# additive bases of abelian groups of rank 2

Weidong Gao  Center for Applied Mathematics
Tianjin University
Tianjin, 300072, P.R. China<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>,  Yuanlin Li  Department of Mathematics and Statistics
Brock University
St. Catharines, ON L2S 3A1, Canada<EMAIL_ADDRESS>,  Yongke Qu  Department of
Mathematics
Luoyang Normal University
LuoYang 471934, P.R. China<EMAIL_ADDRESS>and  Qinghong Wang  College of
Science
Tianjin University of Technology
Tianjin, 300384, P.R. China<EMAIL_ADDRESS>
###### Abstract.

Let $G$ be a finite abelian group and $p$ be the smallest prime dividing
$|G|$. Let $S$ be a sequence over $G$. We say that $S$ is regular if for every
proper subgroup $H\subsetneq G$, $S$ contains at most $|H|-1$ terms from $H$.
Let $\mathsf{c}_{0}(G)$ be the smallest integer $t$ such that every regular
sequence $S$ over $G$ of length $|S|\geq t$ forms an additive basis of $G$,
i.e., $\sum(S)=G$. The invariant $\mathsf{c}_{0}(G)$ was first studied by
Olson and Peng in 1980’s, and since then it has been determined for all finite
abelian groups except for the groups with rank 2 and a few groups of rank 3 or
4 with order less than $10^{8}$. In this paper, we focus on the remaining case
concerning groups of rank 2. It was conjectured by Gao et al. (Acta Arith. 168
(2015) 247-267) that $\mathsf{c}_{0}(G)=m(G)$. We confirm the conjecture for
the case when $G=C_{n_{1}}\oplus C_{n_{2}}$ with $n_{1}|n_{2}$, $n_{1}\geq
2p$, $p\geq 3$ and $n_{1}n_{2}\geq 72p^{6}$.

Keywords: Finite abelian group; Additive basis; Regular sequence.
2020 Mathematics Subject Classifications: 11B75, 11P70

##  1\. Introduction and main results

Through the paper, let $G$ be a finite abelian group, written additively, $p$
be the smallest prime dividing $|G|$ and $\mathsf{r}(G)$ denote the rank of
$G$. Let $S$ be a sequence over $G$. We say that $S$ is an additive basis of
$G$ if every element of $G$ can be expressed as the sum over a nonempty
subsequence of $S$. For every subgroup $H$ of $G$, let $S_{H}$ denote the
subsequence of $S$ consisting of all terms of $S$ contained in $H$. We say
that $S$ is a regular sequence over $G$ if $|S_{H}|\leq|H|-1$ holds for every
proper subgroup $H\subsetneq G$. Let $\mathsf{c}_{0}(G)$ denote the smallest
integer $t$ such that every regular sequence over $G$ of length at least $t$
is an additive basis of $G$. The problem of determining $\mathsf{c}_{0}(G)$
was first proposed by Olson and it was conjectured that
$\mathsf{c}_{0}(C_{p}\oplus C_{p})=2p-1$. In 1987, Peng proved this conjecture
and further determined $\mathsf{c}_{0}(G)$ for all the finite elementary
abelian $p$-groups ([8, 9]). Those results and techniques used for determining
$\mathsf{c}_{0}(G)$ have been successfully applied to the study of nonunique
factorization ([4]). Recently, the problem related to the additive basis of a
finite abelian group has been investigated by several authors ([1, 5, 10, 11,
12]). Let

$m(G)=\left\\{\begin{array}[]{ll}&|G|,\mbox{ if }G\mbox{ is cyclic},\\\
&2p-1,\mbox{ if }G=C_{p}\oplus C_{p},\\\ &kp+2p-3,\mbox{ if }G=C_{p}\oplus
C_{pk}\mbox{ and }k\geq 2,\\\ &\frac{|G|}{p}+p-2,\mbox{
otherwise}.\end{array}\right.$

It has been proved that $\mathsf{c}_{0}(G)=m(G)$ for any of the following
finite abelian groups:

  1. (1)

$G$ is cyclic;

  2. (2)

$|G|$ is even;

  3. (3)

$\mathsf{r}(G)\geq 4$ and $G\neq C_{3}^{3}\oplus C_{3n}$ where $n>3$ is odd
and is not a power of $3$ with $|G|<3.72\times 10^{7}$ ;

  4. (4)

$\mathsf{r}(G)=3$ and either $p\geq 11$ or $3\leq p\leq 7$ with $|G|\geq
3.72\times 10^{7}$;

  5. (5)

$\mathsf{r}(G)\geq 2$ and $G$ is a $p$-group;

  6. (6)

$G=C_{3}\oplus C_{3q}$, where $q$ is a prime.

In this paper, we focus our investigation on the remaining case when $G$ is of
rank $2$. It was conjectured in [1] that $\mathsf{c}_{0}(G)=m(G)$. We confirm
this conjecture for the case when $G=C_{n_{1}}\oplus C_{n_{2}}$ with
$n_{1}|n_{2}$, $n_{1}\geq 2p$, $p\geq 3$ and $n_{1}n_{2}\geq 72p^{6}$.

######  Theorem 1.1.

Let $G=C_{n_{1}}\oplus C_{n_{2}}$ with $n_{1}|n_{2}$, and $p\geq 3$ be the
smallest prime divisor of $|G|$. If $n_{1}\geq 2p$ and $|G|\geq 72p^{6}$, then
$\mathsf{c}_{0}(G)=|G|/p+p-2$.

##  2\. Notations and Preliminaries

Suppose that $G_{0}\subseteq G$ is a subset of $G$ and $\mathcal{F}(G_{0})$ is
the multiplicatively written, free abelian monoid with basis $G_{0}$. The
elements of $\mathcal{F}(G_{0})$ are called sequences over $G_{0}$. We denote
multiplication in $\mathcal{F}(G_{0})$ by the bold symbol $\bm{\cdot}$ rather
than by juxtaposition, and use brackets for all exponentiation in
$\mathcal{F}(G_{0})$.

A sequence $S\in\mathcal{F}(G)$ will be written in the form
$S=g_{1}\bm{\cdot}\ldots\bm{\cdot}g_{\ell},$ where $|S|=\ell$ is the length of
$S$. For $g\in G$, let $\mathsf{v}_{g}(S)=|\\{i\in[1,\ell]:g_{i}=g\\}|\,$
denote the multiplicity of $g$ in $S$. We call $\mbox{supp}(S)=\\{g\in
G:\mathsf{v}_{g}(S)>0\\}$ the support of $S$. Let
$\mathsf{h}(S)=\mbox{max}\\{\mathsf{v}_{g}(S):g\in G\\}$. A sequence
$T\in\mathcal{F}(G)$ is called a subsequence of $S$ and is denoted by $T\mid
S$ if $\mathsf{v}_{g}(T)\leq\mathsf{v}_{g}(S)$ for all $g\in G$. Denote by
$S\bm{\cdot}T^{[-1]}$ the subsequence of $S$ obtained by removing the terms of
$T$ from $S$.

If $S_{1},S_{2}\in\mathcal{F}(G)$, then
$S_{1}\bm{\cdot}S_{2}\in\mathcal{F}(G)$ denotes the sequence satisfying that
$\mathsf{v}_{g}(S_{1}\bm{\cdot}S_{2})=\mathsf{v}_{g}(S_{1})+\mathsf{v}_{g}(S_{2})$
for all $g\in G$. For convenience we write

$g^{[k]}=\underbrace{g\bm{\cdot}\ldots\bm{\cdot}g}_{k}\in\mathcal{F}(G)\quad$

for $g\in G$ and $k$ a nonnegative integer.

Let $\sigma(S)=\sum_{g\in G}\mathsf{v}_{g}(S)g\in G$ be the sum of $S$. Define

$\sum(S)=\\{\sigma(T):\ 1\neq T\mid S\\},$

where $1$ is the empty sequence, and

$\sum\nolimits_{0}(S)=\sum(S)\cup\\{0\\}.$

We call a sequence $S$ a zero-sum sequence if $\sigma(S)=0$, and a zero-sum
free sequence if $0\notin\sum(S)$. We note that a subset of $G$ can be
regarded as a sequence, so all above-mentioned concepts and notations
regarding a sequence of $G$ are valid for a subset of $G$. We say that a
subset $A\subset G\setminus\\{0\\}$ is $2$-zero-sum free if $A$ contains no
two distinct elements with sum zero.

