# On almost perfect linear Lee codes of packing radius $2$

Xiaodong Xu 1 1Guangxi Academy of Sciences, Nanning, 530007, P.R.China.
<EMAIL_ADDRESS>and  Yue Zhou 2, $\dagger$ 2Department of Mathematics,
National University of Defense Technology, 410073 Changsha, China
$\dagger$Corresponding author<EMAIL_ADDRESS>
###### Abstract.

More than 50 years ago, Golomb and Welch conjectured that there is no perfect
Lee codes $C$ of packing radius $r$ in $\mathbb{Z}^{n}$ for $r\geq 2$ and
$n\geq 3$. Recently, Leung and the second author proved that if $C$ is linear,
then the Golomb-Welch conjecture is valid for $r=2$ and $n\geq 3$. In this
paper, we consider the classification of linear Lee codes with the second-best
possibility, that is the density of the lattice packing of $\mathbb{Z}^{n}$ by
Lee spheres $S(n,r)$ equals $\frac{|S(n,r)|}{|S(n,r)|+1}$. We show that, for
$r=2$ and $n\equiv 0,3,4\pmod{6}$, this packing density can never be achieved.

###### Key words and phrases:

Lee metric, perfect code, Golomb-Welch conjecture, Cayley graph

The extended abstract [23] of an earlier version of this paper was presented
in the 12th International Workshop on Coding and Cryptography (WCC) 2022.

##  1\. Introduction

Let $\mathbb{Z}$ denote the ring of integers. For two words
$x=(x_{1},\cdots,x_{n})$ and $y=(y_{1},\cdots,y_{n})\in\mathbb{Z}^{n}$, the
_Lee distance_ (also known as $\ell_{1}$-norm, taxicab metric, rectilinear
distance or Manhattan distance) between them is defined by

$d_{L}(x,y)=\sum_{i=1}^{n}|x_{i}-y_{i}|\text{ for }x,y\in\mathbb{Z}^{n}.$

A _Lee code_ $C$ is just a subset of $\mathbb{Z}^{n}$ endowed with the Lee
distance. If $C$ further has the structure of an additive group, i.e. $C$ is a
lattice in $\mathbb{Z}^{n}$, then we call $C$ a _linear Lee code_. Lee codes
have many practical applications, for example, constrained and partial-
response channels [20], flash memory [21] and interleaving schemes [3].

The minimum distance between any two distinct elements in $C$ is called the
_minimum distance_ of $C$. Given a Lee code of minimum distance $2r+1$, for
any $x\in\mathbb{Z}^{n}$, if one can always find a unique $c\in C$ such that
$d_{L}(x,c)\leq r$, then $\mathcal{C}$ is called a _perfect code_. This is
equivalent to

$\mathbb{Z}^{n}=\dot{\bigcup}_{c\in C}(S(n,r)+c),$

where

$S(n,r):=\left\\{(x_{1},\cdots,x_{n})\in\mathbb{Z}^{n}:\sum_{i=1}^{n}|x_{i}|\leq
r\right\\}$

and $S(n,r)+c:=\\{v+c:v\in S(n,r)\\}$. Thus, the existence of a perfect Lee
code implies a tiling of $\mathbb{Z}^{n}$ by Lee spheres of radius $r$.

Perfect Lee codes exist for $n=1,2$ and any $r$, and for $n\geq 3$ and $r=1$.
Golomb and Welch [7] conjectured that there are no more perfect Lee codes for
other choices of $n$ and $r$. This conjecture is still far from being solved,
despite many efforts and various approaches applied on it. We refer the reader
to the recent survey [11] and the references therein. For perfect codes with
respect to other metrics, see the very recent monograph [6] by Etzion.

Figure 1. Tiling of $\mathbb{R}^{2}$ by $L(2,2)$

Let $L(n,r)$ denote the union of $n$-cubes centered at each point of $S(n,r)$
in $\mathbb{R}^{n}$. It is not difficult to see that there is a tiling of
$\mathbb{Z}^{n}$ by $S(n,r)$ if and only if there is a tiling of
$\mathbb{R}^{n}$ by $L(n,r)$. Figure 1 shows a (lattice) tiling of
$\mathbb{R}^{2}$ by $L(2,2)$. As the shape of $L(n,r)$ is close to a cross-
polytope when $r$ is large enough, one can use the cross-polytope packing
density to prove the Golomb-Welch conjecture provided that $r$ is large enough
compared with $n$. In fact, this idea was first applied by Golomb and Welch
themselves in [7]. There are some other geometric approaches, including the
analysis of some local configurations of the boundary of Lee spheres in a
tiling by Post [17], and the density trick by Astola [2] and Lepistö [13].

However, it seems that the geometric approaches do not work for small $r$ and
large $n$. In the past several years, algebraic approaches have been proposed
and applied on the existence of perfect linear Lee code for small $r$. In
[12], Kim introduced a symmetric polynomial method to study this problem for
sphere radius $r=2$. This approach has been extended by Zhang and Ge [24], and
Qureshi [18] for $r\geq 3$. See [10, 11, 19, 25] for other related results. In
particular, Leung and the second author [14] succeeded in getting a complete
solution to the case with $r=2$: there is no perfect linear Lee code of
minimum distance $5$ in $\mathbb{Z}^{n}$ for $n\geq 3$.

It is worth pointing out that the existence of a perfect linear Lee code of
minimum distance $2r+1$ in $\mathbb{Z}^{n}$ is equivalent to an abelian Cayley
graph of degree $2n$ and diameter $r$ whose number of vertices meets the so-
called _abelian Cayley Moore bound_ ; see [4, 25]. For more results about the
degree-diameter problems in graph theory, we refer to the survey [16].

It is clear that a perfect Lee code $C$ means the packing density of
$\mathbb{Z}^{n}$ by $S(n,r)$ with centers consisting of all the elements in
$C$ is $1$. If there is no perfect linear Lee code for $r\geq 2$ and $n\geq
3$, one may wonder whether the second best is possible, which is about the
existence of a lattice packing of $\mathbb{Z}^{n}$ by $S(n,r)$ with density
$\frac{|S(n,r)|}{|S(n,r)|+1}$. We call such a linear Lee code _almost
perfect_. In Figure 2, we present a lattice packing of $\mathbb{Z}^{2}$ by
$S(2,2)$, and its packing density is
$\frac{|S(2,2)|}{|S(2,2)|+1}=\frac{13}{14}$. This example and the numbers
labeled on the cubes will be explained later in Example 2.4. For convenience,
we abbreviate the term almost perfect linear Lee code to _APLL_ code.

Figure 2. An almost perfect linear Lee code in $\mathbb{Z}^{2}$

The _packing radius_ of a Lee code $C$ is defined to be the largest integer
$r^{\prime}$ such that for any element $w\in\mathbb{Z}^{n}$ there exists at
most one codeword $c\in C$ with $d_{L}(w,c)\leq r^{\prime}$. The _covering
radius_ of a Lee code $C$ is the smallest integer $r^{\prime\prime}$ such that
for any element $w\in\mathbb{Z}^{n}$ there exists at least one codeword $c\in
C$ with $d_{L}(w,c)\leq r^{\prime\prime}$. It is clear that a Lee code is
perfect if and only if its packing radius and covering radius are the same.
For an APLL code with packing density $\frac{|S(n,r)|}{|S(n,r)|+1}$, its
packing radius and covering radius are $r$ and $r+1$, respectively, and its
minimum distance is $2r+1$ for $n>1$ and $2r+2$ for $n=1$.

A Lee code is called _$t$ -quasi-perfect_ if its packing radius and covering
radius are $t$ and $t+1$, respectively. Note that an APLL code is a quasi-
perfect Lee code, but a quasi-perfect Lee code is not necessarily almost
perfect. For instance, two families of $2$-quasi-perfect codes in
$\mathbb{Z}^{n}$ are constructed in [4] and the packing density equals
$\frac{2n^{2}+2n+1}{(2n\pm 1)^{2}}$, where $n=2[\frac{p}{4}]$ for a prime $p$
satisfying $p\geq 7$ and $p\equiv\pm 5\pmod{12}$. In particular, the packing
density equals $\frac{41}{49}$ when $n=4$, and it tends to $\frac{1}{2}$ for
big $n$. We refer to [1, 4, 15] for more constructions and other results on
quasi-perfect Lee codes.

About the existence of APLL codes of packing radius $2$, one can apply the
symmetric polynomial method in [12] and the algebraic number theory approach
in [25] to derive some partial results. For $n\leq 10^{5}$, one can exclude
the existence of APLL codes of minimum distance $5$ for $76,573$ choices of
$n$. For more details, see [8].

The main result of this paper is the following one.

######  Theorem 1.1.

Let $n$ be a positive integer. If $n\equiv 0,3,4\pmod{6}$, then there exists
no almost perfect linear Lee code of packing radius $2$.

As in [14], our proof is given by investigating group ring equations in
$\mathbb{Z}[G]$ where $G$ is of order $|S(n,2)|+1=2(n^{2}+n+1)$. However, the
situation here is different from the one appeared in [14]. We consider a
special subset $T\subseteq G$ and show that $T$ splits into two disjoint
subsets $T_{0}\subseteq H$ and $fT_{1}\subseteq fH$ where $H$ is the unique
subgroup of $G$ of index $2$ and $f$ is the unique involution in $G$. Then, we
analyze the elements appearing in $(T_{0}^{(2)}+T_{1}^{(2)})T_{0}$ and
$(T_{0}^{(2)}+T_{1}^{(2)})T_{1}$. Unfortunately, we could not get any
contradiction when $n\equiv 1,2,5\pmod{6}$. But we conjecture that there is no
APLL code for this case when $n>2$.

The rest part of this paper consists of three sections. In Section 2, we
convert the existence of an almost perfect linear Lee code of packing radius
$2$ into some conditions in group ring. Then we prove Theorem 1.1 in Section
3. In Section 4, we conclude this paper with several remarks and research
problems.

##  2\. Necessary and sufficient conditions in group ring

Let $\mathbb{Z}[G]$ denote the set of formal sums $\sum_{g\in G}a_{g}g$, where
$a_{g}\in\mathbb{Z}$ and $G$ is a finite multiplicative group. The addition
and the multiplication on $\mathbb{Z}[G]$ are defined by

$\sum_{g\in G}a_{g}g+\sum_{g\in G}b_{g}g:=\sum_{g\in G}(a_{g}+b_{g})g,$

and

$(\sum_{g\in G}a_{g}g)\cdot(\sum_{g\in G}b_{g}g):=\sum_{g\in G}(\sum_{h\in
G}a_{h}b_{h^{-1}g})\cdot g,$

for $\sum_{g\in G}a_{g}g,\sum_{g\in G}b_{g}g\in\mathbb{Z}[G]$. Moreover,

$\lambda\cdot(\sum_{g\in G}a_{g}g):=\sum_{g\in G}(\lambda a_{g})g$

for $\lambda\in\mathbb{Z}$ and $\sum_{g\in G}a_{g}g\in\mathbb{Z}[G]$.

For an element $A=\sum_{g\in G}a_{g}g\in\mathbb{Z}[G]$ and $t\in\mathbb{Z}$,
we define

$A^{(t)}:=\sum_{g\in G}a_{g}g^{t}.$

A multi-subset $D$ of $G$ can be viewed as an element $\sum_{g\in
D}g\in\mathbb{Z}[G]$. In the rest of this paper, by abuse of notation, we will
use the same symbol to denote a multi-subset of $G$ and the associated element
in $\mathbb{Z}[G]$. Moreover, $|D|$ denotes the number of distinct elements in
$D$.

The following result converts the existence of APLL codes into an algebraic
combinatorics problem on abelian groups. Its proof is not difficult, and more
or less the same as the proof of Theorem 6 in [9].

######  Lemma 2.1.

There is an APLL code of packing radius $r$ in $\mathbb{Z}^{n}$ if and only if
there is an abelian group $G$ and a homomorphism
$\varphi:\mathbb{Z}^{n}\rightarrow G$ such that the restriction of $\varphi$
to $S(n,r)$ is injective and $G\setminus\varphi(S(n,r))$ has only one element.

