumsets with small cardinality.

For a zero-sum free sequence $S$, we say that a term $x\in \supp(S)$ is a
{\it 1-term} if $|\sum(S)|=|\sum(S
x^{-1})|+1$.


\begin{definition} {\rm (\cite{Ge}, Definition 5.1.3)} A sequence $S\in \mathcal{F}(G)$ is called
$smooth$ if $S=(n_1g)(n_2g)\cdot \, \ldots\,\cdot (n_lg)$, where
$|S|\in \mathbb{N}, \, g\in G, \, 1=n_1\leq \cdots\leq n_l,
\,n=n_1+\cdots+n_l<\mbox{ord}(g)$ and $\sum(S)=\{g,  \ldots, \,ng\}$
( in this case we say more precisely that $S$ is $g$-smooth).
\end{definition}

Let $S$ be a sequence in an Abelian group $G$. Note that
$\sum(S)\cup \{0\}=\{0,g_1\}+\cdots +\{0, g_l\}$, where
$S=g_1\cdot  \ldots \cdot g_l$. In particular,
$\sum(ST)\cup \{0\}=(\sum(T)\cup \{0\})+\{0,g_1\}+\cdots +\{0, g_l\}$
for any nontrivial sequence $T\in \mathcal {F}(G)$. With this observation in hand, the
following lemma is easily verified.

\begin{lemma}\label{l1} Let $A\subseteq G\setminus  \{0\}$ be a finite nonempty subset of an
Abelian group $G$. Let $b\in G\setminus \{0\}$ and $|\{0, b\}+(\{0\}\cup A)|=|A|+2$.
Then $A\cup\{0\}$ is the union of some arithmetic progression with difference $b$ and (possibly)
some disjoint $H-cosets$, where $H=\langle b\rangle$.
\end{lemma}


Let $G$ be an Abelian group. If $A$ and $B$ are subsets of $G$ and $g\in G$, then
$r_g(A,B)$  denotes the number of different representations of $g$ as a sum
of $g=a+b$ $(a\in A, b\in B)$. The following results will be needed.

\begin{theorem}{\bf (Addition theorem of Kemperman-Scherk)} Let $G$ be an Abelian
group, and $A, B$ be nonempty subsets of $G$. Then for each element $g \in A + B$,
$$|A +B| \geq |A| + |B|- r_g(A,B).$$
\end{theorem}

The following result characterizes all zero-sum free sequences $S$
with $|\sum(S)|\le 2|S|-1$. Moreover,  if the equality holds, then
$S$ is one of the forms of (ii), (iii), (iv) and (v).

\begin{theorem}{\rm (\cite{yu08}, Theorem 1.1)} \label{t2}
Let $G$ be a finite Abelian group and let $S$ be a zero-sum free sequence over
$G$ with $|\sum(S)|\le  2|S|-1$. Then $S$ is one of the following:
\begin{enumerate}[(i)]
\item $S$ is $a$-smooth for some $a\in G$.
\item $S=a^kb$, where $k\in\mathbb{N}$ and $a,  b\in G$ are distinct.
\item $S=a^kb^l$,  where $k\ge l \ge 1$ and $a, b \in G$ are
distinct with $2a = 2b$.
\item $S=a^kb^l(a-b) $,  where $k\ge l \ge 1$ and $a, b \in G$ are
distinct with $2a = 2b$.
\item $S=a^kbc$, where $\ord(a)=k+2$ and $b,c\in G\setminus \langle a\rangle $
are distinct with $b+c=a$ and $b-c\in \langle a\rangle $. In this case,
$\sum(S)=\langle a,b,c\rangle\setminus \{0\}$.
\end{enumerate}
\end{theorem}

\begin{lemma}\label{l2} {\rm (\cite{SC07}, Proposition 2)}  Suppose a sequence $S=Ta$ is zero-sum free
over the Abelian group $G$ and $|\sum(S)|=|\sum(T)|+1$, let
$H=\langle a\rangle$ denote the subgroup of $G$ generated by $a$,
then:
\begin{enumerate}[(i)]
\item $\sum(T)$ is the union of a progression $\{a,2a,3a,\ldots,
ka\}$ and some cosets (maybe empty) of $H=\langle a\rangle$ where
$1\le k<\ord (a)-1$.
\item $\sigma(T)=ka$.
\item $a$ is the unique element of $G$ with such property.
\end{enumerate}
\end{lemma}

\begin{lemma}\label{l3} A sequence of $n-1$ integers in the
interval $[0,n-1]$, assuming two distinct values, has a nonempty
subsequence with $sum\equiv 0$ (mod $n$).
\end{lemma}

\begin{proof}
See \cite{BDL}.
\end{proof}

\begin{theorem}[Savchev-Chen Structure Theorem \cite{SC07}]\label{sc3} 
Let $S\in \mathcal{F}(C_n)$
be a zero-sum free sequence of length $|S|=l> \frac{n}{2}$. Then there
exists a sequence $a_1\cdot \ldots \cdot a_l\in \mathcal{F}(\mathbb{Z})$, with
$a_i\in [1, n-1]$, such that
\begin{enumerate}[(i)]
\item $S=(a_1g)\cdot \ldots \cdot (a_lg)$ for some $g\in \supp(S)$, and
\item $\sum(S)=\{g, 2g, \ldots, (a_1+\ldots +a_l)g\}$.
\end{enumerate}
\end{theorem}

\begin{lemma}\label{l4} A sequence of $n-2$ integers in the
interval $[0,n-1]$, assuming more than two distinct values, has a
nonempty subsequence with $sum\equiv 0$ (mod $n$). Furthermore, if
$S$ is a zero-sum free sequence with length $n-2$, then $S=x^{n-2}$
or $S=x^{n-3}\cdot {2x}$ with $x\equiv 0 \pmod n$ and $n\geq 4$.
\end{lemma}

\begin{proof}
See \cite{BDL}.
\end{proof}

Let $G$ be a group and
 $D(G)$ be the Davenport's constant of $G$, i.e., the smallest
integer $d$ such that every sequence $S$  over $G$ with
$|S|\geq d$ satisfies $0\in\sum(S)$. In the following, we give some
properties of $D(G)$.

\begin{lemma}[\cite{Ge-HK06a}, Proposition 5.1.4
 and Theorem 5.8.3] \label{l5}\leavevmode
 \begin{enumerate}[(i)]
\item Let $G$ be a finite Abelian group and $S$ a zero-sum free
sequence over $G$. If $|S|=D(G)-1$, then $\sum(S)=G\setminus \{0\}$.

\item Let $G=C_{n_1}\oplus C_{n_2}$ with $1\leq n_1\leq n_2$. Then
$D(G)=n_1+n_2-1$.
\end{enumerate}
\end{lemma}

\begin{lemma}[\cite{P{09}}, Lemma 2.2, Theorem 2.4 and Theorem 5.1]\label{l6}\leavevmode
\begin{enumerate}[(i)]
\item Let $G$ be a noncyclic finite Abelian
group. Then $D(G)\leq \frac{|G|}{2}+1$, and equality holds when
$G\cong C_2\oplus C_{2m}$.
\item Let $G$ be a finite Abelian group of rank greater than 2. Then
$D(G)\leq \frac{|G|}{4}+2$.
\item $D((\mathbb{Z}/p\mathbb{Z})^r)=r(p-1)+1$ for prime $p$ and $r\geq 1$.
\end{enumerate}
\end{lemma}


Next we discuss the shortest length of zero-sum subsequence  of a
sequence $S$ with length $n-1$ over a finite Abelian group $G$ with
$|G|=n$, and the result will be used throughout the proof of our
main results.

Before stating the main result explicitly, we first introduce the important
concept of setpartitions, which will be used throughout this thesis. Let $S$
be a sequence. A \\{\it n-setpartition} of $S$ is a partition of sequence $S$ into
$n$ nonempty subsequences $A_1, \ldots, A_n$, such that terms in each subsequence
$A_i$ are all distinct, allowing the $A_i$ to be regarded as sets.



\begin{lemma}\label{l7} Let $G$ be an Abelian group of order $|G| = n$
and let $S$ be a sequence of length $n-1$. If $0 \not\in \sum_{\leq
h(S)+1}(S)$, then $S$ is one of the following:
\begin{enumerate}[(i)]
\item $G \cong C_{n}, S = g^{n-1}$ and $ord(g) = n$.
\item $G = C^{r}_{2}, \supp(S) = G\backslash \{0 \}$ and  $h(S) = 1$.
\end{enumerate}
\end{lemma}

\begin{proof} Obviously, $0 \not\in \supp(S)$. First we consider the case $h(S) = n-1$.
Then $S = g^{n-1},\ g\ne0$ and $g\in G$. If $\ord(g)\neq n$, then $\ord(g) \leq h(S)$ and $\ord(g)g =
0$, which is a contradiction. Hence we have $\ord(g)= n$, $G \cong
C_n$ and $S=g^{n-1}$.

For the case $h(S) = 1$, we have $\supp(S) = G\setminus\{0\}$. If there
exists some $g \in G$ such that $2g\neq 0$, then $g\neq 0,\ (-g)g|S$
and $-g+g = 0$, which is a contradiction. Therefore we have $2g =0$ for all
$g \in G$, and  thus $G \cong C_2^r,\ r\in\mathbb{N}$.

 Suppose  that $2 \le h(S) \le n-2$.  We choose $b \in \supp(S)$ with
$\mathsf{v}_{b}(S) = h(S)$. Since $h(S)\ge 2$, we have $\ord(b)\ge 3$.
By Bialostocki's result in \cite{BDL}, the existence of a $(h+1)-$setpartition is
straightforward, where $h=h(S)$. For any $(h+1)-$setpartition
$\mathcal {A}=\mathcal {A}_{0}\mathcal {A}_1 \cdots\mathcal {A}_{h+1}$ of $S$,
we set $\mathcal{B}_i=\mathcal {A}_i \cup \{0\}$
 for $i \in [0,h]$.
 It is clear that $0$
is a unique expression element in $\sum_{i=0}^h\mathcal {B}_i$ in view of
$S$ being zero-sum free.
 If there exist
two distinct integers $j, k\in [0,h]$ such that $|\mathcal {B}_j+
\mathcal {B}_k| \geq |\mathcal {B}_j| + |\mathcal {B}_k|$, then by
the addition theorem of Kemperman-Scherk, it follows that
\begin{eqnarray*}
|\sum(\mathcal {B}_0\cdots \mathcal{B}_h)| & = & |\mathcal
{B}_{j}+\mathcal {B}_{k}+ \sum_{i\neq j,k}{\mathcal {B}_i}|\\ & \geq
& |\mathcal {B}_{j}+\mathcal {B}_{k}|+ |\sum_{i\neq j,k}{\mathcal
{B}_i}|-1 \\
& \geq & |\mathcal {B}_j| + |\mathcal {B}_k|+ \sum_{i\neq
j,k}|{\mathcal {B}_i}|-(h-2)\\
& \geq & \sum_{i=0}^{h}|\mathcal {B}_i|-h+1\\
& \geq & n+1.
\end{eqnarray*}
By the assumptions, we have a contradiction. By the addition theorem of Kemperman-Scherk,
it follows that
 $|\mathcal {B}_j+
\mathcal {B}_k| < |\mathcal {B}_j| + |\mathcal {B}_k|-1$ cannot hold.
 Therefore
we have $|\mathcal {B}_j +\mathcal {B}_k| = |\mathcal
{B}_j|+|\mathcal {B}_k|-1$ for any $j, k \in [0,h], j\neq
k\frac{}{}$ and $|\sum(\mathcal {B}_0\cdots \mathcal{B}_h)| = n$.