Let $\mathsf{D}(G)$ denote the Davenport constant of $G$, which is defined as
the smallest integer $t$ such that every sequence $S$ over $G$ of length
$|S|\geq t$ contains a nonempty zero-sum subsequence. Note that
$\mathsf{D}(C_{n_{1}}\oplus C_{n_{2}})=n_{1}+n_{2}-1$, where $1\leq
n_{1}|n_{2}$ ([6, Theorem 5.8.3]).

For each subset $A$ of $G$, denote by $\langle A\rangle$ the subgroup
generated by $A$. Let ${\rm st}(A)=\\{g\in G:g+A=A\\}$. Then ${\rm st}(A)$ is
the maximal subgroup $H$ of $G$ such that $H+A=A$. The following is the well
known Kneser’s theorem and a proof of it can be found in [7].

######  Lemma 2.1.

(Kneser) [7, Theorem 4.4] Let $A_{1},\ldots,A_{r}$ be nonempty finite subsets
of an abelian group $G$, and let $H={\rm st}(A_{1}+\cdots+A_{r})$. Then,

$|A_{1}+\cdots+A_{r}|\geq|A_{1}+H|+\cdots+|A_{r}+H|-(r-1)|H|.$

######  Lemma 2.2.

[5, Lemma 2.2] $\mathsf{c}_{0}(G)\geq m(G)$ for every finite abelian group
$G$.

######  Lemma 2.3.

[5, Lemma 2.3] Let $S$ be a regular sequence over $G$ of length
$|S|\geq\mbox{max}\\{|G|/p+p-2,\mathsf{D}(G)\\}$. Let $T$ be a nonempty
subsequence of $S$. If $\sum(S)\neq G$, then

  1. (1)

${\rm st}(\sum_{0}(T))=\\{0\\}$;

  2. (2)

$|\sum_{0}(T))|\geq|T|+1$.

We remark that since $|S|\geq\mathsf{D}(G)$, we have $0\in\sum(S)$. Thus
$\sum(S)=\sum_{0}(S)$, and the above lemma follows immediately from [5, Lemma
2.3].

For a finite abelian group $G$ and a positive integer $k$, let
$f(G,k)=\mbox{min}\\{|\sum(S)|:S\mbox{ is a subset of }G\mbox{ with
}|S|=k\mbox{ and }0\notin\sum(S)\\}$. If $0\in\sum(S)$ for every $S\subseteq
G$ of $|S|=k$, we set $f(G,k)=\infty$. Let $f(k)=\mbox{min}\\{f(G,k):G\mbox{
is a finite abelian group}\\}$.

######  Lemma 2.4.

[2, Theorem 1.1] $f(k)\geq\frac{k^{2}}{6}$ holds for every positive integer
$k$.

##  3\. Proof of Theorem 1.1

We first prove a few technical results.

######  Lemma 3.1.

Let $G$ be an abelian group and $A\subseteq G\setminus\\{0\\}$. If $|A|\geq
6p(p+1)+1$, then there is a zero-sum free subset $B\subseteq A$ such that
$|\sum_{0}(B)|\geq p|B|+2$ with $|B|\leq 6p+1$.

###### Proof.

It is sufficient to prove that there is a zero-sum free subset $B\subseteq A$
such that $|\sum(B)|\geq p|B|+1$ with $|B|\leq 6p+1$. We first prove that for
every $k\leq 6p+1$ there is a zero-sum free subset $B\subseteq A$ such that
either $|B|=k$ or $|B|<k$ and $|\sum(B)|\geq p|B|+1$.

We proceed by induction on $k$. If $k=1$, then the result holds trivially.
Assume that the result is true for each $k$ ($k<6p+1$). We next prove it is
also true for $k+1$. By the induction hypothesis, there is a zero-sum free
subset $B_{0}\subseteq A$ such that either $|B_{0}|=k$ or $|B_{0}|<k$ and
$|\sum(B_{0})|\geq p|B_{0}|+1$. If $|\sum(B_{0})|\geq p|B_{0}|+1$, then let
$B=B_{0}$. Hence $|B|<k+1$ and $|\sum(B)|\geq p|B|+1$, so we are done. Thus we
may assume that $|B_{0}|=k$ and $|\sum(B_{0})|\leq p|B_{0}|$. It follows that
$|-\sum(B_{0})|=|\sum(B_{0})|\leq pk\leq
6p^{2}\leq|A|-(6p+1)<|A|-|B_{0}|=|A\setminus B_{0}|$. Hence, there is an
element $g\in A\setminus B_{0}$ such that $g\not\in-\sum(B_{0})$. So,
$B_{0}\cup\\{g\\}$ is a zero-sum free subset of $A$. Set $B=B_{0}\cup\\{g\\}$,
and then the result holds for $k+1$.

Next, let $k=6p+1$. We have just proved that there is a zero-sum free subset
$B$ of $A$ such that either $|B|=6p+1$ or $|\sum(B)|\geq p|B|+1$ and
$|B|<6p+1$. If the latter is true, then we are done. So, we may assume that
$|B|=6p+1$. By Lemma 2.4, $|\sum(B)|\geq f(6p+1)\geq(6p+1)^{2}/6\geq
p(6p+1)+1$ and again we are done. ∎

######  Lemma 3.2.

Let $G$ be a finite abelian group with smallest prime divisor $p$ of $|G|$.
Let $T$ be a sequence over $G\setminus\\{0\\}$ and
$H=\langle\mbox{supp}(T)\rangle$. Then $|\sum_{0}(T)|\geq|T|+1$ if one of the
following holds

  1. (1)

$|T|\leq p-1$;

  2. (2)

$|T|\leq 2p-1$ and $|H|\geq 2p$.

###### Proof.

Let $T=g_{1}\bm{\cdot}\ldots\bm{\cdot}g_{k}$. Then,

$\sum\nolimits_{0}(T)=\\{{g_{1}},0\\}+\cdots+\\{{g_{k}},0\\}.$

Let $M=\mbox{st}(\\{{g_{1}},0\\}+\cdots+\\{{g_{k}},0\\})$. If $M=\\{0\\}$,
then by Lemma 2.1, $|\sum_{0}(T)|=|\\{{g_{1}},0\\}+\cdots+\\{{g_{k}},0\\}|\geq
k+1=|T|+1$ as desired.

Next we assume $M\neq\\{0\\}$, so $|M|\geq p$.

(1) Given that $|T|\leq p-1$. Clearly,
$|\sum_{0}(T)|=|\\{{g_{1}},0\\}+\cdots+\\{{g_{k}},0\\}|=|\\{{g_{1}},0\\}+\cdots+\\{{g_{k}},0\\}+M|\geq|M|\geq
k+1$ as desired.

(2) Given that $|T|\leq 2p-1$ and $|H|\geq 2p$. If $M\supseteq H$, then as
above $|\sum_{0}(T)|\geq|M|\geq|H|\geq 2p\geq|T|+1$. Next we assume that
$\\{0\\}\neq M\nsupseteq H$. Clearly,
$\sum_{0}(T)=\\{{g_{1}},0\\}+\cdots+\\{{g_{k}},0\\}\subsetneq M$. Thus
$g_{i}\notin M$ for some $1\leq i\leq k$, and so $|\\{{g_{i}},0\\}+M|=2|M|$.
By Lemma 2.1, $|\sum_{0}(T)|=|\\{{g_{1}},0\\}+\cdots+\\{{g_{k}},0\\}|\geq
2|M|\geq 2p\geq|T|+1$ as desired. ∎

As a consequence, we obtain the following corollary, which will be used in the
proof for our main result.

######  Corollary 3.3.

Let $G$ be a finite abelian group with smallest prime divisor $p$ of $|G|$.
Let $S$ and $T$ be sequences over $G$. Suppose
$g\notin\langle\mbox{supp}(S)\rangle$ for every $g|T$. Then
$|\sum_{0}(T\bm{\cdot}S)|\geq(|T|+1)|\sum_{0}(S)|$ if one of the following
holds

  1. (1)

$|T|\leq p-1$;

  2. (2)

$|T|\leq 2p-1$ and
$|\langle\mbox{supp}(T\bm{\cdot}S)\rangle/\langle\mbox{supp}(S)\rangle|\geq
2p$.