###### Proof.

First, let us prove the necessary part of the condition. Assume that
$\mathcal{C}$ is an APLL code of packing radius $r$ and $w\in\mathbb{Z}^{n}$
is an arbitrary element which is not covered by the lattice packing
$\dot{\bigcup}_{c\in C}(S(n,r)+c)$. Then, by the definition of APLL codes, $C$
is a subgroup of $\mathbb{Z}^{n}$ and

(1) $\mathbb{Z}^{n}=\left(\dot{\bigcup}_{c\in
C}(S(n,r)+c)\right)\dot{\cup}(w+C).$

Consider the quotient group $G=\mathbb{Z}^{n}/C$ and take $\varphi$ to be the
canonical homomorphism from $\mathbb{Z}^{n}$ to $G$. If there are $x,y\in
S(n,r)$, such that $\varphi(x)=\varphi(y)$, then $z=x-y\in C$. Since $S(n,r)$
contains $y$, $x\in S(n,r)+z$. This means $x$ is covered by both $S(n,r)+z$
and $S(n,r)$. By (1), $z$ must be $0$ which implies $x=y$. Therefore,
$\varphi$ is injective and $\varphi(w)$ is the unique element not in
$\varphi(S(n,r))$.

To prove the sufficient part, one only has to define $C=\ker(\varphi)$. The
verification of the almost perfect property of $C$ is straightforward. ∎

The number of elements in a Lee sphere is

$|S(n,r)|=\sum_{i=0}^{\text{min}\\{n,r\\}}2^{i}\binom{n}{i}\binom{r}{i},$

which can be found in [7] and [22]. In particular, when $r=2$,
$|S(n,r)|=2n^{2}+2n+1$ and the group $G$ considered in Lemma 2.1 is of order
$2n^{2}+2n+2$.

The next result translates the existence of an APLL code into a group ring
condition. The same result has been proved in [8] in the context of the
existence of abelian Moore Cayley graphs with excess one. For completeness, we
include a proof here.

######  Lemma 2.2.

There exists an APLL code of packing radius $2$ in $\mathbb{Z}^{n}$ if and
only if there is an abelian group $G$ of order $2(n^{2}+n+1)$ and an inverse-
closed subset $T\subseteq G$ containing $e$ with $|T|=2n+1$ such that

(2) $T^{2}=2(G-f)-T^{(2)}+2ne,$

where $e$ is the identity element in $G$ and $f$ is the unique element of
order $2$ in $G$.

###### Proof.

The proof of (2) is similar to the proof of Lemma 2.3 in [25]. Suppose that
there exists an APLL code of packing radius $2$. By Lemma 2.1, we have a
homomorphism $\varphi:\mathbb{Z}^{n}\rightarrow G$. Define
$a_{i}=\varphi(e_{i})$ for $i=1,\cdots,n$. As the radius equals $2$, by
definition, $G$ has the following partition

$G=\\{e,f\\}\dot{\bigcup}\\{a_{i}^{\pm 1},a_{i}^{\pm
2}:i=1,\cdots,n\\}\dot{\bigcup}\\{a_{i}^{\pm 1}a_{j}^{\pm 1}:1\leq i<j\leq
n\\},$

where $f$ is the unique element of $G$ not in $\varphi(S(n,2))$.

Let $T=e+\sum_{i=1}^{n}a_{i}+\sum_{i=1}^{n}a_{i}^{-1}$. Hence $|T|=2n+1$.
Moreover,

$\displaystyle T^{2}$
$\displaystyle=e+2\sum_{i=1}^{n}a_{i}+2\sum_{i=1}^{n}a_{i}^{-1}+\left(\sum_{i=1}^{n}a_{i}\right)^{2}+\left(\sum_{i=1}^{n}a_{i}^{-1}\right)^{2}+2\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n}a_{i}^{-1}\right)$
$\displaystyle=(2n+1)e+2\left(\sum_{i=1}^{n}a_{i}+\sum_{i=1}^{n}a_{i}^{-1}\right)+\sum_{i=1}^{n}(a_{i}^{2}+a_{i}^{-2})+2\sum_{1\leq
i<j\leq n}a_{i}^{\pm 1}a_{j}^{\pm 1}$ $\displaystyle=2ne+2(G-f)-T^{(2)}.$

Therefore (2) is proved.

Applying the map $x\mapsto x^{(-1)}$ on the both sides of (2), we get

$T^{2}=2(G-f^{-1})-T^{(2)}+2ne,$

because $T$ is inverse-closed. By comparing this equation with (2), we see
$f=f^{-1}$. As $n^{2}+n+1$ is always odd, there is only one involution in $G$
which is $f$.

Next we prove the sufficiency part. Given an inverse-closed subset $T\subseteq
G$ containing $e$ with $|T|=2n+1$ satisfying $\eqref{eq:main_T^2}$, we define
a homomorphism $\varphi:\mathbb{Z}^{n}\rightarrow G$ by
$\varphi(e_{i})=a_{i}$. It is clear that $\varphi$ is completely determined by
the value of all $a_{i}$’s. As $T$ is inverse-closed and $|T|=2n+1$, the
restriction of $\varphi$ to $S(n,1)$ is injective. Furthermore, by checking
the coefficients of elements in (2) as in the necessity part of the proof, we
see that

$G\setminus\\{f\\}=\\{e\\}\dot{\bigcup}\\{a_{i}^{\pm 1},a_{i}^{\pm
2}:i=1,\cdots,n\\}\dot{\bigcup}\\{a_{i}^{\pm 1}a_{j}^{\pm 1}:1\leq i<j\leq
n\\},$

that is $G\setminus\\{f\\}=\varphi(S(n,2))$. By Lemma 2.1, $C=\ker(\varphi)$
is an APLL code of packing radius $2$. ∎

######  Remark 2.3.

Similar to the alternative proof of the uniqueness of $f\in G$ in Lemma 2.2,
we can show that if there is an APLL code of packing radius $r$, then there is
a group $G$ of even order with a unique involution in $G$ by checking
$a_{i_{1}}\cdots a_{i_{j}}=a_{i_{1}}^{-1}\cdots a_{i_{j}}^{-1}$ for $j\leq r$.
However, the corresponding group ring equations become more complicated, and
we are not going to investigate them in this paper.

The following two examples show that for $n=1,2$, there do exist APLL codes of
packing radius $2$. The corresponding result of Example 2.4 (b) has been
already depicted in Figure 2.

######  Example 2.4.

Let $C_{m}$ denote the cyclic group generated by $g$ of order $m$.

  1. (a)

$n=1$, $G=C_{6}$ and $T=\\{e,g^{\pm 1}\\}$.

  2. (b)

$n=2$, $G=C_{14}$ and $T=\\{e,g^{\pm 2},g^{\pm 3}\\}$.

In Figure 2, we label each element in $\mathbb{Z}^{2}$ which is mapped to
$\varphi(e_{1})=g^{2}$ by $2$ and those mapped to $\varphi(e_{2})=g^{3}$ by
$3$. The center of every Lee sphere is labeled by $0$. The holes are all
labeled by $7$ which corresponds to the unique involution $g^{7}$ in $G$.

In (a) and (b), $G$ has a unique subgroup $H$ of order $n^{2}+n+1$. We will
obtain further necessary and sufficient conditions in terms of subsets in $H$
in the later part, and the corresponding results for (a) and (b) will be given
in Example 2.7.

######  Remark 2.5.

It is easy to check that $\\{e,g^{\pm 3},g^{\pm 2}\\}\subseteq C_{13}$ defines
a perfect linear Lee code in $\mathbb{Z}^{2}$ as a lattice with basis
$\\{(2,3),(-3,2)\\}$; see Figure 1. Accordingly the basis of the APLL code in
Figure 2 generated by $\\{e,g^{\pm 3},g^{\pm 2}\\}\subseteq C_{14}$ is
$\\{(4,2),(3,-2)\\}$.

The isometry group of $\mathbb{Z}^{n}$ is generated by permutations of axes
and reflections; see [11]. Consequently, all APLL codes of packing radius $2$
in $\mathbb{Z}^{2}$ are isometric for the following reasons. By Lemma 2.2, $T$
must contain a primitive element of $C_{14}$. Moreover, it is easy to check
that if we assume that $e,g^{\pm 1}\in T$, then $T$ must be $\\{e,g^{\pm
1},g^{\pm 4}\\}$. Thus, all $T$ satisfying Lemma 2.2 must be of the shape
$\\{e,g^{\pm 1},g^{\pm 4}\\}^{\psi}$ for some group isomorphism $\psi$ of
$C_{14}$. Finally, note that $T$ and $T^{\psi}$ provide the same Lee code in
$\mathbb{Z}^{2}$ for any group isomorphism $\psi$. Therefore, all APLL codes
of packing radius $2$ in $\mathbb{Z}^{2}$ must be isometric.

In Lemma 2.2, we have obtained some necessary and sufficient condition for the
existence of APLL codes of packing radius $2$ in terms of a subset $T$ in an
abelian group $G$ of order $2(n^{2}+n+1)$. We can further split $T$ into
subsets in a subgroup $H$ and its coset $fH$.

Let $H$ denote the unique subgroup of order $n^{2}+n+1$ in $G$. Define
$T_{0}=T\cap H$ and $T_{1}=fT\cap H$. Thus

$T=T_{0}+fT_{1}.$

By (2),

$T_{0}^{2}+T_{1}^{2}+2fT_{0}T_{1}=2(G-f)-T_{0}^{(2)}-T_{1}^{(2)}+2n\cdot e.$

By considering the above equation intersecting $H$ and $fH$, respectively, we
obtain

(3) $\displaystyle T_{0}T_{1}$ $\displaystyle=$ $\displaystyle H-e,$ (4)
$\displaystyle T_{0}^{2}+T_{1}^{2}$ $\displaystyle=$ $\displaystyle
2H-T_{0}^{(2)}-T_{1}^{(2)}+2ne.$

Therefore, we have proved the following result.

######  Lemma 2.6.

Let $T=T_{0}+fT_{1}\subseteq G$ with $T_{0}$ and $T_{1}\subseteq H$. The
subset $T$ satisfies that $e\in T$, $T^{(-1)}=T$ and (2) if and only if $e\in
T_{0}$, $T_{0}^{(-1)}=T_{0}$, $T_{1}^{(-1)}=T_{1}$, (3) and (4) hold.

######  Example 2.7.

The following examples of $T_{0}$ and $T_{1}$ are derived from $T$ in Example
2.4 (a) and (b), respectively. It is easy to check that $T_{0}$ and $T_{1}$
satisfy (3) and (4).

  1. (a)

For $n=1$, $G=C_{6}=\langle g\rangle$ and $H=2C_{6}=\langle g^{2}\rangle\cong
C_{3}$. Let $T=\\{e,g^{\pm 1}\\}$. Then $T_{0}=\\{e\\}$ and
$T_{1}=\\{g^{2},g^{4}\\}$.

  2. (b)

For $n=2$, $G=C_{14}=\langle g\rangle$ and $H=2C_{14}=\langle
g^{2}\rangle\cong C_{7}$. From $T=\\{e,g^{\pm 2},g^{\pm 3}\\}$, one gets

$T_{0}=\\{e,g^{2},g^{12}\\}\text{ and }T_{1}=\\{g^{4},g^{10}\\}.$

######  Remark 2.8.

A _near-factorization_ of a group $H$ is a pair of (not necessarily inverse-
closed) subsets $T_{0}$ and $T_{1}$ such that (3) holds. There are very little
known results on them. However, it has been proved that, up to translate,
$T_{0}$ and $T_{1}$ must be inverse-closed; see [5].

The following result is a collection of obvious necessary conditions for
$T_{0}$ and $T_{1}$, which will be intensively used in Section 3.

######  Lemma 2.9.

Suppose that $T=T_{0}+fT_{1}\subseteq G$ satisfying $e\in T$, $T^{(-1)}=T$,
$|T|=2n+1$, (3) and (4). Then the following statements hold.

  1. (a)

$e\in T_{0}$, $e\notin T_{1}$;

  2. (b)

$T_{0}\cap T_{1}=\emptyset$ and $T_{0}^{(2)}\cap T_{1}^{(2)}=\emptyset$;

  3. (c)

$T_{0}\cap(T_{0}^{(2)}\setminus\\{e\\})=T_{0}\cap T_{1}^{(2)}=\emptyset$;

  4. (d)

$\\{ab:a\neq b,a,b\in T_{0}\\}\cap T_{0}^{(2)}=\\{e\\}$;

  5. (e)

When $n$ is odd, $|T_{0}|=n$ and $|T_{1}|=n+1$;

  6. (f)

When $n$ is even, $|T_{0}|=n+1$ and $|T_{1}|=n$;

  7. (g)

There is no common non-identity element in $T_{0}^{2}$ and $T_{1}^{2}$;

  8. (h)

$T_{0}\cap T_{0}^{(3)}=\\{e\\}$.