Now we choose a special setpartition $\mathcal {A}=\mathcal
{A}_{0}\mathcal {A}_1 \cdots \mathcal {A}_h$ such that $\mathcal
{A}_0 = \{b\}$ and $ b \notin \mathcal {A}_h$. By Bialostocki's result in \cite{BDL},
the special $(h+1)-$setpartition exists.
If there exists an integer $k \in [1,h]$ such that $\langle b\rangle
\subseteq \mathcal {B}_{k}$, then we have $0 \in b + \mathcal {A}_k
\subseteq \sum_{\leq h(S)+1}(S)$, which is impossible. Let $k\in [1,h]$
be arbitrary. Since $|\{0, b\}+\mathcal{B}_k|=|\mathcal{B}_k|+2-1=|\mathcal{A}_k|+2$
and $\langle b\rangle\not\subseteq B_k$, Lemma \ref{l1} implies that
$\mathcal{B}_k$ is the union of
$\{0,b,2b,\ldots,u_{k}b\}$ and some $\langle b\rangle$-cosets, where $0 \leq u_k \leq
\ord(b)-2$. Moreover, since $b \notin \mathcal {B}_h$, we have $u_h
=0$.


For any $g \in \mathcal {B}_k\setminus \{0, b\}$, $k \in [1,h]$, if
$g \notin \langle b\rangle$ and $g \not\in \mathcal {B}_t$ for some
$t \in [1,h]\backslash\{k\}$, then removing $g$ from $\mathcal
{B}_k$ and appending $g$ to $\mathcal {B}_t$, it yields a new
setpartition $\mathcal {B}^{'}=\mathcal {B}_0^{'}\mathcal {B}_1^{'}
\cdots \mathcal {B}_h^{'}$. By the structure of $\mathcal{B}_k$, it is easy to
see that  $\mathcal {B}_k^{'}$ is not the union of
$\{0,b,2b,\ldots,u_{k}b\}$ and some $\langle b\rangle$- cosets, which
is impossible. If $g \in \langle b\rangle$, then  $g \not\in
\mathcal {B}_h$. Removing $g$ from $\mathcal {B}_k$ and appending
$g$ to $\mathcal {B}_{h}$, it  leads to a contradiction as above
($\mathcal {B}_{h}\cup \{g\}$ is not of that form). Therefore we
obtain that $\mathcal {B}_0 = \{0,b\}, \mathcal {B}_{h}$ is the
union of $\{0\}$ and some $\langle b \rangle$-cosets and $\mathcal
{B}_{i}=\mathcal {B}_{h} \cup \{b\}$ for  $i \in [1,h-1]$.

If there exists an element  $g \in \mathcal {B}_{h}\setminus \{0\}$
such that $g+\mathcal {B}_1 \nsubseteq \mathcal {B}_1+\langle
b\rangle$, then, since $g
\in \mathcal{B}_1\setminus \langle b\rangle$ by the description of the
$\mathcal{B}_i$ given above, it follows that there exists some $b_1\in
\mathcal{B}_1\setminus \langle b\rangle$ such that
$ g+b_1\notin \mathcal{B}_1+ \langle b\rangle$. Since the
description of $\mathcal{B}_1$ ensures that
$b_1+\langle b \rangle\subseteq \mathcal{B}_1$, we have that the entire coset
$b_1+\langle b \rangle$ is disjoint from $\mathcal{B}_1+\langle b \rangle$.
Thus removing $g$ from $\mathcal {B}_h$ and appending $g$
to $\mathcal {B}_0$, it yields a new setpartition $\mathcal
{B}^{'}=\mathcal {B}_0^{'}\mathcal {B}_1^{'}\cdots \mathcal
{B}_h^{'}$ such that $(\mathcal {B}_1^{'}+\mathcal {B}_0^{'}
)\setminus (\mathcal B_1+\langle b\rangle)$ contains some $\langle
b\rangle$-cosets. Since $\ord(b) \geq 3$ and $h\ge 2$, it follows that $|\mathcal
{B}_1^{'} + \mathcal {B}_0^{'}| \geq |\mathcal {B}_1^{'}| + |\langle
b\rangle| \geq |\mathcal {B}_1^{'}| + 3 = |\mathcal {B}_1^{'}| +
|\mathcal {B}_0^{'}|$, which is a contradiction.

If every $g \in \mathcal {B}_{h}\setminus \{0\}$ satisfies
$g+\mathcal {B}_1 \subseteq \mathcal {B}_1+\langle b\rangle$,  we
have $\mathcal {B}_1 + \mathcal {B}_h=\mathcal{B}_1+\langle
b\rangle$, which implies $|\mathcal{B}_h|\ge |H|+1$, where
$H=\langle b\rangle$. Recall that $|\mathcal {B}_1 + \mathcal {B}_h| = |\mathcal
{B}_1| + |\mathcal {B}_h|-1\ge |\mathcal
{B}_1|+|H|$, where
$H=\langle b\rangle$; the latter inequality follows in view of the
description of $\mathcal{B}_h$. However,
$|\mathcal {B}_1 + \mathcal {B}_h|\ge |\mathcal
{B}_1|+|H|$ is not possible in view of $\mathcal {B}_1 + \mathcal {B}_h = \mathcal{B}_1+\langle
b\rangle= \mathcal {B}_1 + H$ and the description of $\mathcal{B}_1$ given above, which makes
a contradiction. This completes the proof.
\end{proof}





\section{The proof of the main results}

In order to prove our main results, we need to state a property.


{\bf Property.} {\it Let $G$ be a finite Abelian  group of order $n$ and
 $S$ be a sequence of length $n-1$ over $G$. Let $T$ be a
nonempty zero-sum subsequence of $S$. Setting $U=ST^{-1}$. If any
nonempty zero-sum subsequence of $S$ has length $r\in [1,n-1]$, then the
following statements hold:
\begin{enumerate}[(i)]
\item $|T|=r\le D(G)$ and $|U|=n-1-r\le D(G)-1$, where $D(G)$ is
 Davenport's constant.
\item $T$ is a minimal zero-sum subsequence, and $U$ is zero-sum
free.
\item For every $x|T$, $x\notin \sum_{\geq 2}(U)$. 
\end{enumerate}}


Next we give the proof of Theorem \ref{1}.


\begin{proof}[ {\bf Proof of Theorem \ref{1}}]
First we observe that if the rank of $G$ is greater than 2,
then, by Lemma \ref{l6}(ii), we have $D(G)\leq \frac{|G|}{4}+2$. If
$|G|\geq 9$, then we have $2D(G)\leq |G|-1$, and so there exist two
disjoint nonempty subsequences $S_1$ and $S_2$ of $S$ such that
$\sigma(S_1)=\sigma(S_2)=0$, which is impossible. If $|G|\leq 8$,
then $G\cong C_2\oplus C_2\oplus C_2, r=D(G)=4$ and $S=\prod_{g\in G\setminus
\{0\}}g$, which is also impossible.

Now we assume that the rank of $G$ is 2 and let $G\cong C_k\oplus
C_{km}$ with $k\geq 2$ and $m\geq 1$. Then by Lemma \ref{l5}(ii), we have
$D(G)=km+k-1$. By Property(i), we obtain $2D(G)-1\geq |G|-1$. It
follows that we have $k=2, G\cong C_2\oplus C_{2m}$ and $D(G)=2m+1$,
or $G\cong C_3^2$ and $D(G)=5$.

Suppose that $G\cong C_3^2$ and $D(G)=5$. By Property, we have $4\le r\le 5$.
By Lemma \ref{l7}, we have $h(S)\ge r-1\ge 3$. Let $g|S$ with
$\mathsf{v}_g(S)=h(S)$. Then $\sigma(g^3)=0$, which contradicts that $r\ge 4$.
Therefore we have $G\cong C_2\oplus C_{2m}$, in which
case $D(G)=2m+1$ by Lemma \ref{l5}.

If $0\in \supp(S)$, then $S0^{-1}$ is a zero-free and $4m-2=|G|-2=|S|-1\le D(G)-1=2m$,
and so $m=1$. It follows that $G\cong C_2^2$ and $S=0g_1g_2$, where
$g_1, g_2\in G\setminus \{0\}$ are distinct. In the following
argument, we consider $0\notin \supp(S)$, and so $r\ge 2$.
We shall show $2m-1\leq r\leq 2m+1$. Let $T$ be a zero-sum
subsequence of $S$. If $r\geq 2m+2$, then $|T|=r> D(G)$, and
there exists a zero-sum subsequence $T_1$ of $T$ with length $< r$,
which is a contradiction. If $r\leq 2m-2$, then $n-r\geq 2m+1=D(G)$,
and so there exists a zero-sum subsequence of $ST^{-1}$, which is
also a contradiction. Next we divide into three cases to discuss the result.

{\bf Case 1:} $r=2m+1$. If $0\notin \sum_{\le h(S)+1}(S)$, then by
Lemma \ref{l7}, we have $r=3, m=1$ and $S=\prod_{g\in G\setminus \{0\}}g$. Next we
consider $0\in \sum_{\le h(S)+1}(S)$, then $h(S)\geq r-1=2m$. Take
$a|S$ satisfying $\mathsf{v}_{a}(S)=h(S)\ge r-1=2m$. Then $a^{\ord(a)}|S$
shows that $r=\ord(a)\le 2m$, contrary to case hypothesis.

{\bf Case 2:} $r=2m-1$. Let $T|S$ be a zero-sum subsequence of
length $r$. Let $U=ST^{-1}$. Then $|U|= 2m=D(G)-1$, and so, by
Lemma \ref{l5}(i), $\sum(U)=G\setminus \{0\}$ and
$|\sum(U)|=2|U|-1$. By Property(iii), for any $a\in \supp(T)$, we
have $a\in \supp(U)$, i.e., $$\supp(T)\subseteq \supp(U).$$
By Theorem \ref{t2}, we distinguish four subcases.


(i) $U=a^{x}b$, where $x=2m-1, \ord(a)=2m$ and $b \notin \langle a
\rangle$. If $a\notin \supp(T)$, then $T=b^{2m-1}$, which is not
possible since $T$ is zero-sum free. If $a\in \supp(T)$, then $a^{2m}|S$,
and $\sigma(a^{2m})=0$, which contradicts $r=2m-1\neq 2m$.

(ii) $U=a^{x}(a+g)^{y}$, where $x+y=2m, \ord(a)=2m, \ord(g)=2$ and
$y$ is odd. Then $T=a^{s}(a+g)^{t}$ with $s+t=2m-1$. However,
$\sigma(T)\neq 0$, which contradicts $\sigma(T)=0$.

(iii) $U=a^{x}(a+g)^{y}g$, where $x+y=2m-1, \ord(a)=2m$, and
$\ord(g)=2$. Obviously, $g\notin \supp(T)$. Then $T=a^{s}(a+g)^{t}$
with $s+t=2m-1$, and so $\sigma(T)\neq 0$, which contradicts
$\sigma(T)=0$.

(iv) $U=a^xbc$, where $x=2m-2, \ord(a)=x+2=2m$, and
$b, c\in G\setminus\langle a\rangle$
are distinct with $b+c=a$ and $b-c\in \langle a\rangle$. Obviously,
$\langle a\rangle\cap \supp(T)=\varnothing $. Otherwise, there exists
$sa\in \supp(S)$, with $s\in [1, 2m-2]$, such that either $a^{2m-s}\cdot sa$
is a zero-sum subsequence with length $2m-s+1< r$, where $s\ge 3$, or
$a^{2m-s-1}\cdot sa\cdot b\cdot c$ is a zero-sum subsequence with
length $2m-s+2> r$, where $s=1, 2$. By Property(iii) and
$\sum(U)=G\setminus \{0\}$, we have $T=b^s$ or $T=c^s$ with $s=2m-1$.
If $\sigma(T)=0$, we have $(2m-1)u=4m$. That is to say, $u=4, m=1$, and
$r=1$, which contradicts $\sigma(T)=0$ since $\gcd(2m-1, 2m)=1$.


