# Subsequence Sums of Zero-sum free Sequences II

Pingzhi Yuan
School of Mathematics, South China Normal University , Guangzhou 510631,
P.R.CHINA
e-mail<EMAIL_ADDRESS>
###### Abstract

Let $G$ be a finite abelian group, and let $S$ be a sequence of elements in
$G$. Let $f(S)$ denote the number of elements in $G$ which can be expressed as
the sum over a nonempty subsequence of $S$. In this paper, we determine all
the sequences $S$ that contains no zero-sum subsequences and $f(S)\leq
2|S|-1$.

MSC: Primary 11B75; Secondary 11B50.

Key words: Zero-sum problems, Davenport’s constant, zero-sum free sequences.

00footnotetext:  Supported by NSF of China (No. 10571180).

##  1 Introduction

Let $G$ be a finite abelian group (written additively)throughout the present
paper. $\mathcal{F}(G)$ denotes the free abelian monoid with basis $G$, the
elements of which are called $sequences$ (over $G$). A sequence of not
necessarily distinct elements from $G$ will be written in the form
$S=g_{1}\cdot\,\cdots\,\cdot g_{k}=\prod_{i=1}^{k}g_{i}=\prod_{g\in
G}g^{\mathsf{v}_{g}(S)}\in\mathcal{F}(G)$, where $\mathsf{v}_{g}(S)\geq 0$ is
called the $multiplicity$ of $g$ in $S$. Denote by $|S|=k$ the number of
elements in $S$ (or the $length$ of $S$) and let ${\rm supp}(S)=\\{g\in
G:\,\mathsf{v}_{g}(S)>0\\}$ 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 by $T|S$.
If $T|S$, then let $ST^{-1}$ denote the sequence obtained by deleting the
terms of $T$ from $S$. Furthermore, by $\sigma(S)$ we denote the sum of $S$,
(i.e. $\sigma(S)=\sum_{i=1}^{k}g_{i}=\sum_{g\in G}\mathsf{v}_{g}(S)g\in G$).
By $\sum(S)$ we denote the set consisting of all elements which can be
expressed as a sum over a nonempty subsequence of $S$, i.e.

$\sum(S)=\\{\sigma(T):T\,{\rm is\,a\,nonempty\,subsequence\,of\,}S\\}.$

We write $f(S)=|\sum(S)|$, $<S>$ for the subgroup of $G$ generated by all the
elements of $S$.

Let $S$ be a sequence over $G$. We call $S$ a $zero-sum$ $sequence$ if
$\sigma(S)=0$, a $zero-sum\,free$ $sequence$ if $\sigma(W)\neq 0$ for any
subsequence $W$ of $S$, and $squarefree$ if $\mathsf{v}_{g}(S)\leq 1$ for
every $g\in G$. We denote by $\mathcal{A}^{\star}(G)$ the set of all zero-sum
free sequences in $\mathcal{F}(G)$.

Let $D(G)$ 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
$0\in\sum(S)$. For every positive integer $r$ in the interval
$\\{1,\,\ldots,\,D(G)-1\\}$, let

$f_{G}(r)=\min_{S,\,|S|=r}f(S),$ (1)

where $S$ runs over all zero-sum free sequences of $r$ elements in $G$. How
does the function $f_{G}$ behave?

In 2006, Gao and Leader proved the following result.

Theorem A [5] Let $G$ be a finite abelian group of exponent $m$. Then

(i) If $1\leq r\leq m-1$ then $f_{G}(r)=r$.

(ii) If $\gcd(6,\,m)=1$ and $G$ is not cyclic then $f_{G}(m)=2m-1$.

Recently, Sun[10] showed that $f_{G}(m)=2m-1$ still holds without the
restriction that $\gcd(6,\,m)=1$.

Using some techniques from the author [11], the author [12] proved the
following two theorems.

Theorem B[12, 8] Let $S$ be a zero-sum free sequence over $G$ such that $<S>$
is not a cyclic group, then $f(S)\geq 2|S|-1$.

Theorem C [12] Let $S$ be a zero-sum free sequence over $G$ such that $<S>$ is
not a cyclic group and $f(S)=2|S|-1$. Then $S$ is one of the following forms

(i) $S=a^{x}(a+g)^{y},\,x\geq y\geq 1$, where $g$ is an element of order 2.

(ii) $S=a^{x}(a+g)^{y}g,\,x\geq y\geq 1$, where $g$ is an element of order 2.

(iii) $S=a^{x}b,\,x\geq 1$.

However, Theorem B is an old theorem of Olson and White [8] which has been
overlooked by the author. For more recent progress on this topic, see [4, 9,
13].

The main purpose of the present paper is to determine all the sequences $S$
over a finite abelian group such that $S$ contains no zero-sum subsequences
and $f(S)\leq 2|S|-1$. To begin with, we need the notation of $g$-smooth.