###### Proof.

By the definition of $T$ and $T_{0}$, we have (a). From (3) and the inverse-
closed property of $T_{1}$, we derive $T_{0}\cap T_{1}=\emptyset$. As
$\gcd(2,|H|)=1$, the map $x\mapsto x^{2}$ on $H$ is a group automorphism. Thus
$T_{0}^{(2)}\cap T_{1}^{(2)}=\emptyset$ and (b) is proved.

Note that $T_{0}$, $T_{0}^{(2)}\setminus\\{e\\}$ and $T_{1}^{(2)}$ are all
subsets in $H$. By checking (4), we know that they cannot share any common
element. By the same reason, we also prove (d).

From (3), we get $|T_{0}|\cdot|T_{1}|=n(n+1)$. Notice that $|T_{0}|$ must be
odd because it has an involution $x\mapsto x^{-1}$ with a unique fixed element
$e$. Together with $|T_{0}|+|T_{1}|=2n+1$, we derive (e) and (f). By checking
the elements appeared in $T_{0}^{2}+T_{1}^{2}$ in (4), we get (g).

Finally, we prove (h). It is clear that $e\in T_{0}\cap T_{0}^{(3)}$. Assume
to the contrary that there exists $b\in T_{0}\setminus\\{e\\}$ such that
$b=a^{3}$ for some $a\in T_{0}$. Then $a^{2}=a^{-1}b\in T_{0}^{2}$. As
$a^{-1}\neq b$, this is a contradiction to (d). ∎

In [8], several different approaches have been applied to derive various
nonexistence results. In particular, we need the following one in the next
section.

######  Proposition 2.10 (Corollary 3.1 in [8]).

Suppose that $8n-7$ is not a square in $\mathbb{Z}$ and one of the following
conditions are satisfied:

  * •

$3,7,19$ or $31$ divides $n^{2}+n+1$;

  * •

$13\mid n^{2}+n+1$ and $8n-11\notin\\{13k^{2}:k\in\mathbb{Z}\\}$.

Then there is no inverse-closed subset $T\subseteq G$ satisfying Lemma 2.2.

Although Proposition 2.10 does not work for $n=3$, this situation can be
excluded by a direct verification because the group $G$ is small.

The symmetric polynomial method applied by Kim [12] on the existence of
perfect linear Lee codes can also be used directly here to derive strong
nonexistence results; see Theorem 3.1 in [8]. In particular, it leads to the
following results appeared in [8].

######  Result 2.11.

For $3<n\leq 10^{5}$, if $n^{2}+n+1$ is a prime, then there is no inverse-
closed subset $T\subseteq G$ satisfying Lemma 2.2.

##  3\. Proof of the main result

In this section, we are going to prove Theorem 1.1 by showing the nonexistence
of inverse-closed subsets $T_{0},T_{1}\subseteq H$ satisfying $e\in T_{0}$,
$|T_{0}|+|T_{1}|=2n+1$, (3) and (4).

Define $\hat{T}=T_{0}+T_{1}\in\mathbb{Z}[H]$. Write
$\hat{T}^{(2)}=\sum_{i=0}^{2n}a_{i}$ where $a_{0}=e$, $a_{i}\in T_{0}^{(2)}$
for $i=0,\cdots,|T_{0}|-1$ and $a_{i}\in T_{1}^{(2)}$ for
$i=|T_{0}|,\cdots,2n$. Let $k_{0}=|T_{0}|$ and $k_{1}=|T_{1}|$.

By multiplying $T_{0}$ and $T_{1}$ on the both sides of (4), respectively, and
rearranging the terms, we get

$\displaystyle\hat{T}^{(2)}T_{0}$
$\displaystyle=(2k_{0}-k_{1})H+T_{1}-T_{0}^{3}+2nT_{0},$
$\displaystyle\hat{T}^{(2)}T_{1}$
$\displaystyle=(2k_{1}-k_{0})H+T_{0}-T_{1}^{3}+2nT_{1}.$

Consider the above two equations modulo $3$

$\displaystyle\hat{T}^{(2)}T_{0}$
$\displaystyle\equiv(2k_{0}-k_{1})H+T_{1}-T_{0}^{(3)}+2nT_{0}\pmod{3},$
$\displaystyle\hat{T}^{(2)}T_{1}$
$\displaystyle\equiv(2k_{1}-k_{0})H+T_{0}-T_{1}^{(3)}+2nT_{1}\pmod{3}.$

Note that $T_{0}^{3}\equiv T_{0}^{(3)}\pmod{3}$. If $3\nmid|H|$, then
$T_{0}^{(3)}$ is a subset; for $3\mid|H|$, we will show that there are only
very few elements in $T_{0}^{(3)}$ appearing more than once in Lemma 3.4.
Thus, we can derive some strong conditions on the coefficients of most of the
elements in the right-hand side of the above two equations. For instance, when
$n\equiv 1\pmod{3}$ and $n$ is odd, the first equation becomes

$\hat{T}^{(2)}T_{0}\equiv T_{1}-T_{0}^{(3)}+2T_{0}\pmod{3}.$

As $T_{1}$, $T_{0}^{(3)}$ and $T_{0}$ are approximately of size $n$, most of
the elements in $H$ appear in $\hat{T}^{(2)}T_{0}$ for $3k$ times,
$k=0,1,\cdots$.

Let $X_{i}$ ($Y_{i}$, resp.) be the subset of elements of $H$ appearing in
$\hat{T}^{(2)}T_{0}$ ($\hat{T}^{(2)}T_{1}$, resp.) exactly $i$ times for
$i=0,1,\cdots,M_{0}$ ($M_{1}$, resp.), which means $X_{i}$’s ($Y_{i}$’s,
resp.) form a partition of the group $H$. In particular, we define $M_{0}$ and
$M_{1}$ to be the largest integers such that $X_{M_{0}}$ and $Y_{M_{1}}$ are
non-empty sets. Then

(5) $\displaystyle\hat{T}^{(2)}T_{0}$
$\displaystyle=\sum_{i=0}^{M_{0}}iX_{i},$ (6)
$\displaystyle\hat{T}^{(2)}T_{1}$ $\displaystyle=\sum_{i=0}^{M_{1}}iY_{i}.$

By (5) and (6), we have the following three conditions on the value of
$|X_{i}|$’s and $|Y_{i}|$’s:

(7) $\displaystyle\sum_{i=1}^{M_{0}}i|X_{i}|$ $\displaystyle=(2n+1)k_{0},$ (8)
$\displaystyle\sum_{i=1}^{M_{1}}i|Y_{i}|$ $\displaystyle=(2n+1)k_{1},$ (9)
$\displaystyle\sum_{i=0}^{M_{0}}|X_{i}|=\sum_{i=0}^{M_{1}}|Y_{i}|$
$\displaystyle=n^{2}+n+1.$

Some extra conditions are given by the following Lemma.

######  Lemma 3.1.

For $X_{i}$ and $Y_{i}$ defined in (5) and (6),

(10) $\displaystyle\sum_{i=1}^{M_{0}}|X_{i}|=$
$\displaystyle(2n+1)k_{0}-2(k_{0}-1)k_{0}+\theta_{0}+\sum_{s\geq
3}\frac{(s-1)(s-2)}{2}|X_{s}|,$ (11) $\displaystyle\sum_{i=1}^{M_{1}}|Y_{i}|=$
$\displaystyle(2n+1)k_{1}-2(k_{1}-1)k_{1}+\theta_{1}+\sum_{s\geq
3}\frac{(s-1)(s-2)}{2}|Y_{s}|,$

where $\theta_{0}=|(T_{0}^{2}\setminus T_{0}^{(2)})\cap\hat{T}^{(4)}|$ and
$\theta_{1}=\frac{|T_{1}\cap\hat{T}^{(2)}|}{2}+|(T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\}))\cap\hat{T}^{(4)}|$.

Note that $|X_{i}|$’s and $|Y_{i}|$’s are nonnegative integers. Our main idea
is to use (7) , (8) , (9), (10) and (11) together with $\hat{T}^{(2)}T_{0}$
and $\hat{T}^{(2)}T_{1}\pmod{3}$ to determine $|X_{i}|$’s and $|Y_{i}|$’s. For
some cases, we will end up with a contradiction which means there is no
$T_{0}$ and $T_{1}$ satisfying (3) and (4), for example, see Theorem 3.5; for
the other cases, the value of $|X_{i}|$’s and $|Y_{i}|$’s together with a
careful analysis of (3) and (4) also lead to contradictions, for instance, see
Theorems 3.6 and 3.7.

To prove Lemma 3.1, we need the following fact.

######  Lemma 3.2.

For $X_{i}$’s defined by (5), $e\in X_{1}$, $|X_{i}|$ is even for $i\neq 1$
and $|X_{1}|$ is odd .

###### Proof.

The identity element $e$ is in $X_{1}$ for the following reasons: First,
$e\notin T_{1}^{(2)}T_{0}$, otherwise $T_{0}\cap T_{1}^{(2)}\neq\emptyset$
which contradicts Lemma 2.9 (c); second, $e$ appears in $T_{0}^{(2)}T_{0}$
exactly once which also follows from $T_{0}^{(2)}\cap T_{0}=\\{e\\}$ by Lemma
2.9 (c).

Note that $X_{i}=X_{i}^{(-1)}$ for $i\geq 1$. Hence $|X_{i}|$ is even for
$i\geq 2$, and $|X_{1}|$ is odd because $e\in X_{1}$. Moreover, as
$|H|=n^{2}+n+1$ is odd and $X_{i}$’s form a partition of $H$, $|X_{0}|$ is
even. ∎

Recall that we have defined $k_{0}=|T_{0}|$ and $k_{1}=|T_{1}|$ at the
beginning of this section.

######  Lemma 3.3.

Denote the elements in $T_{0}^{(2)}$ by $a_{i}$ with $i=0,\cdots,k_{0}-1$ and
the elements in $T_{1}^{(2)}$ by $a_{i}$ with $i=k_{0},\cdots,2n$. In
particular, let $a_{0}=e$. Then

$|a_{i}T_{k}\cap a_{j}T_{k}|\in\\{0,1,2\\},$

for any $i<j$ and $k=0,1$. For $\ell=0,1,2$, define

$C_{k,\ell}=\\#\left\\{(a_{i},a_{j}):i<j,|a_{i}T_{k}\cap
a_{j}T_{k}|=\ell\right\\}.$

Then

$C_{k,1}=\begin{cases}2k_{0}-2,&k=0;\\\\[5.0pt]
2k_{1}-\frac{|T_{1}\cap\hat{T}^{(2)}|}{2},&k=1,\end{cases}$

and

$C_{k,2}=\begin{cases}(k_{0}-1)^{2}-\frac{\left|(T_{0}^{2}\setminus
T_{0}^{(2)})\cap\hat{T}^{(4)}\right|}{2},&k=0;\\\\[5.0pt]
k_{1}^{2}-2k_{1}-\frac{\left|(T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\}))\cap\hat{T}^{(4)}\right|}{2},&k=1.\end{cases}$

###### Proof.

An element $a_{i}x=a_{j}y\in a_{i}T_{k}\cap a_{j}T_{k}$ for some $x,y\in
T_{k}$ if and only if

$a_{i}a_{j}^{-1}=x^{-1}y\in T_{k}^{2}.$

By (4), there exist at most two possible choices of $(x,y)$. Moreover, if
$(x,y)$ is such that $a_{i}x=a_{j}y$, then $a_{i}y^{-1}=a_{j}x^{-1}$ which
also provides a solution $(y^{-1},x^{-1})$. Thus $|a_{i}T_{k}\cap
a_{j}T_{k}|=1$ if and only if $x=y^{-1}$, which is equivalent to
$a_{i}a_{j}^{-1}\in T_{k}^{(2)}$.

Next let us determine the value of $C_{k,\ell}$ for $\ell=1,2$.