{\bf Case 3:} $r=2m$. By Lemma \ref{l7}, we have $h(S)\geq r-1=2m-1$.
We set $g\in \supp(S)$ with $h(S)=\mathsf{v}_{g}(S)$. Let $T^{'}=Sg^{-h(S)}$.
Then $\ord(g)=2m$ and $\supp(T^{'})\cap \langle g\rangle =\varnothing$. In
particular, for any $a, b \in \supp(T)$, we must have $2a, 2b,
a+b\in \langle g \rangle$.

If $|T^{'}|=0$, then $S=g^{4m-1}$.

If $|T^{'}|=1$, then $S=g^{4m-2}\cdot g^{'}$ with $g^{'}\notin \langle g
\rangle$.

If $|T^{'}|=2$, then $T^{'}=h_1h_2$ with $h_1,h_2\notin \langle g
\rangle$, and $h_1+h_2=lg$, where $l\in [1, 2m-2]$. But then
$g^{2m-l}\cdot h_1\cdot h_2$ is a zero-sum subsequence, which force
$h_1+h_2=2g$.

If $|T^{'}|\geq 3$, choose any subsequence $h_1h_2h_3|T^{'}$.
Then as shown in the previous case, $h_1+h_2=h_1+h_3=h_2+h_3=2g$,
which implies $h_1=h_2=h_3$, and
so $|\supp(T^{'})|=1$. In addition, $2h_1=2g$ implies $h_1=g+h$
with $\ord(h)=2$. Therefore, by $h(S)=\mathsf{v}_g(S)$, we have
$S=g^{2m+x}(g+h)^{2m-1-x}$, where $x\in [0, 2m-4]$.
This completes the proof.
\end{proof}

In what follows, we prove the theorem \ref{2}.

\begin{proof}[ {\bf Proof of Theorem \ref{2}.}]
First we consider $r=1$, then, by
Property (ii), $0\in \supp(S)$, and $S0^{-1}$ is
zero-sum free.  By Lemma \ref{l4}, we have  $S=0\cdot x^{n-2}$ or
$S=0\cdot x^{n-3}\cdot 2x$ with $ord(x)=n$.

If $|\supp(S)|=1$, then $S=a^{n-1}$, where $ord(a)\ge \frac
{n}{2}$.

In the following argument, we assume that $n\geq 4, |\supp(S)|\geq
2$ and $2 \leq r \leq n-1$. By Lemma \ref{l7}, we obtain $h(S)\geq r-1$.
Next we divide into six cases to prove the result.


Suppose that $r\ge \frac{n+1}{2}$ and choose $a|S$ satisfying $\mathsf{v}_{a}(S)=h(S)\geq r-1 \geq
\frac{n-1}{2}$, it follows that  $\ord(a)
> \mathsf{v}_{a}(S)=h(S)\geq \frac{n-1}{2}$. Therefore,
$\ord(a)=n$, and without loss of generality, we set $a=1$. Let $T$ be
a zero-sum free subsequence of $S$ with $|\sum(T)|\geq 2|T|-1$ and
$|T|$ maximum. Then we have $1\le |T|\le \lfloor\frac{|\sum(T)|+1}{2}\rfloor \le
\lfloor\frac{n}{2}\rfloor \le r-1$. Setting $U=ST^{-1}$, then we have
the following result.

{\bf Claim 1.} Suppose that $r\geq \frac{n+1}{2}$. Let all notation be as above,
 then the following statements hold:
\begin{enumerate}[(i)]
\item If $|T|\leq r-2$, then $U=1^{n-1-|T|}$ and $\supp(T)\subseteq
\{1, n-r, n-r+1\}$.
\item If $|T|=r-1$ and $|\sum(T)|=n-1$, then $U=1^{n-1-|T|}$ and
$\supp(T)\subseteq \{1, n-r, n-r+1\}$.
\end{enumerate}

{\bf Proof of Claim 1:} (i) Since $h(S)\geq r-1>|T|$, we have $1|U$.
Since for any $a|U$, $Ta$ is zero-sum free and
$a$ is a $1-term$ for $Ta$ because of the maximality of $T$, it
follows in view of Lemma \ref{l2} that $U=a^{n-1-|T|}$, and so
$U=1^{n-1-|T|}$ because of $1\in \supp(U)$. But now Lemma \ref{l2}
further implies that $\sum(T)=\{1, 2,
\cdots, k\}$ and $U=1^{n-1-|T|}$ with $k\ge 2|T|-1$. If $|T|=1$, then
$|\supp(S)|=1$, contrary to what we assumed above. So $|T|\ge 2$.
Therefore, there exists a subsequence $T_{1}$ such that
$\sigma(T_1)=0$ and $T|T_1$. By Property(i), we have $|T_1|=r$ and
$|ST_{1}^{-1}|=n-1-r$. Let $x |T_{1}$ with $\bar{x} \neq 1$. If $\bar{x} \geq n-r+2$, we
can obtain a zero-sum subsequence $x\cdot 1^{n-\bar{x}}$ of length less
than $r$. If $\bar{x}\leq n-r-1$, then $\bar{x}=\sigma(1^{\bar{x}})$, and so
$T_{1}\cdot x^{-1}\cdot1^{\bar{x}}$ is a zero-sum subsequence with length $r-1+\bar{x}$,
which forces $\bar{x}=1$. Hence for any $x|T$, either $n-r \leq \bar{x} \leq
n-r+1$ or $\bar{x}=1$.

(ii) For $|T|=r-1$ and $|\sum(T)|=n-1$, we have
$r\leq \frac{n}{2}+1$ and for every $a|U$, $Ta$ is a zero-sum
subsequence. Thus $U=a^{n-1-|T|}$. If $\bar{a}\neq 1$, then
$T=1^{r-1}$ and $|\sum(T)|=n-1=r-1$ implies $r=n$, which is a contradiction. It
follows that $U=1^{n-1-|T|}$.

For any $x|T$, by the similar argument as above, we have either $n-r
\leq \bar{x} \leq n-r+1$ or $\bar{x}=1$. This proves Claim 1.\\

{\bf Case 1. $r\geq \frac{n}{2}+1$.}

Choose $a|S$ satisfying $\mathsf{v}_{a}(S)=h(S)\geq r-1 \geq
\frac{n}{2}$. Since $r\geq \frac{n}{2}+1$, it follows that  $ord(a)
> \mathsf{v}_{a}(S)=h(S)\geq \frac{n}{2}$. Therefore,
$ord(a)=n$, and without loss of generality, we set $a=1$. Let $T$ be
a zero-sum free subsequence of $S$ with $|\Sigma(T)|\geq 2|T|-1$ and
$|T|$ maximum. Then we have $1\le |T|\le \frac{|\sum(T)|+1}{2} \le
\frac{n}{2} \le r-1$.


{\bf Subcase 1.1.} $|T|\leq r-2$.

By Claim 1(i), we have $U=1^{n-1-|T|}$ and $\supp(T)\subseteq \{1,
n-r, n-r+1\}$. There are three options need to be considered.


(i) If $n-r=1$, then $S=1^{n-2}\cdot 2$.

(ii) If $r=\frac{n}{2}+1$, that is, $n-r+1=\frac{n}{2}$, then
$(n-r)+(n-r+1)=n-1$. If $(n-r)\cdot(n-r+1)|S$, then $r=3$ and $n=4$,
whence $S=1^{2}\cdot 2$. So we can assume $\supp(S)\subseteq \{1,n-r\}$
or $\supp(S)\subseteq \{1,n-r+1\}$. In the second case, let $S=1^{s}\cdot
(n-r+1)^{t}=1^{s}\cdot (\frac{n}{2})^{t}$. If $t\ge 2$, then $n=2$, which makes a
contradiction.
Therefore, $S=1^{n-2}\cdot \frac{n}{2}$. In the first case, let $S=1^{s}\cdot
(n-r)^{t}=1^{s}\cdot (\frac{n}{2}-1)^{t}$ with $s\ge r-1\ge
\frac{n}{2}$. We can assume that $t\ge 2$, else $S=1^{n-2}\cdot g$, as desired.
But  $\mathsf{v}_1(S)\ge r-1\ge \frac{n}{2}\ge 2$, so the zero-sum
subsequence $1^2\cdot (\frac{n}{2}-1)^2$ shows that $r=4$, in which case
$n=6$, and now we must have $S\cong 1^3\cdot 2^2$(else $2^3$ would contradict
that $r=4$).

(iii) If $3\le n-r+1<\frac{n}{2}$, then $$n-r+2 \leq 2(n-r)<
(n-r)+(n-r+1)< 2(n-r+1)< n,$$ and  there exists an integer $s< r-2$
such that $(n-r)+(n-r+1)+s*1=n$. If $(n-r)\cdot (n-r+1)|S$, then
$1^s\cdot (n-r)\cdot (n-r+1)$ is a zero-sum subsequence with length
$s+2< r$, which is a contradiction. So we can assume $\supp(S)=\{1,x \}$,
where $\bar{x}=n-r$ or $\bar{x}=n-r+1$.

Suppose that $S=1^{u}\cdot (n-r)^v$. It is easy to show that $u \leq
r-1$, and so $u=r-1$ and $v=n-r\ge 2$. If $v=n-r=2$, then
$S=1^{n-3}\cdot 2^{2}$. If $v=n-r \geq 3$, then $0<n-2(n-r)<r-2$,
that is, $n-2(n-r)< r-2$. Therefore there exists a positive integer $t\le
r-3$ such that $(n-r)+(n-r)+t*1=n$, which is a contradiction.

Suppose $S=1^{u}\cdot (n-r+1)^v$, then $u \geq r-1$. Since $n-r+2 <
2(n-r+1)< n$, it follows that there exists a positive integer $t\le r-3$
such that $2(n-r+1)+t*1=n$. Therefore,  $v=1$ and $S=1^{n-2}\cdot
(n-r+1)$.


{\bf Subcase 1.2 :} $|T|=r-1 \geq \frac{n}{2}$.

Since $|\Sigma(T)|\geq 2|T|-1$,  it follows that $|\Sigma(T)|=n-1,
|T|=\frac{n}{2}$ and $r=\frac{n}{2}+1$, and so, by Claim 1(ii), we
have $U=1^{\frac{n}{2}-1}$ and $\supp(T)\subseteq \{1,
\frac{n}{2}-1, \frac{n}{2}\}$. There are two options need to be considered.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


(i) If $\frac{n}{2}|T$, then we have $\mathsf{v}_{\frac{n}{2}}(S)=1$.
If $\frac{n}{2}-1 |T$, then $\frac{n}{2}+(\frac{n}{2}-1)+1=n$
implies that $r=3$ and $n=4$, and so $S=1^{2}\cdot 2$. Therefore,
$S=1^{n-2}\cdot \frac{n}{2}$ for $n \geq 4$.

(ii) If $\frac{n}{2}-1|T$, then we must have $\mathsf{v}_{1}(S)\leq
\frac{n}{2}$, and $\frac{n}{2}\nmid S$ in view of the previous paragraph. Let
$S=1^{s}\cdot(\frac{n}{2}-1)^{t}$ with $s\geq t$ for $\mathsf{v}_1(S)=h(S)$. Since
$\frac{n}{2}\geq s\geq r-1=\frac{n}{2}$, it follows that
$s=\frac{n}{2}$ and $t=\frac{n}{2}-1$. If $\frac{n}{2}-1\geq 4$, then
$\sigma((\frac{n}{2}-1)^{2}\cdot1^2)=0$ and
$\sigma((\frac{n}{2}-1)^{4}\cdot1^{4})=0$, which is a contradiction.
If $\frac{n}{2}-1=3$, then $n=8, r=5$ and $S=1^{4}\cdot3^{3}$, but
the subsequence $1^2\cdot 3^2$ shows $r=4$, which
is a contradiction. If $\frac{n}{2}-1=2$, then we have $n=6$ and
$S=1^{3}\cdot 2^{2}$. If $\frac{n}{2}-1=1$, then we have $n=4$ and
$S=1^3$, which contradicts our assumption $|\supp(S)|\ge 2$.

