# On characterizing nilpotent Lie algebra by their multiplier, $s(L)=6,7$

afsaneh shamsaki  School of Mathematics and Computer Science
Damghan University, Damghan, Iran<EMAIL_ADDRESS>and  Peyman
Niroomand<EMAIL_ADDRESS><EMAIL_ADDRESS>School of Mathematics
and Computer Science
Damghan University, Damghan, Iran

###### Abstract.

Let $L$ be an $n$-dimensional non-abelian nilpotent Lie algebra and
$s(L)=\frac{1}{2}(n-1)(n-2)+1-\dim\mathcal{M}(L)$ where $\mathcal{M}(L)$ is
the Schur multiplier of a Lie algebra $L.$ The structures of nilpotent Lie
algebras $L$ when $s(L)\in\\{0,1,2,3,4,5\\}$ are determined. In this paper, we
classify all non-abelian nilpotent Lie algebras $L$ when $s(L)=6,7.$

###### Key words and phrases:

Schur multiplier; nilpotent Lie algebra, capable Lie algebra, triple tensor
product

###### 1991 Mathematics Subject Classification:

17B30, 17B05, 17B99

Corresponding Author: Peyman Niroomand

##  1\.  Introduction and preliminaries

Let $L$ be a Lie algebra on a fixed field $\mathbb{F}$ and $L\cong F/R$ for a
free Lie algebra. Then the Schur multiplier $\mathcal{M}(L)$ of $L$ is
isomorphic to $R\cap F^{2}/[R,F]$. Classification of finite dimensional
nilpotent Lie algebras according to the dimension of its Schur multiplier has
been a line of investigation by many authors in [1, 2, 10]. One of the main
themes of the Lie theoretical part of research on the Schur multipliers of Lie
algebras has been to determine all nilpotent Lie algebras according to their
Schur multipliers. It is know that from [10, Lemma 2.2] that for a nilpotent
Lie algebra $L$, we have $\dim\mathcal{M}(L)=1/2n(n-1)-t(L)$ for an integer
$t(L)\geq 0$. All finite dimensional nilpotent Lie algebras $L$ with
$t(L)=0,1,\dots,7,8$ have been classified by several papers in [1, 5, 6]. The
second author, as a joint paper, showed that the dimension of Schur multiplier
of a non-abelian nilpotent Lie algebra is equal to $1/2(n-1)(n-2)+1-s(L)$ for
an integer $s(L)\geq 0$. Similarly, we may classify nilpotent Lie algebras $L$
rely on $s(L)$. The classification nilpotent Lie algebras in term of
$s(L)=0,\ldots,5$ are given in [12, 14, 16, 21, 22]. This classification
simplify the problem of determining a Lie algebra in terms of $t(L)$ and
recently are used to answer the same question for an $n$-Lie super algebra and
Leibniz $n$-algebras in [17, 18]. In this paper, we are going to give a
classification of all nilpotent Lie algebras $L$ for $t(L)=6,7$.
The following theorems and lemmas are useful in the next sections.

######  Theorem 1.1.

[12, Theorem 3.1] Let $L$ be an $n$-dimensional nilpotent Lie algebra and
$\dim L^{2}=m\geq 1$. Then

$\displaystyle\dim\mathcal{M}(L)\leq\dfrac{1}{2}(n+m-2)(n-m-1)+1.$

Moreover, if $m=1$, then the equality holds if and only if $L\cong H(1)\oplus
A(n-3)$.

For a Lie algebra $X,$ let $X^{ab}$ be used to denote $X/X^{2}.$

######  Theorem 1.2.

[12, Corollary 2.3] Let $L$ be a finite dimensional Lie algebra and $K$ be a
central ideal of $L$. Then

$\displaystyle\dim\mathcal{M}(L)+\dim(L^{2}\cap
K)\leq\dim\mathcal{M}(L/K)+\dim\mathcal{M}(K)+\dim((L/K)^{ab}\otimes K).$

For a Lie algebra $L$, the concept epicenter $Z^{*}(L)$ is introduced in [19].
The importance of $Z^{*}(L)$ is due to notion of the fact that $L$ is capable
if and only if $Z^{*}(L)=0$. Another notion having relation to the capability
is the concept of the exterior center of a Lie algebra $Z^{\wedge}(L)$ which
is introduced in [13]. It is known that from [13, Lemma 3.1],
$Z^{*}(L)=Z^{\wedge}(L)$.

######  Lemma 1.3.

[21, Corollary 2.3] Let $L$ be a non-capable nilpotent Lie algebra of
dimension $n$ such that $\dim L^{2}\geq 2$. Then

$n-3<s(L).$

The following Lemma determines the structure of all nilpotent Lie algebras $L$
when $\dim L^{2}=1.$

######  Lemma 1.4.

[14, Lemma 3.3] Let $L$ be an $n$-dimensional Lie algebra with derived
subalgebra of dimension $1.$ Then $L$ is isomorphic to $H(m)\oplus A(n-2m-1)$
for some $m\geq 1.$

The Lie algebra $L$ is a central product of $A$ and $B,$ if $L=A+B,$ where $A$
and $B$ are ideals of $L$ such that $[A,B]=0$ and $A\cap B\subseteq Z(L).$ We
denote the central product of two Lie algebras $A$ and $B$ by $A\dotplus B.$
We say a Lie algebra $L$ is semidirect sum of an ideal $I$ and subalgebra $K$
if $L=I+K$ and $I\cap K=0.$ The semidirect sum of an ideal $I$ and a
subalgebra $K$ is denoted by $I\rtimes K.$

######  Lemma 1.5.

[24, Lemma 1] Let $L$ be a nilpotent Lie algebra and $H$ be a subalgebra of
$L$ such that $L^{2}=H^{2}+L^{3}.$ Then $L^{i}=H^{i}$ for all $i\geq 2.$
Moreover, $H$ is an ideal of $L.$

Let $L$ be a Lie algebra we assume that the reader is familiar with the basic
definitions and properties of the tensor square $L\otimes L$ in [3]. By a
terminology of [3], let $L\square L$ be the submodule of tensor square
$L\otimes L$ generated by the elements $l\otimes l$ for all $l\in L$. One can
check that $L\square L$ lies in the center of $L\otimes L$ and the exterior
square $L\wedge L$ of $L$ is defined as $L\otimes L/L\square L$.
By looking at the classification of all nilpotent Lie algebras in [4, 10], the
nilpotent Lie algebras with derived subalgebra of dimension $m$ such that
$2\leq m\leq 4$ are listed in the following tables.