Calculation of $C_{0,1}$. By the previous analysis, for $i<j$,
$|a_{i}T_{0}\cap a_{j}T_{0}|=1$ if and only if $a_{i}a_{j}^{-1}\in
T_{0}^{(2)}\setminus\\{e\\}$, which is equivalent to
$\sqrt{a_{i}}\sqrt{a_{j}^{-1}}\in T_{0}\setminus\\{e\\}$.

If $0\leq i<j\leq k_{0}$, i.e. $\sqrt{a_{i}},\sqrt{a_{j}}\in T_{0}$, then
$\sqrt{a_{i}}$ must be $e$ because $e\in T_{0}=e\cdot T_{0}$ which is a subset
of $T_{0}^{2}$, and Lemma 2.9 (c) and (4) tell us that
$T_{0}\cap\left\\{ab:a,b\in T_{0}\setminus\\{e\\},a\neq b\right\\}=\emptyset$.
If $k_{0}\leq i<j\leq 2n$, then by Lemma 2.9 (g) and $T_{0}\subseteq
T_{0}^{2}$, $\sqrt{a_{i}},\sqrt{a_{j}}$ cannot belong to $T_{1}$. Therefore,
we have proved that for $i<j$, $|a_{i}T_{0}\cap a_{j}T_{0}|=1$ if and only if
one of the following cases happens:

  1. (a)

$a_{i}=e$ and $\sqrt{a_{j}}\in T_{0}\setminus\\{e\\}$;

  2. (b)

$\sqrt{a_{i}}\in T_{0}$ and $\sqrt{a_{j}}\in T_{1}$ such that
$\sqrt{a_{i}a_{j}^{-1}}\in T_{0}$.

By (3), there are totally $k_{0}-1$ ordered pairs $(a_{i},a_{j})$ satisfying
(b). Thus, $C_{0,1}=2(k_{0}-1)$.

Calculation of $C_{1,1}$. Similarly, for $i<j$, $|a_{i}T_{1}\cap
a_{j}T_{1}|=1$ if and only if $a_{i}a_{j}^{-1}\in T_{1}^{(2)}$, which is
equivalent to one of the following cases:

  1. (a)

$a_{i}=e$ and $\sqrt{a_{j}}\in T_{1}$;

  2. (b)

$\sqrt{a_{i}},\sqrt{a_{j}}\in T_{k}$ for the same $k=0$ or $1$ such that
$\sqrt{a_{i}a_{j}^{-1}}\in T_{1}$.

The first case is obtained by checking the coefficients of elements in
$T_{0}T_{1}$ via (3) and the second case is from (4). It is clear that Case
(a) provides $k_{1}$ ordered pairs $(a_{i},a_{j})$ with $i<j$.

To count this number for Case (b), without loss of generality, we order the
elements in $T_{0}$ and $T_{1}$ such that $a_{i}^{-1}=a_{k_{0}-i}$ for
$i=1,\cdots,k_{0}$ and $a_{k_{0}+i}^{-1}=a_{2n-i}$ for $i=0,\cdots,k_{1}-1$.

Given an element $c\in T_{1}$, by (4), there always exists exactly one
unordered pair $\\{a,b\\}\subseteq T_{0}$ or $T_{1}$ such that $ab=c$. Suppose
that $a=\sqrt{a_{u}}$, $b=\sqrt{a_{v}^{-1}}$ and $a,b\in T_{0}$. As $c\neq e$,
$u$ and $v$ must be different. Consequently,

$c=ab=\sqrt{a_{u}}\sqrt{a_{v}^{-1}}=\sqrt{a_{k_{0}-v}}\sqrt{a_{k_{0}-u}^{-1}}.$

Note that $u<v$ implies $k_{0}-v<k_{0}-u$. Thus, if $a_{u}\neq a_{k_{0}-v}$,

$\\#\left\\{(\sqrt{a_{i}},\sqrt{a_{j}})\in T_{0}\times
T_{0}:c=\sqrt{a_{i}a_{j}^{-1}},i<j\right\\}=\begin{cases}2,&u<v;\\\
0,&u>v.\end{cases}$

If $a_{u}=a_{k_{0}-v}$, then $a_{u}=a_{v}^{-1}$ and $c$ is represented by
exactly one ordered pair which implies $c=a_{u}\in T_{0}^{(2)}$.

Therefore, there are totally

$2\cdot\frac{|T_{1}\cap(T_{0}^{2}\setminus T_{0}^{(2)})|}{2}+\frac{|T_{1}\cap
T_{0}^{(2)}|}{2}=|T_{1}\cap T_{0}^{2}|-|T_{1}\cap
T_{0}^{(2)}|+\frac{|T_{1}\cap T_{0}^{(2)}|}{2}$

ordered pair $(\sqrt{a_{i}},\sqrt{a_{j}})\in T_{0}\times T_{0}$ with $i<j$
such that $\sqrt{a_{i}a_{j}^{-1}}\in T_{1}$. A similar analysis for $a,b\in
T_{1}$ leads to

$2\cdot\frac{|T_{1}\cap(T_{1}^{2}\setminus T_{1}^{(2)})|}{2}+\frac{|T_{1}\cap
T_{1}^{(2)}|}{2}=|T_{1}\cap T_{1}^{2}|-|T_{1}\cap
T_{1}^{(2)}|+\frac{|T_{1}\cap T_{1}^{(2)}|}{2}$

ordered pair $(\sqrt{a_{i}},\sqrt{a_{j}})\in T_{1}\times T_{1}$ with $i<j$
such that $\sqrt{a_{i}a_{j}^{-1}}\in T_{1}$. As (4) tells us that $T_{1}$ is
covered by the elements in $T_{0}^{2}$ and $T_{1}^{2}$, by the above counting
results,

$C_{1,1}=k_{1}+\left(|T_{1}|-\frac{|T_{1}\cap
T_{0}^{(2)}|}{2}-\frac{|T_{1}\cap
T_{1}^{(2)}|}{2}\right)=2k_{1}-\frac{|T_{1}\cap\hat{T}^{(2)}|}{2}.$

Calculation of $C_{0,2}$. For $i<j$, $|a_{i}T_{0}\cap a_{j}T_{0}|=2$ if and
only if $a_{i}a_{j}^{-1}\in T_{0}^{2}\setminus T_{0}^{(2)}$, which is
equivalent to one of the following cases:

  1. (a)

$\sqrt{a_{i}}\in T_{0}$ and $\sqrt{a_{j}}\in T_{1}$ such that
$a_{i}a_{j}^{-1}\in T_{0}^{2}\setminus T_{0}^{(2)}$;

  2. (b)

$\sqrt{a_{i}},\sqrt{a_{j}}\in T_{k}$ for the same $k=0$ or $1$ such that
$a_{i}a_{j}^{-1}\in T_{0}^{2}\setminus T_{0}^{(2)}$.

By (3), $T_{0}^{(2)}T_{1}^{(2)}=H-e$. Thus the cardinality of the ordered
pairs $(a_{i},a_{j})$ satisfying (a) and $i<j$ is

$\displaystyle|T_{0}^{2}\setminus T_{0}^{(2)}|$
$\displaystyle=\frac{1}{2}\left(\\#\\{(a,b):a,b\in
T_{0}\\}-\\#\\{(a,a^{-1}):a\in T_{0}\\}-\\#\\{(a,a):a\in T_{0},a\neq
e\\}\right)$
$\displaystyle=\frac{1}{2}(k_{0}^{2}-k_{0}-(k_{0}-1))=\frac{(k_{0}-1)^{2}}{2}.$

For (b), we follow the argument in the calculation of $C_{1,1}$. For any $c\in
T_{0}^{2}\setminus T_{0}^{(2)}$, suppose that $a=\sqrt{a_{u}}$,
$b=\sqrt{a_{v}^{-1}}$ and $a,b\in T_{0}$ such that
$c=a_{u}a_{v}^{-1}=(ab)^{2}\in(T_{0}^{(2)})^{2}$. As $c\neq e$, $u$ and $v$
must be different. Consequently,

$c={a_{u}}{a_{v}^{-1}}={a_{k_{0}-v}}{a_{k_{0}-u}^{-1}}.$

It follows that

$\\#\left\\{(\sqrt{a_{i}},\sqrt{a_{j}})\in T_{0}\times
T_{0}:c=a_{i}a_{j}^{-1},i<j\right\\}=\begin{cases}2,&a_{u}\neq
a_{k_{0}-v},u<v;\\\ 0,&a_{u}\neq a_{k_{0}-v},u>v;\\\
1,&a_{u}=a_{k_{0}-v}.\end{cases}$

Therefore, there are totally

$\displaystyle 2\cdot\frac{|(T_{0}^{2}\setminus
T_{0}^{(2)})\cap((T_{0}^{(2)})^{2}\setminus
T_{0}^{(4)})|}{2}+\frac{|(T_{0}^{2}\setminus T_{0}^{(2)})\cap
T_{0}^{(4)}|}{2}$ $\displaystyle=$ $\displaystyle
2\cdot\frac{|(T_{0}^{2}\setminus
T_{0}^{(2)})\cap(T_{0}^{(2)})^{2}|}{2}-\frac{|(T_{0}^{2}\setminus
T_{0}^{(2)})\cap T_{0}^{(4)}|}{2}$

ordered pair $(\sqrt{a_{i}},\sqrt{a_{j}})\in T_{0}\times T_{0}$ with $i<j$
such that $a_{i}a_{j}^{-1}\in T_{0}^{2}\setminus T_{0}^{(2)}$.

A similar analysis for $a,b\in T_{1}$ leads to

$2\cdot\frac{|(T_{0}^{2}\setminus
T_{0}^{(2)})\cap(T_{1}^{(2)})^{2}|}{2}-\frac{|(T_{0}^{2}\setminus
T_{0}^{(2)})\cap T_{1}^{(4)}|}{2}$

ordered pair $(\sqrt{a_{i}},\sqrt{a_{j}})\in T_{1}\times T_{1}$ with $i<j$
such that $a_{i}a_{j}^{-1}\in T_{0}^{2}\setminus T_{0}^{(2)}$. As (4) tells us
that

$(T_{0}^{(2)})^{2}+(T_{1}^{(2)})^{2}=2H-T_{0}^{(4)}-T_{1}^{(4)}+2ne,$

which means that $T_{0}^{2}\setminus T_{0}^{(2)}$ is covered by the elements
in $(T_{0}^{(2)})^{2}$ and $(T_{1}^{(2)})^{2}$. By the counting results for
cases (a) and (b),

$\displaystyle C_{0,2}$ $\displaystyle=|T_{0}^{2}\setminus
T_{0}^{(2)}|+\left(|T_{0}^{2}\setminus T_{0}^{(2)}|-\frac{|(T_{0}^{2}\setminus
T_{0}^{(2)})\cap\hat{T}^{(4)}|}{2}\right)$
$\displaystyle=(k_{0}-1)^{2}-\frac{|(T_{0}^{2}\setminus
T_{0}^{(2)})\cap\hat{T}^{(4)}|}{2}.$

Calculation of $C_{1,2}$. For $i<j$, $|a_{i}T_{1}\cap a_{j}T_{1}|=2$ if and
only if $a_{i}a_{j}^{-1}\in T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\})$, which
is equivalent to one of the following cases:

  1. (a)

$\sqrt{a_{i}}\in T_{0}$ and $\sqrt{a_{j}}\in T_{1}$ such that
$a_{i}a_{j}^{-1}\in T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\})$;

  2. (b)

$\sqrt{a_{i}},\sqrt{a_{j}}\in T_{k}$ for the same $k=0$ or $1$ such that
$a_{i}a_{j}^{-1}\in T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\})$.