Therefore, by the above description, $S$ is one of the following
in this case:

(i) $S\cong1^{n-2}\cdot (n-r+1)$ with $n\geq 3$.

(ii) $S\cong1^{n-3}\cdot (n-r)^2$ with $r=n-2$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%55
{\bf Case 2. $r < \frac{n}{2}-1$.}

Let $T|S$ be a nontrivial zero-sum subsequence.  Setting $U=ST^{-1}$,  by
Property(i) and (ii), we have $|T|=r$ and $U$ is  zero-sum free. It
follows that $$|U|=n-1-r>\frac{n}{2}.$$
Applying the Savchev-Chen Structure Theorem, we can suppose that
 $U=1^{v}x_{1}\cdot \ldots \cdot x_t$ with $v+t=n-1-r \geq
\frac{n+1}{2}$ and $2 \leq \overline{x_1} \leq \ldots \leq \overline{x_t}\leq
v+\sum_{i=1}^{t}\overline{x_{i}}\leq n-1$ and
$\sum(U)=\{1,2,\ldots,v+\sum_{i=1}^{t}\overline{x_{i}}\}$. Consequently,
since $r-2\le n-1-r\le v+t\le v+\sum_{i=1}^{t}\overline{x_{i}}$,
it follows that $\{1,2,\cdots, r-2\}\subseteq
\sum_{\le r-2}(U)$, which implies that $\bar{x}\leq n-(r-2)-1=n-r+1$
for every $x|T$. Therefore, for any
$x|T$, we have either $\bar{x}=1$ or $$n-r+1\geq \bar{x}\geq
1+v+\sum_{i=1}^{t}\overline{{x_i}}\geq 1+\sigma(\bar{U})\geq 1+|U|=n-r.$$
Moreover, since the zero-sum subsequence $T$ cannot be all $1$'s,
there exists $x|T$ with $\bar{x}\neq 1$, which means that the above estimate
must hold for some $x|T$. In consequence, either $U=1^{n-1-r}$ or
$U=1^{n-r-2}\cdot 2$(as otherwise the above estimate can be improved to
shows no much $x|T$ exists). Moreover, $n-r \nmid
T$. Otherwise, $1^r\cdot (n-r)$ gives a zero-sum subsequence of length
$r+1$ in view of $r\le n-r-2$. Thus $T=1^{t}\cdot (n-r+1)^s$.

Suppose that $U=1^{n-1-r}$.  Since $r\geq 2$, we have
$n-2r+2< n$. If $n-r \leq n-2r+2$, then
$r=2$, and so $T=1\cdot(n-1)$ and $U=1^{n-3}$, that is, $S=1^{n-2}\cdot
(n-1)$. If $n-2r+2 \leq n-r-1$ and $s\geq 2$, then
$$T\cdot(n-r+1)^{-2}\cdot 1^{(n-2r+2)}$$ is a zero-sum
subsequence, which forces $n-2r+2=2$ and $r=\frac{n}{2}$. It arises  a
contradiction. Therefore, $T=(n-r+1)\cdot 1^{r-1}$ and
$S=1^{n-2}\cdot (n-r+1)$.

Suppose that $U=1^{n-r-2}\cdot 2$. If $n-r+1 \leq n-2$, then $r\geq
3$. However $\sigma((n-r+1)\cdot 1^{r-1})=0$ and
$\sigma((n-r+1)\cdot 2 \cdot 1^{r-3})=0$, which is a contradiction.
If $n-r+1
> n-2$, then $r=2$, and so $S=1^{n-3}\cdot 2\cdot (n-1)$.

Therefore, in this case $S$ is one of the following:

(i) $S\cong1^{n-2}\cdot (n-r+1)$ with $n\geq 3$.

(ii) $S\cong1^{n-3}\cdot 2 \cdot (n-1)=1^{n-3}\cdot 2 \cdot (n-r+1)$
with $r=2$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

{\bf Case 3. $r = \frac{n}{2}-1$.}

Let $T$ be a nontrivial zero-sum subsequence of $S$, and set $U=ST^{-1}$. By
Property(i) and (ii), we have $|T|=\frac{n}{2}-1, |U|=\frac{n}{2}$ and $U$ is
zero-sum free. Then $$\frac{n}{2}=|U|\leq |\sum(U)|\leq n-1= 2|U|-1.$$
Since $r \geq 2$, we have $n\geq 6$. By Theorem \ref{t2}, we divide the
proof into five subcases.


{\bf Subcase 3.1.} $U$ is $a$-smooth for some $a\in G$. Since
$|U|=\frac{n}{2}$, without loss
of generality set, we may set $a=1$. In the following, we prove that either $S=1^{n-2}\cdot
(\frac{n}{2}+2)$ with $n\geq 6$ or $S=1^{2}\cdot 3^{3}$ or
$S=1^3\cdot 2\cdot 5$ with $n=6$.

Let $U=1^{v}x_1\cdot \ldots \cdot x_t,  2\leq
\overline{x_1}\leq \cdots \leq \overline{x_t}$ be
$1$-smooth with $v+\sum_{i=1}^{t}\overline{x_i}\leq
n-1$ and $v\geq 1$. Since
\begin{eqnarray*}\label{eq:1}
n-1 &\ge& v+\overline{x_1} + \cdots + \overline{x_t} \\
& \ge & v+2(t-1)+\overline{x_t}\\
 & \ge & v+ 2t-2+\overline{x_t}\\
  & \ge &
2(v+t)-v-2+\overline{x_t}\\
&=& n-v-2+\overline{x_t}, \end{eqnarray*}
 we have $\overline{x_t}\leq v+1$. Let $\mathsf{v}_{x_t}(U)=b$.


(i) If  $\overline{x_t}=v+1$, then equality holds in (\ref{eq:1}), which implies
that either $b=1$ or $v=1$. If $v=1$, then $U=1\cdot 2^{\frac{n}{2}-1}$ and
$|\sum(U)|=n-1$. Thus for any $x|T$, we have $x\in \sum(U)$, and so,
by Property(iii), we have either $\bar{x}=1$ or $\bar{x}=2$. This contradicts
that $T$ is a minimal zero-sum subsequence with length $
r=\frac{n}{2}-1$. If $b=1$ with $v\ge 2$, then we have $U=1^{v}\cdot
2^{\frac{n}{2}-v-1}\cdot (v+1)$ and $|\sum(U)|=n-1$ with $v\geq 2$.
Similarly, for any $x|T$, we have $x\in \sum(U)$, and so $\bar{x}=1$ or
$\bar{x}=2$ or $\bar{x}=v+1$. Since $v\geq 2$, we have $2\nmid T$, so
$T=1^{s}\cdot (v+1)^t$ with $s+t=r$. Because the zero-sum subsequence
$T$ cannot be all $1$'s, there exists $x|T$ such that $\bar{x}=v+1$. If $2|U$,
then by $v+1=(v-1)+2, 1^s\cdot 2\cdot (v+1)^{t-1}\cdot U^{'}$ is a
zero-sum subsequence with length $>r$, where $U^{'}$ is a subsequence of
$U\cdot 2^{-1}$ satisfying $\sigma(U^{'})=v-1$ and $|U^{'}|\ge 1$. So we
can assume
$U=1^{v}\cdot (v+1)=1^{\frac{n}{2}-1}\cdot \frac{n}{2}$ and
$T=1^{s}\cdot (v+1)^t=1^{s}\cdot (\frac{n}{2})^{t}$. Therefore, $r=2$,
and so $n=6$ and $S=1^{2}\cdot 3^{3}$.

(ii) If $\overline{x_t}\leq v$, then $\sum(U)=\{1, 2, \cdots,
v+\sum_{i=1}^{t}{\overline{x_i}}\}$ and $$\frac{n}{2}=|U|\leq |\sum(U)|=
v+\sum_{i=1}^{t}\overline{{x_i}}\le n-1.$$
For any $x\in \sum(U)$ and $\bar{x}\neq 1$, if $2\le \bar{x}\le v$, then there
exists a subsequence $U^{'}=1^{\bar{x}}$ of $U$ such that $\sigma(U^{'})=\bar{x}$
and $|U^{'}|=\bar{x}\ge 2$. If $\bar{x}\ge v+1$, then there exists a subsequence
$U^{''}$ of $U$ such that $\sigma(\overline{U^{''}})=\bar{x}$ and $|U^{''}|\ge 2$.
So by Property (iii), for any $x|T$, we have either $\bar{x}=1$ or
$\bar{x}\ge \sigma(\bar{U})+1\ge \frac{n}{2}+1$. If there is an element $x|T$ such that
$\bar{x}>n-(r-1)=\frac{n}{2}+2$, then there exists a subsequence $U_2$ of
$U$ such that $x\cdot U_2$ is a zero-sum subsequence of length less than
$r$. Therefore, for any $x|T$, we have either $\bar{x}=1$ or
$\frac{n}{2}+1\le \sigma(\bar{U})+1\le \bar{x} \le \frac{n}{2}+2$, and so we
have either $\sigma(\bar{U})=\frac{n}{2}$ or $\sigma(\bar{U})=\frac{n}{2}+1$.
Since $|U|=\frac{n}{2}$ and $U$ is $1$-smooth, we have either
$U=1^{\frac{n}{2}}$ or $U=1^{\frac{n}{2}-1}\cdot 2$.

Since $(\frac{n}{2}+1)\cdot 1^{\frac{n}{2}-1}$ is a zero-sum
subsequence of length $r+1$, it follows that $\frac{n}{2}+1\nmid T$.
Thus $T=1^{s}\cdot (\frac{n}{2}+2)^t$ with $s+t=\frac{n}{2}-1$, and
so $s+(\frac{n}{2}+2)t\equiv 0(\mbox{mod}  n)$. It follows that we
have $t=1$ and $T=1^{\frac{n}{2}-2}\cdot (\frac{n}{2}+2)$.

If $U=1^{\frac{n}{2}}$, then $S=1^{n-2}\cdot (\frac{n}{2}+2)$. If
$U=1^{\frac{n}{2}-1}\cdot 2$, then we must have $\frac{n}{2}-2< 2$(otherwise,
$1^{\frac{n}{2}-4}\cdot 2\cdot (\frac{n}{2}+2)$ is a zero-sum subsequence
of $S$ with length $\frac{n}{2}-2<r$),
 and so
$n=6$ and $S=1^{3}\cdot 2\cdot 5$.


{\bf Subcase 3.2.} $U=a^{\frac{n}{2}-1}\cdot b$ is not smooth with
$\ord(a)\geq \frac{n}{2}$.

(i) $\ord(a)=\frac{n}{2}$. Without loss of generality, we set $a=2$.
Then $U=2^{\frac{n}{2}-1}\cdot b$, where $b$ is odd. By
Property(iii), for any $x|T$, we have either $\bar{x}=2$ or $x=b$.
Clearly, $2 \nmid T$. Hence we have $x=b$ and $T=b^{\frac{n}{2}-1}$,
and so $\sigma(b^{\frac{n}{2}-1})=0$. Therefore,
$\ord(b)=\frac{n}{2}-1$ and $\frac{n}{2}-1 \mid n$, and so $n=6$ and
$S=2^{3}\cdot 3^2$.