###### Proof.

Let
$\varphi:\langle\mbox{supp}(T\bm{\cdot}S)\rangle\rightarrow\langle\mbox{supp}(T\bm{\cdot}S)\rangle/\langle\mbox{supp}(S)\rangle$
be the canonical epimorphism. Let $T=g_{1}\bm{\cdot}\ldots\bm{\cdot}g_{k}$ and
$\varphi(T)=\varphi(g_{1})\bm{\cdot}\ldots\bm{\cdot}\varphi(g_{k})$. Since
$g_{i}\notin\langle\mbox{supp}(S)\rangle$ for every $i\in[1,k]$, we have
$\varphi(g_{i})\neq 0$ for $i\in[1,k]$. Thus $\varphi(T)$ is a sequence over
$\varphi(G)\setminus\\{\varphi(0)\\}$. By Lemma 3.2, we have
$|\sum_{0}(\varphi(T))|\geq|\varphi(T)|+1=|T|+1$. Therefore,
$|\sum_{0}(T\bm{\cdot}S)|\geq|\sum_{0}(\varphi(T))|\cdot|\sum_{0}(S)|\geq(|T|+1)|\sum_{0}(S)|$.
∎

######  Lemma 3.4.

[1, Lemma 3.1] If $A$ is a $2$-zero-sum free subset of three elements in an
abelian group, then either $|\sum_{0}(A)|\geq 7$ or $A$ contains some element
of order two.

We now prove our main result.

Proof of Theorem 1.1.

By Lemma 2.2, it suffices to prove $\mathsf{c}_{0}(G)\leq n/p+p-2$. Let $S$ be
a regular sequence over $G$ of length $|S|=n/p+p-2$. We only need to show that
$\sum(S)=G$. Since $2p\leq n_{1}|n_{2}$, we have $n/p+p-2\geq
n_{1}+n_{2}-1=\mathsf{D}(G)$. Hence $0\in\sum(S)$ and thus
$\sum(S)=\sum_{0}(S)$.

Assume to the contrary that $\sum\nolimits_{0}(S)\neq G$, that is
$|\sum\nolimits_{0}(S)|<n$. Thus, by Lemma 2.3 (1), we have ${\rm
st}(\sum\nolimits_{0}(S))=\\{0\\}$.

Suppose that $\mathsf{h}(S)\leq 2p-2$. If $|\operatorname{supp}(S)|\leq
6p(p+1)$, then $|S|\leq|\operatorname{supp}(S)|\cdot\mathsf{h}(S)\leq
12p(p^{2}-1)$. Since $|S|=\frac{n}{p}+p-2\geq\frac{72p^{6}}{p}+p-2\geq
72p^{5}>12p(p^{2}-1)$, we get a contradiction. Hence
$|\operatorname{supp}(S)|\geq 6p(p+1)+1$. By Lemma 3.1, we may choose
$A_{1}\subseteq\operatorname{supp}(S)\subseteq G$, such that $|A_{1}|\leq
6p+1$ and $|\sum_{0}(A_{1})|\geq p|A_{1}|+2$. Let $t\geq 1$ be the maximal
integer such that $S$ has a factorization

$S=A_{1}\bm{\cdot}\ldots\bm{\cdot}A_{t}\bm{\cdot}T$

with $A_{i}\subseteq G$, $|A_{i}|\leq 6p+1$ and $|\sum_{0}(A_{i})|\geq
p|A_{i}|+2$ for every $i\in[1,t]$.

By the maximality of $t$ and Lemma 3.1, we have $|\mbox{supp}(T)|\leq
6p(p+1)$. Thus $|T|\leq|\mbox{supp}(T)|\cdot\mathsf{h}(S)\leq 12p(p^{2}-1)$.
Note that

$\sum(S)=\sum\nolimits_{0}(S)=\sum\nolimits_{0}(A_{1})+\cdots+\sum\nolimits_{0}(A_{t})+\sum\nolimits_{0}(T).$

Since $|A_{i}|\leq 6p+1$ for each $i\in[1,t]$, we have
$t\geq\frac{|S|-|T|}{6p+1}$. By Lemma 2.3 (2),
$|\sum\nolimits_{0}(T)|\geq|T|+1$. Since ${\rm st}(\sum(S))=\\{0\\}$, by Lemma
2.1, we have

$\displaystyle|\sum\nolimits_{0}(S)|=$
$\displaystyle|\sum\nolimits_{0}(A_{1})+\cdots+\sum\nolimits_{0}(A_{t})+\sum\nolimits_{0}(T)|$
$\displaystyle\geq$
$\displaystyle(|\sum\nolimits_{0}(A_{1})|-1)+\cdots+(|\sum\nolimits_{0}(A_{t})|-1)+|\sum\nolimits_{0}(T)|$
$\displaystyle\geq$ $\displaystyle\sum_{i=1}^{t}(p|A_{i}|+1)+|T|+1$
$\displaystyle=$ $\displaystyle p(|S|-|T|)+t+|T|+1$ (3.1) $\displaystyle\geq$
$\displaystyle(p+\frac{1}{6p+1})(|S|-|T|)+|T|+1$ $\displaystyle\geq$
$\displaystyle n+\frac{n}{p(6p+1)}+p^{2}-2p-(p-\frac{6p}{6p+1})|T|+1\
(\mbox{as }|S|=\frac{n}{p}+p-2)$ $\displaystyle\geq$ $\displaystyle
n+\frac{n}{p(6p+1)}+p^{2}-2p-\frac{12p^{2}(6p-5)(p^{2}-1)}{6p+1}+1\ (\mbox{as
}|T|\leq 12p(p^{2}-1))$ $\displaystyle\geq$ $\displaystyle
n+\frac{n}{p(6p+1)}-\frac{12p^{2}\cdot 6p\cdot p^{2}}{6p+1}$
$\displaystyle\geq$ $\displaystyle n\ \ \ \ \ \ \ \ \ (\mbox{as }n\geq
72p^{6}),$

yielding a contradiction. Therefore, we must have $\mathsf{h}(S)\geq 2p-1$. We
choose $g_{1}\in\mbox{supp}(S)$ such that

$|S_{\langle g_{1}\rangle}|=\mbox{max}\\{|S_{\langle
g\rangle}|:g\in\mbox{supp}(S)\mbox{ and }\mathsf{v}_{g}(S)\geq 2p-1\\}.$

Let $S_{1}=S_{\langle g_{1}\rangle}$ and $\lambda$ be the maximal integer such
that $S$ has a factorization

$S=(T_{1}\bm{\cdot}S_{1})\bm{\cdot}(T_{2}\bm{\cdot}S_{2})\bm{\cdot}\ldots\bm{\cdot}(T_{\lambda}\bm{\cdot}S_{\lambda})\bm{\cdot}S^{\prime}$

with the following properties:

  * (1)

$|T_{i}|=2p-1$ for each $i\in[1,\lambda]$;

  * (2)

for each $i\in[1,\lambda]$, $\mathsf{v}_{g_{i}}(S_{i})\geq 2p-1$ for some
$g_{i}\in G$; moreover, $S_{i}$ contains all terms from
$S\bm{\cdot}((T_{1}\bm{\cdot}S_{1})\bm{\cdot}(T_{2}\bm{\cdot}S_{2})\bm{\cdot}\ldots\bm{\cdot}(T_{i-1}\bm{\cdot}S_{i-1}))^{[-1]}$
contained in $\langle g_{i}\rangle$;

  * (3)

$|\sum_{0}(T_{i}\bm{\cdot}S_{i}^{\prime})|\geq 2p(|S_{i}^{\prime}|+1)$ for all
subsequences $S_{i}^{\prime}|S_{i}$, where $i\in[1,\lambda]$.

Clearly, $S_{1}$ satisfies Property $(2)$.

Claim.