For Case (a), (3) tells us that each element in
$T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\})$ is covered by elements in
$T_{0}T_{1}$ exactly once. As a consequence, the cardinality of ordered pairs
$(a_{i},a_{j})$ satisfying (a) and $i<j$ is

$\displaystyle|T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\})|$
$\displaystyle=\frac{1}{2}\left(\\#\\{(a,b):a,b\in
T_{1}\\}-\\#\\{(a,a^{-1}):a\in T_{1}\\}-\\#\\{(a,a):a\in T_{1}\\}\right)$
$\displaystyle=\frac{1}{2}(k_{1}^{2}-2k_{1}).$

For Case (b), we follow the same argument for Case (b) of $C_{0,2}$.
Consequently,

$\displaystyle C_{1,2}$
$\displaystyle=2|T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\})|-\frac{|(T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\}))\cap\hat{T}^{(4)}|}{2}$
$\displaystyle=k_{1}^{2}-2k_{1}-\frac{|(T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\}))\cap\hat{T}^{(4)}|}{2}.\qed$

By Lemma 3.3, it is straightforward to get

(12) $\sum_{i<j}|a_{i}T_{k}\cap
a_{j}T_{k}|=\begin{cases}2k_{0}(k_{0}-1)-\left|(T_{0}^{2}\setminus
T_{0}^{(2)})\cap\hat{T}^{(4)}\right|,&k=0;\\\\[5.0pt]
2k_{1}(k_{1}-1)-\frac{|T_{1}\cap\hat{T}^{(2)}|}{2}-\left|(T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\}))\cap\hat{T}^{(4)}\right|,&k=1.\end{cases}$

Recall that $\theta_{0}=|(T_{0}^{2}\setminus T_{0}^{(2)})\cap\hat{T}^{(4)}|$.
Let
$\theta_{1,1}=|(T_{1}^{2}\setminus(T_{1}^{(2)}\cup\\{e\\}))\cap\hat{T}^{(4)}|$
and $\theta_{1,2}=|T_{1}\cap\hat{T}^{(2)}|$ which equals
$|T_{1}^{(2)}\cap\hat{T}^{(4)}|$. By (4) and $\hat{T}^{(4)}\cap
T_{0}^{(2)}=\\{e\\}$,

(13)
$\theta_{0}+\theta_{1,1}+\theta_{1,2}=|\hat{T}^{(4)}\setminus\\{e\\}|=2n.$

Hence

$\theta_{0}+\theta_{1}=\theta_{0}+\theta_{1,1}+\frac{\theta_{1,2}}{2}\leq 2n.$

Now we are ready to prove Lemma 3.1.

###### Proof of Lemma 3.1..

We only prove the result for $T_{0}$. The proof of (11) is the same.

By the inclusion–exclusion principle, we count the distinct elements in
$\hat{T}^{(2)}T_{0}$,

(14)
$|\hat{T}^{(2)}T_{0}|=\sum_{i=0}^{2n}|a_{i}T_{0}|-\sum_{i<j}|a_{i}T_{0}\cap
a_{j}T_{0}|+\sum_{r\geq 3}(-1)^{r-1}|a_{i_{1}}T_{0}\cap
a_{i_{2}}T_{0}\cap\cdots\cap a_{i_{r}}T_{0}|,$

where $i_{1}<i_{2}<\cdots<i_{r}$ cover all the possible values. By the
definition of $X_{i}$’s, the left-hand side of (14) equals
$\sum_{i=1}^{M_{0}}|X_{i}|$.

Now we determine the value of the different sums in (14). First,
$\sum_{i=0}^{2n}|a_{i}T_{0}|=(2n+1)k_{0}$. Second, (12) tells us the value of
$\sum_{i<j}|a_{i}T_{0}\cap a_{j}T_{0}|$.

Finally, suppose that $g\in a_{i_{1}}T_{0}\cap a_{i_{2}}T_{0}\cap\cdots\cap
a_{i_{r}}T_{0}$ with $r\geq 3$. It means that $g\in X_{s}$ for some $s\geq 3$.
Then the contribution for $g$ in the sum $\sum_{r\geq
3}(-1)^{r-1}|a_{i_{1}}T_{0}\cap a_{i_{2}}T_{0}\cap\cdots\cap a_{i_{r}}T_{0}|$
is

$(-1)^{3-1}\binom{s}{3}+(-1)^{4-1}\binom{s}{4}+\cdots=\binom{s}{2}-\binom{s}{1}+\binom{s}{0}=\frac{(s-1)(s-2)}{2}.$

Therefore,

$\sum_{r\geq 3}(-1)^{r-1}|a_{i_{1}}T_{0}\cap a_{i_{2}}T_{0}\cap\cdots\cap
a_{i_{r}}T_{0}|=\sum_{s\geq 3}|X_{s}|\frac{(s-1)(s-2)}{2}.$

Plugging them all into (14), we get (10). ∎

Recall that at the very beginning of this section we have obtained

(15) $\displaystyle\hat{T}^{(2)}T_{0}$
$\displaystyle=(2k_{0}-k_{1})H+T_{1}-T_{0}^{3}+2nT_{0},$ (16)
$\displaystyle\hat{T}^{(2)}T_{1}$
$\displaystyle=(2k_{1}-k_{0})H+T_{0}-T_{1}^{3}+2nT_{1}.$

In the following, we investigate (15) and (16) separately in different cases
depending on the value of $n$ modulo $3$.

###  3.1. Case: $n\equiv 1\pmod{3}$

First we need to prove the following observation.

######  Lemma 3.4.

When $n\equiv 1\pmod{3}$,

  1. (a)

there is no element in $T_{j}^{(3)}$ appearing more than $2$ times for
$j=0,1$,

  2. (b)

$e$ appears only once in $T_{0}^{(3)}$, and

  3. (c)

there are $0$ or $2$ elements appearing twice in $T_{0}^{(3)}$, and there are
at most $2$ elements appearing twice in $T_{1}^{(3)}$.

###### Proof.

Recall that $|H|=n^{2}+n+1$. By the assumption, $3\mid|H|$ but $9\nmid|H|$.
Thus there are exactly two elements in $H$ of order $3$. We denote one of them
by $h$, which means the other one is $h^{-1}$.

(a). For fixed $j=0$ or $1$, suppose that there exist distinct elements $a,b$
in $T_{j}$ such that $a^{3}=b^{3}$. Then $ab^{-1}$ must be of order $3$ in
$H$. Without loss of generality, assume $ab^{-1}=h$.

Toward a contradiction, suppose that there exists $c\neq a,b$ in $T_{j}$ such
that $c^{3}=a^{3}=b^{3}$ which means $ac^{-1}$ or $bc^{-1}$ equals $h$. By
(4), there are at most one unordered pair $\\{u,v\\}$ with $u,v\in T_{j}$ such
that $uv=h$. It follows that the unordered pairs $\\{a,b^{-1}\\}$ and
$\\{a,c^{-1}\\}$ (or $\\{b,c^{-1}\\}$) must be equal. However, this is
impossible.

(b). Note that $e=a^{3}$ for some $a\in T_{0}$ if and only if $a=e,h$ or
$h^{-1}$. Hence, if $e$ appears more than once in $T_{0}^{(3)}$, then it
appears for at least three times in $T_{0}^{(3)}$ which contradicts (a).

(c). For $j=1$ or $2$, if $g=t_{1}^{3}=t_{2}^{3}$ appears twice in
$T_{j}^{(3)}$, then $t_{1}t_{2}^{-1}$ has to be of order $3$. By (4), the
unordered pair $\\{t_{1},t_{2}^{-1}\\}$ must be $\\{a,b^{-1}\\}$ or
$\\{b,a^{-1}\\}$ which are given in (a). Therefore, there is no element
different from $a^{3}$ and $a^{-3}$ appearing twice in $T_{j}^{(3)}$.

In particular, for $j=0$, (b) tells us that $a^{3}\neq e$. Hence $a^{3}\neq
a^{-3}\in T_{0}^{(3)}$. ∎

By Lemma 2.9 (e) and (f), $2k_{0}-k_{1}\equiv 2k_{1}-k_{0}\equiv 0\pmod{3}$.
Using (15), (16) and $3\mid(n-1)$,

(17) $\displaystyle\hat{T}^{(2)}T_{0}$ $\displaystyle\equiv
T_{1}-T_{0}^{(3)}+2T_{0}\pmod{3},$ (18) $\displaystyle\hat{T}^{(2)}T_{1}$
$\displaystyle\equiv T_{0}-T_{1}^{(3)}+2T_{1}\pmod{3}.$

By Lemma 2.9 (b) and (h),

$T_{0}\cap T_{1}=\emptyset,\text{ and }T_{0}\cap T_{0}^{(3)}=\\{e\\}.$

Let $\Delta_{i}$ be the set of elements appearing twice in $T_{i}^{(3)}$ for
$i=1,2$. By Lemma 3.4, $|\Delta_{0}|,|\Delta_{1}|\leq 2$ and there is no
element appearing more than 2 times in $T_{i}^{(3)}$ for $i=1,2$. Note that

$|T_{0}^{(3)}|=k_{0}-|\Delta_{0}|,\text{ and
}|T_{1}^{(3)}|=k_{1}-|\Delta_{1}|.$

Depending on the parity of $n$, we investigate (18) and (17), respectively.

######  Theorem 3.5.

For $n\equiv 4\pmod{6}$ and $n>2$, there is no inverse-closed subsets $T_{0}$
and $T_{1}\subseteq H$ with $e\in T_{0}$ and $k_{0}+k_{1}=2n+1$ satisfying (3)
and (4).

###### Proof.

Suppose to the contrary that there exist $T_{0}$ and $T_{1}$ satisfying (3)
and (4).

As $2\mid n$, $k_{0}=n+1$ and $k_{1}=n$. We concentrate on (18). Let
$u_{0}=|T_{1}^{(3)}\cap T_{0}|$ and $u_{1}=|T_{1}^{(3)}\cap T_{1}|$. By Lemma
2.9, $T_{0}\cap T_{1}=\emptyset$. Furthermore, we separate $\Delta_{1}$ into
three disjoint parts $\Delta_{1}^{0}=\Delta_{1}\cap T_{0}$,
$\Delta_{1}^{1}=\Delta_{1}\cap T_{1}$ and
$\Delta_{1}^{2}=\Delta_{1}\setminus(T_{0}\cup T_{1})$.

By comparing the coefficients of elements in (18), we get

$\bigcup_{i\geq 0}Y_{3i+2}=(T_{1}\setminus
T_{1}^{(3)})~{}\dot{\cup}~{}\left(T_{1}^{(3)}\setminus(T_{0}~{}\dot{\cup}~{}T_{1}\cup\Delta_{1})\right)~{}\dot{\cup}~{}\Delta_{1}^{0}.$

Note that $|T_{1}^{(3)}|=k_{1}-|\Delta_{1}|$. It follows that

$\sum_{i\geq
0}|Y_{3i+2}|=(n-u_{1})+(n-|\Delta_{1}|-(u_{0}+u_{1})-|\Delta_{1}^{2}|)+|\Delta_{1}^{0}|.$

Similarly,

$\bigcup_{i\geq 0}Y_{3i+1}=(T_{0}\setminus
T_{1}^{(3)})~{}\dot{\cup}~{}\Delta_{1}^{2}~{}\dot{\cup}~{}(T_{1}\cap
T_{1}^{(3)}\setminus\Delta_{1}^{1}).$

Thus

$\sum_{i\geq
0}|Y_{3i+1}|=(n+1-u_{0})+|\Delta_{1}^{2}|+u_{1}-|\Delta_{1}^{1}|.$

To summarize, we have proved

(19) $\sum_{i\geq
0}|Y_{3i+2}|=2n-2u_{1}-u_{0}-2|\Delta_{1}^{2}|-|\Delta_{1}^{1}|,~{}\sum_{i\geq
0}|Y_{3i+1}|=n+1-u_{0}+u_{1}+|\Delta_{1}^{2}|-|\Delta_{1}^{1}|,$

Now (11) becomes

$\displaystyle\sum_{i=1}^{M_{1}}|Y_{i}|$
$\displaystyle=(2n+1)n-2(n-1)n+\theta_{1}+\sum_{s\geq
3}\frac{(s-1)(s-2)}{2}|Y_{s}|$ (20) $\displaystyle=3n+\theta_{1}+\sum_{s\geq
3}\frac{(s-1)(s-2)}{2}|Y_{s}|.$

Plugging the two equations in (19) into it to replace $3n$, we get

$\displaystyle\sum_{i=1}^{M_{1}}|Y_{i}|=$ $\displaystyle\sum_{i\geq
0}|Y_{3i+1}|+\sum_{i\geq
0}|Y_{3i+2}|+2u_{0}+u_{1}+|\Delta_{1}^{2}|+2|\Delta_{1}^{1}|-1$
$\displaystyle+\theta_{1}+\sum_{s\geq 3}\frac{(s-1)(s-2)}{2}|Y_{s}|,$

which implies

$\displaystyle 1=$ $\displaystyle
2u_{0}+u_{1}+|\Delta_{1}^{2}|+2|\Delta_{1}^{1}|+\theta_{1}$ (21)
$\displaystyle+\sum_{s>3,3\mid
s}\left(\frac{(s-1)(s-2)}{2}-1\right)|Y_{s}|+\sum_{s\geq 3,3\nmid
s}\left(\frac{(s-1)(s-2)}{2}\right)|Y_{s}|.$

As $|Y_{i}|$, $u_{0}$, $u_{1}$,$|\Delta_{1}^{2}|$, $|\Delta_{1}^{1}|$ and
$\theta$ are all nonnegative integers, (21) implies that $u_{0}=0$ and each
$|Y_{i}|=0$ for $i\geq 4$. Recall that $u_{1}$ is the cardinality of the
inverse-closed subset $T_{1}^{(3)}\cap T_{1}$ which does not contain the
identity element $e$ of $H$. Hence $u_{1}$ must be even. Similarly,
$|\Delta_{1}^{1}|$ and $|\Delta_{1}^{2}|$ are also even. Thus, by (21),

$u_{1}=|\Delta_{1}^{1}|=|\Delta_{1}^{2}|=0,$

and $\theta_{1}=1$. Plugging them into (19), we get

$|Y_{1}|=n+1,~{}|Y_{2}|=2n.$

By (8),

$3|Y_{3}|=(2n+1)n-(n+1)-4n=2n^{2}-4n-1\equiv 3\pmod{6}.$

Consequently, $|Y_{1}|$ and $|Y_{3}|$ are both odd. As $Y_{i}$’s form a
partition of the group $H$ and each $Y_{i}$ is inverse-closed, there exists
only one of them of odd size. We get a contradiction.