(ii) $\ord(a)=n$. Without loss of generality, we set $a=1$. Then
$U=1^{\frac{n}{2}-1}\cdot b$ with $\bar{b} \geq \frac{n}{2}+1$ because
$U$ is not $1$-smooth. However,
$0\in \sum(U)$, which is a contradiction.

{\bf Subcase 3.3.} $U=a^{k}\cdot (a+g)^{l}$ is not smooth with $|U|=\frac{n}{2},
\ord(g)=2$ and $k\geq l\geq 1$. If $l$ is even, then since $k\ge 1$
and $\ord(g)=2$, we must have
$$\sum(U)=\{a,\cdots, (k+l)a,a+g,\cdots,(k+l-1)a+g\}.$$
If $l$ is odd, then, similarly, we must have
$$\sum(U)=\{a,\cdots, (k+l-1)a,a+g,\cdots,(k+l)a+g\}.$$
Thus $|\sum(U)|=(k+l-1)+(k+l)=n-1$
and $\ord(a)\geq \frac{n}{3}$.

(i) $\ord(a)=\frac{n}{3}$. Since $r=\frac{n}{2}-1$ and
$\{a,2a,\cdots, (k+l-1)a\}\subseteq \sum(U)$ with $k+l-1=\frac{n}{2}-1$,
we have $\frac{n}{3}>\frac{n}{2}-1$, which contradicting that $n\ge 6$.

(ii) $\ord(a)=\frac{n}{2}$. Without loss of generality, we set $a=2$.
Then $U=2^{k}\cdot (\frac{n}{2}+2)^l$ with $l$  odd. For any $x|T$,
by Property(iii), we have either $\bar{x}=2$ or $\bar{x}=\frac{n}{2}+2$. If
$2|T$, then $2^{k}\cdot (\frac{n}{2}+2)^{l-1}\cdot 2$ is a zero-sum
subsequence of length $\frac{n}{2}> r$, which is a contradiction.
Therefore, $T=(\frac{n}{2}+2)^{\frac{n}{2}-1}$, and so $n\le 4$,
which is also a contradiction.

(iii) $ord(a)=n$. Without loss of generality, we set $a=1$. Then
$U=1^{k}\cdot (\frac{n}{2}+1)^l$ with $k\geq l \geq 1$. Similarly,
for any $x|T$, we have either $\bar{x}=1$ or $\bar{x}=\frac{n}{2}+1$. Thus
$T=1^{s}(\frac{n}{2}+1)^t$ with $s+t=\frac{n}{2}-1$. However
$\sigma(T)\neq 0$.

{\bf Subcase 3.4.} $U=a^{k}\cdot (a+g)^{l}\cdot g$ is not smooth with
$|U|=\frac{n}{2}, \ord(g)=2$ and $k\geq l\geq 1$. By a similar argument
of subcase 3.3, we have $|\sum(U)|=n-1$ and $\ord(a)\geq \frac{n}{3}$.

(i) $\ord(a)=\frac{n}{3}$. Since $r=\frac{n}{2}-1$ and
$\{a,2a,\cdots, (k+l)a\}\subseteq \sum(U)$ with $k+l=\frac{n}{2}-1$,
we have $\frac{n}{3}>\frac{n}{2}-1$, which contradicting that $n\ge 6$.

(ii) $\ord(a)=\frac{n}{2}$. Without loss of generality, we set $a=2$ and
$g=\frac{n}{2}$. Then $U=2^{k}\cdot (\frac{n}{2}+2)^l \cdot
\frac{n}{2}$. By Property(iii), for any $x|T$, we have $\bar{x}=2,
\frac{n}{2}$ or $\frac{n}{2}+2$. If $\frac{n}{2}|T$, then $r=2,
n=6$, and $S=2\cdot 3^3\cdot 5$, which is equivalent to $S=1\cdot
3^3\cdot 4$. If $2|T$, then we have either
$\sigma(2^k(\frac{n}{2}+2)^{l}2)=0$ or
$\sigma(2^k(\frac{n}{2}+2)^{l}\frac{n}{2}2)=0$. However these length
$\geq \frac{n}{2}>r$, which is a contradiction. Therefore,
$T=(\frac{n}{2}+2)^{\frac{n}{2}-1}$, and so $(\frac{n}{2}+2)^{\frac{n}{2}}$
or $(\frac{n}{2}+2)^{\frac{n}{2}}\cdot \frac{n}{2}$ is a zero-sum
subsequence of length at least $\frac{n}{2}>r$, which is a contradiction.

(iii) $\ord(a)=n$. Without loss of generality, we set $a=1$. Then
$U=1^{k}\cdot (\frac{n}{2}+1)^{l}\cdot {\frac{n}{2}}$. Similarly,
for any $x|T$, we have $\bar{x}=1, \frac{n}{2}$ or $\frac{n}{2}+1$. If
$\frac{n}{2}|T$, then we have $r=2, n=6$ and $S=1\cdot 3^3\cdot 4$.
If $T=1^{s}\cdot (\frac{n}{2}+1)^l$, then $s+(\frac{n}{2}+1)l\neq0\pmod n$,
which is a contradiction.

{\bf Subcase 3.5.} $U=a^x\cdot b\cdot c$ is not smooth, where
$\ord(a)=x+2=\frac{n}{2}$ and $b,c\in G\setminus \langle a\rangle$
are distinct with $b+c=a$ and $b-c\in \langle a\rangle$. Then
$\sum(U)=\langle a,b,c\rangle\setminus \{0\}=G\setminus \{0\}$.

By Property(iii), $\langle a\rangle\cap \supp(T)=\varnothing $. If
$bc|T$, then $T(bc)^{-1}a$ is a zero-sum subsequence with length
$r-1$. So by Property(iii) and $\sum(U)=G\setminus \{0\}$, we
have $T=b^s$ or $T=c^s$ with $s=\frac{n}{2}-1$. If $n> 6$, we have
$\sigma(T)\neq 0$, which contradicts $\sigma(T)=0$. If $n=6$,
without loss of generality, we set $a=2$. Then $U=2\cdot 3\cdot 5$
and $T=3^2$. Hence, $S=2\cdot 3^3\cdot 5\cong 1\cdot 3^3\cdot 4$.

Therefore, in this case $S$ is one of the following:

(i) $S\cong1^{n-2}\cdot (n-r+1)$ with $n\geq 3$.

(ii) $S\cong1^{2}\cdot 3^{3}$ or $S\cong2^{2}\cdot 3^{3}$ or
$S\cong1\cdot 3^3\cdot 4$ or $S\cong1^3\cdot
2\cdot 5$ with $n=6$ and $r=2$.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

{\bf Case 4. $r=\frac{n+1}{2}$}.

By Lemma \ref{l7}, we have $h(S) \geq r-1=\frac{n-1}{2}$. Then there
exists some element $a|S$ such that $\mathsf{v}_a(S)=h(S)\geq
r-1=\frac{n-1}{2}$ and $\ord(a)=n$. Without loss of generality, we
set $a=1$. Let $T$ be a zero-sum free subsequence of $S$ such that
$|\sum(T)|\geq 2|T|-1$ with $|T|$ maximal. Set $U=ST^{-1}$. Since
$n-1\ge |\sum(T)|\ge 2|T|-1$, we have $|T|\leq \lfloor\frac{n}{2}\rfloor=r-1$.

{\bf Subcase 4.1.}   $|T| \leq r-2$.

By Claim 1(i), we have $U=1^{n-1-|T|}$ and $$\supp(T)\subseteq \{1,
n-r, n-r+1\}=\{1, \frac{n-1}{2}, \frac{n+1}{2}\}.$$ Since $n-1-|T|
\geq n-r+1=\frac{n+1}{2}$, it follows that $\frac{n-1}{2}\cdot
1^{\frac{n+1}{2}}$ is a zero-sum sequence with length $r+1$.
Hence $\frac{n-1}{2}\nmid T$, and so $T=1^t\cdot (\frac{n+1}{2})^s$.
If $s\geq 3$, then we obtain a zero-sum subsequence
$(\frac{n+1}{2})^3\cdot 1^{\frac{n-3}{2}}$ of length $r+1$. If
$s=2$, then $S\cong1^{n-3}(\frac{n+1}{2})^2$. If $s=1$, then we have
$S\cong1^{n-2}\cdot \frac{n+1}{2}$.

{\bf Subcase 4.2.}  $|T|=r-1=\frac{n-1}{2}$.

Obviously, $|\sum(T)|\geq n-2$. There are three options consider.

(i) If $Ta$ is a zero-sum free for every $a|U$, then
$|\sum(Ta)|=|\sum(T)|+1$. By Lemma \ref{l2}, we have
$U=a^{n-1-|T|}=a^{\frac{n-1}{2}}$, $\sum(T)=\{a, 2a, \cdots,
(\ord(a)-2)a\}\cup (H-cosets)$ and $\sigma(T)=(\ord(a)-2)a$, where
$H=\langle a\rangle$. If $a\neq 1$, then
$T=1^{r-1}=1^{\frac{n-1}{2}}$ because of
$\mathsf{v}_1(S)=h(S)\ge \frac{n-1}{2}$. It follows from $|\sum(T)|\geq
2|T|-1$ that $|T|=1$ and $n=3$, contradicting that $n\ge 4$.
Thus $a=1$ and $\sum(T)=\{a, 2a, \cdots,(n-2)a\}$.
However, $T\cdot 1^2$ is a zero-sum subsequence of length $r+1$.

(ii) If $Ta$ is a zero-sum subsequence for every $a|U$, then
$U=a^{n-1-|T|}=a^{\frac{n-1}{2}}$. Similarly, we have
$U=1^{\frac{n-1}{2}}$, and so $\sigma(T)=n-1$. For any $x|T$, since
$\sigma(T)=n-1$, if $2\le \bar{x}\le \frac{n-3}{2}$, then
$\sigma(Tx^{-1})=\bar{y}\in[\frac{n+1}{2},n-3]$, and then
$1^{n-\bar{y}}\cdot T\cdot x^{-1}$ is zero-sum subsequence of length
distinct from $r$. If $\bar{x}> \frac{n+1}{2}$, then $x\cdot 1^{\bar{x}}$
is a zero-sum subsequence of length $n-\bar{x}+1<\frac{n+1}{2}=r$, a
contradiction. Thus we have either $\bar{x}=1$ or $\frac{n-1}{2}\leq \bar{x}\leq
\frac{n+1}{2}$. Because the zero-sum subsequence $T\cdot a$ cannot be
all $1$'s, there exists $x|T$ such that $\bar{x}=\frac{n-1}{2}$ or
$\bar{x}=\frac{n+1}{2}$. Since $\sigma(\frac{n-1}{2}\cdot \frac{n+1}{2})=0$
and $T$ is zero-sum free,
it follows that $\frac{n-1}{2}$ and $\frac{n+1}{2}$ are not in $T$
simultaneously. So we can assume $\supp(T)\subseteq \{1,\frac{n-1}{2}\}$
or $\supp(T)\subseteq \{1,\frac{n+1}{2}\}$.

If $\supp(T)\subseteq \{1,\frac{n-1}{2}\}$, then $T=1^{t}\cdot(\frac{n-1}{2})^{s}$ with
$t+s=\frac{n-1}{2}$. If $s=1$, then $T=1^{r-2}\cdot
\frac{n-1}{2}=1^{\frac{n-3}{2}}\cdot \frac{n-1}{2}$, and
$\sigma(T)=n-2$, which is a contradiction. Thus $s \geq 2$. Since
$\frac{n-1}{2}+\frac{n-1}{2}+1 =n$, it follows that $r=3, n=5$ and
$T=(\frac{n-1}{2})^2$, and so $S=1^{2}\cdot 2^{2}$.