###### Definition 1.1

[7, Definition 5.1.3] A sequence $S\in\mathcal{F}(G)$ is called $smooth$ if
$S=(n_{1}g)(n_{2}g)\cdot\,\cdots\,\cdot(n_{l}g)$, 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).

We have

###### Theorem 1.1

Let $G$ be a finite abelian group and let $S$ be a zero-sum free sequence over
$G$ with $f(S)\leq 2|S|-1$. Then $S$ has one of the following forms:

(i) $S$ is $a$-smooth for some $a\in G$.

(ii) $S=a^{k}b$, where $k\in\mathbb{N}$ and $a,b\in G$ are distinct.

(iii) $S=a^{k}b^{l}$, where $k\geq l\geq 2$ and $a,b\in G$ are distinct with
$2a=2b$.

(iv) $S=a^{k}b^{l}(a-b)$, where $k\geq l\geq 2$ and $a,b\in G$ are distinct
with $2a=2b$.

For a sequence $S$ over $G$ we call

$\mathsf{h}(S)=\max\\{\mathsf{v}_{g}(S)|g\in G\\}\in[0,|S|]$
$the\,maximum\,of\,the\,multiplicities\,of\,S.$

Let $S=a^{x}b^{y}T$ with $x\geq y\geq\mathsf{h}(T)$, then Theorem 1.1(i) can
be stated more precisely as that $S$ is $a$-smooth or $b$-smooth.

##  2 Some Lemmas

Let $\emptyset\neq G_{0}\subseteq G$ be a subset of $G$ and $k\in\mathbb{N}$.
Define

$\mathsf{f}(G_{0},\,k)=\min\\{f(S):\,S\in\mathcal{F}(G_{0})\,\,{\rm~{}zero-
sumfree,\,squarefree~{}and~{}}\,|S|=k\\}$

and set $\mathsf{f}(G_{0},\,k)=\infty$, if there are no sequences over $G_{0}$
of the above form.

###### Lemma 2.1

Let $G$ be a finite abelian group.

  1. 1.

If $k\in\mathbb{N}$ and $S=S_{1}\cdot\,\cdots\,\cdot
S_{k}\in\mathcal{A}^{\star}(G)$, then

$f(S)\geq f(S_{1})+\cdots+f(S_{k})\,.$

  2. 2.

If $G_{0}\subset G$, $k\in\mathbb{N}$ and $\mathsf{f}(G_{0},\,k)>0$, then

$\mathsf{f}(G_{0},\,k)\ \left\\{\begin{array}[]{ll}=1\,,&\mbox{if}\quad
k=1\,,\\\ =3\,,&\mbox{if}\quad k=2\,,\\\ \geq 5\,,&\mbox{if}\quad k=3\,,\\\
\geq 6\,,&\mbox{if}\quad k=3\quad\mbox{and}\quad 2g\neq 0\quad\mbox{for
all}\quad g\in G_{0}\,,\\\ \geq 2k\,,&\mbox{if}\quad k\geq
4\,.\end{array}\right.$

  * Proof.

1\. See [6, Theorem 5.3.1].

2\. See [6, Corollary 5.3.4].

$\Box$

###### Lemma 2.2

Let $a,\,b$ be two distinct elements in an abelian group $G$ such that
$a^{2}b^{2}\in\mathcal{A}^{\star}(G),\,2a\neq 2b,a\neq 2b$, and $b\neq 2a$.
Then $f(a^{2}b^{2})=8$.

  * Proof.

It is easy to see that $a,\,2a,\,b,\,2b,\,a+b,\,a+2b,\,2a+b,\,2a+2b$ are all
the distinct elements in $\sum(a^{2}b^{2})$. We are done. $\Box$

###### Lemma 2.3

Let $S=a^{k}b$ be a zero-sum free sequence over $G$. If $S=a^{k}b$ is not
$a$-smooth, then $f(S)=2k+1$.

  * Proof.

The assertion follows from the fact that
$a,\,\ldots,\,ka,\,b,\,a+b,\,\ldots,\,ka+b$ are all the distinct elements in
$\sum(a^{k}b)$.$\Box$

###### Lemma 2.4

[10, Lemma 4] Let $S$ be a zero-sum free sequence over $G$. If there is some
element $g$ in $S$ with order $2$, then $f(S)\geq 2|S|-1$.

###### Lemma 2.5

Let $k\geq l\geq 2$ be two integers, and let $a$ and $b$ be two distinct
elements of $G$ such that $a^{k}b^{l}\in\mathcal{A}^{\star}(G)$ and
$a^{k}b^{l}$ is not smooth. Then we have

(i) If $2a\neq 2b$, then $f(a^{k}b^{l})\geq 2(k+l)$.

(ii) If $2a=2b$, then $f(a^{k}b^{l})=2(k+l)-1$.

  * Proof.