$\lambda\geq 1,\ \ \ \mbox{i.e., $T_{1}$ always exists}.$

We first show that $\langle\mbox{supp}(S)\rangle=G$. For otherwise, if
$H=\langle\mbox{supp}(S)\rangle<G$, then $|H|\leq\frac{n}{p}$ where $p$ is the
smallest prime divisor of $|G|$. Since $S$ is regular, we have
$|H|-1\geq|S_{H}|=|S|=\frac{n}{p}+p-2\geq|H|$, yielding a contradiction. Thus
$\langle\mbox{supp}(S)\rangle=G$. It follows that

$G/\langle\mbox{supp}(S_{1})\rangle=\langle\mbox{supp}(S\bm{\cdot}S_{1}^{[-1]}),\mbox{supp}(S_{1})\rangle/\langle\mbox{supp}(S_{1})\rangle.$

Note that

$|S\bm{\cdot}S_{1}^{[-1]}|\geq\frac{n}{p}+p-2-\frac{n}{p^{2}}\geq 2p.$

Next we choose a suitable subsequence $T_{1}|S\bm{\cdot}S_{1}^{[-1]}$. If
$G/\langle\mbox{supp}(S_{1})\rangle$ is a cyclic group of prime order, then
let $T_{1}$ be an arbitrary subsequence of $S\bm{\cdot}S_{1}^{[-1]}$ with
length $|T_{1}|=2p-1$. Since
$\langle\operatorname{supp}(T_{1}),\operatorname{supp}(S_{1})\rangle/\langle\mbox{supp}(S_{1})\rangle$
is not trivial, we must have
$\langle\operatorname{supp}(T_{1}),\operatorname{supp}(S_{1})\rangle/\langle\mbox{supp}(S_{1})\rangle=G/\langle\mbox{supp}(S_{1})\rangle$.
Therefore,

$|\langle\operatorname{supp}(T_{1}),\operatorname{supp}(S_{1})\rangle/\langle\mbox{supp}(S_{1})\rangle|=|G/\langle\mbox{supp}(S_{1})\rangle|\geq
n_{1}\geq 2p,$

as $|\langle\mbox{supp}(S_{1})\rangle|\leq|\langle g_{1}\rangle|\leq n_{2}$.
Next we assume that $|G/\langle\mbox{supp}(S_{1})\rangle|$ is a composite
number. Since

$G/\langle\mbox{supp}(S_{1})\rangle=\langle\mbox{supp}(S\bm{\cdot}S_{1}^{[-1]}),\mbox{supp}(S_{1})\rangle/\langle\mbox{supp}(S_{1})\rangle,$

there are two elements $h_{1},h_{2}\in\mbox{supp}(S\bm{\cdot}S_{1}^{[-1]})$
such that $|\langle
h_{1},h_{2},\operatorname{supp}(S_{1})\rangle/\langle\mbox{supp}(S_{1})\rangle|$
is a composite number. It follows that

$|\langle
h_{1},h_{2},\operatorname{supp}(S_{1})\rangle/\langle\mbox{supp}(S_{1})\rangle|\geq
2p,$

as $p$ is the smallest prime divisor of $n$. Let $T_{1}$ be a subsequence of
$S\bm{\cdot}S_{1}^{[-1]}$ with $h_{1}\bm{\cdot}h_{2}|T_{1}$. Then,

$|\langle\operatorname{supp}(T_{1}),\operatorname{supp}(S_{1})\rangle/\langle\mbox{supp}(S_{1})\rangle|\geq
2p.$

So we can always choose a subsequence $T_{1}|S\bm{\cdot}S_{1}^{[-1]}$ such
that
$|\langle\operatorname{supp}(T_{1}),\operatorname{supp}(S_{1})\rangle/\langle\mbox{supp}(S_{1})\rangle|\geq
2p$ and $|T_{1}|=2p-1$.

Finally, we verify Property $(3)$ holds. Let $S_{1}^{\prime}|S_{1}$ be a
subsequence of $S_{1}$. Then
$\langle\mbox{supp}(S_{1}^{\prime})\rangle\leq\langle\operatorname{supp}(S_{1})\rangle$
and thus
$|\langle\operatorname{supp}(T_{1}),\operatorname{supp}(S_{1}^{\prime})\rangle/\langle\mbox{supp}(S_{1}^{\prime})|\geq|\langle\operatorname{supp}(T_{1}),\operatorname{supp}(S_{1})\rangle/\langle\mbox{supp}(S_{1})\rangle|\geq
2p$.
Note that $g\notin\langle\mbox{supp}(S_{1})\rangle$ for every $g|T_{1}$. Thus
$g\notin\langle\mbox{supp}(S_{1}^{\prime})\rangle$ for every $g|T_{1}$. By
Corollary 3.3 (2) and Lemma 2.3 (2), we have
$|\sum_{0}(T_{1}\bm{\cdot}S_{1}^{\prime})|\geq(|T_{1}|+1)|\sum_{0}(S_{1}^{\prime})|\geq
2p(|S_{1}^{\prime}|+1)$, implying Property $(3)$. In summary, we have found a
subsequence $T_{1}$ (together with $S_{1}$) satisfying all three properties.
This completes the proof of Claim.

Next we distinguish the remaining proof into the following two cases according
to the value of $\mathsf{h}(S^{\prime})$.

Case 1. $\mathsf{h}(S^{\prime})\leq 2p-2$.

Let $t\geq 0$ be the maximal integer such that $S^{\prime}$ has a
factorization

$S^{\prime}=A_{1}\bm{\cdot}\ldots\bm{\cdot}A_{t}\bm{\cdot}T$

with $A_{i}\subseteq G$, $|A_{i}|\leq 6p+1$ and $|\sum_{0}(A_{i})|\geq
p|A_{i}|+2$. By the maximality of $t$ and Lemma 3.1, we have
$|\mbox{supp}(T)|\leq 6p(p+1)$. Thus
$|T|\leq|\mbox{supp}(T)|\cdot\mathsf{h}(S^{\prime})\leq 12p(p^{2}-1)$. Since
$|T_{i}\bm{\cdot}S_{i}|\geq 4p-2$ for each $i\in[1,\lambda]$, we have
$\lambda\leq\frac{|S|-|S^{\prime}|}{4p-2}$. Similarly,
$t\geq\frac{|S^{\prime}|-|T|}{6p+1}$. By Lemma 2.3 (2),
$|\sum\nolimits_{0}(T)|\geq|T|+1$. Since ${\rm
st}(\sum\nolimits_{0}(S))=\\{0\\}$, by Lemma 2.1, we have

$\displaystyle|\sum\nolimits_{0}(S)|=$
$\displaystyle|\sum_{i=1}^{\lambda}\sum\nolimits_{0}(T_{i}\bm{\cdot}S_{i})+\sum_{j=1}^{t}\sum\nolimits_{0}(A_{j})+\sum\nolimits_{0}(T)|$
$\displaystyle\geq$
$\displaystyle\sum_{i=1}^{\lambda}(|\sum\nolimits_{0}(T_{i}\bm{\cdot}S_{i})|-1)+\sum_{j=1}^{t}(|\sum\nolimits_{0}(A_{j})|-1)+|\sum\nolimits_{0}(T)|$
$\displaystyle\geq$
$\displaystyle\sum_{i=1}^{\lambda}(2p|S_{i}|+2p-1)+\sum_{j=1}^{t}(p|A_{j}|+1)+|T|+1$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{\lambda}(2p|T_{i}\bm{\cdot}S_{i}|-(2p-1)^{2})+p(|S^{\prime}|-|T|)+t+|T|+1$
$\displaystyle\geq$ $\displaystyle
2p(|S|-|S^{\prime}|)-(2p-1)^{2}\lambda+(p+\frac{1}{6p+1})(|S^{\prime}|-|T|)+|T|+1$
$\displaystyle\geq$
$\displaystyle(p+1/2)(|S|-|S^{\prime}|)+(p+\frac{1}{6p+1})(|S^{\prime}|-|T|)+|T|+1$
$\displaystyle\geq$ $\displaystyle(p+\frac{1}{6p+1})(|S|-|T|)+|T|+1$
$\displaystyle\geq$ $\displaystyle n\ \ \ \ \ \ \ \ (\mbox{as }|T|\leq
12p(p^{2}-1)),$

yielding a contradiction. We note that the last inequality was obtained by
using a similar calculation as in (3.1).

Case 2. $\mathsf{h}(S^{\prime})\geq 2p-1$.

There exists $g_{\lambda+1}\in G$ such that
$\mathsf{v}_{g_{\lambda+1}}(S^{\prime})=\mathsf{h}(S^{\prime})\geq 2p-1$. Let
$S_{\lambda+1}=S_{\langle g_{\lambda+1}\rangle}^{\prime}$ and
$U=S^{\prime}\bm{\cdot}S_{\lambda+1}^{[-1]}$. Then

$S=(T_{1}\bm{\cdot}S_{1})\bm{\cdot}(T_{2}\bm{\cdot}S_{2})\bm{\cdot}\ldots\bm{\cdot}(T_{\lambda}\bm{\cdot}S_{\lambda})\bm{\cdot}S_{\lambda+1}\bm{\cdot}U.$

We next distinguish the rest proof into the following two subcases.

Subcase 2.1. $|U|\leq 2p-2$.