Therefore, we have excluded the existence of $|Y_{i}|$’s which means there do
not exist inverse-closed subsets $T_{0}$ and $T_{1}\subseteq H$ with $e\in
T_{0}$ and $k_{0}+k_{1}=2n+1$ satisfying (3) and (4). ∎

For $n$ odd and $n\equiv 1\pmod{3}$, we could not derive any contradiction by
analyzing the value of $|X_{i}|$’s and $|Y_{i}|$’s.

###  3.2. Case: $n\equiv 0\pmod{3}$

We separate the proof into two cases according to the parity of $n$.

######  Theorem 3.6.

For $n\equiv 3\pmod{6}$, there is no inverse-closed subsets $T_{0}$ and
$T_{1}\subseteq H$ with $e\in T_{0}$ and $k_{0}+k_{1}=2n+1$ satisfying (3) and
(4).

###### Proof.

As $n$ is odd, $|T_{0}|=k_{0}=n$ and $|T_{1}|=k_{1}=n+1$. Toward a
contradiction, suppose that $T_{0}$ and $T_{1}$ exist such that (3) and (4)
hold.

First Lemma 2.9 tells us that $2k_{0}-k_{1}\equiv 2k_{1}-k_{0}\equiv
2\pmod{3}$. Under the assumption that $3\mid n$,

(22) $\displaystyle\hat{T}^{(2)}T_{0}$ $\displaystyle\equiv
2H+T_{1}-T_{0}^{(3)}\pmod{3},$ $\displaystyle\hat{T}^{(2)}T_{1}$
$\displaystyle\equiv 2H+T_{0}-T_{1}^{(3)}\pmod{3}.$

Moreover, as $3\nmid|H|$, there is no repeating element in $T_{0}^{(3)}$ and
$T_{1}^{(3)}$.

Assume that there are $\ell_{0}$ elements in $T_{1}\cap T_{0}^{(3)}$ and
$\ell_{1}$ elements in $T_{0}\cap T_{1}^{(3)}$. It follows that

$\displaystyle\sum_{i\geq 0}|X_{3i+1}|=k_{0}-\ell_{0},\quad$
$\displaystyle\sum_{i\geq 0}|X_{3i+2}|=n^{2}-n+2\ell_{0},$
$\displaystyle\sum_{i\geq 0}|X_{3i}|=k_{1}-\ell_{0},$
$\displaystyle\sum_{i\geq 0}|Y_{3i+1}|=k_{1}-\ell_{1},\quad$
$\displaystyle\sum_{i\geq 0}|Y_{3i+2}|=n^{2}-n+2\ell_{1},$
$\displaystyle\sum_{i\geq 0}|Y_{3i}|=k_{0}-\ell_{1}.$

In particular, for $2\nmid n$,

(23) $\sum_{i\geq 0}|X_{3i+1}|=n-\ell_{0},\quad\sum_{i\geq
0}|X_{3i+2}|=n^{2}-n+2\ell_{0},\quad\sum_{i\geq 0}|X_{3i}|=n+1-\ell_{0}.\\\ $

The strategy of our proof is as follows: First we consider the possible
nonnegative integer solutions for $|X_{i}|$’s satisfying (23). Several
inequalities will provide us

(24) $\begin{cases}|X_{0}|=n+1,\\\ \ell_{0}=0,\end{cases}\text{ or
}\quad\begin{cases}|X_{0}|=n-1,\\\ \ell_{0}=0\text{ or }2.\end{cases}$

Then, we will show that such $X_{0}$ cannot exist.

Next we provide the details of the proof of (24). Now (10) becomes

$\sum_{i=1}^{M_{0}}|X_{i}|=3n+\theta_{0}+\sum_{s\geq
3}\frac{(s-1)(s-2)}{2}|X_{s}|.$

Subtract (9) from (10),

$-|X_{0}|=-n^{2}+2n-1+\theta_{0}+\sum_{s\geq 3}\frac{(s-1)(s-2)}{2}|X_{s}|,$

which means

(25) $n^{2}-|X_{0}|-2n-\theta_{0}+1=\sum_{s\geq
3}\frac{(s-1)(s-2)}{2}|X_{s}|\leq\frac{M_{0}-1}{2}\sum_{s\geq 3}(s-2)|X_{s}|.$

On the other hand, by (7), (9) and (23),

$\displaystyle\sum_{i\geq 1}3i(|X_{3i+1}|+|X_{3i+2}|)+\sum_{i\geq
1}(3i-2)|X_{3i}|$ (26) $\displaystyle=$ $\displaystyle\sum_{i\geq
1}i|X_{i}|-\sum_{i\geq 0}|X_{3i+1}|-2\sum_{i\geq 0}|X_{3i+2}|-\sum_{i\geq
0}2|X_{3i}|+2|X_{0}|$ $\displaystyle=$ $\displaystyle 2|X_{0}|-\ell_{0}-2.$

As $M_{0}>1$, by Lemma 3.2, $|X_{M_{0}}|$ is even, and $|X_{M_{0}}|\geq 2$. By
(26),

$(M_{0}-2)2\leq 2|X_{0}|-\ell_{0}-2,$

which means

(27) $M_{0}\leq|X_{0}|+1-\frac{\ell_{0}}{2}.$

Moreover, by (26), we also have

$\sum_{s\geq 3}(s-2)|X_{s}|\leq 2|X_{0}|-\ell_{0}-2.$

Plugging it and (27) into (25), we have

$\displaystyle n^{2}-|X_{0}|-2n-\theta_{0}+1$
$\displaystyle\leq\frac{M_{0}-1}{2}(2|X_{0}|-\ell_{0}-2)$ (28)
$\displaystyle\leq(|X_{0}|-\ell_{0}/2)(|X_{0}|-\ell_{0}/2-1)$
$\displaystyle\leq|X_{0}|(|X_{0}|-1).$

By $|X_{0}|\leq\sum_{i\geq 0}|X_{3i}|=k_{1}-\ell_{0}\leq n+1$ and a direct
computation with the above inequality using $\theta_{0}\leq 2n$, we derive
that $|X_{0}|=n+1,n$, $n-1$ or $n-2$. Moreover, $|X_{1}|$ is odd by Lemma 3.2.
Consequently, by (23), $\ell_{0}$ must be even for odd $n$. Taking (28) and
$\sum_{i\geq 0}|X_{3i}|=n+1-\ell_{0}$ into account, we get four possible cases

  1. (I)

$|X_{0}|=n+1$ and $\ell_{0}=0$;

  2. (II)

$|X_{0}|=n-1$ and $\ell_{0}=0$ or $2$;

  3. (III)

$|X_{0}|=n$;

  4. (IV)

$|X_{0}|=n-2$.

By Lemma 3.2, $|X_{0}|$ must be even. Consequently Case (III) and Case (IV)
are impossible, because $n$ is assumed to be odd.

Hence, we have proved (24).

Determination of $X_{0}$. Next we show that

(29) $X_{0}=\begin{cases}T_{1},&|X_{0}|=n+1;\\\
T_{1}\setminus\\{\gamma,\gamma^{-1}\\},&|X_{0}|=n-1,\end{cases}$

where $\gamma$ is an element in $T_{1}$. Afterwards we will finish the whole
proof by getting some contradiction from (29).

Recall

(15) $\hat{T}^{(2)}T_{0}=(n-1)H+T_{1}-T_{0}^{3}+2nT_{0}.$

First we consider the case with $\ell_{0}=0$. For an arbitrary element $t\in
T_{1}$, since $\ell_{0}=|T_{1}\cap T_{0}^{(3)}|=0$, $t$ can only be
represented as

$abc\text{ for some pairwise distinct }a,b,c\in T_{0},\quad\text{or}\quad
aab\text{ for some distinct }a,b\in T_{0},$

in $T_{0}^{3}$.

Taking account of the ordering of the elements in the representations, the
coefficient of $t$ in $T_{0}^{3}\in\mathbb{Z}[H]$ is divisible by $3$.
Together with $n\equiv 0\pmod{3}$ and $T_{1}\cap T_{0}=\emptyset$, the
coefficient of $t$ in the right-hand side of (15) must be congruent to $0$
modulo $3$. Since $\sum_{i\geq 0}|X_{3i}|=n+1$ and $|T_{1}|=n+1$,

$X_{0}=\begin{cases}T_{1},&|X_{0}|=n+1;\\\
T_{1}\setminus\\{\gamma,\gamma^{-1}\\},&|X_{0}|=n-1,\end{cases}$

for some $\gamma\in T_{1}$.

When $\ell_{0}=2$, by (24), $|X_{0}|=n-1$. By our definition of $\ell_{0}$,
there exist exactly two elements $\gamma$ and $\gamma^{-1}\in T_{1}$ such that
$\gamma,\gamma^{-1}\in T_{0}^{(3)}$. The coefficients of them in the right-
hand side of (15) must be congruent to $1$ modulo $3$, and the coefficient of
any element in $T_{1}\setminus\\{\gamma,\gamma^{-1}\\}$ is still congruent to
$0$ modulo $3$. Thus $X_{0}=T_{1}\setminus\\{\gamma,\gamma^{-1}\\}$.

Therefore, we have obtained (29).

In the final step, we show that (29) cannot hold which concludes the proof of
Theorem 3.6.

If $T_{1}=X_{0}$, then (15) tells us

$(T_{0}^{(2)}+T_{1}^{(2)})T_{0}\cap T_{1}=\emptyset,$

which cannot hold, because $T_{0}T_{1}=H-e$ means for any non-identity $a\in
T_{0}^{(2)}+T_{1}^{(2)}$ there always exist $t_{0}\in T_{0}$ and $t_{1}\in
T_{1}$ such that $t_{0}t_{1}=a$.

If $T_{1}\setminus\\{\gamma,\gamma^{-1}\\}=X_{0}$, then (15) tells us
$t\notin(T_{0}^{(2)}+T_{1}^{(2)})T_{0}$ for any $t\in
T_{1}\setminus\\{\gamma,\gamma^{-1}\\}$. By (3),

$(T_{1}-\gamma-\gamma^{-1})T_{0}=H-e-(\gamma T_{0}+\gamma^{-1}T_{0}).$

For any $x\in T_{0}$, if $tx=y$, then $t=x^{-1}y$ which means $y\notin
T_{0}^{(2)}+T_{1}^{(2)}$. Thus the above equation implies

(30) $\gamma T_{0}+\gamma^{-1}T_{0}+e=T_{0}^{(2)}+T_{1}^{(2)}.$

As $\gamma^{2}\in T_{1}^{(2)}$,

$\gamma^{2}=\gamma a\quad\text{or}\quad\gamma^{2}=\gamma^{-1}a$

for some $a\in T_{0}\setminus\\{e\\}$. The first case is impossible because
$T_{0}\cap T_{1}=\emptyset$. The second case implies that $a=\gamma^{3}\in
T_{0}\cap T_{1}^{(3)}$. By (30), $\gamma a=\gamma^{4}$ belongs to
$T_{0}^{(2)}$ or $T_{1}^{(2)}$.