If $nb\neq sa$ for any $n$ and $s$ with $1\leq n\leq l$ and $1\leq s\leq k$,
then $ra+sb,\,r+s\neq 0,\,0\leq r\leq k,\,0\leq s\leq b$ are all the distinct
elements in $\sum(a^{k}b^{l})$, and so

$f(a^{k}b^{l})=kl+k+l\geq 2(k+l).$

Now we assume that $nb=sa$ for some $n$ and $s$ with $1\leq n\leq l$ and
$1\leq s\leq k$. Let $n$ be the least positive integer with $nb=sa,\,1\leq
n\leq l,\,1\leq s\leq k$ . Then $n\geq 2$ and $s\geq 2$ by our assumptions. It
is easy to see that

$a,\,\ldots,\,ka,\,\ldots,\,(k+[\frac{l}{n}]s)a,$
$b,\,a+b,\,\ldots,\,b+ka,\,\ldots,\,b+(k+[\frac{l-1}{n}]s)a,$ $\ldots\ldots$
$(n-1)b,\,\ldots,\,(n-1)b+ka,\,\ldots,\,(n-1)b+(k+[\frac{l-n+1}{n}]s)a$

are all the distinct elements in $\sum(a^{k}b^{l})$, and so

$f(a^{k}b^{l})=k+[\frac{l}{n}]s+1+k+[\frac{l-1}{n}]s+\cdots+1+k+[\frac{l-n+1}{n}]s$
$=n(k-s+1)+ls+s-1.$

Since $n(k-s+1)+ls+s-1-2(k+l)=(n-2)(k-s)+(l-1)(s-2)+n-3$, we have
$f(a^{k}b^{l})\geq 2(k+l)-1$ and the equality holds if and only if $n=s=2$,
that is $2a=2b$. This completes the proof.

$\Box$

Remark: Note that if $a^{k}b^{l}\in\mathcal{A}^{\star}(G),\,k\geq l\geq 2$,
then the conditions that $a^{k}b^{l}$ is smooth and $2a=2b$ cannot hold
simultaneously. Otherwise, we may suppose that $2a=2b$ and $a^{k}b^{l}$ is
$a$-smooth (the case that $a^{k}b^{l}$ is $b$-smooth is similar), then
$b=ta,\,2\leq t\leq(k+1)$. It follows that
$b+(t-2)a=2(t-1)a=2b-2a=0,\,0<t-2\leq k-1$, which contradicts the fact that
$a^{k}b^{l}\in\mathcal{A}^{\star}(G)$.

###### Lemma 2.6

[12, Lemma 2.9]Let $S=a^{k}b^{l}g,\,k\geq l\geq 1$ be a zero-sum free sequence
over $G$ with $b-a=g$ and ${\rm ord}(g)=2$, then $f(S)=2(k+l)+1$.

###### Lemma 2.7

Let $S_{1}\in\mathcal{F}(G)$ and $a,\,g\in G$ such that
$S=S_{1}a\in\mathcal{A}^{\star}(G)$, $S_{1}$ is $g$-smooth and $S$ is not
$g$-smooth. Then $f(S)=2f(S_{1})+1$.

  * Proof.

If $a\not\in<g>$, then $\sum(S)=\sum(S_{1})\cup\\{a\\}\cup(\sum(S_{1})+a)$,
and so $f(S)=2f(S_{1})+1$.

If $a\in<g>$, we let $\sum(S_{1})=\\{g,\,\ldots,ng\\}$,
$a=tg,\,t\in\mathbb{N}$, then $t\geq n+2$ by our assumptions. It follows that
$\sum(S)=\\{g,\,\ldots,\,ng,\,tg,\,(t+1)g,\,\ldots,\,(t+n)g\\}$, and so
$f(S)=2f(S_{1})+1$.$\Box$

###### Lemma 2.8

Let $k\geq 2$ be a positive integer and $a,\,b,\,c$ three distinct elements in
$G$ such that $a^{k}bc\in\mathcal{A}^{\star}(G)$ and $a^{k}bc$ is not
$a$-smooth. Then $f(a^{k}bc)\geq 2k+4$.

  * Proof.

Observe that $f(a^{k}bc)\geq 2k+4$ when $a^{k}bc$ is $b$ or $c$-smooth. We
consider first the case that $a^{k}b$ is $a$-smooth (the case that $a^{k}c$ is
$a$-smooth is similar). It is easy to see $f(a^{k}b)\geq k+2$, and so
$f(a^{k}bc)=2f(a^{k}b)+1\geq 2k+5$ by Lemma 2.7. Therefore we may assume that
both $a^{k}b$ and $a^{k}c$ are not $a$-smooth in the remaining arguments. We
divide the proof into three cases.