By the choice of $S_{1}$, we have $|S_{\lambda+1}|\leq|S_{1}|$. Since
$|T_{i}\bm{\cdot}S_{i}|\geq 4p-2$ for each $i\in[2,\lambda]$, we have
$\lambda-1\leq\frac{|S|-|S_{1}\bm{\cdot}T_{1}|-|S_{\lambda+1}|-|U|}{4p-2}$. By
Lemma 2.3 (2),
$|\sum\nolimits_{0}(S_{\lambda+1}\bm{\cdot}U)|\geq|S_{\lambda+1}\bm{\cdot}U|+1$.
Since ${\rm st}(\sum_{0}(S))=\\{0\\}$, by Lemma 2.1, we have

$\displaystyle|\sum\nolimits_{0}(S)|=$
$\displaystyle|\sum\nolimits_{0}(T_{1}\bm{\cdot}S_{1})+\sum_{i=2}^{\lambda}\sum\nolimits_{0}(T_{i}\bm{\cdot}S_{i})+\sum\nolimits_{0}(S_{\lambda+1}\bm{\cdot}U)|$
$\displaystyle\geq$
$\displaystyle(|\sum\nolimits_{0}(T_{1}\bm{\cdot}S_{1})|-1)+\sum_{i=2}^{\lambda}(|\sum\nolimits_{0}(T_{i}\bm{\cdot}S_{i})|-1)+|\sum\nolimits_{0}(S_{\lambda+1}\bm{\cdot}U)|$
$\displaystyle\geq$
$\displaystyle(2p|S_{1}|+2p-1)+\sum_{i=2}^{\lambda}(2p|S_{i}|+2p-1)+|S_{\lambda+1}\bm{\cdot}U|+1$
$\displaystyle\geq$
$\displaystyle(p+1/2)|S_{1}|+\sum_{i=2}^{\lambda}(2p|T_{i}\bm{\cdot}S_{i}|-(2p-1)^{2})+(p+1/2)|S_{\lambda+1}|+2p$
$\displaystyle=$
$\displaystyle(p+1/2)(|S_{1}|+|S_{\lambda+1}|)+2p(|S|-|T_{1}\bm{\cdot}S_{1}|-|S_{\lambda+1}|-|U|)-(2p-1)^{2}({\lambda}-1)+2p$
$\displaystyle\geq$
$\displaystyle(p+1/2)(|S_{1}|+|S_{\lambda+1}|)+(p+1/2)(|S|-|T_{1}\bm{\cdot}S_{1}|-|S_{\lambda+1}|-|U|)+2p$
$\displaystyle=$ $\displaystyle(p+1/2)(|S|-|T_{1}|-|U|)+2p$
$\displaystyle\geq$ $\displaystyle(p+1/2)(n/p+p-2-(4p-3))+2p$
$\displaystyle\geq$ $\displaystyle n\ \ \ \ \ \ \ \ \ (\mbox{as }n\geq
6p^{3}-3p^{2}-p),$

yielding a contradiction.

Subcase 2.2. $|U|\geq 2p-1$.

Let $H=\langle\mbox{supp}(U),g_{\lambda+1}\rangle$. If $|H/\langle
g_{\lambda+1}\rangle|\geq 2p$, then by a similar method as used in the proof
of Claim, we can find a subsequence $T_{\lambda+1}|U$ with
$|T_{\lambda+1}|=2p-1$ such that $S_{\lambda+1}$ and $T_{\lambda+1}$ satisfy
Properties $(1)-(3)$, yielding a contradiction to the maximality of $\lambda$.
Therefore, we must have $|H/\langle g_{\lambda+1}\rangle|\leq 2p-1$. Since
$\langle g_{\lambda+1}\rangle$ is a cyclic subgroup of $G$, we have
$|G/\langle g_{\lambda+1}\rangle|\geq n_{1}\geq 2p$. Thus $H\neq G$.
Therefore,

$|H|\leq n/p\mbox{ \ \ \ \ and \ \ \ \ }|S^{\prime}|\leq|S_{H}|\leq n/p-1\
(\mbox{as }S^{\prime}|S_{H}\mbox{ and }S\mbox{ is regular}).$

Suppose that $\mbox{supp}(S_{i})\nsubseteq H$ for some $i\in[1,\lambda]$. Then
let $S_{i}^{\prime}=S_{i}\bm{\cdot}g_{i}^{[-(p-1)]}$. We have

$S=\left(\prod_{1\leq j\neq
i\leq\lambda}(T_{j}\bm{\cdot}S_{j})\right)\bm{\cdot}(T_{i}\bm{\cdot}S_{i}^{\prime})\bm{\cdot}(g_{i}^{[p-1]}\bm{\cdot}S^{\prime}).$

Let $U^{\prime}|U$ with $|U^{\prime}|=p-1$. By Corollary 3.3 (1) and Lemma 2.3
(2),
$|\sum_{0}(U^{\prime}\bm{\cdot}g_{\lambda+1}^{[2p-1]})|\geq(|U^{\prime}|+1)|\sum_{0}(g_{\lambda+1}^{[2p-1]})|=2p^{2}$.
By Lemma 2.3 (1), ${\rm st}(\sum_{0}(S^{\prime}))=\\{0\\}$. Note that

$\sum\nolimits_{0}(S^{\prime})=\sum\nolimits_{0}(S^{\prime}\bm{\cdot}(U^{\prime}\bm{\cdot}g_{\lambda+1}^{[2p-1]})^{[-1]})+\sum\nolimits_{0}(U^{\prime}\bm{\cdot}g_{\lambda+1}^{[2p-1]}).$

By Lemma 2.1 and Lemma 2.3 (2),
$|\sum_{0}(S^{\prime})|\geq|\sum_{0}(S^{\prime}\bm{\cdot}(U^{\prime}\bm{\cdot}g_{\lambda+1}^{[2p-1]})^{[-1]})|+|\sum_{0}(U^{\prime}\bm{\cdot}g_{\lambda+1}^{[2p-1]})|-1\geq|S^{\prime}\bm{\cdot}(U^{\prime}\bm{\cdot}g_{\lambda+1}^{[2p-1]})^{[-1]}|+2p^{2}=|S^{\prime}|+2p^{2}-3p+2$.
Therefore, by Corollary 3.3 (1),

$|\sum\nolimits_{0}(g_{i}^{[p-1]}\bm{\cdot}S^{\prime})|\geq(|g_{i}^{[p-1]}|+1)|\sum\nolimits_{0}(S^{\prime})|\geq
p|\sum\nolimits_{0}(S^{\prime})|\geq p|S^{\prime}|+2p^{3}-3p^{2}+2p.$

As in Subcase 2.1, we have
$\lambda-1\leq\frac{|S|-|T_{i}\bm{\cdot}S_{i}|-|S^{\prime}|}{4p-2}$. Since
${\rm st}(\sum_{0}(S))=\\{0\\}$, by Lemma 2.1,

$\displaystyle|\sum\nolimits_{0}(S)|=$ $\displaystyle|\sum_{1\leq j\neq
i\leq\lambda}\sum\nolimits_{0}(T_{j}\bm{\cdot}S_{j})+\sum\nolimits_{0}(T_{i}\bm{\cdot}S_{i}^{\prime})+\sum\nolimits_{0}(g_{i}^{[p-1]}\bm{\cdot}S^{\prime})|$
$\displaystyle\geq$ $\displaystyle\sum_{1\leq j\neq
i\leq\lambda}(|\sum\nolimits_{0}(T_{j}\bm{\cdot}S_{j})|-1)+(|\sum\nolimits_{0}(T_{i}\bm{\cdot}S_{i}^{\prime})|-1)+|\sum\nolimits_{0}(g_{i}^{[p-1]}\bm{\cdot}S^{\prime})|$
$\displaystyle\geq$ $\displaystyle\sum_{1\leq j\neq
i\leq\lambda}(2p|T_{j}\bm{\cdot}S_{j}|-(2p-1)^{2})+(2p|S_{i}^{\prime}|+2p-1)+p|S^{\prime}|+2p^{3}-3p^{2}+2p$
$\displaystyle=$ $\displaystyle
2p(|S|-|S_{i}\bm{\cdot}T_{i}|-|S^{\prime}|)-(2p-1)^{2}(\lambda-1)+2p(|S_{i}|-p+1)$
$\displaystyle+p|S^{\prime}|+2p^{3}-3p^{2}+4p-1$ $\displaystyle\geq$
$\displaystyle(p+1/2)(|S|-|S_{i}\bm{\cdot}T_{i}|-|S^{\prime}|)+2p|S_{i}|+p|S^{\prime}|+2p^{3}-5p^{2}+6p-1$
$\displaystyle=$
$\displaystyle(p+1/2)|S|+(p-1/2)|S_{i}|-(p+1/2)|T_{i}|-|S^{\prime}|/2+2p^{3}-5p^{2}+6p-1$
$\displaystyle\geq$
$\displaystyle(p+1/2)(n/p+p-2)-(2p-1)-(n/p-1)/2+2p^{3}-5p^{2}+6p-1$
$\displaystyle\geq$ $\displaystyle n,$

yielding a contradiction.