Table 1. : nilpotent Lie algebras of dimension at most $6$ with $\dim L^{2}=2$ Name | Nonzero multiplication
---|---
$L_{4,3}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4}$
$L_{5,3}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4}$
$L_{5,5}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{5},[x_{2},x_{4}]=x_{5}$
$L_{5,8}$ | $[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5}$
$L_{6,3}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4}$
$L_{6,5}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{5},[x_{2},x_{4}]=x_{5}$
$L_{6,8}$ | $[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5}$
$L_{6,10}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{6},[x_{4},x_{5}]=x_{6}$
$L_{6,22}(\varepsilon)$ | $[x_{1},x_{2}]=x_{5},[x_{1},x_{3}]=x_{6},[x_{2},x_{4}]=\varepsilon x_{6},[x_{3},x_{4}]=x_{5}$
Table 2. : $7$-dimensional nilpotent Lie algebras with $\dim L^{2}=2$ Name | Nonzero multiplication
---|---
$L_{6,3}\oplus A(1)$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4}$
$L_{6,5}\oplus A(1)$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{5},[x_{2},x_{4}]=x_{5}$
$L_{6,8}\oplus A(1)$ | $[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5}$
$L_{6,22}(\varepsilon)\oplus A(1)$ | $[x_{1},x_{2}]=x_{5},[x_{1},x_{3}]=x_{6},[x_{2},x_{4}]=\varepsilon x_{6},[x_{3},x_{4}]=x_{5}$
$L_{6,10}\oplus A(1)$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{6},[x_{4},x_{5}]=x_{6}$
$27A$ | $[x_{1},x_{2}]=x_{6},[x_{1},x_{4}]=x_{7},[x_{3},x_{5}]=x_{7}$
$27B$ | $[x_{1},x_{2}]=[x_{3},x_{4}]=x_{6},[x_{1},x_{5}]=[x_{2},x_{3}]=x_{7}$
$157$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{7},[x_{2},x_{4}]=x_{7},[x_{5},x_{6}]=x_{7}$
Table 3. : nilpotent Lie algebras of dimension at most $6$ with $\dim L^{2}=3$ Name | Nonzero multiplication
---|---
$L_{5,6}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},[x_{1},x_{4}]=[x_{2},x_{3}]=x_{5}$
$L_{5,7}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},[x_{1},x_{4}]=x_{5}$
$L_{5,9}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},[x_{2},x_{3}]=x_{5}$
$L_{6,6}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},[x_{1},x_{4}]=[x_{2},x_{3}]=x_{5}$
$L_{6,7}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},[x_{1},x_{4}]=x_{5}$
$L_{6,9}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},[x_{2},x_{3}]=x_{5}$
$L_{6,11}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},$
| $[x_{1},x_{4}]=[x_{2},x_{3}]=[x_{2},x_{5}]=x_{6}$
$L_{6,12}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},[x_{1},x_{4}]=[x_{2},x_{5}]=x_{6}$
$L_{6,13}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{2},x_{4}]=x_{5},$
| $[x_{1},x_{5}]=[x_{3},x_{4}]=x_{6}$
$L_{6,19}(\epsilon)$ | $[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5},[x_{1},x_{5}]=[x_{2},x_{4}]=x_{6},$
| $[x_{3},x_{5}]=\epsilon x_{6}$
$L_{6,20}$ | $[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5},[x_{1},x_{5}]=[x_{2},x_{4}]=x_{6}$
$L_{6,23}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{2},x_{4}]=x_{5},[x_{1},x_{4}]=x_{6}$
$L_{6,24}(\epsilon)$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{2},x_{4}]=x_{5},$
| $[x_{1},x_{4}]=\varepsilon x_{6},[x_{2},x_{3}]=x_{6}$
$L_{6,25}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{5},[x_{1},x_{4}]=x_{6}$
$L_{6,26}$ | $[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5},[x_{2},x_{3}]=x_{6}$
Table 4. : $7$-dimensional nilpotent Lie algebras with $\dim L^{2}=3$ Name | Nonzero multiplication
---|---
$37A$ | $[x_{1},x_{2}]=x_{5},[x_{2},x_{3}]=x_{6},[x_{2},x_{4}]=x_{7}$
$37B$ | $[x_{1},x_{2}]=x_{5},[x_{2},x_{3}]=x_{6},[x_{3},x_{4}]=x_{7}$
$37C$ | $[x_{1},x_{2}]=[x_{3},x_{4}]=x_{5},[x_{2},x_{3}]=x_{6},[x_{2},x_{4}]=x_{7}$
$37D$ | $[x_{1},x_{2}]=[x_{3},x_{4}]=x_{5},[x_{1},x_{3}]=x_{6},[x_{2},x_{4}]=x_{7}$
$257A$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{2},x_{4}]=x_{6},[x_{1},x_{5}]=x_{7}$
$257B$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{6},[x_{1},x_{4}]=[x_{2},x_{5}]=x_{7}$
$257C$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{2},x_{4}]=x_{6},[x_{2},x_{5}]=x_{7}$
$257D$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{2},x_{4}]=x_{6},$
| $[x_{1},x_{4}]=[x_{2},x_{5}]=x_{7}$
$257E$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{4},x_{5}]=x_{6},[x_{2},x_{4}]=x_{7}$
$257F$ | $[x_{1},x_{2}]=x_{3},[x_{2},x_{3}]=[x_{4},x_{5}]=x_{6},[x_{2},x_{4}]=x_{7}$
$257G$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{4},x_{5}]=x_{6},$
| $[x_{1},x_{5}]=[x_{2},x_{4}]=x_{7}$
$257H$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{2},x_{4}]=x_{6},[x_{4},x_{5}]=x_{7}$
$257I$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{1},x_{4}]=x_{6},$
| $[x_{1},x_{5}]=[x_{2},x_{3}]=x_{7}$
$257J$ |  $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{2},x_{4}]=x_{6},$ $[x_{1},x_{5}]=[x_{2},x_{3}]=x_{7}$
$257K$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{6},[x_{2},x_{3}]=[x_{4},x_{5}]=x_{7}$
$257L$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=[x_{2},x_{4}]=x_{6},$
| $[x_{2},x_{3}]=[x_{4},x_{5}]=x_{7}$
$147A$ | $[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5},$
| $[x_{1},x_{6}]=[x_{2},x_{5}]=[x_{3},x_{4}]=x_{7}$
$147B$ | $[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5},$
| $[x_{1},x_{4}]=[x_{2},x_{6}]=[x_{3},x_{5}]=x_{7}$
$1457A$ | $[x_{1},x_{i}]=x_{i+1}~{}~{}i=2,3,~{}~{}[x_{1},x_{4}]=[x_{5},x_{6}]=x_{7}$
$1457B$ | $[x_{1},x_{i}]=x_{i+1}~{}~{}i=2,3,$
| $[x_{1},x_{4}]=[x_{2},x_{3}]=[x_{5},x_{6}]=x_{7}$
$137A$ | $[x_{1},x_{2}]=x_{5},[x_{1},x_{5}]=[x_{3},x_{6}]=x_{7},[x_{3},x_{4}]=x_{6}$
$137B$ | $[x_{1},x_{2}]=x_{5},[x_{3},x_{4}]=x_{6},$
| $[x_{1},x_{5}]=[x_{2},x_{4}]=[x_{3},x_{6}]=x_{7}$
$137C$ | $[x_{1},x_{2}]=x_{5},[x_{1},x_{4}]=[x_{2},x_{3}]=x_{6},$
| $[x_{1},x_{6}]=x_{7},[x_{3},x_{5}]=-x_{7}$
$137D$ | $[x_{1},x_{2}]=x_{5},[x_{1},x_{4}]=[x_{2},x_{3}]=x_{6},$
| $[x_{1},x_{6}]=[x_{2},x_{4}]=x_{7},[x_{3},x_{5}]=-x_{7}$
$1357A$ | $[x_{1},x_{2}]=x_{4},[x_{1},x_{4}]=[x_{2},x_{3}]=x_{5},$
| $[x_{1},x_{5}]=[x_{2},x_{6}]=x_{7},[x_{3},x_{4}]=-x_{7}$
$1357B$ | $[x_{1},x_{2}]=x_{4},[x_{1},x_{4}]=[x_{2},x_{3}]=x_{5},$
| $[x_{1},x_{5}]=[x_{3},x_{6}]=x_{7},[x_{3},x_{4}]=-x_{7}$
$1357C$ | $[x_{1},x_{2}]=x_{4},[x_{1},x_{4}]=[x_{2},x_{3}]=x_{5},$
| $[x_{1},x_{5}]=[x_{2},x_{4}]=x_{7},[x_{3},x_{4}]=-x_{7}$
Table 5. : $7$-dimensional decomposable nilpotent Lie algebras with $\dim L^{2}=3$ Name | Name
---|---
| $L_{4,3}\oplus H(1)$ | $L_{6,19}(\epsilon)\oplus A(1)$
| $L_{5,6}\oplus A(2)$ | $L_{6,20}\oplus A(1)$
| $L_{5,7}\oplus A(2)$ | $L_{6,23}\oplus A(1)$
| $L_{5,9}\oplus A(2)$ | $L_{6,24}(\epsilon)\oplus A(1)$
| $L_{6,11}\oplus A(1)$ | $L_{6,25}\oplus A(1)$
| $L_{6,12}\oplus A(1)$ | $L_{6,26}\oplus A(1)$
| $L_{6,13}\oplus A(1)$ |
Table 6. : nilpotent Lie algebras of dimension at most $6$ with $\dim L^{2}=4$ Name | Nonzero multiplication
---|---
$L_{6,14}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},[x_{1},x_{4}]=x_{5},$
| $[x_{2},x_{3}]=x_{5},[x_{2},x_{5}]=x_{6},[x_{3},x_{4}]=-x_{6},$
$L_{6,15}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},[x_{1},x_{4}]=x_{5},$
| $[x_{2},x_{5}]=x_{6},[x_{3},x_{4}]=-x_{6},$
$L_{6,16}$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},[x_{1},x_{4}]=x_{5}$
|  $[x_{2},x_{5}]=x_{6},$ $[x_{3},x_{4}]=-x_{6},$
$L_{6,17}$ |  $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},$ $[x_{1},x_{4}]=x_{5},$
|  $[x_{1},x_{3}]=x_{4},[x_{1},x_{5}]=x_{6},$ $[x_{2},x_{3}]=x_{6},$
$L_{6,18}$ |  $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},$ $[x_{1},x_{4}]=x_{5},$
| $[x_{1},x_{5}]=x_{6},$
$L_{6,21}(\varepsilon)$ | $[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{4},[x_{2},x_{3}]=x_{5}$
|  $[x_{1},x_{4}]=x_{6},$ $[x_{1},x_{3}]=x_{4},$ $[x_{2},x_{5}]=\varepsilon
x_{6}.$

By using [12, Theorem 3.1], [14, Theorems 3.9 and 4.5], [16, Main Theorem],
[21, Main Theorem] and [22, Theorem 2.3] the structures of all nilpotent Lie
algebras $L$ with $s(L)\in\\{0,1,2,3,4,5\\}$ are summarized as following.

######  Theorem 1.6.

Let $L$ be an $n$-dimensional nilpotent Lie algebra.

  * (i).

If $s(L)=0,$ then $L$ is isomorphic to $H(1)\oplus A(n-3).$

  * (ii).

If $s(L)=1,$ then $L$ is isomorphic to $L_{5,8}.$

  * (iii).

If $s(L)=2,$ then $L$ is isomorphic to one of the Lie algebras $L_{5,8}\oplus
A(1),$ $L_{4,3}$ or $H(m)\oplus A(n-2m-1)$ for all $m\geq 2.$

  * (iv).

If $s(L)=3,$ then $L$ is isomorphic to one of the Lie algebras $L_{5,8}\oplus
A(2),$ $L_{4,3}\oplus A(1),$ $L_{5,5},$ $L_{6,22}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}})$ or $L_{6,26}.$

  * (v).

If $s(L)=4,$ then $L$ is isomorphic to one of the Lie algebras $L_{5,8}\oplus
A(3),$ $L_{4,3}\oplus A(2),$ $L_{5,5}\oplus A(1),$
$L_{6,22}(\varepsilon)\oplus A(1)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}),$ $L_{5,6},$
$L_{5,7},$ $L_{5,9}$ or $37A.$

  * (vi).

If $s(L)=5,$ then $L$ is isomorphic to one of the Lie algebras $L_{5,8}\oplus
A(4),$ $L_{4,3}\oplus A(3),$ $L_{5,5}\oplus A(2),$
$L_{6,22}(\varepsilon)\oplus A(2)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}),$
$L_{6,26}\oplus A(1),$ $L_{6,10},$ $L_{6,23},$ $L_{6,25},$ $37B,$ $37C$ or
$37D.$

The following lemma and propositions are used in the next sections.

######  Lemma 1.7.

Let $L$ be a nilpotent Lie algebra of dimension at most $6$ with $\dim
L^{2}=3.$ Then $s(L),$ the structure and the Schur multiplier of $L$ are given
in the following table.

Table 7.  Name | $\dim\mathcal{M}(L)$ | $s(L)$
---|---|---
$L_{5,6},$ $L_{5,7},$ $L_{5,9}$ | $3$ | $4$
$L_{6,6},$ $L_{6,7},$ $L_{6,9},$ $L_{6,11},$ $L_{6,12},$ $L_{6,19}(\varepsilon),$ $L_{6,20},$ $L_{6,24}(\varepsilon)$ | $5$ | $6$
$L_{6,13}$ | $4$ | $7$
$L_{6,23},$ $L_{6,25}$ | $6$ | $5$
$L_{6,26}$ | $8$ | $3$

###### Proof.

The proof is obtained by a modification of the proof [21, Lemma 2.5]. ∎

######  Proposition 1.8.

Let $L$ be an $7$-dimensional nilpotent Lie algebra with $\dim L^{2}=3.$ Then
the structure, the Schur multiplier of $L$ and $s(L)$ are presented in the
following table.

Table 8.  Name | $\dim\mathcal{M}(L)$ | $s(L)$
---|---|---
$(37A)$ | $12$ | $4$
$(37B),$ $(37C),$ $(37D)$ | $11$ | $5$
$(257A),$ $257C,$ $257F$ | $9$ | $7$
$(257B),$ $257D,$ $257E,$ $(257G),$ $(257H),$ $(257I),$ $(257J)$ | $8$ | $8$
$147A,$ $147B,$ $L_{4,3}\oplus H(1),$ $L_{5,6}\oplus A(2),$ $L_{5,7}\oplus A(2),$ $L_{5,9}\oplus A(2)$ | $8$ | $8$
$L_{6,11}\oplus A(1),$ $L_{6,12}\oplus A(1),$ $L_{6,19}(\varepsilon)\oplus A(1),$ $L_{6,20}\oplus A(1)$ | $8$ | $8$
$L_{6,24}(\varepsilon)\oplus A(1),$ $(257K),$ $(257L),$ | $8$ | $8$
$1457A,$ $1457B,$ $(1357B),$ $(1357C)$ | $6$ | $10$
$(137A),$ $(137B),$ $(137C),$ $(137D),$ $(1357A),$ $L_{6,13}\oplus A(1)$ | $7$ | $9$
$L_{6,23}\oplus A(1),$ $L_{6,25}\oplus A(1)$ | $9$ | $7$
$L_{6,26}\oplus A(1)$ | $11$ | $5$

###### Proof.

The proof is obtained similar to the proof of [21, Lemma 2.5]. ∎

######  Proposition 1.9.

Let $L$ be a nilpotent Lie algebra of dimension at most $6$ with $\dim
L^{2}=4.$ Then the structure, the Schur multiplier of $L$ and $s(L)$ are
presented in the following table.