(i) If $a^{k}(b+c)$ is not $a$-smooth, then
$a,\,\ldots,\,ka,\,b,\,b+a,\,\ldots,\,b+c,\,b+c+a,\,\ldots,\,b+c+ka$ are
distinct elements in $\sum(a^{k}bc)$, and so

$f(a^{k}bc)\geq k+k+1+k+1\geq 2k+4.$

(ii) If neither $a^{k}(b-c)$ nor $a^{k}(c-b)$ is $a$-smooth, then
$a,\,\ldots,\,ka,\,b,\,b+a,\,\ldots,\,c,\,c+a,\,\ldots,\,c+ka,\,b+c+ka$ are
distinct elements in $\sum(a^{k}bc)$, and so

$f(a^{k}bc)\geq k+k+1+k+1+1\geq 2k+5.$

(iii) If $a^{k}(b+c)$ is $a$-smooth and $a^{k}(b-c)$ (or $a^{k}(c-b)$ ) is
$a$-smooth, then we have

$b+c=sa,\quad b-c=ta,\quad 1\leq s,\,t\leq k+1,\quad s\neq t.$

It is easy to see that $a,\,\ldots,\,ka,\,(k+1)a,\,\ldots,\,(k+s)a,$
$c,\,c+a,\,\ldots,\,c+(k+t)a$ are all distinct elements in $\sum(a^{k}bc)$,
and so

$f(a^{k}bc)=k+s+k+t+1\geq 2k+4.$

The second equality holds if and only if $(s,\,t)=(1,\,2)$ or $(2,\,1)$. We
are done. $\Box$

The following corollary follows immediately from Lemmas 2.1, and 2.7 and the
proof of Lemma 2.9.

###### Corollary 2.1

Let $k\geq 1$ be a positive integer and $a,\,b,\,c,\,d$ four distinct elements
in $G$ such that $a^{k}bcd\in\mathcal{A}^{\star}(G)$ and $a^{k}bcd$ is not
$a$-smooth. Then $f(a^{k}bcd)\geq 2k+6$.

###### Lemma 2.9

Let $a,b,x$ be three distinct elements in $G$ such that
$a^{k}b^{l}x\in\mathcal{A}^{\star}(G),\,k\geq l\geq 1$, $2a=2b$, and $x\neq
a-b$, then $f(a^{k}b^{l}x)\geq 2(k+l+1)+1$.

  * Proof.

If there are no distinct pairs
$(m,\,n)\neq(0,\,0),(m_{1},\,n_{1})\neq(0,\,0),\,0\leq m,\,m_{1}\leq k,\,0\leq
n,\,n_{1}\leq l$ such that $ma+nb=m_{1}a+n_{1}b+x$, then
$\sum(a^{k}b^{l}x)=\sum(a^{k}b^{l})\cup\\{x\\}\cup(\sum(a^{k}b^{l})+x)$, and
so $f(a^{k}b^{l}x)=2f(a^{k}b^{l})+1=4(k+l)-1\geq 2(k+l+1)+1$.

If there are two distinct pairs
$(m,\,n)\neq(0,\,0),(m_{1},\,n_{1})\neq(0,\,0),\,0\leq m,\,m_{1}\leq k,\,0\leq
n,\,n_{1}\leq l$ such that $ma+nb=m_{1}a+n_{1}b+x$, then $x=a-b$ or
$x=ua+b,\,1\leq u\leq(k+l-1)$ or $x=vb,\,v\geq 2$ or $x=ta,\,t\geq 2$.

Let $x=ua+b,1\leq u\leq(k+l-1)$, then
$a,\,\ldots,\,(k+l+u)a,\,b,\,\cdots,\,b+(k+l+u)a$ are all distinct elements in
$\sum(a^{k}b^{l}x)$, and so $f(a^{k}b^{l}x)=2(k+l+u)+1\geq 2(k+l+1)+1$.

Let $x=vb,\,2\leq v\leq(k+l)$ (the case that $x=ta,\,t\geq 2$ is similar). If
$k$ is even, then $b,\,\ldots,\,(k+l+v)b,\,a,\,a+b,\,\ldots,a+(k+l-2+v)b$ are
all distinct elements in $\sum(a^{k}b^{l}x)$, and so
$f(a^{k}b^{l}x)=2(k+l+v-1)+1\geq 2(k+l+1)+1$. If $k$ is odd, then
$b,\,\ldots,\,(k+l+v-1)b,\,a,\,a+b,\,\ldots,a+(k+l-1+v)b$ are all distinct
elements in $\sum(a^{k}b^{l}x)$, and so $f(a^{k}b^{l}x)=2(k+l+v-1)+1\geq
2(k+l+1)+1$. We are done. $\Box$

###### Lemma 2.10

Let $a,b,x$ be three distinct elements in $G$ such that
$a^{k}b^{2}x\in\mathcal{A}^{\star}(G),\,k\geq 2$ and $a^{k}b^{2}x$ is not
$a$-smooth or $b$-smooth, then $f(a^{k}b^{2}x)=2k+5$ if and only if $2a=2b$
and $x=b-a$.