Next assume that $\mbox{supp}(S_{i})\subseteq H$ for every $i\in[1,\lambda]$.
Let

$S^{\prime\prime}=S_{1}\bm{\cdot}\ldots\bm{\cdot}S_{\lambda}\bm{\cdot}S^{\prime}.$

Then $S^{\prime\prime}|S_{H}$. Let

$n^{\prime}=|H|\leq n/p.$

Since
$S^{\prime\prime}=S_{1}\bm{\cdot}\ldots\bm{\cdot}S_{\lambda}\bm{\cdot}S^{\prime}$
with $|S_{i}|\geq|T_{i}|$ for $i\in[1,\lambda]$, we have
$|S^{\prime\prime}|\geq(|S|+|S^{\prime}|)/2\geq((n/p+p-2)+|U|+|S_{\lambda+1}|)/2\geq(n^{\prime}+|U|+|S_{\lambda+1}|+1)/2$.

Let $T=h_{1}\bm{\cdot}h_{2}\bm{\cdot}S_{\lambda+1}$ where
$h_{1}\bm{\cdot}h_{2}|U$. Let $t\geq 0$ be the maximal integer such that
$S^{\prime\prime}$ has a new factorization

$S^{\prime\prime}=A_{1}\bm{\cdot}\ldots\bm{\cdot}A_{t}\bm{\cdot}W\bm{\cdot}T$

where each $A_{i}$ is a $2$-zero-sum free $3$-subset of $G$, and $W$ is a
subsequence of $S^{\prime\prime}$ which contains no $2$-zero-sum free
$3$-subset of $G$. Then

(3.2)
$3t+|W|+|S_{\lambda+1}|+2=|S^{\prime\prime}|\geq(n^{\prime}+|U|+|S_{\lambda+1}|+1)/2$

and

$\mbox{supp}(W)\subseteq\\{u_{1},-u_{1},u_{2},-u_{2}\\}$

for some distinct elements $u_{1},u_{2}\in H$. By Lemma 3.4,
$|\sum_{0}(A_{i})|\geq 7$ for $i\in[1,t]$.

We now show that $|W|>|U|$. Since $S^{\prime\prime}\subseteq H$ and
$|H|=n^{\prime}$, we must have

$\sum\nolimits_{0}(S^{\prime\prime})\subsetneq H,\mbox{\ \ \ and thus\ \ \
}|\sum\nolimits_{0}(S^{\prime\prime})|\leq n^{\prime}-1.$

For otherwise, $\sum_{0}(S^{\prime\prime})=H$ and thus ${\rm
st}(\sum_{0}(S^{\prime\prime}))\supseteq H\neq\\{0\\}$. However, by Lemma 2.3
(1), ${\rm st}(\sum_{0}(S^{\prime\prime}))=\\{0\\}$, yielding a contradiction.
By Corollary 3.3 (1) and Lemma 2.3 (2), $|\sum\nolimits_{0}(T)|\geq
3(|S_{\lambda+1}|+1)$. It follows from Lemma 2.1 and 2.3 (2) that

$\displaystyle n^{\prime}-1\geq$
$\displaystyle|\sum\nolimits_{0}(S^{\prime\prime})|=|\sum_{i=1}^{t}\sum\nolimits_{0}(A_{i})+\sum\nolimits_{0}(W)+\sum\nolimits_{0}(T)|$
$\displaystyle\geq$
$\displaystyle\sum_{i=1}^{t}(|\sum\nolimits_{0}(A_{i})|-1)+(|\sum\nolimits_{0}(W)|-1)+|\sum\nolimits_{0}(T)|$
$\displaystyle\geq$ $\displaystyle\sum_{i=1}^{t}6+|W|+3(|S_{\lambda+1}|+1)$
$\displaystyle=$ $\displaystyle 6t+|W|+3|S_{\lambda+1}|+3$ $\displaystyle=$
$\displaystyle 2|S^{\prime\prime}|-|W|+|S_{\lambda+1}|-1$ $\displaystyle\geq$
$\displaystyle n^{\prime}+|U|+2|S_{\lambda+1}|-|W|\ \ \ \ \ \ \ \ (\mbox{by
}\eqref{S''}).$

This gives

$|W|\geq|U|+2|S_{\lambda+1}|+1>|U|.$

Note that
$W|S^{\prime\prime}\bm{\cdot}S_{\lambda+1}^{[-1]}=S_{1}\bm{\cdot}\ldots\bm{\cdot}S_{\lambda}\bm{\cdot}U$.
In view of Property (2), if $u_{1}$ lies in $\operatorname{supp}(S_{j})$ for
some $j\leq\lambda+1$, then $\mathsf{v}_{\pm
u_{1}}(S^{\prime\prime})=\mathsf{v}_{\pm u_{1}}(S_{j})$. Therefore,
$u_{1}\in\operatorname{supp}(U)$ if and only if $\mathsf{v}_{\pm
u_{1}}(U)=\mathsf{v}_{\pm u_{1}}(S^{\prime\prime})$, meaning all terms equal
to $u_{1}$ or $-u_{1}$ in $S^{\prime\prime}$ occur in $U$ provided that
$u_{1}\in\operatorname{supp}(U)$; and likewise for $-u_{1}$, $u_{2}$ and
$-u_{2}$. Since $W|S^{\prime\prime}$ and
$\mbox{supp}(W)\subseteq\\{u_{1},-u_{1},u_{2},-u_{2}\\}$, we have
$W|u_{1}^{\mathsf{v}_{u_{1}}(S^{\prime\prime})}\bm{\cdot}(-u_{1})^{\mathsf{v}_{-u_{1}}(S^{\prime\prime})}\bm{\cdot}u_{2}^{\mathsf{v}_{u_{2}}(S^{\prime\prime})}\bm{\cdot}(-u_{2})^{\mathsf{v}_{-u_{2}}(S^{\prime\prime})}$.
If $u_{1}\bm{\cdot}u_{2}|W$ and $u_{1}\bm{\cdot}u_{2}|U$, then by what we just
proved,
$u_{1}^{\mathsf{v}_{u_{1}}(S^{\prime\prime})}\bm{\cdot}(-u_{1})^{\mathsf{v}_{-u_{1}}(S^{\prime\prime})}\bm{\cdot}u_{2}^{\mathsf{v}_{u_{2}}(S^{\prime\prime})}\bm{\cdot}(-u_{2})^{\mathsf{v}_{-u_{2}}(S^{\prime\prime})}|U$,
so $W|U$, implying $|W|\leq|U|$, yielding a contradiction to $|W|>|U|$.
Therefore, without loss of generality, we may assume that $u_{1}|W$ and
$u_{1}\notin\operatorname{supp}(U)$. Thus $u_{1}\in\mbox{supp}(S_{j})$ for
some $j\leq\lambda+1$. Since the sequence $W$ is disjoint from $S_{\lambda+1}$
by construction, it follows that $u_{1}\in\operatorname{supp}(S_{j})$ for some
$j\leq\lambda$. In view of Property (2), we conclude that
$-u_{1}\in\mbox{supp}(S_{j})$ and $v\notin\langle u_{1}\rangle\leq\langle
g_{j}\rangle$ for every $v\in\mbox{supp}(S_{\lambda+1})$. Write
$W=W_{1}\bm{\cdot}W_{2}$ with

$W_{1}|u_{1}^{[\mathsf{v}_{u_{1}}(S^{\prime\prime})]}\bm{\cdot}(-u_{1})^{[\mathsf{v}_{-u_{1}}(S^{\prime\prime})]}\mbox{
\ \ and \ \
}W_{2}|u_{2}^{[\mathsf{v}_{u_{2}}(S^{\prime\prime})]}\bm{\cdot}(-u_{2})^{[\mathsf{v}_{-u_{2}}(S^{\prime\prime})]}.$

Now fix $v|S_{\lambda+1}$. Let $V=W_{1}\bm{\cdot}v$ and
$T^{\prime}=T\bm{\cdot}v^{[-1]}$. We obtain another factorization

$S^{\prime\prime}=A_{1}\bm{\cdot}\ldots\bm{\cdot}A_{t}\bm{\cdot}V\bm{\cdot}W_{2}\bm{\cdot}T^{\prime}.$

Since $v\notin\langle u_{1}\rangle$, by Corollary 3.3 (1) and Lemma 2.3 (2),
$|\sum\nolimits_{0}(V)|\geq 2(|W_{1}|+1)$. Note that