Table 9.  Name | $\dim\mathcal{M}(L)$ | $s(L)$
---|---|---
$L_{6,14},$ $L_{6,16}$ | $2$ | $9$
$L_{6,15},$ $L_{6,17},$ $L_{6,18}$ | $3$ | $8$
$L_{6,21}(\varepsilon)$ | $4$ | $7$

###### Proof.

The proof is obtained by a similar argument as in the proof of [21, Lemma
2.5]. ∎

##  2\. the structures of a nilpotent Lie algebra $L$ with $s(L)=6$

In this section, we determine the structures of a all nilpotent Lie algebras
$L$ when $s(L)=6.$ First, we begin by the following propositions which are
needed in the proof of the next theorems.

######  Proposition 2.1.

There is no $n$-dimensional nilpotent Lie algebra with $s(L)=6,$ when

  * (i).

$\dim L^{2}\geq 5,$

  * (ii).

$\dim L^{2}=1.$

###### Proof.

The proof is obtained in the same way as the proof [21, Proposition 2.4]. ∎

######  Proposition 2.2.

Let $L$ be a non-capable nilpotent Lie algebra of dimension at most $7$ with
$\dim L^{2}=2.$ Then $s(L),$ the structure and the Schur multiplier of $L$ are
given in the following table.

Table 10.  Name | $\dim\mathcal{M}(L)$ | $s(L)$
---|---|---
$L_{6,10}$ | $6$ | $5$
$27A,$ $L_{6,10}\oplus A(1),$ $157$ | $10$ | $6$

###### Proof.

By using [20, Theorem 1.1], $L_{6,10}$ is only non-capable Lie algebra with
derived subalgebra of dimension $2$ such that $n\leq 6.$ From [20, Proposition
3.1], we have $\dim\mathcal{M}(L_{6,10})=6$ and so $s(L_{6,10})=5.$ Also, by
using Table LABEL:ta2, we have $L_{6,3}\oplus A(1),$ $L_{6,5}\oplus A(1),$
$L_{6,8}\oplus A(1),$ $L_{6,22}(\varepsilon)\oplus A(1)$ and $L_{6,10}\oplus
A(1)$ are decomposable $7$-dimensional nilpotent Lie algebra and only
$L_{6,10}\oplus A(1)$ is non-capable by using [7, Proposition 3.1]. From [1,
Theorem 1], $\dim\mathcal{M}(L_{6,10}\oplus A(1))=10$ and so $s(L_{6,10}\oplus
A(1))=6.$
From Table LABEL:ta2, we know that $27A,$ $27B$ and $157$ are $7$-dimensional
Lie algebra with derived subalgebra of dimension $2.$ On the other hand, [11,
Proposition 2.10 and Lemma 2.11] show $27A$ and $27B$ are non-capable and
capable, respectively and $\dim\mathcal{M}(27A)=10$ and
$\dim\mathcal{M}(27B)=9,$ so $s(27A)=6$ and $s(27B)=7.$ By a similar way and
using [11, Proposition 2.10 and Lemma 2.11], the Lie algebra $157$ is non-
capable and $\dim\mathcal{M}(157)=10,$ so $s(157)=6.$ Therefore $L$ is
isomorphic to one of the Lie algebras $L_{6,10},$ $L_{6,10}\oplus A(1),$ $27A$
or $157.$ ∎

######  Theorem 2.3.

Let $L$ be an $n$-dimensional nilpotent Lie algebra with $s(L)=6$ and $\dim
L^{2}=2.$ Then $L$ is isomorphic to one of the nilpotent Lie algebras
$L_{5,8}\oplus A(5),L_{4,3}\oplus A(4),$ $L_{5,5}\oplus A(3)$,
$L_{6,22}(\varepsilon)\oplus A(3)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}),$
$L_{6.10}\oplus A(1),$ $27A$ or $157.$

###### Proof.

If $L$ is a capable Lie algebra, then the proof is similar to [21, Theorem
2.6] and $L$ is isomorphic to one of the nilpotent Lie algebras $L_{5,8}\oplus
A(5),L_{4,3}\oplus A(4)$, $L_{5,5}\oplus A(3)$ or $L_{6,22}(\varepsilon)\oplus
A(3)$ for all $\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}).$
If $L$ is a non-capable Lie algebra, then $n\leq 8$ by using Lemma 1.3. Let
$n\leq 7.$ Then $L$ is isomorphic to one of the non-capable Lie algebras
$27A,$ $L_{6,10}\oplus A(1)$ or $157$ by using Proposition 2.2. Let now $n=8.$
Since $Z^{\wedge}(L)\subseteq L_{2}$ and $n=8,$ $L/Z^{\wedge}(L)$ is
isomorphic to $A(6)$ or $H(1)\oplus A(4).$ On the other hand, $L\wedge L\cong
L/Z^{\wedge}(L)\wedge L/Z^{\wedge}(L)$ by using [13, Corollary 2.3]. Thus
$\dim L\wedge L$ is equal to $15$ or $17$ and so $\mathcal{M}(L)$ is
isomorphic to $A(13)$ or $A(15)$ by using [3, Theorem 35 (iii)]. Hence
$s(L)=9$ or $s(L)=7,$ respectively. Therefore there is no $8$-dimensional non-
capable nilpotent Lie algebra with $\dim L^{2}=2$ and $s(L)=6.$ ∎

Recall that a Lie algebra $L$ is called generalized Heisenberg of rank $n$ if
$L^{2}=Z(L)$ and $\dim L^{2}=n.$

######  Lemma 2.4.

Let $L$ be an $n$-dimensional generalized Heisenberg of rank $3$ with
$s(L)=6.$ Then $n\leq 7.$

###### Proof.

The proof is obtained by a modification of the proof of [22, Lemma 2.8]. ∎

######  Theorem 2.5.

Let $L$ be an $n$-dimensional nilpotent Lie algebra with $s(L)=6$ and $\dim
L^{2}=3.$ Then $L$ is isomorphic to one of the nilpotent Lie algebras
$37A\oplus A(1),$ $L_{6,6},$ $L_{6,7},$ $L_{6,9},$ $L_{6,11},$ $L_{6,12},$
$L_{6,19}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F^{*}}/(\stackrel{{\scriptstyle*}}{{\sim}}),$
$L_{6,20}$ or $L_{6,24}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}).$

###### Proof.

If $Z(L)\not\subseteq L^{2},$ then there is a central one dimensional ideal
$I$ of $L$ such that $I\cap L^{2}=0.$ Since $\dim(L/I)^{2}=3,$ by using
Theorem 1.1, we have

$\displaystyle\dim\mathcal{M}(L/I)\leq\frac{1}{2}n(n-5)+1.$

If the equality holds, then

$\frac{1}{2}(n-2)(n-3)+1-s(L/I)=\dim\mathcal{M}(L/I)=\frac{1}{2}n(n-5)+1$

and so $s(L/I)=3.$ By using Table LABEL:ta6, $L/I\cong L_{6,26}.$ On the other
hand, $Z(L)\not\subseteq L^{2}$ and $I\not\subset L^{2}$ thus $L\cong
L_{6,26}\oplus A(1).$ Hence $s(L_{6,26}\oplus A(1))=5$ by using Table
LABEL:ta7. It is a contradiction. Thus

$\displaystyle\dim\mathcal{M}(L/I)\leq\frac{1}{2}n(n-5).$

If the equality holds, then

$\displaystyle\frac{1}{2}(n-2)(n-3)+1-s(L/I)=\dim\mathcal{M}(L/I)=\frac{1}{2}n(n-5)$

and $s(L/I)=4.$ Then $L/I$ is isomorphic to one of the Lie algebras $L_{5,6},$
$L_{5,7},$ $L_{5,9}$ or $37A.$ Since $Z(L)\not\subseteq L^{2}$ and
$I\not\subset L^{2},$ we have $L$ is isomorphic to one of the Lie algebras
$L_{6,6}\cong L_{5,6}\oplus A(1),$ $L_{6,7}\cong L_{5,7}\oplus A(1),$
$L_{6,9}\cong L_{5,9}\oplus A(1)$ or $37A\oplus A(1).$ By using Table
LABEL:ta6, we have $s(L_{6,6})=s(L_{6,7})=s(L_{6,9})=6$ and looking at Table
LABEL:ta7 and [1, Theorem 1] we have $s(37A\oplus A(1))=6.$
If $\dim\mathcal{M}(L/I)\leq\frac{1}{2}n(n-5)-1,$ then Theorem 1.1 implies

$\dim\mathcal{M}(L)=\frac{1}{2}(n-1)(n-2)-5\leq\frac{1}{2}n(n-5)-1+n-4=\frac{1}{2}(n^{2}-3n)-5$

which is a contradiction.
If $Z(L)\subseteq L^{2},$ then there is a one dimensional ideal $I$ of $L$
such that $I\subseteq Z(L)\cap L^{2}.$ Hence by using Theorem 1.2 we have

$\displaystyle\frac{1}{2}(n-1)(n-2)-4\leq\frac{1}{2}(n-2)(n-3)+1-s(L/I)+n-3$

and so $s(L/I)\leq 4.$ Since $\dim(L/I)^{2}=2,$ by using Theorem 1.6 $L/I$ is
isomorphic to one of the Lie algebras $L_{5,8},$ $L_{4,3},$ $L_{5,8}\oplus
A(1),$ $L_{4,3}\oplus A(1),$ $L_{5,8}\oplus A(2),$ $L_{5,5},$
$L_{6,22}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}),$
$L_{4,3}\oplus A(2),$ $L_{5,8}\oplus A(3),$ $L_{5,5}\oplus A(1)$ or
$L_{6,22}(\varepsilon)\oplus A(1)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}).$ Hence $n\leq
9.$
If $n\leq 7,$ then $L$ is isomorphic to $L_{6,6},$ $L_{6,7},$ $L_{6,9},$
$L_{6,11},$ $L_{6,12},$ $L_{6,19}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F^{*}}/(\stackrel{{\scriptstyle*}}{{\sim}}),$
$L_{6,20}$ or $L_{6,24}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}})$ by looking at
Tables LABEL:ta6, LABEL:ta7.
Let $L/I\cong L_{5,8}\oplus A(2),$ then $n=8.$ If $L$ is of nilpotency class
$2,$ then there is no such a Lie algebra by using Lemma 2.4. Let $L$ is of
nilpotency class $3$. Then $\dim\gamma_{3}(L)=1.$ On the other hand,
$\mathcal{M}(L_{5,8})=9$ by using [20, Lemma 2.10] and $\mathcal{M}(A(2))=1$
via [10, Lemma 23]. Hence $\dim\mathcal{M}(L/I)=13$ by invoking [1, Theorem
1]. Using [23, proof of Theorem 1.1], we have