  * Proof.

We divide the proof into four cases.

Case 1 $a^{k}b^{2}$ is not smooth and $2b=sa,\,2\leq s\leq k$. If $x=b-a$,
then $a,\,\ldots,\,(k+s)a,\,b-a,\,b,\,\ldots,\,b+(k+s-1)a$ are all the
distinct elements in $\sum(a^{k}b^{2}x)$, and so $f(a^{k}b^{2}x)=2(k+s)+1$. If
$x=ta,\,2\leq t\leq k$, then $a,\,\ldots,\,(k+s+t)a,\,b,\,\ldots,\,b+(k+t)a$
are all the distinct elements in $\sum(a^{k}b^{2}x)$, and so
$f(a^{k}b^{2}x)=2(k+t)+s+1$. If $x=ta+b,\,1\leq t\leq k$, then
$a,\,\ldots,\,(k+s+t)a,\,b,\,\ldots,\,b+(k+t+s)a$ are all the distinct
elements in $\sum(a^{k}b^{2}x)$, and so $f(a^{k}b^{2}x)=2(k+t+s)+1$. Therefore
$f(a^{k}b^{2}x)=2k+5$ if and only if $2a=2b$ and $x=b-a$ in this case.

Case 2 $a^{k}b^{2}$ is not smooth and $2b=sa,\,s>k$ or $2b\not\in<a>$, then
$f(a^{k}b^{2})=3k+2$. If $k\geq 3$, then $f(a^{k}b^{2}x)\geq
f(a^{k}b^{2})+1=3k+3>2k+5$. If $k=2$ and $f(abx)=7$, then $f(a^{2}b^{2}x)\geq
f(abx)+f(ab)=7+3>2k+5$. If $k=2$ and $f(abx)=6$ (i.e., $x=a+b$ or $x=a-b$ or
$x=b-a$), then it is easy to check that $f(a^{2}b^{2}x)>2k+5$.

Case 3 $a^{k}b^{2}$ is smooth and $a^{k}b^{2}x$ is not smooth. If $a^{k}b^{2}$
is $a$-smooth, then $f(a^{k}b^{2}x)=2f(a^{k}b^{2})+1\geq 2(k+2\times
2)+1>2k+5$. If $a^{k}b^{2}$ is $b$-smooth, then
$f(a^{k}b^{2}x)=2f(a^{k}b^{2})+1\geq 2(2+2k)+1>2k+5$.

Case 4 $a^{k}b^{2}x$ is $x$-smooth. We have $f(a^{k}b^{2}x)\geq 1+2k+2\times
3>2k+5$.

This completes the proof of the lemma.

$\Box$

##  3 Proofs of the Main Theorems

To prove the main theorem of the present paper, we still need the following
two obviously facts on smooth sequences.

Fact 1 Let $r$ be a positive integer and $a\in G$. If
$WT_{i}\in\mathcal{A}^{\star}(G)$ is $a$-smooth for all $i=1,\ldots,r$, then
$S=T_{1}\cdot\,\cdots\,\cdot T_{r}W$ is $a$-smooth.

Fact 2 Let $r,\,k,\,l$ be three positive integers and $a,\,b$ two distinct
elements in $G$. If $S\in\mathcal{A}^{\star}(G)$ is $a$-smooth and
$a^{k}b^{l}T_{i}\in\mathcal{A}^{\star}(G)$ is $a$-smooth or $b$-smooth for all
$i=1,\ldots,r$, then the sequence $Sa^{k}b^{l}T_{1}\cdot\,\cdots\,\cdot T_{r}$
is $a$-smooth or $b$-smooth.

Proof of Theorem 1.1:

We start with the trivial case that $S=a^{k}$ with $k\in\mathbb{N}$ and $a\in
G$. Then $\sum(S)=\\{a,\ldots,ka\\}$, and since $S$ is zero-sum free, it
follows that $k<ord(a)$. Thus $S$ is $a$-smooth.

If $S=S_{1}g\in\mathcal{A}^{\star}(G)$, where $g$ is an element of order 2,
then $f(S)\geq 2|S|-1$ by Lemma 2.4, and $f(S)\geq f(S_{1})+2$ since
$\sum(S)\supseteq\sum(S_{1})\cup\\{g,\,g+\sigma(S_{1})\\}$. If
$S=S_{1}g_{1}g_{2}\in\mathcal{A}^{\star}(G)$, where $g_{1}$ and $g_{2}$ are
two elements of order 2, then $f(S)\geq 2|S|$ since
$\sum(S)\supseteq\sum(S_{1}g_{1})\cup\\{g_{2},\,g_{1}+g_{2},\,g_{1}+g_{2}+\sigma(S_{1})\\}$.
Therefore it suffices to determine all $S\in\mathcal{A}^{\star}(G)$ such that
$S$ does not contain any element of order 2 and $f(S)\leq 2|S|-1$, and when
$f(S)\leq 2|S|-1$, determine all $Sg\in\mathcal{A}^{\star}(G)$ such that $g$
is an element of order 2 and $f(Sg)=2|S|+1$.