(3.3)
$|S^{\prime\prime}|=3t+|V|+|W_{2}|+|T|-1=3t+|W_{1}|+|W_{2}|+|S_{\lambda+1}|+2.$

As before, we obtain

$\displaystyle n^{\prime}-1\geq$
$\displaystyle|\sum\nolimits_{0}(S^{\prime\prime})|=|\sum_{i=1}^{t}\sum\nolimits_{0}(A_{i})+\sum\nolimits_{0}(V)+\sum\nolimits_{0}(W_{2})+\sum\nolimits_{0}(T^{\prime})|$
$\displaystyle\geq$
$\displaystyle\sum_{i=1}^{t}(|\sum\nolimits_{0}(A_{i})|-1)+(|\sum\nolimits_{0}(V)|-1)+(|\sum\nolimits_{0}(W_{2})|-1)+|\sum\nolimits_{0}(T^{\prime})|$
$\displaystyle\geq$
$\displaystyle\sum_{i=1}^{t}6+2|W_{1}|+1+|W_{2}|+3|S_{\lambda+1}|$
$\displaystyle=$ $\displaystyle 6t+2|W_{1}|+|W_{2}|+3|S_{\lambda+1}|+1$
$\displaystyle=$ $\displaystyle 2|S^{\prime\prime}|-|W_{2}|+|S_{\lambda+1}|-3\
\ \ \ \ \ \ (\mbox{by }\eqref{W_1})$ $\displaystyle\geq$ $\displaystyle
n^{\prime}+|U|+2|S_{\lambda+1}|-|W_{2}|-2\ \ \ \ \ \ \ \ (\mbox{by
}\eqref{S''}),$

implying

$|W_{2}|\geq|U|+2|S_{\lambda+1}|-1>|U|.$

Next we consider $W_{2}$. Since
$\operatorname{supp}(W_{2})\subseteq\\{u_{2},-u_{2}\\}$, if
$u_{2}\in\operatorname{supp}(U)$, then as above, we obtain $W_{2}|U$, implying
$|W_{2}|\leq|U|$, giving a contradiction. Thus
$u_{2}\notin\operatorname{supp}(U)$. As above, we conclude that $\\{\pm
u_{2}\\}\subseteq\mbox{supp}(S_{j^{\prime}})$ for some $j^{\prime}\leq\lambda$
and $z\notin\langle u_{2}\rangle\leq\langle g_{j^{\prime}}\rangle$ for every
$z\in\mbox{supp}(S_{\lambda+1})$. Since $|S_{\lambda+1}|\geq 2p-1\geq 5$, we
can choose $z|S_{\lambda+1}\bm{\cdot}v^{[-1]}$. Define $V=W_{1}\bm{\cdot}v$,
$Z=W_{2}\bm{\cdot}z$ and
$T^{\prime\prime}=T\bm{\cdot}v^{[-1]}\bm{\cdot}z^{[-1]}$. Thus we obtain
another factorization

$S^{\prime\prime}=A_{1}\bm{\cdot}\ldots\bm{\cdot}A_{t}\bm{\cdot}V\bm{\cdot}Z\bm{\cdot}T^{\prime\prime}.$

As before, we obtain

$\displaystyle n^{\prime}-1\geq$
$\displaystyle|\sum\nolimits_{0}(S^{\prime\prime})|=|\sum_{i=1}^{t}\sum\nolimits_{0}(A_{i})+\sum\nolimits_{0}(V)+\sum\nolimits_{0}(Z)+\sum\nolimits_{0}(T^{\prime\prime})|$
$\displaystyle\geq$
$\displaystyle\sum_{i=1}^{t}(|\sum\nolimits_{0}(A_{i})|-1)+(|\sum\nolimits_{0}(V)|-1)+(|\sum\nolimits_{0}(Z)|-1)+|\sum\nolimits_{0}(T^{\prime\prime})|$
$\displaystyle\geq$
$\displaystyle\sum_{i=1}^{t}6+2|W_{1}|+1+2|W_{2}|+1+3(|S_{\lambda+1}|-1)$
$\displaystyle=$ $\displaystyle 6t+2|W_{1}|+2|W_{2}|+3|S_{\lambda+1}|-1$
$\displaystyle=$ $\displaystyle 2|S^{\prime\prime}|+|S_{\lambda+1}|-5\ \ \ \ \
\ \ (\mbox{by }\eqref{W_1})$ $\displaystyle\geq$ $\displaystyle
n^{\prime}+|U|+2|S_{\lambda+1}|-4\ \ \ \ \ \ \ \ (\mbox{by }\eqref{S''})$
$\displaystyle\geq$ $\displaystyle n^{\prime}\ \ \ \ \ \ (\mbox{as }|U|\geq
2p-1\geq 5),$

yielding a contradiction. In all cases we have found contradictions. Thus we
must have $\sum(S)=G$. This completes the proof. ∎

We end this paper with the following remark.

######  Remark 3.5.

It was proved that $f(6)=19$ in [3, Theorem 1.2]. Thus Lemma 3.1 can be
improved as follows: if $A\subseteq G\setminus\\{0\\}$ is a subset with
$|A|\geq 21$, then there is a subset $B\subset A$ such that $|\sum_{0}(B)|\geq
3|B|+2$ with $|B|\leq 6$. Therefore, when $p=3$, we may relax the condition
$|G|\geq 72p^{6}=52488$ to $|G|\geq 3045$, and the same proof as in Theorem
1.1 still works.

Acknowledgements. Part of this work was carried out during a visit of the
first author to Brock University as an international visiting scholar. He
would like to gratefully thank the host institution for its hospitality and
for providing an excellent atmosphere for research. This work was supported in
part by the National Natural Science Foundation of China (No. 11701256,
11871258, 12071344), the Youth Backbone Teacher Foundation of Henan’s
University (No. 2019GGJS196), the China Scholarship Council (Grant No.
201908410132), and it was also supported in part by a Discovery Grant from the
Natural Sciences and Engineering Research Council of Canada (Grant No. RGPIN
2017-03903).

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# PandaGPT:
One Model To Instruction-Follow Them All

Yixuan Su♠, , &Tian Lan∗ &Huayang Li♢,11footnotemark: 1 ,22footnotemark: 2
&Jialu Xu &Yan Wang &Deng Cai♣,∗

♠University of Cambridge ♢Nara Institute of Science and Technology
♣Tencent AI Lab
https://panda-gpt.github.io/
Major contributors. Contact<EMAIL_ADDRESS>and<EMAIL_ADDRESS>done
during internship at Tencent AI Lab.

###### Abstract

We present PandaGPT, an approach to emPower large lANguage moDels with visual
and Auditory instruction-following capabilities. Our pilot experiments show
that PandaGPT can perform complex tasks such as detailed image description
generation, writing stories inspired by videos, and answering questions about
audios. More interestingly, PandaGPT can take multimodal inputs simultaneously
and compose their semantics naturally. For example, PandaGPT can connect how
objects look in an image/video and how they sound in an audio. To do so,
PandaGPT combines the multimodal encoders from ImageBind and the large
language models from Vicuna. Notably, only aligned image-text pairs are
required for the training of PandaGPT. Thanks to the strong capability of
ImageBind in embedding data from different modalities into the same space,
PandaGPT displays emergent, i.e. zero-shot, cross-modal behaviors for data
other than image and text (e.g., video, audio, depth, thermal, and IMU). We
hope that PandaGPT serves as an initial step toward building AGI that can
perceive and understand inputs in different modalities holistically, as we
humans do.

##  1 Introduction

Humans possess remarkable abilities to perceive and understand information
from diverse sensory modalities, such as seeing a painting and hearing an
audio guide. Analogously, to learn simultaneously, holistically, and directly
from many different forms of information holds great promise for enabling
machines to have a more comprehensive and better understanding of the world.
To this end, there has been an emergent interest in developing artificial
intelligence (AI) systems capable of perceiving and understanding information
from multiple modalities simultaneously in a manner similar to humans.

However, much of the prior research has focused on tackling individual
modalities in isolation. For instance, while significant progress has been
made in text-to-image retrieval and generation [18], visually-grounded
instruction following [12, 31], and speech understanding and generation [29],
these advances have largely been confined to separate combinations of text and
other modalities or, at best, a few visual modalities (e.g., image and video).
These models are limited in their ability to connect information from
different modalities and lack the capacity to perceive and understand
multimodal inputs holistically, thereby neglecting the inherent richness and
complementary nature of multimodal data.