(2.1) $2=n-m-c\leq\dim\ker\lambda_{3},~{}~{}\text{and}$ (2.2)
$\dim\mathcal{M}(L)=\dim\mathcal{M}(L/\gamma_{3}(L))-\dim\gamma_{3}(L)+\dim
L/\gamma_{2}(L)\otimes\gamma_{3}(L)-\dim\ker\lambda_{3},$

so

$\displaystyle 16=\dim\mathcal{M}(L)$
$\displaystyle\leq\dim\mathcal{M}(L/\gamma_{3}(L))-\dim\gamma_{3}(L)+\dim
L/\gamma_{2}(L)\otimes\gamma_{3}(L)-2$ $\displaystyle\leq 15.$

It is a contradiction.
If $L/I\cong L_{6,22}(\varepsilon)\oplus A(1),$ then $n=8$ and we get
similarly contradiction.
If $L/I\cong L_{5,8}\oplus A(3),$ then $n=9$ and we have a contradiction. The
proof is completed. ∎

From [10], a Lie algebra $L$ is called a stem Lie algebra if $Z(L)\subseteq
L^{2}.$

######  Lemma 2.6.

There is no stem $7$-dimensional Lie algebra with $s(L)=6$ or $7$ such that
$L/I\cong L_{6,26},$ $\dim I=1$ and $I\subseteq Z(L).$

###### Proof.

Let $L$ be a nilpoten Lie algebra of class two. By contrary, let there is a
stem $7$-dimensional Lie algebra $L$ of nilpotency class two such that
$L/I\cong L_{6,26}$ and $I\subseteq Z(L).$ Let $I=\langle x_{7}\rangle.$ Then

$\displaystyle[x_{1},x_{2}]=x_{4}+\alpha_{1}x_{7},$
$\displaystyle[x_{1},x_{3}]=x_{5}+\alpha_{2}x_{7},$
$\displaystyle[x_{1},x_{4}]=\alpha_{3}x_{7},$
$\displaystyle[x_{1},x_{5}]=\alpha_{4}x_{7},$
$\displaystyle[x_{1},x_{6}]=\alpha_{5}x_{7},$
$\displaystyle[x_{2},x_{3}]=x_{6}+\alpha_{6}x_{7},$
$\displaystyle[x_{2},x_{4}]=\alpha_{7}x_{7},$
$\displaystyle[x_{2},x_{5}]=\alpha_{8}x_{7},$
$\displaystyle[x_{2},x_{6}]=\alpha_{9}x_{7},$
$\displaystyle[x_{3},x_{4}]=\alpha_{10}x_{7},$
$\displaystyle[x_{3},x_{5}]=\alpha_{11}x_{7},$
$\displaystyle[x_{3},x_{6}]=\alpha_{12}x_{7},$
$\displaystyle[x_{4},x_{5}]=\alpha_{13}x_{7},$
$\displaystyle[x_{4},x_{6}]=\alpha_{14}x_{7},$
$\displaystyle[x_{5},x_{6}]=\alpha_{15}x_{7}.$

Since $L$ is of nilpotency class two, a change of variables
$x^{\prime}_{4}=x_{4}+\alpha_{1}x_{7},$ $x^{\prime}_{5}=x_{5}+\alpha_{2}x_{7}$
and $x^{\prime}_{6}=x_{6}+\alpha_{6}x_{7}$ and relabeling, we have

$L=\langle
x_{1},\dots,x_{7}\mid[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5},[x_{2},x_{3}]=x_{6}\rangle\cong
L_{6,26}\oplus A(1).$

On the other hand, $L$ is stem, thus it is a contradiction.
Let $L$ be of nilotency class $3.$ Since $L_{6,26}$ is of nilpotency class two
and $L/I\cong L_{6,26},$ we have $\dim L^{3}=1.$ Now, by looking at the
classification all $7$-dimensional nilpotent Lie algebras in [9], $L$ should
be isomorphic to one of the Lie algebras

$357A=\langle
x_{1},\dots,x_{7}\mid[x_{1},x_{2}]=x_{3},[x_{1},x_{3}]=x_{5},[x_{1},x_{4}]=x_{7},[x_{2},x_{4}]=x_{6}\rangle,$
$247N=\langle
x_{1},\dots,x_{7}\mid[x_{1},x_{i}]=x_{i+2},~{}~{}i=2,3,[x_{1},x_{5}]=x_{6},[x_{2},x_{3}]=x_{7},[x_{2},x_{4}]=x_{6}\rangle,$
$147F=\langle
x_{1},\dots,x_{7}\mid[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=-x_{6},[x_{1},x_{5}]=x_{7},[x_{1},x_{6}]=x_{7}$
$[x_{2},x_{3}]=x_{5},[x_{2},x_{4}]=x_{7},[x_{2},x_{6}]=x_{7},[x_{3},x_{4}]=x_{7}\rangle.$
$147D=\langle
x_{1},\dots,x_{7}\mid[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=-x_{6},[x_{1},x_{5}]=x_{7},[x_{1},x_{6}]=x_{7}$
$[x_{2},x_{3}]=x_{5},[x_{2},x_{6}]=x_{7},[x_{3},x_{4}]=-2x_{7}\rangle,$
$\text{or}~{}~{}147E=\langle
x_{1},\dots,x_{7}\mid[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=-x_{6},[x_{1},x_{5}]=-x_{7},[x_{2},x_{3}]=x_{5}$
$[x_{2},x_{6}]=\lambda x_{7},[x_{3},x_{4}]=(1-\lambda)x_{7}\rangle$

One parameter family, with an invariant
$I(\lambda)=\frac{(1-\lambda+\lambda^{2})^{3}}{\lambda^{2}(\lambda-1)^{2}},\lambda\neq
0,1.$ When $\lambda=0$ or $1,$ it is isomorphic to $247P.$
By using the method of Hardy and Stitzinger in [1], we can see that
$\dim\mathcal{M}(357A)=8,$ $\dim\mathcal{M}(247N)=7,$ and
$\dim\mathcal{M}(147D)=\dim\mathcal{M}(147E)=\dim\mathcal{M}(147F)=7,$ so
$s(357A)=14$ and $s(247N)=s(147D)=s(147E)=s(147F)=15.$ Therefore there is no
$7$-dimensional Lie algebra such that $L/I\cong L_{6,26},$ $\dim I=1$ and
$I\subseteq Z(L).$ ∎

######  Theorem 2.7.

There is no $n$-dimensional nilpotent Lie algebra with $s(L)=6$ and $\dim
L^{2}=4.$

###### Proof.

Let $L$ be a Lie algebra such that $Z(L)\not\subset L^{2}.$ Then there exist a
central one dimension ideal $I$ of $L$ such that $I\cap L^{2}=0.$ Since
$\dim\mathcal{M}(L)=\frac{1}{2}(n-1)(n-2)-5,$ we have

$\displaystyle\frac{1}{2}(n-1)(n-2)-5\leq\frac{1}{2}(n-2)(n-3)+1-s(L/I)+n-5$

by using Theorem 1.2. Hence $s(L/I)\leq 3.$ On the other hand, $(L/I)^{2}=4.$
Thus by using Theorem 1.6 for $L/I,$ there is no Lie algebra with $s(L/I)\leq
3$ and $(L/I)^{2}=4.$
Assume that $Z(L)\subseteq L^{2}.$ Then there exists a one dimension ideal $I$
of $L$ such that $I\subseteq Z(L)\cap L^{2}.$ By using Theorem 1.2, we have

$\displaystyle\frac{1}{2}(n-1)(n-2)-4\leq\frac{1}{2}(n-2)(n-3)+1-s(L/I)+n-4$

and so $s(L/I)\leq 3.$ Since $s(L/I)\leq 3$ and $\dim(L/I)^{2}=3,$ by using
Theorem 1.6, we have $L/I\cong L_{6,26}$ and so $n=7.$ Thus there is no such a
Lie algebra by using Lemma 2.6, as required. ∎

######  Theorem 2.8.

Let $L$ be an $n$-dimensional nilpotent Lie algebra with $s(L)=6.$ Then $L$ is
isomorphic to one of the Lie algebras $L_{5,8}\oplus A(5),L_{4,3}\oplus A(4)$,
$L_{5,5}\oplus A(3)$, $L_{6,22}(\varepsilon)\oplus A(3)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}),$
$L_{6.10}\oplus A(1),$ $27A,$ $157,$ $37A\oplus A(1),$ $L_{6,6},$ $L_{6,7},$
$L_{6,9},$ $L_{6,11},$ $L_{6,12},$ $L_{6,19}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F^{*}}/(\stackrel{{\scriptstyle*}}{{\sim}}),$
$L_{6,20}$ or $L_{6,24}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}).$

###### Proof.

The proof is obtained by using Proposition 2.1 Theorems 2.3, 2.5 and 2.7. ∎

##  3\.  the structures of a nilpotent Lie algebra $L$ with $s(L)=7$

In this section, we classify all $n$-dimensional nilpotent Lie algebras $L$
with $s(L)=7.$ We begin by the following proposition.

######  Proposition 3.1.

There is no $n$-dimensional nilpotent Lie algebra with $s(L)=7,$ when

  * (i).

$\dim L^{2}\geq 5,$

  * (ii).