If $\gamma^{4}\in T_{0}^{(2)}$, then $\gamma^{2}\in T_{0}$ which means
$T_{0}\cap T_{1}^{(2)}\neq\emptyset$. It contradicts Lemma 2.9 (c).

For $\gamma^{4}\in T_{1}^{(2)}$, we need more detailed analysis. As $a^{2}\in
T_{0}^{(2)}$, by (30) there exists $b\in T_{0}$ such that

$a^{2}=\gamma^{6}=\gamma b,\text{ or }\gamma^{-1}b,$

which means $b=\gamma^{5}$ or $\gamma^{7}\in T_{0}$. If $b=\gamma^{5}\in
T_{0}$, then

$T_{1}^{(2)}\ni\gamma^{2}=\gamma^{5}\cdot\gamma^{-3}=b\cdot a^{-1}\in
T_{0}^{2}$

whence $\gamma^{2}\in T_{0}^{2}\cap T_{1}^{(2)}$ which is not possible by
Lemma 2.9 (g); if $b=\gamma^{7}\in T_{0}$, then
$\gamma^{4}=b\gamma^{-3}=ba^{-1}\in T_{0}^{2}$. Recall that $\gamma^{4}\in
T_{1}^{(2)}$. Hence $\gamma^{4}\in T_{0}^{2}\cap T_{1}^{(2)}$ which
contradicts Lemma 2.9 (g) again.

To summarize, we have proved that there exist no $T_{0}$ and $T_{1}$
satisfying (3) and (4). ∎

The proof for even $n$ which is congruent to $0$ modulo $3$ is similar, but we
have to handle some extra problems.

######  Theorem 3.7.

For $n\equiv 0\pmod{6}$, there is no inverse-closed subsets $T_{0}$ and
$T_{1}\subseteq H$ with $e\in T_{0}$ and $k_{0}+k_{1}=2n+1$ satisfying (3) and
(4).

###### Proof.

Now $n$ is divisible by $2$ and $3$, $|T_{0}|=k_{0}=n+1$ and
$|T_{1}|=k_{1}=n$. Toward a contradiction, suppose that $T_{0}$ and $T_{1}$
exist such that (3) and (4) hold. The smallest possible value of $n$ equals
$6$. However, throughout the rest part of the proof, we can always assume that

(31) $n\geq 30.$

The reason is as follows: for $n=6,12$ and $24$, $n^{2}+n+1$ is prime which
means the nonexistence result is already covered by Result 2.11. For $n=18$,
it can be excluded by Proposition 2.10.

Instead of looking at $X_{i}$’s as in the proof of Theorem 3.6, we consider
$Y_{i}$’s. By (16),

(32) $\hat{T}^{(2)}T_{1}\equiv 2H+T_{0}-T_{1}^{(3)}\pmod{3}.$

It follows that

(33) $\displaystyle\sum_{i\geq 0}|Y_{3i+1}|=n-\ell_{1},\quad$
$\displaystyle\sum_{i\geq 0}|Y_{3i+2}|=n^{2}-n+2\ell_{1},$
$\displaystyle\sum_{i\geq 0}|Y_{3i}|=n+1-\ell_{1},$

where $\ell_{1}$ is defined to be the cardinality of $T_{0}\cap T_{1}^{(3)}$,
and (11) becomes

$\sum_{i=1}^{M_{1}}|Y_{i}|=3n+\theta_{1}+\sum_{s\geq
3}\frac{(s-1)(s-2)}{2}|Y_{s}|.$

By subtracting (9) from (11), we get

(34) $n^{2}-|Y_{0}|-2n+1-\theta_{1}=\sum_{s\geq
3}\frac{(s-1)(s-2)}{2}|Y_{s}|.$

On the other hand, by (8), (9) and (33),

$\displaystyle\sum_{i\geq 1}3i(|Y_{3i+1}|+|Y_{3i+2}|)+\sum_{i\geq
1}(3i-2)|Y_{3i}|$ (35) $\displaystyle=$ $\displaystyle\sum_{i\geq
1}i|Y_{i}|-\sum_{i\geq 0}|Y_{3i+1}|-2\sum_{i\geq 0}|Y_{3i+2}|-\sum_{i\geq
0}2|Y_{3i}|+2|Y_{0}|$ $\displaystyle=$ $\displaystyle 2|Y_{0}|-\ell_{1}-2.$

It is easy to check that (35) is the same as (26) by replacing $X_{i}$ with
$Y_{i}$, and $\ell_{0}$ with $\ell_{1}$.

As we do not know whether $|Y_{M_{1}}|\geq 2$, we cannot immediately follow
the proof of Theorem 3.6 to determine the possible value of $|Y_{i}|$’s. Next
we concentrate on the proof of the following claim.

Claim 1. $|Y_{M_{1}}|\geq 2$.

Toward a contradiction, suppose that $|Y_{M_{1}}|=1$ which is equivalent to
$Y_{M_{1}}=\\{e\\}$. By

$\hat{T}^{(2)}T_{1}=\sum_{i=1}^{M_{1}}iY_{i},$

we know that $M_{1}=|\hat{T}^{(2)}\cap T_{1}|=|T_{0}^{(2)}\cap
T_{1}|+|T_{1}^{(2)}\cap T_{1}|\leq k_{1}=n$, and $2\mid M_{1}$ because
$|T_{0}^{(2)}\cap T_{1}|$ and $|T_{1}^{(2)}\cap T_{1}|$ are both even.
Moreover, by (32), $e\in T_{0}$ and $e\notin T_{1}^{(3)}$, we see that $3\mid
M_{1}$. Therefore, we have obtained the following restriction on $M_{1}$:

(36) $6\leq M_{1}\leq n,\text{ and }6\mid M_{1}.$

By (35), we obtain $M_{1}-2\leq 2|Y_{0}|-\ell_{1}-2$, which means

(37) $M_{1}\leq 2|Y_{0}|-\ell_{1}.$

Let $M_{1}^{\prime}=\max\\{i<M_{1}:|Y_{i}|>0\\}$. If $M^{\prime}_{1}<3$, then
by (33), (36) and $|Y_{M_{1}}|=1$, we have

$\displaystyle|Y_{0}|$ $\displaystyle=n-\ell_{1},$ $\displaystyle|Y_{1}|$
$\displaystyle=n-\ell_{1},$ $\displaystyle|Y_{2}|$
$\displaystyle=n^{2}-n+2\ell_{1}.$

Plugging them in to (34), we get

$n^{2}-3n+\ell_{1}+1-\theta_{1}=\frac{(M_{1}-1)(M_{1}-2)}{2}\leq\frac{n^{2}-3n+2}{2},$

where the last inequality comes from (36). Recall that $\theta_{1}\leq 2n$.
Thus

$n^{2}-3n+\ell_{1}+1-2n\leq\frac{n^{2}-3n+2}{2}.$

It implies that $n\leq 7$ which contradicts our assumption (31). Therefore
$M^{\prime}_{1}\geq 3$.

As $e\notin Y_{M^{\prime}_{1}}$, by the inverse-closed property of
$\hat{T}^{(2)}$ and $T_{1}$, $|Y_{M^{\prime}_{1}}|$ must be even. In
particular, $|Y_{M^{\prime}_{1}}|\geq 2$. By (35), we get

$(M_{1}^{\prime}-2)\cdot 2+(M_{1}-2)\leq 2|Y_{0}|-\ell_{1}-2,$

which means

(38) $M^{\prime}_{1}\leq|Y_{0}|-\frac{\ell_{1}+M_{1}}{2}+2.$

By (35), we also get

(39) $2|Y_{0}|-\ell_{1}-2\geq\sum_{3\leq s\leq
M^{\prime}_{1}}(s-2)|Y_{s}|+(M_{1}-2).$

By (34),

$n^{2}-|Y_{0}|-2n+1-\theta_{1}\leq\frac{M^{\prime}_{1}-1}{2}\sum_{3\leq s\leq
M^{\prime}_{1}}(s-2)|Y_{s}|+\frac{(M_{1}-1)(M_{1}-2)}{2}.$

Plugging (39) and (38) into it, we get

$\displaystyle n^{2}-|Y_{0}|-2n+1-\theta_{1}$
$\displaystyle\leq\frac{M^{\prime}_{1}-1}{2}(2|Y_{0}|-\ell_{1}-M_{1})+\frac{(M_{1}-1)(M_{1}-2)}{2}$
$\displaystyle\leq\left(|Y_{0}|-\frac{M_{1}+\ell_{1}}{2}+1\right)\left(|Y_{0}|-\frac{M_{1}+\ell_{1}}{2}\right)+\frac{(M_{1}-1)(M_{1}-2)}{2}$
$\displaystyle\leq\left(|Y_{0}|-\frac{M_{1}}{2}+1\right)\left(|Y_{0}|-\frac{M_{1}}{2}\right)+\frac{(M_{1}-1)(M_{1}-2)}{2}.$

We add $|Y_{0}|$ on both sides of the above inequality and get

(40) $n^{2}-2n+1-\theta_{1}\leq F(|Y_{0}|,M_{1}),$

where

$F(x,y):=x^{2}+\frac{3}{4}y^{2}-xy+2x-2y+1.$

Recall that $|Y_{0}|\leq n-\ell_{1}$ by the third equation in (33) and (36),
we only have to consider the value of $F(x,y)$ for $(x,y)\in[0,n]\times[6,n]$.
Our goal is to show that the value of $F(x,y)$ is always smaller than
$n^{2}-2n+1-\theta_{1}$ which means (40) cannot hold.

It is easy to see that $F$ defines an elliptic paraboloid and the minimum
value is attained at point $(-1/2,1)$. Hence, the maximum value of $F(x,y)$
for $(x,y)\in[0,n]\times[6,n]$ can be obtained only if $x=n$ or $y=n$.

For $x=n$,

$F(n,y)=\frac{3}{4}y^{2}-(n+2)y+n^{2}+2n+1.$

Its maximum value for $y\in[6,n]$ with $6\mid y$ is

$F(n,6)=n^{2}-4n+16,$

and $F(n,12)=n^{2}-10n+85$ is the second largest value under the assumption
(31).

First, we can simply check that $y=6$ is impossible: Now $M_{1}=6$ and
$|Y_{0}|=n$, which means $\ell_{1}=0$ and $|Y_{3}|=0$. By (33), (11) and (34),

$\displaystyle|Y_{1}|+|Y_{4}|$ $\displaystyle=n,$
$\displaystyle|Y_{2}|+|Y_{5}|$ $\displaystyle=n^{2}-n,$
$\displaystyle|Y_{1}|+2|Y_{2}|+4|Y_{4}|+5|Y_{5}|$ $\displaystyle=2n^{2}+n-6,$
$\displaystyle 3|Y_{4}|+6|Y_{5}|$ $\displaystyle=n^{2}-3n+1-\theta_{1}-10.$

Consequently, $|Y_{1}|=\frac{n^{2}}{3}-\frac{4n}{3}-\frac{\theta_{1}}{3}+1$
which must be smaller than or equal to $n$ by the first equation. Recall that
$\theta_{1}\leq 2n$. Thus

$\frac{n^{2}}{3}-{2n}+1\leq n,$

which implies $n<9$. It contradicts our assumption (31).

For the second largest value $F(n,12)$, as $\theta_{1}\leq 2n$, if (40) holds,
then

$n^{2}-4n+1\leq n^{2}-2n+1-\theta_{1}\leq n^{2}-10n+85,$

which means $n\leq 14$. But these value have already been excluded by (31).
Therefore,

$n^{2}-2n+1-\theta_{1}>F(n,y),$

for any $y\in[6,n]$ with $6\mid y$.

For $y=n$,

$F(x,n)=x^{2}-(n-2)x+\frac{3}{4}n^{2}-2n+1.$

Its maximum value for $x\in[0,n]$ is

$F(n,n)=\frac{3}{4}n^{2}+1.$

It is larger than or equal to $n^{2}-4n+1$ if and only if $n\leq 16$, which
again contradicts (31).