To begin with, we determine all $S\in\mathcal{A}^{\star}(G)$ such that $S$
does not contain any element of order 2 and $f(S)\leq 2|S|-1$. Let
$S=a^{x}b^{y}c^{z}T$ with $x\geq y\geq z\geq\mathsf{h}(T)$ and
$a,\,b,c\not\in{\rm supp}(T)$. The case that $|{\rm supp}(S)|=2$ follows from
Lemmas 2.3 and 2.5 and the remark after Lemma 2.5. Therefore we may assume
that $|{\rm supp}(S)|\geq 3$ and $S$ does not contain any element of order $2$
in the following arguments.

If $x=y=z$, then $S$ allows the product decomposition

$S=S_{1}\cdot\,\cdots\,\cdot S_{x},$

where $S_{i}=abc\cdot\,\cdots,\,i=1,\,\ldots,\,x$ are squarefree of length
$|S_{i}|\geq 3$. By Lemma 2.1, we obtain

$f(S)\geq\sum_{i=1}^{x}f(S_{i})\geq 2\sum_{i=1}^{x}|S_{i}|=2|S|.$

If $x\geq y>z\geq\mathsf{h}(T)$, or $x>y\geq z\geq\mathsf{h}(T)$, then $S$
allows a product decomposition

$S=T_{1}\cdot\,\cdots\,\cdot T_{r}W$

having the following properties:

  * •

$r\geq 1$ and, for every $i\in[2,\,r]$, $S_{i}\in\mathcal{F}(G)$ is squarefree
of length $|S_{i}|=3$.

  * •

$W\in\mathcal{F}(G)$ has the form $W=a^{k},\,k\geq 1$ or $W=a^{k}b,\,k\geq 1$
or $W=a^{k}b^{l},\,k\geq l\geq 2$.

We choose a product decomposition such that $k$ is the largest integer in
$W=a^{k}$ (or $a^{k}b$ or $a^{k}b^{l},\,k\geq l\geq 2$) among all such product
decompositions. We divide the remaining proof into three cases.

Case 1 $W=a^{k},\,k\geq 1$. If $T_{i}=xyz$ with $a\not\in\\{x,\,y,\,z\\}$ for
some $i,\,1\leq i\leq r$ such that $a^{k}xyz$ is not $a$-smooth whenever
$k>1$, then $S$ admits the product decomposition

$S=T_{1}\cdot\,\cdots\,\cdot T_{i-1}T_{i}^{\prime}T_{i+1}\cdot\,\cdots\,\cdot
T_{r},$

where $T_{i},\,i=1,\,\ldots,\,r$ have the properties described above and
$T_{i}^{\prime}=a^{k}xyz$. By Lemma 2.1, and Corollary 2.1, we get

$f(S)\geq\sum_{j\neq i}^{r}f(T_{j})+f(T_{i}^{\prime})\geq\sum_{j\neq
i}^{r}2|T_{j}|+2|T_{i}^{\prime}|=2|S|.$

If $T_{i}=axy$ for some $i,\,1\leq i\leq r$ such that $a^{k+1}xy$ is not
$a$-smooth, then $S$ admits the product decomposition

$S=T_{1}\cdot\,\cdots\,\cdot T_{i-1}T_{i}^{\prime}T_{i+1}\cdot\,\cdots\,\cdot
T_{r},$

where $T_{i},\,i=1,\,\ldots,\,r$ have the properties described above and
$T_{i}^{\prime}=a^{k+1}xy$. By Lemmas 2.1 and 2.8, we get

$f(S)\geq\sum_{j\neq i}^{r}f(T_{j})+f(T_{i}^{\prime})\geq\sum_{j\neq
i}^{r}2|T_{j}|+2|T_{i}^{\prime}|=2|S|.$

Therefore we have proved that if $S$ is not $a$-smooth and $W=a^{k}$, then
$f(S)\geq 2|S|$.

Case 2 $W=a^{k}b,\,k\geq 1$.

Let $T_{i}=xyz$ with $a\not\in\\{x,\,y,\,z\\}$ for some $i,\,1\leq i\leq r$.
If $k=1$, then $T_{i}W=abxyz$. If $k=2$, then $T_{i}W=abx\cdot ayz$. If $k\geq
3$ and one sequence among three sequences $a^{k-1}yz,\,a^{k-1}xz$, and
$a^{k-1}xy$, say, $a^{k-1}yz$ is not $a$-smooth, then $T_{i}W=abx\cdot
a^{k-1}yz$. It follows from Lemmas 2.1 and 2.8 that $f(T_{i}W)\geq
2|T_{i}|+2|W|$, and so $f(S)\geq 2|S|$.