$\dim L^{2}=1.$

###### Proof.

The proof is obtained by the same argument as in the proof of [21, Proposition
2.4]. ∎

######  Proposition 3.2.

Let $L$ be a non-capable $8$-dimensional Lie algebra with derived subalgebra
of dimension $2$ and $s(L)=7$ such that $L/Z^{\wedge}(L)\cong H(1)\oplus
A(4).$ Then $L$ is isomorphic to one of the nilpotent Lie algebras
$L_{6,10}\oplus A(2),$ $27A\oplus A(1),$ $157\oplus A(1),$ $L_{6,10}\dotplus
H(1),$ $H(1)\oplus H(2)$ or

$S_{1}=\langle
x_{1},\dots,x_{8}\mid[x_{1},x_{2}]=x_{6},[x_{1},x_{4}]=x_{8},[x_{3},x_{5}]=x_{8},[x_{2},x_{7}]=x_{8}\rangle.$

###### Proof.

Let $L$ be a Lie algebra and $Z(L)\not\subseteq L^{2}.$ By using [20, Theorem
1.1] and looking at the classification of all nilpotent Lie algebras in [9],
$L_{6,10},$ $27A$ and $157$ are non-capable Lie algebras with derived
subalgebra of dimension $2.$ Since $Z(L)\not\subseteq L^{2},$ $L$ is
isomorphic to one of the nilpotent Lie algebras $L_{6,10}\oplus A(2),$
$27A\oplus A(1)$ or $157\oplus A(1)$ by using [7, Proposition 3.1]. Now, by
using Tables LABEL:ta8, [10, Lemma 23] and [1, Theorem 1] we have
$\dim\mathcal{M}(L_{6,10}\oplus A(2))=\dim\mathcal{M}(27A\oplus
A(1))=\dim\mathcal{M}(157\oplus A(1))=15,$ so $s(L_{6,10}\oplus A(2))=\dim
s(27A\oplus A(1))=s(157\oplus A(1))=7.$
Let $L$ be a Lie algebra and $Z(L)\subseteq L^{2}.$ Consider the following
cases.
Case 1. If $\dim Z(L)=1,$ then $L$ is of nilpotency class $3$ and since $L$ is
non-capable, $L$ is isomorphic to $L_{6,10}\dotplus H(1)$ by using [15,
Theorem 4.6].
Case 2. Let $\dim Z(L)=2.$ Then $L^{2}=Z(L)$ and $L$ is of nilpotency class
$2.$ There exist ideals $I_{1}/Z^{\wedge}(L)$ and $I_{2}/Z^{\wedge}(L)$ of
$L/Z^{\wedge}(L)$ such that $I_{1}/Z^{\wedge}(L)\cong H(1)\oplus A(3)$ and
$I_{2}/I\cong A(1).$ Since $L$ is of nilpotency class $2,$ $I_{1}$ is of
nilpotency class $2.$ Since $L$ is a nilpotent Lie algebra, we have
$Z^{\wedge}(L)\subseteq L^{2}$ by using [15, Corollary 2.3] On the other hand,
$I_{1}\cap I_{2}=Z^{\wedge}(L)\subseteq I_{1}.$ Hence $Z^{\wedge}(L)\subseteq
I_{1}\cap L^{2}=I_{1}^{2}$ and so $Z^{\wedge}(L)\subseteq I_{1}^{2}.$ Since
$L^{2}/Z^{\wedge}(L)=I_{1}^{2}+Z^{\wedge}(L)/Z^{\wedge}(L)$ and
$Z^{\wedge}(L)\subseteq I_{1}^{2},$ we have $I_{1}^{2}=L^{2}$ and
$Z^{\wedge}(L)\subseteq Z^{\wedge}(I_{1}).$ Thus $I_{1}$ is non-capable Lie
algebra. By using the classification of all nilpotent Lie algebras in [4, 9]
we have $I_{1}$ is isomorphic to $27A.$ On the other hand, $I_{2}=A(1)\oplus
Z^{\wedge}(L)$ and $L=I_{1}+A(1)\oplus Z^{\wedge}(L)=I_{1}+A(1).$ Since $L$ is
a stem Lie algebra and $[I_{1},I_{2}]\subseteq Z^{\wedge}(L),$ we have
$[I_{1},A(1)]=Z^{\wedge}(L).$ Otherwise, $L=I_{1}\oplus A(1)$ and since $L$ is
stem Lie algebra, it is a contradiction. Now, we are going to obtain
$[27A,A(1)].$ Let $A(1)\cong\langle x_{7}\rangle$ and

$27A\cong\langle
x_{1},\dots,x_{6},x_{8}\mid[x_{1},x_{2}]=x_{6},[x_{1},x_{4}]=x_{8},[x_{3},x_{5}]=x_{8}\rangle.$

We have

$\displaystyle[x_{1},x_{7}]=\alpha_{1}x_{8},$
$\displaystyle[x_{2},x_{7}]=\alpha_{2}x_{8},$
$\displaystyle[x_{3},x_{7}]=\alpha_{3}x_{8},$
$\displaystyle[x_{4},x_{7}]=\alpha_{4}x_{8},$
$\displaystyle[x_{5},x_{8}]=\alpha_{5}x_{8}.$

Put $x^{\prime}_{7}=x_{7}-\alpha_{3}x_{5}.$ By relabeling we have

$\displaystyle[x_{1},x_{7}]=\alpha_{1}x_{8},$
$\displaystyle[x_{2},x_{7}]=\alpha_{2}x_{8},$ $\displaystyle[x_{3},x_{7}]=0,$
$\displaystyle[x_{4},x_{7}]=\alpha_{4}x_{8},$
$\displaystyle[x_{5},x_{7}]=\alpha_{5}x_{8}.$

A change of variable $x^{\prime}_{7}=x_{7}-\alpha_{5}x_{3}$ and relabeling we
have

$\displaystyle[x_{1},x_{7}]=\alpha_{1}x_{8},$
$\displaystyle[x_{2},x_{7}]=\alpha_{2}x_{8},$ $\displaystyle[x_{3},x_{7}]=0,$
$\displaystyle[x_{4},x_{7}]=\alpha_{4}x_{8},$ $\displaystyle[x_{5},x_{7}]=0.$

Put $x^{\prime}_{7}=x_{7}-\alpha_{1}x_{4},$ we have

$\displaystyle[x_{1},x_{7}]=0,$ $\displaystyle[x_{2},x_{7}]=\alpha_{2}x_{8},$
$\displaystyle[x_{3},x_{7}]=0,$ $\displaystyle[x_{4},x_{7}]=\alpha_{4}x_{8},$
$\displaystyle[x_{5},x_{7}]=0.$

Therefore

$L=\langle
x_{1},\dots,x_{8}\mid[x_{1},x_{2}]=x_{6},[x_{1},x_{4}]=x_{8},[x_{3},x_{5}]=x_{8},[x_{2},x_{7}]=\alpha_{2}x_{8},[x_{4},x_{7}]=\alpha_{4}x_{8}\rangle.$

Consider the following cases.
Case 1. Let $\alpha_{2}\neq 0$ and $\alpha_{4}=0.$ Then by using a change of
variable $x^{\prime}_{7}=\alpha_{2}x_{7}$ and relabeling, we have

$S_{1}=\langle
x_{1},\dots,x_{8}\mid[x_{1},x_{2}]=x_{6},[x_{1},x_{4}]=x_{8},[x_{3},x_{5}]=x_{8},[x_{2},x_{7}]=x_{8}\rangle.$

Case 2. Let $\alpha_{2}=0$ and $\alpha_{4}\neq 0.$ A change of variable
$x^{\prime}_{7}=\alpha_{4}x_{7}$ and relabeling we have

$[x_{1},x_{2}]=x_{6},[x_{1},x_{4}]=x_{8},[x_{3},x_{5}]=x_{8},[x_{4},x_{7}]=x_{8}.$

Put $x^{\prime}_{1}=x_{1}+x_{7}.$ By relabeling we have

$\langle
x_{1},\dots,x_{8}\mid[x_{1},x_{2}]=x_{6},[x_{3},x_{5}]=x_{8},[x_{4},x_{7}]=x_{8}\rangle\cong
H(1)\oplus H(2).$

Case 3. Let $\alpha_{2}\neq 0$ and $\alpha_{4}\neq 0.$ Then by using
$x^{\prime}_{7}=\alpha_{4}^{-1}x_{7}$ and relabeling, we have

$[x_{1},x_{2}]=x_{6},[x_{1},x_{4}]=x_{8},[x_{3},x_{5}]=x_{8},[x_{2},x_{7}]=\alpha_{2}x_{8},[x_{4},x_{7}]=x_{8}.$

Therefore by a change of variable $x^{\prime}_{2}=\alpha_{2}^{-1}x_{2},$
$x^{\prime}_{6}=\alpha_{2}^{-1}x_{6}$ and relabeling we have

$[x_{1},x_{2}]=x_{6},[x_{1},x_{4}]=x_{8},[x_{3},x_{5}]=x_{8},[x_{2},x_{7}]=x_{8},[x_{4},x_{7}]=x_{8}.$

Finally, appropriate changes of variables $x^{\prime}_{1}=x_{1}+x_{7},$
$x^{\prime}_{2}=x_{2}-x_{4}$ and $x^{\prime}_{6}=x_{6}-x_{8}$ and relabeling
show that

$\langle
x_{1},\dots,x_{8}\mid[x_{1},x_{2}]=x_{6},[x_{3},x_{5}]=x_{8},[x_{4},x_{7}]=x_{8}\rangle\cong
H(1)\oplus H(2).$

Therefore $L$ is isomorphic to $H(1)\oplus H(2)$ or $S_{1}$ in this case, as
required. ∎

######  Theorem 3.3.

Let $L$ be an $n$-dimensional nilpotent Lie algebra with $s(L)=7$ and $\dim
L^{2}=2.$ Then $L$ is isomorphic to $L_{5,8}\oplus A(6)$ $L_{4,3}\oplus A(5),$
$L_{5,5}\oplus A(4),$ $L_{6,22}(\varepsilon)\oplus A(4)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}),$ $27B,$
$L_{6,10}\oplus A(2),$ $27A\oplus A(1),$ $157\oplus A(1),$ $L_{6,10}\dotplus
H(1),$ $H(1)\oplus H(2)$ or $S_{1}.$

###### Proof.

If $L$ is a capable Lie algebra, by using a proof similar that be used in [21,
Theorem 2.6], $L$ is isomorphic to one of the Lie algebras $L_{5,8}\oplus
A(6)$ $L_{4,3}\oplus A(5),$ $L_{5,5}\oplus A(4),$ $L_{6,22}(\varepsilon)\oplus
A(4)$ for all $\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}})$
or $27B.$
If $L$ is a non-capable Lie algebra, then $n\leq 9$ by using Lemma 1.3. Let
$n\leq 7,$ then there is no such a Lie algebra by using Proposition 2.2. Let
$n=8.$ Similar to the proof of Theorem 2.3, we have $L/Z^{\wedge}(L)\cong
H(1)\oplus A(4)$ and $s(L)=7.$ Now, by using Proposition 3.2, $L$ is
isomorphic to one of the Lie algebras $L_{6,10}\oplus A(2),$ $27A\oplus A(1),$
$157\oplus A(1),$ $L_{6,10}\dotplus H(1),$ $H(1)\oplus H(2)$ or $S_{1}.$
If $n=9,$ then $L/Z^{\wedge}(L)$ is isomorphic to $A(6)$ or $H(1)\oplus A(5).$
Since by using [13, Corollary 2.2] $L\wedge L\cong L/Z^{\wedge}(L)\wedge
L/Z^{\wedge}(L),$ we have $\dim L\wedge L$ is equal to $15$ or $23,$ so
$\mathcal{M}(L)$ is isomorphic to $A(13)$ or $A(21)$ by using [3, Theorem 35
(iii)]. Hence $s(L)=16$ or $s(L)=8,$ respectively. Therefore by our assumption
there is no such a Lie algebra. ∎