If $\supp(T)\subseteq \{1,\frac{n+1}{2}\}$, then $T=1^{t}\cdot(\frac{n+1}{2})^{s}$ with
$t+s=\frac{n-1}{2}$. If $s\geq 3$, then
$1^{\frac{n-3}{2}}\cdot(\frac{n+1}{2})^3$ is z zero-sum subsequence of length $>r$.
If $s=2$, then $T=1^{r-3}\cdot (\frac{n+1}{2})^2$. However,
$\sigma(1\cdot T)\neq 0$. Therefore, $T=1^{r-2}\cdot \frac{n+1}{2}$
and $S=1^{n-2}\cdot \frac{n+1}{2}$.

(iii) Suppose that there exist $a, b\in\supp(U)$ such that $Ta$ is zero-sum
free and $Tb$ is zero-sum. Then by Lemma \ref{l2}, we have $\sum(T)=\{a,
2a, \cdots, (\ord(a)-2)a\}\cup(H-cosets)$ with $H=\langle a\rangle$ and
$\sigma(T)=(\ord(a)-2)a$, and so $b=2a$. Since $T\cdot a^2$ and $T\cdot b$
are zero-sum, by Lemma \ref{l2}(iii), we must have
$U=a\cdot (2a)^{n-r-1}=a\cdot (2a)^{\frac{n-3}{2}}$. If $a\neq 1$ and
$2a\neq  1\pmod n$, then we have $T=1^{r-1}=1^{\frac{n-1}{2}}$ and
$2a\equiv 1\pmod n$ because of $\mathsf{v}_1(S)=h(S)\ge \frac{n-1}{2}$.
So $U=(\frac{n+1}{2})^{\frac{n-3}{2}}\cdot a$. Since
$1^{\frac{n-3}{2}}\cdot (\frac{n+1}{2})^3$ is zero-sum of length $r+1$,
we have $\frac{n-3}{2}\le 2$, and so $n\le 7$. If $\bar{a}\ge \frac{n+3}{2}$,
then $1^{n-\bar{a}}\cdot a$ is zero-sum of length $n-\bar{a}+1<r$.
If $2\le \bar{a}\le \frac{n-1}{2}$, then
$1^{\frac{n-1}{2}-\bar{a}}\cdot \frac{n+1}{2}\cdot a$ is zero-sum of length
$\frac{n-1}{2}-\bar{a}+2<r$. Then we have $\bar{a}=\frac{n+1}{2}$, so
$2a\equiv 1\pmod n$, which is a contradiction. Thus $a=1$ or $2a\equiv 1\pmod n$,
so $U=1 \cdot 2^{\frac{n-3}{2}}$, or $U=1^{\frac{n-3}{2}}\cdot \frac{n+1}{2}$.

If $U=1 \cdot 2^{\frac{n-3}{2}}$, then
$\sum(U)=\{1, 2, \cdots, n-2\}$. Applying Property(iii) to
$T\cdot (2a)=T\cdot 2$ and recalling that $|T|=r-1$ and
$\sum(T)=\{1,2,\cdots,n-2\}$(since $a=1$),  for any
$x|T$, we have $\bar{x}=1$ or $\bar{x}=2$ or $\bar{x}= n-1$. If $n-1|T$, then the
zero-sum subsequence $-1\cdot 1$ implies that $r=2$, in which case
$n=3$, contrary to $n\ge 4$. So for any $x|T$, we have $\bar{x}=1$ or $\bar{x}=2$.
We can assume $T=1^s\cdot 2^t$ with $s+t=\frac{n-1}{2}$ and
$s+2t=n-2$, and so $t=\frac{n-3}{2}$ and $s=1$. It follows that
$T=1\cdot 2^{\frac{n-3}{2}}$ and $S=1^2\cdot 2^{n-3}$,
which is equivalent to $S\cong1^{n-3}\cdot (\frac{n+1}{2})^2$.

Suppose $U=1^{\frac{n-3}{2}}\cdot \frac{n+1}{2}$. If there is a term
$x|T$ satisfying $2\le \bar{x}\le \frac{n-1}{2}$, then
$1^{\frac{n-1}{2}-\bar{x}}\cdot \frac{n+1}{2}\cdot x$ is zero-sum of
length $\frac{n+3}{2}-\bar{x}<r$. If there is a term $x|T$ satisfying
$\bar{x}\ge \frac{n+3}{2}$, then
$1^{n-\bar{x}}\cdot x$ is zero-sum of
length $n-\bar{x}+1\le\frac{n-1}{2}<r$. Thus for any $x|T$, we have
$\bar{x}=1$ or $\bar{x}=\frac{n+1}{2}$. So we can assume $T=1^s\cdot (\frac{n+1}{2})^t$.
If $t\ge 2$, then $1^{\frac{n-3}{2}}\cdot(\frac{n+1}{2})^3$ is zero-sum
of length $\frac{n+3}{2}>r$. Thus we have $t\le 1$. Since
$|\sum(T)|\ge 2|T|-1$, we have
$T=1^{r-2}\cdot (\frac{n+1}{2})=1^{\frac{n-3}{2}}\cdot \frac{n+1}{2}$,
and so $S=1^{n-3}\cdot (\frac{n+1}{2})^2$.


Therefore, in this case $S$ is one of the following:
\begin{enumerate}[(i)]
\item $S\cong1^{n-2}\cdot (n-r+1)$ with $n\geq 3$.
\item $S\cong1^{n-3}\cdot (\frac{n+1}{2})^2=1^{n-3}\cdot (n-r+1)^2$ with
$r=\frac{n+1}{2}$ and $n\geq 3$.
\item $S\cong 1^2\cdot 2^2$ with $n=5$ and $r=3$.
\end{enumerate}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Now we give two results. Suppose $\frac{n-1}{2}\leq r\leq \frac{n}{2}$
and $\mathsf{v}_a(S)=\mathsf{v}_1(S)=h(S)\ge r-1$. Let $T=S\cdot 1^{-h(S)}$.
Assuming $\sigma(\overline{T})\ge n-r+1$, define $T_0$ as follows. Let
$T=x_1\cdot \ldots \cdot x_l$ with $2\le \overline{x_1}\le \cdots \le \overline{x_l}$.
Then there must be a minimal index $u\ge 1$ such that
$\overline{x_1}+ \cdots + \overline{x_u}\ge n-r+1$. Set $T_0=x_1\cdot \ldots \cdot x_u$.
Then we have the following results.


{\bf Claim 2} If $\bar{x}< r$
for every $x|T$, then either $\sigma(\overline{T})\leq n-r$
or else $\sigma(\overline{T})\ge n-r+1$ and
$\overline{\sigma(T_0)}=\sigma(\overline{T_0})=n-r+|T_0|$ with $2\le |T_0|\le r-1$.

{\bf Proof of Claim 2:} We assume $\sigma(\overline{T})\ge n-r+1$ and
proceed to show $\sigma(\overline{T_0})$ has the desired value.
Since $\overline{x_1}\le r\le n-r$(in view of $r\le \frac{n}{2}$), we
must have $|T_0|\ge 2$. Let $T_0^{'}=T_0\cdot x_u^{-1}$. Then the
minimality of $u$ ensures that $\sigma(\overline{T_0^{'}})\le n-r$,
which together with $\overline{x_u}\le r$ gives $\sigma(\overline{T_0})\le n$.

Suppose $\sigma(\overline{T_0})=n$. Then we must have $\overline{x_u}=r$
and $T_0$ is a zero-sum subsequence. Thus $|T_0|=r$ and
$n=\sigma(\overline{T_0})\ge r+2(r-1)$, which implies $r\le \lfloor \frac{n+2}{3}\rfloor$.
In view of $r\ge \lceil \frac{n-1}{2}\rceil$ and $n\ge 4$, this is only
possible if $n=7$ or $n=5$ or $n=4$, with
$r=\lfloor \frac{n+2}{3}\rfloor=\lceil \frac{n-1}{2}\rceil$. If
$n=4$ or $n=5$, then $r=2$, in which case $\supp(T_0)=\{2\}$.
Thus, if $n=4$, then $T_0=2\cdot 2$, so that $\mathsf{v}_1(S)\le |S|-2=1$,
contradicting that $\mathsf{v}_1(S)=h(S)$, while if $n=5$,
then $\sigma(\overline{T_0})=n=5$ is not possible as $\sigma(\overline{T_0})$
must be even. Finally, if $n=7$, then $r=3$ and $T_0=2\cdot 2\cdot 3$.
But then, since $\mathsf{v}_1(S)=h(S)\ge r-1=2$, the zero-sum
$1^2\cdot 2\cdot 3$ contradicts that $r=3$. So we have obtained a
contradiction in all cases and can instead assume
$\sigma(\overline{T_0})\le n-1$.

Thus $n-r+1\le \overline{\sigma(T_0)}\le\sigma(\overline{T_0})\le n-1$(
with the first inequality in view of the definition of $T_0$). But
now $T_0\cdot 1^{n-\sigma(\overline{T_0})}$ is a zero-sum sequence of
length $|T_0|+n-\sigma(\overline{T_0})$. Moreover, it is a subsequence
in view of $n-\sigma(\overline{T_0})\le r-1\le h(S)\le \mathsf{v}_1(S)$,
which forces $\sigma(\overline{T_0})=n-r+|T_0|$. Since
$\sigma(\overline{T_0})\le n-1$, it  follows that $|T_0|\le r-1$.\\



{\bf Claim 3} Suppose that  $\sigma(\overline{T})\ge n-r+1$ and
$\bar{x}\le r$ for every $x|T$. Let $T_{1}=T_0\cdot 1^{r-|T_0|}$(which
is a zero-sum subsequence of length $r$). Then $\supp(T)=\{x\}$ for
some $\bar{x}\in[2,n-1]$, so that $T_1=1^{v}\cdot x^{u}$ and
$ST_{1}^{-1}=1^{t}\cdot x^s$ with $s\ge 1$. Moreover,
$\bar{x}> v$ and $\bar{x}> t$.

{\bf Proof of Claim 3:} First, let us show that $T_0\neq T$.
Suppose to the contrary that $T_0=T$. Then Claim 2 implies
$\sigma(\overline{T})=n-r+|T|$ with $2\le |T|\le r-1$.
Consequently, letting $T^{'}=T\cdot x_1^{-1}$, we have
$\sigma(\overline{T^{'}})=n-r+|T|-\overline{x_1}$ with
$\overline{x_1}\in [2,r]$, in which case
$$3\le \overline{x_1}+1\le n-\sigma(\overline{T^{'}})=
r-|T|+\overline{x_1}\le 2r-|T|\le n-|T|=\mathsf{v}_1(S)+1.$$
If the above upper bound were tight, then we must have $\overline{x_1}=r$
and $r=\frac{n}{2}$. Hence $\supp(T)=\{r\}=\{\frac{n}{2}\}$
with $|T|\ge |T_0|\ge 2$, whence the zero-sum subsequence
$\frac{n}{2}\cdot \frac{n}{2}$ force $\frac{n}{2}=r=2$. But now
$\mathsf{v}_2(S)\ge 2$ has greater multiplicity than $1$,
contradicting that $\mathsf{v}_1(S)=h(S)$. Therefore, we conclude
that the above upper bound is not right. As a result, the zero-sum
subsequence $1^{n-\sigma(\overline{T^{'}})}\cdot T^{'}$ forces
$$r=n-\sigma(\overline{T^{'}})+|T^{'}|=n-(n-r+|T|-\overline{x_1})+|T|-1,$$
implying $\overline{x_1}=1$, which contradicts that $\overline{x_1}\ge 2$.
So we instead conclude that $T_0\neq T$. Thus let
$x\in \supp(T\cdot T_0^{-1})$.