Let $T_{i}=bxy$ for some $i,\,1\leq i\leq r$, then $k\geq 2$. If $k=2$, then
$T_{i}W=abx\cdot aby$. If $k>2$ and $a^{k-1}by$ (or $a^{k-1}bx$) is not
$a$-smooth, then $T_{i}W=abx\cdot a^{k-1}by$ (or $T_{i}W=aby\cdot a^{k-1}bx$).
It follows from Lemmas 2.1 and 2.8 that $f(T_{i}W)\geq 2|T_{i}|+2|W|$, and so
$f(S)\geq 2|S|$.

Let $T_{i}=abx$ for some $i,\,1\leq i\leq r$, then $T_{i}W=a^{k+1}b^{2}x$. If
$a^{k+1}b^{2}x$ is not $a$-smooth or $b$-smooth, then by Lemma 2.10 we have
$f(T_{i}W)\geq 2|T_{i}|+2|W|$, and so $f(S)\geq 2|S|$.

Therefore we have proved that if $S$ is not $a$-smooth or $b$-smooth, then
$f(S)\geq 2|S|$ in this case.

Case 3 $W=a^{k}b^{l},\,k\geq l\geq 2$. If $2a\neq 2b$ and $a^{k}b^{l}$ is not
smooth, then by Lemma 2.5 we have $f(W)\geq 2|W|$ and we are done. Note that
the conditions that $2a=2b$ and $a^{k}b^{l}$ is smooth cannot hold
simultaneous. Here we omit the similar arguments as we have done in Case 1.

Subcase 1 $2a=2b$.

Let $T_{i}=xyz$ with $a\not\in\\{x,\,y,\,z\\}$ for some $i,\,1\leq i\leq r$,
then $T_{i}W=abxy\cdot a^{k-1}b^{l-1}z$. It follows from Lemmas 2.1 and 2.9
that $f(T_{i}W)\geq 2|T_{i}|+2|W|$, and so $f(S)\geq 2|S|$.

Let $T_{i}=byz$ for some $i,\,1\leq i\leq r$, then $k\geq l+1$,
$T_{i}W=aby\cdot a^{k-1}b^{l}z$. It follows from Lemmas 2.1 and 2.9 that
$f(T_{i}W)\geq 2|T_{i}|+2|W|$, and so $f(S)\geq 2|S|$.

Let $T_{i}=abx$ for some $i,\,1\leq i\leq r$, then $T_{i}W=a^{k+1}b^{l+1}x$.
If $a^{k+1}b^{2}x$ is not $a$-smooth or $b$-smooth, then by Lemma 2.10 we have
$f(T_{i}W)\geq 2|T_{i}|+2|W|$, and so $f(S)\geq 2|S|$.

Subcase 2 $a^{k}b^{l}$ is smooth, $a\neq 2b$, and $b\neq 2a$. Then
$W=(a^{2}b^{2})^{s}W_{1}$, $W_{1}=a^{k_{1}}$ or $W_{1}=a^{k_{1}}b$. If
$S_{1}=SW^{-1}W_{1}$ is not $a$-smooth or $b$-smooth, then $f(S_{1})\geq
2|S_{1}|$, and so by Lemmas 2.1 and 2.2 $f(S)\geq sf(a^{2}b^{2})+f(S_{1})\geq
8s+2|S_{1}|=2|S|$. If $S_{1}=SW^{-1}W_{1}$ is $a$-smooth or $b$-smooth, then
$S$ is $a$-smooth or $b$-smooth.

Subcase 3 $a=2b$.

Let $T_{i}=xyz$ with $a,\,b\not\in\\{x,\,y,\,z\\}$ for some $i,\,1\leq i\leq
r$, then it is easy to see that $f(T_{i}W)=f(a^{k}b^{l}xyz)=f(b^{2k+l}xyz)$.
It follows from Corollary 2.1 that $b^{2k+l}xyz$ is $b$-smooth or
$f(T_{i}W)\geq 2(|T_{i}|+|W|)$.

Let $T_{i}=bxy$ with $a,\,b\not\in\\{x,\,y\\}$ for some $i,\,1\leq i\leq r$,
then $f(T_{i}W)=f(a^{k}b^{l+1}xy)=f(b^{2k+l+1}xyz)$. It follows from Lemma 2.8
that $b^{2k+l+1}xy$ is $b$-smooth or $f(T_{i}W)\geq 2(|T_{i}|+|W|)$.

Let $T_{i}=abx$ with $a\neq x,\,b\neq x$ for some $i,\,1\leq i\leq r$, then
$f(T_{i}W)=f(a^{k+1}b^{l+1}x)=f(b^{2k+l+3}xyz)$. It follows from Lemma 2.3
that $b^{2k+l+3}x$ is $b$-smooth or $f(T_{i}W)\geq 2(|T_{i}|+|W|)$.