######  Lemma 3.4.

Let $L$ be an $n$-dimensional generalized Heisenberg of rank $3$ with
$s(L)=7.$ Then $n\leq 8.$

###### Proof.

The proof is obtained by the similar procedure as in [22, Lemma 2.8] ∎

######  Proposition 3.5.

There is no stem $8$-dimensional Lie algebra $L$ of nilpotency class two with
derived subalgebra of dimension $3$ and $s(L)=7$ such that $L/I\cong
L_{6,22}(\varepsilon)\oplus A(1)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}})$ and
$I\subseteq Z(L).$

###### Proof.

There exist ideals $I_{1}/I$ and $I_{2}/I$ of $L/I$ such that $I_{1}/I\cong
L_{6,22}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}})$ and
$I_{2}/I\cong A(1).$ Since $L^{2}/I\cong(I_{1}^{2}+I)/I,$ we have
$L^{2}=I_{1}^{2}+I.$ Hence $\dim I_{1}^{2}=2$ or $3.$ On the other hand, $L$
is of nilpotecy class two thus $I_{1}$ is of nilpotency class two. Let $\dim
I_{1}^{2}=3.$ Since $I_{1}$ is of nilpotency class two and $\dim I_{1}^{2}=3,$
$I_{1}$ is isomorphic to one of the Lie algebras $37A,$ $37B,$ $37C$ or $37D$
by looking at Table LABEL:ta4. If $\dim I_{1}^{2}=2,$ then $I_{1}$ is
isomorphic to $L_{6,22}(\varepsilon)\oplus A(1)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}})$ by looking at
Table LABEL:ta2. On the other hand, $I_{2}/I\cong A(1),$ we have $I_{2}\cong
A(2).$ We claim that and $[I_{1},I_{2}]=I.$ Otherwise $[I_{1},I_{2}]=0$ and so
$I_{2}\subseteq Z(L).$ Hence $\dim Z(L)\geq 4.$ It is a contradiction.
Therefore $L=I_{1}+I_{2}=I_{1}\rtimes A(1)$ such that $I_{1}$ is isomorphic to
one of the Lie algebras $37A,$ $37B,$ $37C,$ $37D$ or
$L_{6,22}(\varepsilon)\oplus A(1)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}})$ and
$[I_{1},A(1)]=\langle x_{7}\rangle.$
Let $L\cong I_{1}\rtimes A(1)$ such that $I_{1}\cong 37A=\langle
x_{1},\dots,x_{6},x_{8}\mid[x_{1},x_{2}]=x_{5},[x_{2},x_{3}]=x_{6},[x_{2},x_{4}]=x_{8}\rangle$
and $A(1)=\langle x_{7}\rangle.$ Then the relations of $L$ are as follow.

$\displaystyle[x_{1},x_{2}]=x_{5},$ $\displaystyle[x_{2},x_{3}]=x_{6},$
$\displaystyle[x_{2},x_{4}]=x_{8},$
$\displaystyle[x_{1},x_{7}]=\alpha_{1}x_{8},$
$\displaystyle[x_{2},x_{7}]=\alpha_{2}x_{8},$
$\displaystyle[x_{3},x_{7}]=\alpha_{3}x_{8},$
$\displaystyle[x_{4},x_{7}]=\alpha_{4}x_{8}.$

By using a change of variable $x^{\prime}_{7}=x_{7}-\alpha_{2}x_{4}$ and
relabiling, we have

$\displaystyle[x_{1},x_{2}]=x_{5},$ $\displaystyle[x_{2},x_{3}]=x_{6},$
$\displaystyle[x_{2},x_{4}]=x_{8},$
$\displaystyle[x_{1},x_{7}]=\alpha_{1}x_{8},$ $\displaystyle[x_{2},x_{7}]=0,$
$\displaystyle[x_{3},x_{7}]=\alpha_{3}x_{8},$
$\displaystyle[x_{4},x_{7}]=\alpha_{4}x_{8}.$

Since $[I_{1},A(1)]=\langle x_{8}\rangle,$ at least one of the $\alpha_{1},$
$\alpha_{3}$ or $\alpha_{4}$ is non-zero. Let $\alpha_{1}\neq 0.$ Then we are
going to obtain the Schur multiplier of $L.$ By using the method of Hardy and
Stitzinger in [1]. Start with

$\displaystyle[x_{1},x_{2}]=x_{5}+s_{1},$ $\displaystyle[x_{1},x_{3}]=s_{2},$
$\displaystyle[x_{1},x_{4}]=s_{3},$ $\displaystyle[x_{1},x_{5}]=s_{4},$
$\displaystyle[x_{1},x_{6}]=s_{5},$
$\displaystyle[x_{1},x_{7}]=\alpha_{1}x_{8}+s_{6},$
$\displaystyle[x_{1},x_{8}]=s_{7},$ $\displaystyle[x_{2},x_{3}]=x_{6}+s_{8},$
$\displaystyle[x_{2},x_{4}]=x_{8}+s_{9},$ $\displaystyle[x_{2},x_{5}]=s_{10},$
$\displaystyle[x_{2},x_{6}]=s_{11},$ $\displaystyle[x_{2},x_{7}]=s_{12},$
$\displaystyle[x_{2},x_{8}]=s_{13},$ $\displaystyle[x_{3},x_{4}]=s_{14},$
$\displaystyle[x_{3},x_{5}]=s_{15},$ $\displaystyle[x_{3},x_{6}]=s_{16},$
$\displaystyle[x_{3},x_{7}]=\alpha_{3}x_{8}+s_{17},$
$\displaystyle[x_{3},x_{8}]=s_{18},$ $\displaystyle[x_{4},x_{5}]=s_{19},$
$\displaystyle[x_{4},x_{6}]=s_{20},$
$\displaystyle[x_{4},x_{7}]=\alpha_{4}x_{8}+s_{21},$
$\displaystyle[x_{4},x_{8}]=s_{22},$ $\displaystyle[x_{5},x_{6}]=s_{23},$
$\displaystyle[x_{5},x_{7}]=s_{24},$ $\displaystyle[x_{5},x_{8}]=s_{25},$
$\displaystyle[x_{6},x_{7}]=s_{26},$ $\displaystyle[x_{6},x_{8}]=s_{27},$
$\displaystyle[x_{7},x_{8}]=s_{28},$

where $s_{1},...,s_{28}$ generate $\mathcal{M}(L)$. Put
$x^{\prime}_{5}=x_{5}+s_{1},$ $x^{\prime}_{6}=x_{6}+s_{8}$ and
$x^{\prime}_{8}=x_{8}+s_{9}.$ A change of variables allows that
$s_{1}=s_{8}=s_{9}=0.$ Now, by using the Jocobi identity we have

$\displaystyle
s_{5}=[x_{1},x_{6}]=[x_{1},[x_{2},x_{3}]]=-([x_{3},[x_{1},x_{2}]]+[x_{2},[x_{3},x_{1}]])=-s_{15},$
$\displaystyle
s_{7}=[x_{1},x_{8}]=[x_{1},[x_{2},x_{4}]]=-([x_{4},[x_{1},x_{2}]]+[x_{2},[x_{4},x_{1}]])=-s_{19},$
$\displaystyle
s_{18}=[x_{3},x_{8}]=[x_{3},[x_{2},x_{4}]]=-([x_{4},[x_{3},x_{2}]]+[x_{2},[x_{4},x_{3}]]=s_{20},$
$\displaystyle
s_{18}=[x_{3},x_{8}]=[x_{3},\frac{1}{\alpha_{1}}[x_{1},x_{7}]]=-\frac{1}{\alpha_{1}}([x_{7},[x_{3},x_{1}]]+[x_{1},[x_{7},x_{3}]])=\frac{\alpha_{3}}{\alpha_{1}}s_{7},$
$\displaystyle
s_{22}=[x_{4},x_{8}]=[x_{4},\frac{1}{\alpha_{1}}[x_{1},x_{7}]]=-\frac{1}{\alpha_{1}}([x_{7},[x_{4},x_{1}]]+[x_{1},[x_{7},x_{4}]])=\frac{\alpha_{4}}{\alpha_{1}}s_{7},$
$\displaystyle
s_{23}=[x_{5},x_{6}]=[x_{5},[x_{2},x_{3}]]=-([x_{3},[x_{5},x_{2}]]+[x_{2},[x_{3},x_{5}]]=0,$
$\displaystyle
s_{24}=[x_{5},x_{7}]=-[x_{7},[x_{1},x_{2}]]=[x_{2},[x_{7},x_{1}]]+[x_{1},[x_{2},x_{7}]]=-\alpha_{1}s_{13},$
$\displaystyle
s_{25}=[x_{5},x_{8}]=[x_{5},[x_{2},x_{4}]]=-([x_{4},[x_{5},x_{2}]]+[x_{2},[x_{4},x_{5}]])=0,$
$\displaystyle
s_{26}=[x_{6},x_{7}]=-[x_{7},[x_{2},x_{3}]]=[x_{3},[x_{7},x_{2}]]+[x_{2},[x_{3},x_{7}]]=\alpha_{3}s_{13},$
$\displaystyle
s_{27}=[x_{6},x_{8}]=[x_{6},[x_{2},x_{4}]]=-([x_{4},[x_{6},x_{2}]]+[x_{2},[x_{4},x_{6}]])=0,$
$\displaystyle
s_{28}=[x_{7},x_{8}]=[x_{7},[x_{2},x_{4}]]=-([x_{4},[x_{7},x_{2}]]+[x_{2},[x_{4},x_{7}]])=\alpha_{4}s_{13}.$

Therefore

$\mathcal{M}(L)=\langle
s_{2},s_{3},s_{4},s_{5},s_{6},s_{7},s_{10},s_{11},s_{12},s_{13},s_{14},s_{16},s_{17},s_{21}\rangle$

and so $\dim\mathcal{M}(L)=14.$ Hence $s(L)\neq 7.$ Also, if $\alpha_{3}$ or
$\alpha_{4}$ are non-zero, then $\dim\mathcal{M}(L)=14$ and so $s(L)\neq 7.$
By using a similar way if $L=I_{1}\rtimes A(1)$ such that $I_{1}$ is
isomorphic to one of the Lie algebras $37B,$ $37C,$ $37D$ or
$L_{6,22}(\varepsilon)\oplus A(1)$ and $[I_{1},A(1)]=\langle x_{8}\rangle,$
then $s(L)\neq 7.$ Therefore by our assumption there is no such a Lie algebra.
∎