Let $x_i\in \supp(T_0)$ and let $T_0^{'}=T_0\cdot x_i^{-1}$. If
$\sigma(\overline{T_0^{'}})\ge n-r+1$, then
$1^{n-\sigma(\overline{T_0^{'}})}\cdot T_0^{'}$ will be a  zero-sum
subsequence (in view of $\sigma(\overline{T_0^{'}})\le
\sigma(\overline{T_0})\le n-1$) of length (in view of Claim 2)
$$n-\sigma(\overline{T_0^{'}})+|T_0^{'}|=
n-(n-r+|T_0|-\overline{x_i})+|T_0|-1=r+\overline{x_i}-1,$$
which forces $\overline{x_i}=1$, contradicting that $\overline{x_i}\ge 2$.
Therefore we may instead assume $\sigma(\overline{T_0^{'}})\le n-r$.
Thus, since $\bar{x}\le r$ for every $x|T$, it follows that
$\sigma(\overline{T_0^{'}\cdot x})\le n$. However, if
$\sigma(\overline{T_0^{'}\cdot x})=n$, then $T_0^{'}\cdot x$ is a zero-sum
subsequence of length $|T_0^{'}\cdot x|=|T_0|\le r-1$(in view of Claim 2),
which is not possible. Therefore
$\overline{\sigma(T_0^{'}\cdot x)}\le n-1$. By
the definitions of $T_0$ and $x$, we have $\bar{x}\ge \overline{x_u}\ge \overline{x_i}$.
Thus $\sigma(\overline{T_0^{'}\cdot x})\ge \sigma(\overline{T_0})\ge n-r+1$.
Consequently, as in the proof of Claim 2, the zero-sum subsequence
$1^{n-\sigma(\overline{T_0^{'}\cdot x})}\cdot T_0^{'}\cdot x$ forces
$$\sigma(\overline{T_0})-\overline{x_i}+\bar{x}=\sigma(\overline{T_0^{'}}\cdot x)
=n-r+|T_0^{'}\cdot x|=n-r+|T_0|,$$
whence Claim 2 implies $x=x_i$. Since $x_i\in \supp(T_0)$ and
$x\in \supp(T\cdot T_0^{-1})$ were arbitrary, we have now established
that $\supp(T)=\{x\}$, which clearly implies that $T_1=1^v\cdot x^u$ and
$S\cdot T_1^{-1}=1^t\cdot x^s$.

We have $s\ge 1$ since $T\neq T_0$. That $T_1$ is zero-sum follows by a simple calculation
using Claim 2. If $\bar{x}\le t$, then we could replace a single term
$x$ in the zero-sum $T_1$ with $\bar{x}\ge 2$ terms equal to $1$ from
$S\cdot T^{-1}$, yielding a zero-sum of length other than $r$. Likewise,
If $\bar{x}\le v$, then we could replace $v\ge \bar{x}\ge 2$ terms equal
to $1$ in $T_1$ with single term equal to $x$ from $S\cdot T_1^{-1}$(which
exists since $s\ge 1$) to yield a zero-sum of length other than $r$. Thus
$\bar{x}> v$ and $\bar{x}> t$.\\






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

{\bf Case 5. $r=\frac{n}{2}$.}

By Lemma \ref{l7}, we have $h(S) \geq r-1=\frac{n}{2}-1$. Then there
exists $a|S$ such that $\mathsf{v}_{a}(S)\geq \frac{n}{2}-1$. There are two subcases to consider.

{\bf Subcase 5.1.} $\ord(a)=\frac{n}{2}$. Without loss of generality,
we set $a=2$. Let $T=S\cdot 2^{-h(S)}$ and $x|T$, which exists else
$S=2^{n-1}$, as desired. If $x$ is even, then $x\neq 2$ as all terms equal
to $2$ have already been removed, and then we obtain a zero-sum subsequence
of length less than $\frac{n}{2}$, which is a contradiction, by combining $x$
with an appropriate number of the other $\mathsf{v}_2(S)=h(S)\ge \frac{n}{2}-1$
terms from $S$ equal to $x=2$. Therefore all terms of $T$ are odd.

If $S$ has only one odd element, then $S=2^{n-2}\cdot b$ with $b$
odd. If $S$ has precisely two odd elements $b_1$ and $b_2$, then $b_1
+b_2$ is even and $S=2^{n-3}\cdot {b_1}\cdot {b_2}$. It is easy to
show that $\overline{{b_1}+{b_2}}=4$. Thus $S=2^{n-3}\cdot b \cdot (4-b)$ with
$b$ odd. If $S$ has more than two odd elements, then, letting
 $x_1, x_2$ and $x_3$
be three odd elements in $S$, we have $\overline{x_1 +x_2}=\overline{x_1 +x_3}
=\overline{x_2 +x_3} =4$,
and so $\overline{x_1} =\overline{x_2} =\overline{x_3}= \frac{n}{2}+2$
with $\frac{n}{2}$ odd. Thus
$S=2^{v}\cdot(\frac{n}{2}+2)^{u}$ with $v \geq \frac{n}{2}-1$ and
$\frac{n}{2}$ odd.

{\bf Subcase 5.2.} $\ord(a)=n$. Without loss of generality, we set
$a=1$. Let $T=S\cdot 1^{-h(S)}$ and $x|T$. Then we must have $\bar{x} \leq
\frac{n}{2}+1$, otherwise $x\cdot 1^{n-\bar{x}}$ will be
zero-sum of length $<\frac{n}{2}$. So for all $x\in \supp(S)$, we
have $\bar{x} \leq \frac{n}{2}+1$. First we give a Claim.

{\bf Claim 4} Let all notation be as above and $n\ge 6$. If there
exists $x|T$ such that
$\bar{x}= \frac{n}{2}+1$, then $S=1^u\cdot (\frac{n}{2}+1)^v$, where
$u+v=n-1, u\geq \frac{n}{2}-1$ and $v\geq 1$.

{\bf Proof of Claim 4:} For $n\geq 6$, we have $r\geq 3$. If
there exists $y|T$ such that
$\bar{y}<\frac{n}{2}$, then $\frac{n}{2}+1+\bar{y}> \frac{n}{2}+2$, and so
$(\frac{n}{2}+1)\cdot y \cdot 1^{\frac{n}{2}-1-\bar{y}}$ is a zero-sum
subsequence of length $\frac{n}{2}+1-\bar{y}< \frac{n}{2}$, which is
a contradiction. Clearly, if
$\frac{n}{2}|T$, then we have
$\mathsf{v}_{1}(S)=\frac{n}{2}-1, |T|=|S|-\mathsf{v}_1(S)=\frac{n}{2}$
and $\mathsf{v}_{\frac{n}{2}}(T)=1$, as $\mathsf{v}_{1}(S)=\frac{n}{2}-1$
and $1^{\frac{n}{2}}\cdot \frac{n}{2}$ and $(\frac{n}{2})^2$ are zero-sum of
length $\neq r$. Thus
$T=\frac{n}{2}\cdot (\frac{n}{2}+1)^{\frac{n}{2}-1}$. However,
$\frac{n}{2}\cdot (\frac{n}{2}+1)^{2}\cdot 1^{\frac{n}{2}-2}$ is a
zero-sum subsequence of length $r+1$, which is also a contradiction.

By the above argument, we have $S=1^u\cdot (\frac{n}{2}+1)^v$, where
$u+v=n-1, u\geq \frac{n}{2}-1$ and $v\geq 1$. This proves the
Claim 4.\\

For $n=4$, we have $r=2$ and there must exist some $x|T$ with
$\bar{x}\ge 2=\frac{n}{2}$. In this case, we have $S=1\cdot 2\cdot 3$
or $S=1^2\cdot 3$ in view of $\mathsf{v}_1(S)=h(S)$. So we can
assume $n\ge 6$. If there is some $x|T$ with $\bar{x}\ge \frac{n}{2}+1$,
then Claim 4 completes the proof. Therefore we may assume
$\bar{x}\le \frac{n}{2}=r$ for all $x|T$, and now we can apply Claim 2.
As a result, either $\sigma(\overline{T})\le \frac{n}{2}$ or there exists a
subsequence $T_0|T$ with $\sigma(\overline{T_0})=\frac{n}{2}+|T_0|\le n-1$.
Let $|T|=u\ge 2$.

Let $T_{1}=T_0\cdot 1^{\frac{n}{2}-u}$ be a zero-sum subsequence. By
Claim 3, we have $T_1=1^{v}x^{u}$ and $ST_{1}^{-1}=1^{t}x^s$ with
$\bar{x}> v, \bar{x}> t$. Since $\sigma(\overline{T_1})=n$, it follows that
$\bar{x}=\frac{n-v}{u}=\frac{n-r}{u}+1=\frac{n}{2u}+1$ with
$\bar{x} >t$ and $\bar{x}>v$. Since $n-1\ge 2\bar{x}\ge v+t+2\geq \frac{n}{2}+1$,
we have $\bar{x} \ge \frac{n}{4}+\frac{1}{2}$. In view of the definition
of $T_0$ and $r=\frac{n}{2}$, we have $|T_0|\ge 2$, so $u=2$.  Thus
$T_1=x^{2}\cdot 1^{\frac{n}{2}-2}$ and $\bar{x}=\frac{n}{4}+1$ with $4|n$.
For $n\ge 10$, $(\frac{n}{4}+1)^3\cdot 1^{\frac{n}{4}-3}$ will be a
zero-sum subsequence of incorrect length unless $\bar{x}=\frac{n}{4}+1$
has multiplicity $2$, in which case $S=(\frac{n}{4}+1)^2\cdot 1^{n-3}$
and $(\frac{n}{4}+1)\cdot 1^{\frac{3n}{4}-1}$ is this time of incorrect
length, which is a contradiction. For $n=8$, then we have $x=3$, so
$T_1=1^2\cdot 3^2$ and $S=1^{4}\cdot 3^{3}$ by $\mathsf{v}_1(S)=h(S)$.


Therefore, in this case, $S$ is one of the following:

(i) $S\cong 2^{n-1}$.

(ii) $S\cong 2^{n-2}\cdot b$, where $b$ is odd.

(iii) $S\cong 2^{n-3}\cdot b \cdot(4-b)$, where $b$ is odd.

(iv) $S\cong 2^u \cdot (\frac{n}{2}+2)^v$, where $\frac{n}{2}$ is odd and $u\geq \frac{n}{2}-1$.

(v) $S\cong 1^u \cdot (\frac{n}{2}+1)^v$ with $u\geq \frac{n}{2}-1$ and $n\geq 5$.

(v) $S\cong 1^{4}\cdot 3^{3}$ with $n=8$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

{\bf Case 6. $r=\frac{n-1}{2}$.}

By Lemma \ref{l7}, we have $h(S) \geq r-1=\frac{n-3}{2}$. Then there
exists $a|S$ such that $\mathsf{v}_{a}(S)=h(S)\geq \frac{n-3}{2}$.
Since $n\ge 4$, so that $\frac{n}{3}< \frac{n-1}{2}$, we may
without loss of generality set $a=1$. Let
$T=S\cdot 1^{-h(S)}$. For any $x|T$, it is easy to show that $\bar{x} \leq
\frac{n+3}{2}$. We first give a Claim.

{\bf Claim 5} If there exists $x|T$ such that
$\bar{x}\geq \frac{n+1}{2}$, then we have $S=1^{n-2}\cdot \frac{n+3}{2}$
or $S=1^2\cdot 2\cdot 4$ with $n=5$, or $S=1^3\cdot 5^3\cong 1^3\cdot 3^3$
with $n=7$.