Subcase 4 $b=2a$. Similar to Subcase 3.

Therefore we have proved that if $S$ is not $a$-smooth or $b$-smooth, then
$f(S)\geq 2|S|-1$ and $f(S)=2|S|-1$ if and only if $S=a^{k}b$ or
$S=a^{k}b^{l},\,2a=2b,\,k\geq l\geq 2$.

Finally, when $f(S)\leq 2|S|-1$, we will determine all
$Sg\in\mathcal{A}^{\star}(G)$ such that $g$ is an element of order 2 and
$f(Sg)=2|S|+1$.

(i) If $S$ is $a$-smooth (the case that $S$ is $b$-smooth is similar), we set
$\sum(S)=\\{a,\,\ldots,\,na\\},\,n\leq 2|S|-1$, then $g\not\in\sum(S)$ since
$g$ is an element of order 2 and $Sg\in\mathcal{A}^{\star}(G)$. It follows
that $\sum(Sg)=\sum(S)\cup\\{g\\}\cup\\{g+\sum(S)\\}$, and so $f(Sg)=2n+1$.
Therefore $f(Sg)\leq 2|S|+1$ if and only if $S=a^{k}$.

(ii) $S=a^{k}b$ is not smooth, by Lemma 2.8, $f(a^{k}bg)\leq 2k+1$ only if
$a^{k}bg$ is $a$-smooth, which is impossible since $g$ is an element of order
2 and $a^{k}bg\in\mathcal{A}^{\star}(G)$.

(iii) $S=a^{k}b^{l},\,2a=2b,\,k\geq l\geq 2$. The result follows from Lemmas
2.5 and 2.9.

Therefore we have proved that if
$S=a^{x}b^{y}\cdot\,\cdots\in\mathcal{A}^{\star}(G),\,x\geq y\geq\cdots$,
where $a,\,b,\,\ldots$ are distinct elements of $G$ and $f(S)\leq 2|S|-1$,
then $S$ is $a$-smooth or $b$-smooth or $S=a^{k}b,\,b\not\in\sum(a^{k})$ or
$S=a^{k}b^{l},\,k\geq l\geq 2,2a=2b$ or $S=a^{k}b^{l},\,k\geq l\geq
2,2a=2b,\,g=a-b$. Theorem 1.1 is proved.

$\Box$

Acknowledgement: The author wishes to thank Alfred Geroldinger for sending the
preprint [7] to him. He also thanks the referee for his/her valuable
suggestions.

## References

  * [1] J.D. Bovey, P. Erdős, and I. Niven, _Conditions for zero sum modulo $n$_, Canad. Math. Bull. 18 (1975), 27 – 29.
  * [2] B. Bollob$\acute{a}$s and I. Leader, _The number of $k$-sums modulo $k$_, J. Number Theory 78(1999), 27-35.
  * [3] S.T. Chapman and W.W. Smith, _A characterization of minimal zero-sequences of index one in finite cyclic groups_ , Integers 5(1) (2005), Paper A27, 5pp.
  * [4] W. Gao, Y. Li, J. Peng, and F. Sun, _On subsequence sums of a zero-sum free sequence II_ , the Electronic Journal of Combinatorics 15 (2008), $\sharp$R117.
  * [5] W.D. Gao and I. Leader, _sums and $k$-sums in an abelian groups of order $k$_, J. Number Theory 120(2006), 26-32.
  * [6] A. Geroldinger and F. Halter-Koch, _Non-Unique Factorizations. Algebraic, Combinatorial and Analytic Theory_ , Pure and Applied Mathematics, Vol. 278, Chapman & Hall/CRC, 2006.
  * [7] A. Geroldinger, _Additive group theory and non-unique factorizations_ , to appear.
  * [8] J. E. Olson and E.T.White, _sums from a sequence of group elements_ , in : Number Theory and Algebra, Academic Press, New York, 1977, pp. 215-222.
  * [9] A. Pixton, _Sequences with small subsum sets_ , J. Number Theory 129(2009), 806-817.
  * [10] F. Sun, _On subsequence sums of a zero-sum free sequence_ , the Electronic Journal of Combinatorics 14(2007), $\sharp$R52.
  * [11] P.Z. Yuan, _On the index of minimal zero-sum sequences over finite cyclic groups_ , J. Combin. Theory Ser. A114(2007), 1545-1551.
  * [12] P.Z. Yuan, _Subsequence sums of a zero-sumfree sequence_ , European Journal of Combinatorics,  30(2009), 439-446.
  * [13] P.Z. Yuan, _Subsequence Sums of Zero-sum-free Sequences_ , to appear in the Electronic Journal of Combinatorics.

Pingzhi Yuan
School of Mathematics
South China Normal University
Guangdong, Guangzhou 510631
P.R.CHINA
<EMAIL_ADDRESS>