Let $cl(L)$ be used to show the nilpotency class of $L.$ Then

######  Proposition 3.6.

There is no stem $8$-dimensional Lie algebra $L$ with derived subalgebra of
dimension $3$ and $s(L)=7$ such that $L/I\cong L_{5,8}\oplus A(2)$ and
$I\subseteq Z(L).$

###### Proof.

There are ideals $I_{1}/I$ and $I_{2}/I$ such that $I_{1}/I\cong L_{5,8}\oplus
A(1)$ and $I_{2}/I\cong A(1).$ Consider the following cases.
Case 1. Let $\dim Z(L)=1.$ Then $Z(L)=I.$ Since
$L^{2}/Z(L)=I^{2}_{1}+Z(L)/Z(L),$ we have $L^{2}=I^{2}_{1}+Z(L)$ and by using
Lemma 1.5, $L^{2}=I^{2}_{1},$ $Z(L)=I^{3}_{1}$ and $cl(L)=cl(I_{1})=3.$ Also,
$I_{2}=A(1)\oplus Z(L).$ It is clear that $[I_{1},I_{2}]\subseteq I_{1}\cap
I_{2}$ and $I_{1}\cap I_{2}=Z(L),$ hence $[I_{1},I_{2}]\subseteq Z(L).$ On the
other hand, $L=I_{1}+I_{2}=I_{1}+A(1)$ and so $[I_{1},I_{2}]=[I_{1},A(1)].$ If
$[I_{1},A(1)]=0,$ then $L=I_{1}\oplus A(1)$ and since $L$ is stem, we have a
contradiction. Therefore $[I_{1},A(1)]=Z(L).$ We claim that $Z(I_{1})\cap
I_{1}^{2}=Z(L),$ otherwise there is $a\in Z(I_{1})\cap I_{1}^{2}$ such that
$a\notin Z(L).$ Since $a\in I_{1}^{2},$ there are $b,c\in I_{1}$ such that
$[b,c]=a.$ Now, let $A(1)=\langle x_{8}\rangle.$ Then
$[x_{8},[b,c]]=-([c,[x_{8},b]]+[b,[c,x_{8}]])$ by using Jacobi identity. Since
$[x_{8},b]$ and $[c,x_{8}]$ are central, we have $[x_{8},[b,c]]=0$ and so
$a\in Z(L).$ It is a contradiction. If $I_{1}$ is stem, then
$Z(I_{1})=I_{1}^{3}$ and $\dim Z(I_{1})=1.$ By using the classification of all
$7$-dimensional Lie algebra in [9] $I_{1}$ is isomorphic to one of the
nilpotent Lie algebras $147A$ or $147B.$ If $I_{1}$ is not stem, then $I_{1}$
is isomorphic to one of the nilpotent Lie algebras
$L_{6,19}(\varepsilon)\oplus A(1)$ for all
$\varepsilon\in\mathbb{F}^{*}/(\stackrel{{\scriptstyle*}}{{\sim}})$ or
$L_{6,20}\oplus A(1)$ by using Table LABEL:ta5. Now, we are going to determine
$L.$ It is sufficient to obtain $[I_{1},A(1)].$ Assume that $A(1)=\langle
x_{8}\rangle$ and

$I_{1}\cong 147A=\langle
x_{1},\dots,x_{7}\mid[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5},[x_{1},x_{6}]=x_{7},[x_{2},x_{5}]=x_{7},$
$[x_{3},x_{4}]=x_{7}\rangle.$

We have

$\displaystyle[x_{1},x_{8}]=\alpha_{1}x_{7},$
$\displaystyle[x_{2},x_{8}]=\alpha_{2}x_{7},$
$\displaystyle[x_{3},x_{8}]=\alpha_{3}x_{7},$
$\displaystyle[x_{4},x_{8}]=\alpha_{4}x_{7},$
$\displaystyle[x_{5},x_{8}]=\alpha_{5}x_{7},$
$\displaystyle[x_{6},x_{8}]=\alpha_{6}x_{7}.$

Put $x^{\prime}_{8}=x_{8}-\alpha_{1}x_{6}.$ Then by relabeling

$\displaystyle[x_{1},x_{8}]=0,$ $\displaystyle[x_{2},x_{8}]=\alpha_{2}x_{7},$
$\displaystyle[x_{3},x_{8}]=\alpha_{3}x_{7},$
$\displaystyle[x_{4},x_{8}]=\alpha_{4}x_{7},$
$\displaystyle[x_{5},x_{8}]=\alpha_{5}x_{7},$
$\displaystyle[x_{6},x_{8}]=\alpha_{6}x_{7}.$

A change of variable $x^{\prime}_{8}=x_{8}-\alpha_{1}x_{5}$ and relabeling
allows that

$\displaystyle[x_{1},x_{8}]=0,$ $\displaystyle[x_{2},x_{8}]=0,$
$\displaystyle[x_{3},x_{8}]=\alpha_{3}x_{7},$
$\displaystyle[x_{4},x_{8}]=\alpha_{4}x_{7},$
$\displaystyle[x_{5},x_{8}]=\alpha_{5}x_{7},$
$\displaystyle[x_{6},x_{8}]=\alpha_{6}x_{7}.$

Finally, by putting $x^{\prime}_{8}=x_{8}-\alpha_{3}x_{4}$ and relabeling we
have

$\displaystyle[x_{1},x_{8}]=0,$ $\displaystyle[x_{2},x_{8}]=0,$
$\displaystyle[x_{3},x_{8}]=0,$ $\displaystyle[x_{4},x_{8}]=\alpha_{4}x_{7},$
$\displaystyle[x_{5},x_{8}]=\alpha_{5}x_{7},$
$\displaystyle[x_{6},x_{8}]=\alpha_{6}x_{7}.$

By using Jacobi identity
$[x_{4},x_{8}]=[[x_{1},x_{2}],x_{8}]=[x_{2},[x_{8},x_{1}]]+[x_{1},[x_{2},x_{8}]]$
and since $[x_{8},x_{1}]$ and $[x_{2},x_{8}]$ are central, we have
$[x_{4},x_{8}]=0.$ By using a similar way $[x_{5},x_{8}]=0.$ Therefore

$\displaystyle[x_{1},x_{8}]=0,$ $\displaystyle[x_{2},x_{8}]=0,$
$\displaystyle[x_{3},x_{8}]=0,$ $\displaystyle[x_{4},x_{8}]=0,$
$\displaystyle[x_{5},x_{8}]=0,$ $\displaystyle[x_{6},x_{8}]=\alpha_{6}x_{7}.$

Since $[I_{1},A(1)]\neq 0,$ we have $\alpha_{6}\neq 0$ and so by using
$x^{\prime}_{8}=\alpha_{6}^{-1}x_{8}$ and relabeling
$[I_{1},A(1)]=\langle[x_{6},x_{8}]=x_{7}\rangle.$ Therefore

$L\cong\langle
x_{1},\dots,x_{8}\mid[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5},[x_{1},x_{6}]=x_{7},[x_{2},x_{5}]=x_{7},$
$[x_{3},x_{4}]=x_{7},[x_{6},x_{8}]=x_{7}\rangle.$

By using the method of Hardy and Stitzinger in [1], we can see
$\dim\mathcal{M}(L)=12$ and so $s(L)\neq 7.$ Also, if $I_{1}$ is isomorphic to
one of the nilpotent Lie algebras $147B,$ $L_{6,19}(\varepsilon)\oplus A(1)$
or $L_{6,20},$ then by using a similar way one can see that $s(L)\neq 7.$
Case 2. Let $\dim Z(L)=2$ and $Z(L)\subset L^{2}.$ Then $L/Z(L)\cong
H(1)\oplus A(3)$ and so $\dim Z_{2}(L)=6.$ Also, by our assumption, we know
that $\dim L=8$ and $s(L)=7,$ hence $\dim\mathcal{M}(L)=15.$ On the other
hand, $L^{3}\subseteq Z(L).$ If $L^{3}\neq Z(L),$ then $\dim L^{3}=1$ and
$L/I^{3}\cong L_{5,8}\oplus A(2).$ When $L^{3}=Z(L),$ then $\dim L^{3}=2$ and
$L/I^{3}\cong H(1)\oplus A(3).$ By using [20, Lemmas 2.3 and 2.10], [10, Lemma
23] and [1, Theorem 1] one can see $\dim\mathcal{M}(L_{5,8}\oplus A(2))=10$
and $\dim\mathcal{M}(H(1)\oplus A(3))=13.$ Now, from [16, Theorem 2.6], we
have

(3.1) $\dim
L^{3}+\dim\mathcal{M}(L)\leq\dim\mathcal{M}(L/L^{3})+\dim(L/Z_{2}(L)\otimes
L^{3}),$

which is a contradiction.
Case 3. Let $\dim Z(L)=3$ and $Z(L)=L^{2}.$ Then $L$ and $I_{1}$ are of
nilpotency class two. Since $L^{2}=I^{2}_{1}+I,$ we have $\dim I^{2}_{1}=2$ or
$3.$ If $\dim I^{2}_{1}=2,$ then $\dim I^{2}_{1}\cap I=0$ and so $I\not\subset
I^{2}_{1}.$ On the other hand, $I\subseteq I_{1}$ and $I\subseteq Z(L),$ thus
$I\subseteq Z(I_{1})$ and so $Z(I_{1})\not\subset I^{2}_{1}.$ Now, by looking
at Table LABEL:ta2, $I_{1}$ is isomorphic to $L_{5,8}\oplus A(2).$ Similar to
Case 1, we have

$L\cong\langle
x_{1},\dots,x_{8}\mid[x_{1},x_{2}]=x_{4},[x_{1},x_{3}]=x_{5},[x_{6},x_{8}]=x_{7}\rangle.$

By using the method of Hardy and Stitzinger in [1], one can see that
$\dim\mathcal{M}(L)=14$ and so $s(L)\neq 7.$
If $\dim I^{2}_{1}=3,$ then $I_{1}$ is isomorphic to one of the nilpotent Lie
algebras $37A,$ $37B$ or $37C$ by using Table LABEL:ta7. By using a similar
way we may obtain that $\mathcal{M}(L)=14.$ Hence $s(L)\neq 7.$ Therefore
there is no such a Lie algebra. ∎

######  Theorem 3.7.

Let $L$ be an $n$-dimensional nilpotent Lie algebra with $s(L)=7$ and $\dim
L^{2}=3.$ Then $L$ is isomorphic to one of the nilpotent Lie algebras
$L_{6,23}\oplus A(1),$ $L_{6,25}\oplus A(1),$ $37B\oplus A(1),$ $37C\oplus
A(1),$ $37D\oplus A(1),$ $L_{6,26}\oplus A(2),$ $L_{6,13},$ $257A,$ $257C$ or
$257F.$