In this paper, we present PandaGPT, the first general-purpose model capable of
instruction-following data from six modalities. PandaGPT leverages the power
of multimodal encoders from ImageBind [8] and the expressive language models
from Vicuna [4], demonstrating impressive and emergent cross-modal
capabilities across six modalities: image/video, text, audio, depth, thermal,
and inertial measurement units (IMU). Crucially, PandaGPT achieves these
capabilities despite being only trained on aligned image-text pairs, thanks to
the shared embedding space provided by ImageBind.

This integration of multimodal information enables PandaGPT to perform a wide
range of tasks, including generating detailed descriptions of images,
composing engaging stories inspired by videos, and providing accurate answers
to questions about audio inputs. Most interestingly, the core innovation of
PandaGPT lies in its ability to naturally compose the semantics of multimodal
inputs, which enables a rich set of compositional multimodal tasks across
different modalities. For example, it can seamlessly connect the visual
appearance of objects in a photo with their corresponding sounds in an audio
clip, producing a cohesive and comprehensive understanding of the scene. These
cross-modal capabilities empower the model to go beyond traditional unimodal
analysis. We hope PandaGPT serves as an initial step toward building AGI that
can perceive and understand inputs in different modalities holistically, as
humans do.

##  2 Related Work

### Large Language Models.

Large language models (LLMs) pre-trained over massive unlabeled text have
dominated the field of natural language processing (NLP) today [19, 5, 20, 3].
With alignment techniques such as supervised instruction tuning [21, 28, 13]
and reinforcement learning from human feedback [23, 16], LLMs exhibit
surprisingly effective zero- and few-shot generalization abilities to perform
almost any NLP tasks. The most successful examples could be OpenAI’s ChatGPT
[15] and GPT4 [14], which have made a profound impact on the entire AI
research community and beyond. There also have been enormous open-source
efforts to replicate the success, such as BLOOM [22], LLaMA [27], Alpaca [26],
Vicuna [4], OpenAlpaca [24] among many others.

### Multi-modal Alignment.

Feature alignment among multiple modalities has attracted great interest for
its applications such as cross-modal retrieval [7, 6, 2]. Recently, CLIP [18]
learns a joint embedding space for image and text. Flamingo [1], BLIP-2 [11],
and MAGIC [25] bridge powerful pre-trained vision-only and language-only
models and show strong zero-shot abilities. AudioCLIP [9] adds audio into the
CLIP framework for audio classification. ImageBind [8] learn a joint embedding
across six different modalities (image/video, text, audio, depth, thermal, and
IMU data) using image-paired data only. More recently, there has been a surge
of interest to combine multi-modal alignment and large language models for
multi-modal instruction following. LLaVa [12], Mini-GPT4 [31], and Video-LLaMA
[30] enable visually-grounded instruction following. DetGPT [17] proposes
reasoning-based object detection. SpeechGPT [29] adds speech understanding and
generation abilities to LLMs. However, these advances have largely been
confined to separate combinations of text and other modalities (e.g.,
image/video or audio).

Figure 1: Illustration of PandaGPT. During training, we only train the linear
projection matrix and the additional LoRA weights (as indicated with dashed
boxes) while keeping the parameters of ImageBind and Vicuna frozen.

##  3 Method

PandaGPT combines the multi-modal encoders from ImageBind and the large
language models from Vicuna, achieving impressive capabilities in vision- and
audio-grounded instruction following tasks. To align the feature space of
multimodal encoders from ImageBind and large language models from Vicuna111We
use the version-0 of Vicuna-13B as our base language model., we train PandaGPT
using 160k image-language instruction-following data released by [12] and
[31]. Each training instance consists of an image $\mathcal{I}$ and a multi-
turn conversation data
$(\boldsymbol{x}_{1},\boldsymbol{y}_{1},...,\boldsymbol{x}_{n},\boldsymbol{y}_{n})$,
where $\boldsymbol{x}_{i}$ and $\boldsymbol{y}_{i}$ are the human’s
instruction and the system’s response at the $i$-th turn. To reduce the number
of trainable parameters, we only train (i) a linear projection matrix $f$ to
connect the representation produced by ImageBind to Vicuna; and (ii)
additional LoRA [10] weights on the Vicuna’s attention modules.222The total
number of trainable parameters is around 0.4% of the parameters of Vicuna.
Figure 1 illustrates the architecture of PandaGPT.

The training objective of PandaGPT is defined as

$\mathcal{L}(\theta_{f},\theta_{l})=\prod_{i=1}^{n}p_{\theta}(\boldsymbol{y}_{i}|\boldsymbol{x}_{<i},\boldsymbol{y}_{<i-1},f(h_{\mathcal{I}})),$
(1)

where $\theta_{f}$ and $\theta_{l}$ correspond to the learnable parameters of
the linear projection matrix and LoRA weights. The $h_{\mathcal{I}}$ is the
image representation produced by ImageBind and
$\theta=\\{\theta_{f},\theta_{l},\theta_{1},\theta_{2}\\}$, where $\theta_{1}$
and $\theta_{2}$ are frozen parameters of ImageBind and Vicuna. Note that the
loss is only computed from the part of system responses during training. We
train PandaGPT on the image-language instruction-following dataset for two
epochs using a learning rate of 5e-4 with linear decay. The maximum sequence
length for Vicuna-13B is set to 400 based on our computation resources
(8$\times$A100 40G GPUs). The training takes around 7 hours to complete.

It is worth noting that the current version of PandaGPT is only trained with
aligned image-text data. However, by leveraging the binding property across
six modalities (image/video, text, audio, depth, thermal, and IMU) inherited
from the frozen ImageBind encoders, PandaGPT demonstrates emergent, i.e. zero-
shot, cross-modal capabilities across all of the modalities.

##  4 Capabilities of PandaGPT

Compared to existing multimodal instruction-following models trained
individually for one particular modality, PandaGPT can understand and combine
the information in different forms together, including image/video, text,
audio, depth (3D), thermal (infrared radiation), and inertial measurement
units (IMU) readings. We find that the capabilities of PandaGPT (see concrete
examples in Section 6) include but are not limited to:

  * •

image/video-grounded question answering: see examples of Figure 2 , 3, and 4.

  * •

image/video-inspired creative writing: see examples of Figure 5.

  * •

visual and auditory reasoning: see examples of Figure 6, 7, and 8.

  * •

multimodal arithmetic: PandaGPT is also capable of working with input composed
across modalities. By arithmetically adding information from different
modalities as input, PandaGPT can produce results that reflect concepts from
different parts. See Figure 9 and 10 for examples of image and audio
arithmetic, and see Figure 11 and 12 for examples of video and audio
arithmetic.

##  5 Limitations

Despite the amazing ability in handling multiple modalities and their
combinations. There are multiple ways to further improve PandaGPT.

  1. 1.

The training of PandaGPT can be enriched by using other alignment data, for
instance, other modalities paired with text (e.g., audio-text pairs).

  2. 2.

We only use one embedding vector for the content in other modalities than
text, more research into fine-grained feature extraction such as cross-modal
attention mechanisms could be beneficial to the performance.

  3. 3.

PandaGPT currently only allows multimodal information to be used as input,
future possibilities include generating richer multimedia content (e.g.,
creating images and response in audio).

  4. 4.

New benchmarks to evaluate the composition ability of multimodal inputs is
demanded.

  5. 5.

PandaGPT can also exhibit several common deficiencies of existing language
models, including hallucination, toxicity, and stereotypes.

Lastly, we would like to note that PandaGPT is a research prototype and cannot
be readily used for real-world applications.

##  6 Examples

Figure 2: Example showing PandaGPT’s capability in image-grounded question
answering. Figure 3: Example showing PandaGPT’s capability in image-grounded
question answering. Figure 4: Example showing PandaGPT’s capability in video-
grounded question answering. Figure 5: Example showing PandaGPT’s capability
in image/video-inspired creative writing. Figure 6: Example showing PandaGPT’s
capability in visual reasoning. Figure 7: Example showing PandaGPT’s
capability in auditory reasoning. Figure 8: Example showing PandaGPT’s
capability in auditory reasoning. Figure 9: Example showing PandaGPT’s
capability in multimodal arithmetic (Image and Audio). Figure 10: Example
showing PandaGPT’s capability in multimodal arithmetic (Image and Audio).
Figure 11: Example showing PandaGPT’s capability in multimodal arithmetic
(Video and Audio). Figure 12: Example showing PandaGPT’s capability in
multimodal arithmetic (Video and Audio).

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