Therefore, (40) cannot hold which means that we have proved Claim 1.

Following the proof of Theorem 3.6, our next goal is to show that the only
possible value of $|Y_{0}|$ are as follows:

(41) $\begin{cases}|Y_{0}|=n+1,\\\ \ell_{1}=0,\end{cases}\text{ or
}\quad\begin{cases}|Y_{0}|=n-1,\\\ \ell_{1}=0\text{ or }2,\end{cases}\text{ or
}\quad\begin{cases}|Y_{0}|=n,\\\ \ell_{1}=0.\end{cases}$

By replacing $Y_{i}$’s with $X_{i}$’s, one can get (23) and (26) from (33) and
(35), respectively. Moreover, (25) becomes (34) if we change $X_{i}$’s to
$Y_{i}$’s and change $\theta_{0}$ to $\theta_{1}$.

Hence, by the same argument for (28), we get

$\displaystyle n^{2}-|Y_{0}|-2n+1-\theta_{1}$
$\displaystyle\leq\frac{M_{1}-1}{2}(2|Y_{0}|-\ell_{1}-2)$ (42)
$\displaystyle\leq(|Y_{0}|-\ell_{1}/2)(|Y_{0}|-\ell_{1}/2-1)$
$\displaystyle\leq|Y_{0}|(|Y_{0}|-1).$

Taking account of $\theta_{1}\leq 2n$, we get $|Y_{0}|=n-1,n$ or $n+1$ under
the assumption (31). Note that $\ell_{1}=|T_{0}\cap T_{1}^{(3)}|$ must be
even, because $T_{0}$ and $T_{1}$ are inverse-closed and $e\notin
T_{1}^{(3)}$. Taking (42) and $|Y_{0}|\leq\sum_{i\geq 0}|Y_{3i}|=n+1-\ell_{1}$
into account, we get four possible cases

  1. (I)

$|Y_{0}|=n+1$ and $\ell_{1}=0$;

  2. (II)

$|Y_{0}|=n-1$ and $\ell_{1}=0$ or $2$;

  3. (III)

$|Y_{0}|=n$, $\ell_{1}=0$;

  4. (IV)

$|Y_{0}|=n-2$, $\ell_{1}=0$.

To prove (41), we only have to exclude Case (IV).

Case (IV). By (35) and Claim 1,

$(M_{1}-2)2\leq 2|Y_{0}|-\ell_{1}-2,$

which implies

$M_{1}\leq|Y_{0}|+1-\frac{\ell_{0}}{2}=n-2+1=n-1.$

On the other hand, by (42),

$M_{1}\geq 1+\frac{n^{2}-(n-2)-2n+1-\theta_{1}}{(n-2)-1}\geq n-1.$

Therefore $M_{1}$ must be $n-1$. Plugging $\ell_{1}=0$ into the last equation
of (33), we get

(43) $\sum_{i\geq 0}|Y_{3i}|=n+1.$

By (35) and $M_{1}=n-1\equiv 2\mod 3$,

$2n-6=2|Y_{0}|-\ell_{1}-2=\sum_{i\geq 1}3i|Y_{3i+1}|+\sum_{i\geq 1,3i+2\neq
M_{1}}3i|Y_{3i+2}|+2(n-1-2)+\sum_{i\geq 1}(3i-2)|Y_{3i}|.$

As a consequence, $|Y_{3}|=|Y_{4}|=\cdots=|Y_{M_{1}-1}|=0$. However, it means
$\sum_{i\geq 0}|Y_{3i}|=|Y_{0}|=n-2$ which contradicts (43).

Therefore, we have proved (41).

Determination of $Y_{0}$. We basically follow the argument used in the proof
of Theorem 3.6. Our goal is to prove

(44) $Y_{0}=\begin{cases}T_{0},&\text{ if }|Y_{0}|=n+1;\\\
T_{0}\setminus\\{e\\},&\text{ if }|Y_{0}|=n;\\\
T_{0}\setminus\\{\gamma,\gamma^{-1}\\},&\text{ if }|Y_{0}|=n-1,\end{cases}$

for some $\gamma\in T_{0}$.

Recall

(16) $\hat{T}^{(2)}T_{1}=(n-1)H+T_{0}-T_{1}^{3}+2nT_{1}.$

First let us consider the case with $\ell_{1}=0$. For an arbitrary element
$t\in T_{0}$, as $\ell_{1}=|T_{0}\cap T_{1}^{(3)}|=0$, $t$ can only be
represented as

$abc\text{ for some pairwise distinct }a,b,c\in T_{1},\quad\text{or}\quad
aab\text{ for some distinct }a,b\in T_{1},$

in $T_{1}^{3}$.

Taking account of the ordering of the elements in the representations, the
coefficient of $t$ in $T_{1}^{3}\in\mathbb{Z}[H]$ is divisible by $3$.
Together with $n\equiv 0\pmod{3}$ and $T_{1}\cap T_{0}=\emptyset$, the
coefficient of $t\in T_{0}$ in the right-hand side of (16) must be congruent
to $0$ modulo $3$. Since $\sum_{i\geq 0}|Y_{3i}|=n+1$, $|T_{0}|=n+1$ and
$Y_{0}$ is inverse-closed, we obtain (44).

When $\ell_{1}=2$ which happens only if $|Y_{0}|=n-1$ by (41), there exist
exactly two elements $\gamma,\gamma^{-1}\in T_{0}\cap T_{1}^{(3)}$. The
coefficients of $\gamma$ or $\gamma^{-1}$ in the right-hand side of (16) must
be congruent to $1$ modulo $3$, and the coefficient of any element in
$T_{0}\setminus\\{\gamma,\gamma^{-1}\\}$ is still congruent to $0$ modulo $3$.
Thus $Y_{0}=T_{0}\setminus\\{\gamma,\gamma^{-1}\\}$ which corresponds to the
last case in (44).

The final part of our proof is to show that (44) cannot hold which concludes
the proof of Theorem 3.7.

If $T_{0}=Y_{0}$, then (16) tells us

$(T_{0}^{(2)}+T_{1}^{(2)})T_{1}\cap T_{0}=\emptyset,$

which cannot hold, because $T_{0}T_{1}=H-e$ means for any non-identity $a\in
T_{0}^{(2)}+T_{1}^{(2)}$ there always exist $t_{0}\in T_{0}$ and $t_{1}\in
T_{1}$ such that $t_{0}t_{1}=a$.

If $T_{0}\setminus\\{e\\}=Y_{0}$, then $t\notin(T_{0}^{(2)}+T_{1}^{(2)})T_{1}$
for any $t\in T_{0}\setminus\\{e\\}$. By (3),

$Y_{0}T_{1}=(T_{0}-e)T_{1}=H-e-T_{1}.$

As $Y_{0}\cap(T_{0}^{(2)}+T_{1}^{(2)})T_{1}=\emptyset$ implies
$Y_{0}T_{1}\cap(T_{0}^{(2)}+T_{1}^{(2)})=\emptyset$,

$T_{0}^{(2)}+T_{1}^{(2)}\subseteq e+T_{1},$

which is impossible by the sizes of them.

If $T_{0}\setminus\\{\gamma,\gamma^{-1}\\}=Y_{0}$ for some $\gamma\neq e$,
then $t\notin(T_{0}^{(2)}+T_{1}^{(2)})T_{1}$ for any $t\in
T_{0}\setminus\\{\gamma,\gamma^{-1}\\}$. By (3),

$Y_{0}T_{1}=(T_{0}-\gamma-\gamma^{-1})T_{1}=H-e-(\gamma
T_{1}+\gamma^{-1}T_{1}).$

As $Y_{0}\cap(T_{0}^{(2)}+T_{1}^{(2)})T_{1}=\emptyset$ implies
$Y_{0}T_{1}\cap(T_{0}^{(2)}+T_{1}^{(2)})=\emptyset$, the above equation and
the sizes of $T_{0}$ and $T_{1}$ lead to

(45) $\gamma T_{1}+\gamma^{-1}T_{1}+e=T_{0}^{(2)}+T_{1}^{(2)}.$

Since $\gamma^{2}\in T_{0}^{(2)}$,

$\gamma^{2}=\gamma\alpha\quad\text{or}\quad\gamma^{2}=\gamma^{-1}\alpha$

for some $\alpha\in T_{1}$. The first case is impossible because $T_{0}\cap
T_{1}=\emptyset$. The second case implies that $\alpha=\gamma^{3}\in T_{1}\cap
T_{0}^{(3)}$. By (45), $\gamma\alpha=\gamma^{4}$ belongs to $T_{0}^{(2)}$ or
$T_{1}^{(2)}$.

If $\gamma^{4}\in T_{1}^{(2)}$, then $\gamma^{2}\in T_{1}$ which means
$\gamma^{2}\in T_{1}\cap T_{0}^{(2)}$. It follows that the coefficient of $e$
in $\hat{T}^{(2)}T_{1}$ is larger than $0$. However, $e\in
Y_{0}=T_{0}\setminus\\{\gamma,\gamma^{-1}\\}$. This is a contradiction.

If $\gamma^{4}\in T_{0}^{(2)}$, then $\gamma^{2}\in T_{0}\cap T_{0}^{(2)}$. It
contradicts Lemma 2.9 (c).

Therefore, we have finished the proof that there exist no $T_{0}$ and $T_{1}$
satisfying (3) and (4). ∎

The main result, i.e. Theorem 1.1 is a simple combination of Theorems 3.5, 3.6
and 3.7.

##  4\. Concluding remarks

In this paper, we investigate the classification of linear Lee codes of
packing radius $2$ and packing density $\frac{|S(n,2)|}{|S(n,2)|+1}$ in
$\mathbb{Z}^{n}$. We call linear Lee codes with this density almost perfect.
There are still several cases remaining open.

######  Problem 4.1.

For $n\equiv 1,2,5\pmod{6}$, show that there is an almost perfect linear Lee
code of packing radius $2$ if and only if $n=2$.

If this problem is solved, then we get the complete nonexistence proof of APLL
codes of packing radius $2$, for $n\geq 3$. As a direct consequence, Example
2.4 provides all possible APLL codes of packing radius $2$ up to isometry.

To solve Problem 4.1, one may try to follow the approach in [14] to
investigate the coefficients of the elements in $\hat{T}^{(4)}T_{0}$ and
$\hat{T}^{(4)}T_{1}$ modulo $5$. In a very recent work [26], this approach has
been applied and Problem 4.1 has been solved except for only a finite number
of $n$.

One may naturally expect a similar classification result provided that the
packing density equals $\frac{|S(n,2)|}{|S(n,2)|+\epsilon}$, where $\epsilon$
is a small integer compared with $n$. However, it seems not an easy task,
because the order of the group $G$ equals
$|S(n,2)|+\epsilon=2n^{2}+2n+1+\epsilon$ which is determined by $\epsilon$ and
some key steps of the proof should rely on the structure of $G$. One may
compare the proof for $\epsilon=0$ and for $\epsilon=1$ with $n\equiv
0\pmod{3}$ in [14, Proposition 3.3] and Theorems 3.6, 3.7. The first one
consists of only a half-page proof and the second one is much more
complicated. Therefore, a small change of the order of $G$ could lead to a big
change of the proof.

Another interesting question is about quasi-perfect linear Lee codes. As we
have mentioned in the introduction, the packing density of the 2-quasi-perfect
Lee codes constructed in [4, 15] tends to $\frac{1}{2}$ when
$n\rightarrow\infty$. However, to the best of our knowledge, there is no
construction with packing density tending to $1$ when $n\rightarrow\infty$.
Therefore, we propose the following open problem.

######  Problem 4.2.

Find better upper and lower bounds of the packing density of $2$-quasi-perfect
linear Lee codes in $\mathbb{Z}^{n}$ for infinitely many $n$.

## Acknowledgment

The authors express their gratitude to the anonymous reviewers for their
detailed and constructive comments which are very helpful to the improvement
of the presentation of this paper. Xiaodong Xu is partially supported by the
Natural Science Foundation of China (No. 71471176). Yue Zhou is partially
supported by the Natural Science Foundation of Hunan Province (No. 2019RS2031)
and the Training Program for Excellent Young Innovators of Changsha (No.
kq2106006).

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