{\bf Proof of Claim 5:} Since $n\geq 5$, we have $r\geq 2$ and
$h(S)\geq 1$. If $\frac{n+1}{2}|T$, then obviously, we have
$\mathsf{v}_{1}(S)=\frac{n-3}{2}$. If $\frac{n-1}{2}|T$, then
obviously, we have $\mathsf{v}_{1}(S)\leq \frac{n-1}{2}$. If
$\mathsf{v}_{\frac{n-1}{2}}(S)\ge 2$, then $1\cdot (\frac{n-1}{2})^2$
is a zero-sum subsequence of length $3$, which implies $r=3$ and $n=7$.
Since $\mathsf{v}_1(S)\ge \frac{n-3}{2}=2$, we have
$1^2\cdot (\frac{n-1}{2})^2=1^2\cdot 3^2|S$, and so $4$ is not in $S$.
If $2|S$, then $1^2\cdot 2\cdot 3$ is a zero-sum subsequence of length
$4>r=3$. So we can assume $\supp(S)\subset \{1,3,5\}$. Because
$\mathsf{v}_1(S)=h(S)$ and $1\cdot 3\cdot 5^2, 1^4\cdot 3$ and
$1^3\cdot 3^2\cdot 5$ is a zero-sum sequence of incorrect length,
we must have $S=1^3\cdot 3^3$, which contradicts that
$\bar{x}\ge \frac{n+1}{2}$ foe some $x|T$. So we can assume
$\mathsf{v}_{\frac{n-1}{2}}(S)\le 1$.

If $(\frac{n-1}{2})(\frac{n+1}{2})|T$, then we have $r=2$ and $n=5$,
so $\mathsf{v}_1(S)=1=h(S)$ and $S=1\cdot 2\cdot 3\cdot 4$, which is a
contradiction.

If $(\frac{n+3}{2})(\frac{n+1}{2})|T$, then our above work shows that
$\mathsf{v}_1(S)=\frac{n-3}{2}$, so that $|T|=\frac{n+1}{2}\ge 3$.
If there were some $x|T$ with $\bar{x}\le\frac{n-3}{2}$, then the
zero-sum subsequence $1^{\frac{n-3}{2}-\bar{x}}\cdot x\cdot \frac{n+3}{2}$
must have length $r$, implying $\bar{x}=1$, which contradicts the
definition of $T$. Therefore, in view of the previous paragraph, we have
$\supp(T)=\{\frac{n+3}{2}, \frac{n+1}{2}\}$. We must also have
$\mathsf{v}_{\frac{n+1}{2}}(T)=1$, as otherwise
$1^{\frac{n-3}{2}-1}\cdot (\frac{n+1}{2})^2\cdot \frac{n+3}{2}$
will be a zero-sum subsequence of length $r+1\neq r$. Thus
$T=\frac{n+1}{2}\cdot (\frac{n+3}{2})^{\frac{n-1}{2}}$. But now
$\mathsf{v}_{\frac{n+3}{2}}(S)\geq \frac{n-1}{2}>
\frac{n-3}{2}= \mathsf{v}_1(S)$, contradicting that
$\mathsf{v}_1(S)=h(S)$.

If $(\frac{n+3}{2})(\frac{n-1}{2})|T$, then our above work
shows that $\mathsf{v}_{\frac{n-1}{2}}(S)=1, \mathsf{v}_1(S)\le
\frac{n-1}{2}$ and $\mathsf{v}_{\frac{n+1}{2}}(S)=0$. In view of
$\mathsf{v}_1(S)\le
\frac{n-1}{2}$, we have $|T|\ge \frac{n-1}{2}$. If there were some
$x|T$ with $\bar{x}\le \frac{n-3}{2}$, then the zero-sum subsequence
$1^{\frac{n-3}{2}-\bar{x}}\cdot x\cdot \frac{n+3}{2}$ must have
length $r$, implying $\bar{x}=1$, which contradicts the definition
of $T$. Therefore $\supp(T)= \{\frac{n+1}{2}, \frac{n+3}{2}\}$
with $\mathsf{v}_{\frac{n-1}{2}}(S)=1$, which implies
$T=\frac{n-1}{2}\cdot (\frac{n+3}{2})^{|T|-1}$. Thus
$\mathsf{v}_{\frac{n+3}{2}}(S)\ge |T|-1\ge \frac{n-3}{2}$.
Consequently, if $n> 5$, then the zero-sum subsequence
$1^{\frac{n-5}{2}}\cdot \frac{n-1}{2}\cdot (\frac{n+3}{2})^2$ has
length greater than $r$, which is a contradiction. On the other hand, if
$n=5$, then $S=1^2\cdot 2\cdot 4$ follows in view of
$\mathsf{v}_1(S)=h(S)$. In view of the above work, if $\frac{n+3}{2}|S$,
then we have $\mathsf{v}_{\frac{n-1}{2}}(S)=\mathsf{v}_{\frac{n+1}{2}}(S)=0$.
Moreover, as argued in both of the above paragraphs, we also cannot
have any $x|T$ with $\bar{x}\le \frac{n-3}{2}$. Thus
$\supp(T)=\{\frac{n+3}{2}\}$. If $\mathsf{v}_{\frac{n+3}{2}}(S)\ge 3$,
then we must have $\mathsf{v}_1(S)=h(S)\ge 3$, which implies $n\ge 7$,
and then the zero-sum subsequence $(\frac{n+3}{2})^3\cdot 1^{\frac{n-3}{2}-3}$
has incorrect length for $n> 7$, while for $n=7$, we have
 $S=1^3\cdot 5^3\cong 1^3\cdot 3^3$, as desired. If
 $\mathsf{v}_{\frac{n+3}{2}}(S)=2$, then $S=1^{n-3}\cdot (\frac{n+3}{2})^2$,
 in which case the zero-sum subsequence $1^{n-3}\cdot (\frac{n+3}{2})^2$
 has length $n-1>r$, which is a contradiction. Finally, if
 $\mathsf{v}_{\frac{n+3}{2}}(S)=1$, then $S=1^{n-2}\cdot \frac{n+3}{2}$,
 as desired. So we can now assume $\frac{n+3}{2}\nmid T$.

Since $\frac{n+3}{2}\nmid T$, our hypotheses ensure that $\frac{n+1}{2}|T$.
Then we must have $\supp(T)\subseteq \{2, \frac{n+1}{2}\}$, otherwise
$1^{\frac{n-1}{2}-\bar{x}}\cdot x\cdot \frac{n+1}{2}$ would be a zero-sum
subsequence of incorrect length if $3\le \bar{x}\le \frac{n-1}{2}$.
If $2|T$ and $\mathsf{v}_2(T)\ge 2$, then
$1^{\frac{n-9}{2}}\cdot 2^2\cdot \frac{n+1}{2}$ is zero-sum, and this
implies $n=7$ or $n=5$. If $n=5$, then $\mathsf{v}_1(S)=h(S)=\mathsf{v}_2(S)\ge 2$
and $\mathsf{v}_{\frac{n+1}{2}}(S)\ge 1$ imply $|S|\ge 5\ge n-1$,
contrary to hypothesis. For $n=7$, we have $1^2\cdot 2^2\cdot 4|S$, so
$S=1^2\cdot 2^2\cdot 4^2$ or $S=1^3\cdot 2^2\cdot 4$. The latter has
$1^3\cdot 4$ as a length $4>r$ zero-sum, while the former has
$1^2\cdot 2^2 \cdot 4^2$ as a length $6> r$ zero-sum, both contradictions.
So we have $\mathsf{v}_2(S)\le 1$. Hence
$\mathsf{v}_{\frac{n+1}{2}}(S)\ge \frac{n-1}{2}\ge 2$(as shown at the start
of the proof), so $n=5$ and $\mathsf{v}_2(S)= 1$, which implies
$S=1^2\cdot2\cdot 3$ in view of $\mathsf{v}_1(S=h(S)$. This
proves Claim 5.\\


In view of Claim 5, we may assume $\bar{x}\leq \frac{n-1}{2}=r$
for every $x|T$. Let
$T=S\cdot 1^{-h(S)}$. By Claim 2, we have that either $\sigma(\overline{T})\leq
\frac{n+1}{2}$ or there exists $T_0|T$ such that $\sigma(\overline{T_0})
=\frac{n+1}{2}+|T_0|\le n-1$. If $\sigma(\overline{T})\leq \frac{n+1}{2}$, then either
$0\notin \sum(S)$ or there exists a zero-sum subsequence of length
$>r$, which is impossible. Therefore, there exists a subsequence
$T_0|T$ such that $\sigma(\overline{T_0})=\frac{n+1}{2}+|T_0|\le n-1$
with $2\le |T_0|\le \frac{n-3}{2}$. Let $|T_0|=u\ge 2$.

Let $T_1=T_0\cdot 1^{\frac{n-1}{2}-u}$. Then $\sigma(\overline{T_1})=n$. By
Claim 3, we have $T_1=x^{u}\cdot 1^{v}$ and $ST_{1}^{-1}=x^{s}\cdot
1^{t}$ with $\bar{x}> v$ and $\bar{x}> t$. Since $\sigma(\overline{T_1})=0$,
we conclude that $\bar{x}=\frac{n+1}{2u}+1$. Since $\supp(S)=\{1,x\}$ with
$\mathsf{v}_1(S)=h(S)$, it follows that
$v+t=\mathsf{v}_1(S)\ge \frac{1}{2}|S|=\frac{n-1}{2}$. Thus,
since $\bar{x}> v$ and $\bar{x}> t$, we have
$\bar{x}\ge \frac{n+3}{2}$. Hence $\frac{n+1}{2u}+1=\bar{x}
\ge \frac{n+3}{4}$, which implies $u\le 3$ in view of $n\ge 5$.

If $u=3$, then $\frac{n+1}{2u}+1=\bar{x}\ge \frac{n+3}{4}$
implies $n=5,x=2$ and $r=2$. But  $\supp(S)=\{1,x\}$
together with $\mathsf{v}_1(S)=h(S)$ and $|T_0|\ge 2$
implies $S=1^2\cdot 2^2$, so that the zero-sum subsequence
$1\cdot 2^2$ contradicts that $r=2$. Therefore we may instead
assume $u=2$, in which case  $\bar{x}=\frac{n+1}{4}+1$
with $4 \mid n+1$, and $T_1=x^{2}\cdot 1^{v}$ with
$v=r-2=\frac{n-5}{2}$. Since $\bar{x} > v$ and $4\mid n+1$, we have
$n=11$ or $n=7$. If $n=11$, then $\supp(S)=\{1,4\}, r=5$ and
$\mathsf{v}_1(S)=h(S)$ ensure that $S=1^5\cdot 4^5, S=1^6\cdot 4^4,
S=1^7\cdot 4^3,S=1^8\cdot 4^2$ and $S=1^9\cdot 4$ are the only
possibilities for $S$. However, the zero-sum sequences
$1^2\cdot 4^5, 1^6\cdot 4^4$ and $1^7\cdot 4$ yield the
contradiction $r\neq 5$ in all but the final case. This completes
the case $n=11$.

If $n=7$, then $\supp(S)=\{1,3\}, r=3$ and $\mathsf{v}_1(S)=h(S)$
ensure that $S=1^{3}\cdot 3^{3}, S=1^4\cdot 3^2$ and $S=1^{5}\cdot 3$
are the only possibilities for $S$. However, the zero-sum sequences
$1^4\cdot 3$ and $1^7\cdot 4$ contradicts that $r=3$ in the second
case, leaving only the other two.

Therefore, in this case, $S$ is one of the following:
\begin{enumerate}[(i)]
\item $S\cong1^{n-2}\cdot (n-r+1)$ with $n\geq 5$.
\item $S\cong1^2\cdot 2 \cdot 4$ with $n=5$; $S\cong 1^{3}\cdot 3^{3}$ with
$n=7$. 
\end{enumerate}
This completes the proof.
\end{proof}

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