###### Proof.

If $Z(L)\not\subseteq L^{2},$ then there is a central one dimensional ideal
$I$ of $L$ such that $I\cap L^{2}=0.$ By using a similar method in the proof
of Theorem 2.5, we can see $L$ is isomorphic to $L_{6,23}\oplus A(1),$
$L_{6,25}\oplus A(1),$ $37B\oplus A(1),$ $37C\oplus A(1),$ $37D\oplus A(1)$ or
$L_{6,26}\oplus A(2).$ Also, by using Tables LABEL:ta6, LABEL:ta7, [10, Lemma
23] and [1, Theorem 1] we have $s(L_{6,23}\oplus A(1))=s(L_{6,25}\oplus
A(1))=s(37B\oplus A(1))=s(37C\oplus A(1))=s(37D\oplus A(1))=s(L_{6,26}\oplus
A(2))=7.$
If $Z(L)\subseteq L^{2},$ then there is a one dimensional ideal $I$ of $L$
such that $I\subseteq Z(L)\cap L^{2}.$ Hence by using we have

$\displaystyle\frac{1}{2}(n-1)(n-2)-5\leq\frac{1}{2}(n-2)(n-3)+1-s(L/I)+n-3$

and so $s(L/I)\leq 5.$ Since $\dim(L/I)^{2}=2,$ by using Theorem 1.6, we have
one of the following cases.
Case 1. If $L/I$ is isomorphic to one of the nilpotent Lie algebras $L_{5,8},$
$L_{4,3},$ $L_{5,8}\oplus A(1),$ $L_{4,3}\oplus A(1),$ $L_{5,5},$
$L_{6,22}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}})$ or
$L_{4,3}\oplus A(2),$ then $n\leq 7.$ Hence $L$ is isomorphic to one of the
nilpotent Lie algebras $L_{6,13},$ $257A,$ $257C$ or $257F$ by looking at
Tables LABEL:ta6 and LABEL:ta7.
Case 2. If $L/I\cong L_{5,8}\oplus A(2),$ then there is no Lie algebra $L$
with $s(L)=7$ and $\dim L^{2}=3$ by using Proposition 3.6.
Case 3. Let $L/I\cong L_{6,22}(\varepsilon)\oplus A(1)$ for all
$\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}).$ If $L$ is of
nilpotency class two, then there is no Lie algebra with $s(L)=7$ and $\dim
L^{2}=3$ by using Theorem 3.5. If $L$ is of nilpotency class $3,$ then

$\displaystyle 15=\dim\mathcal{M}(L)$
$\displaystyle\leq\dim\mathcal{M}(L/\gamma_{3}(L))-\dim\gamma_{3}(L)+\dim
L/\gamma_{2}(L)\otimes\gamma_{3}(L)-2$ $\displaystyle\leq 14.$

by using (2.1) and (2.2). It is a contradiction.
Case 4. Let $L/I$ is isomorphic to one of the nilpotent Lie algebras
$L_{5,8}\oplus A(3),$ $L_{5,8}\oplus A(4)$ or $L_{6,22}(\varepsilon)\oplus
A(2)$ for all $\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}).$
If $L$ is of nilpotency class two, then there is no Lie algebra with $s(L)=7$
and $\dim L^{2}=3$ by using Lemma 3.4. If $L$ is of nilpotency class $3,$ then
we have a contradiction similar to the Case $3.$
Case 5. Let $L/I$ is isomorphic to $L_{4,3}\oplus A(3),$ $L_{5,5}\oplus A(2).$
If $L/I\cong L_{4,3}\oplus A(3),$ then $\dim\mathcal{M}(L/I)=11$ by using [20,
Lemma 2.9], [10, Lemma 23] and [1, Theorem 1]. When $cl(L)=3$ and
$\dim\gamma_{3}(L)=1,$ then by using the inequalities (2.1) and (2.2) we have
a contradiction. In the case $cl(L)=3$ and $\dim\gamma_{3}(L)=2,$ by using the
inequality (3.1), we get a contradiction. If $cl(L)=4,$ then $I=\gamma_{4}(L)$
and $\dim\gamma_{4}(L)=1.$ Since $\dim\mathcal{M}(L)=15$ and
$\dim\mathcal{M}(L/\gamma_{3}(L))=11,$ by using [23, proof of Theorem 1.1], we
have

$1=n-m-c\leq\dim\ker\lambda_{4},~{}~{}\text{and}$
$\dim\mathcal{M}(L)=\dim\mathcal{M}(L/\gamma_{4}(L))-\dim\gamma_{4}(L)+\dim
L/\gamma_{2}(L)\otimes\gamma_{4}(L)-\dim\ker\lambda_{4}$

which a contradiction. If $L/I\cong L_{5,5}\oplus A(2),$ by using a similar
way we get a contradiction. ∎

######  Proposition 3.8.

There is no stem $8$-dimensional nilpotent Lie algebra $L$ of nilpotency class
$2$ with derived subalgebra of dimension $4$ and $s(L)=7$ such that $L/I\cong
37A$ and $I\subseteq Z(L).$

###### Proof.

Let $I=\langle x_{8}\rangle$ and $37A=\langle
x_{1},\dots,x_{6},x_{7}\mid[x_{1},x_{2}]=x_{5},[x_{2},x_{3}]=x_{6},[x_{2},x_{4}]=x_{7}\rangle.$
Then

$\displaystyle[x_{1},x_{2}]=x_{5}+\alpha_{1}x_{8},$
$\displaystyle[x_{1},x_{3}]=\alpha_{2}x_{8},$
$\displaystyle[x_{1},x_{4}]=\alpha_{3}x_{8},$
$\displaystyle[x_{1},x_{5}]=\alpha_{4}x_{8},$
$\displaystyle[x_{1},x_{6}]=\alpha_{5}x_{8},$
$\displaystyle[x_{1},x_{7}]=\alpha_{6}x_{8},$
$\displaystyle[x_{2},x_{3}]=x_{6}+\alpha_{7}x_{8},$
$\displaystyle[x_{2},x_{4}]=x_{7}+\alpha_{8}x_{8},$
$\displaystyle[x_{2},x_{5}]=\alpha_{9}x_{8},$
$\displaystyle[x_{2},x_{6}]=\alpha_{10}x_{8},$
$\displaystyle[x_{2},x_{7}]=\alpha_{11}x_{8},$
$\displaystyle[x_{3},x_{4}]=\alpha_{12}x_{8},$
$\displaystyle[x_{3},x_{5}]=\alpha_{13}x_{8},$
$\displaystyle[x_{3},x_{6}]=\alpha_{14}x_{8},$
$\displaystyle[x_{3},x_{7}]=\alpha_{15}x_{8},$
$\displaystyle[x_{4},x_{5}]=\alpha_{16}x_{8},$
$\displaystyle[x_{4},x_{6}]=\alpha_{17}x_{8},$
$\displaystyle[x_{4},x_{7}]=\alpha_{18}x_{8},$
$\displaystyle[x_{5},x_{6}]=\alpha_{19}x_{8},$
$\displaystyle[x_{5},x_{7}]=\alpha_{20}x_{8},$
$\displaystyle[x_{6},x_{7}]=\alpha_{21}x_{8}.$

Since $L$ is of nilpotency class two, we have
$\alpha_{4}=\alpha_{5}=\alpha_{6}=\alpha_{9}=\alpha_{10}=\alpha_{11}=\alpha_{13}=\alpha_{14}=\alpha_{15}=\alpha_{16}=\alpha_{17}=\alpha_{18}=\alpha_{19}=\alpha_{20}=\alpha_{21}=0.$
By using a change of variables $x^{\prime}_{5}=x_{5}+\alpha_{1}x_{8},$
$x^{\prime}_{6}=x_{6}+\alpha_{7}x_{8}$ and
$x^{\prime}_{7}=x_{7}+\alpha_{8}x_{8},$ we have
$\alpha_{1}=\alpha_{7}=\alpha_{8}=0.$ Since $\dim L^{2}=4,$ at least one of
the coefficients $\alpha_{2},$ $\alpha_{3}$ or $\alpha_{12}$ is non-zero.
Let $\alpha_{2}\neq 0.$ Then by using $x^{\prime}_{8}=\alpha_{2}x_{8}$ and
relabeling we have

$L=\langle
x_{1},\dots,x_{7},x_{8}\mid[x_{1},x_{2}]=x_{5},[x_{2},x_{3}]=x_{6},[x_{2},x_{4}]=x_{7},[x_{1},x_{3}]=x_{8},[x_{1},x_{4}]=\alpha_{3}x_{8},$
$[x_{3},x_{4}]=\alpha_{12}x_{8}\rangle.$

In the following, we are going to obtain the Schur multiplier of $L,$ by using
the method of Hardy and Stitzinger in [1]. Start with

$\displaystyle[x_{1},x_{2}]=x_{5}+s_{1},$
$\displaystyle[x_{1},x_{3}]=x_{8}+s_{2},$
$\displaystyle[x_{1},x_{4}]=\alpha_{3}x_{8}+s_{3},$
$\displaystyle[x_{1},x_{5}]=s_{4},$ $\displaystyle[x_{1},x_{6}]=s_{5},$
$\displaystyle[x_{1},x_{7}]=s_{6},$ $\displaystyle[x_{1},x_{8}]=s_{7},$
$\displaystyle[x_{2},x_{3}]=x_{6}+s_{8},$
$\displaystyle[x_{2},x_{4}]=x_{7}+s_{9},$ $\displaystyle[x_{2},x_{5}]=s_{10},$
$\displaystyle[x_{2},x_{6}]=s_{11},$ $\displaystyle[x_{2},x_{7}]=s_{12},$
$\displaystyle[x_{2},x_{8}]=s_{13},$
$\displaystyle[x_{3},x_{4}]=\alpha_{6}x_{8}+s_{14},$
$\displaystyle[x_{3},x_{5}]=s_{15},$ $\displaystyle[x_{3},x_{6}]=s_{16},$
$\displaystyle[x_{3},x_{7}]=s_{17},$ $\displaystyle[x_{3},x_{8}]=s_{18},$
$\displaystyle[x_{4},x_{5}]=s_{19},$ $\displaystyle[x_{4},x_{6}]=s_{20},$
$\displaystyle[x_{4},x_{7}]=s_{21},$ $\displaystyle[x_{4},x_{8}]=s_{22},$
$\displaystyle[x_{5},x_{6}]=s_{23},$ $\displaystyle[x_{5},x_{7}]=s_{24},$
$\displaystyle[x_{5},x_{8}]=s_{25},$ $\displaystyle[x_{6},x_{7}]=s_{26},$
$\displaystyle[x_{6},x_{8}]=s_{27},$ $\displaystyle[x_{7},x_{8}]=s_{28},$

where $s_{1},...,s_{28}$ generate $\mathcal{M}(L)$. By using the Jocobi
identity, we have

$\displaystyle
s_{5}=[x_{1},x_{6}]=[x_{1},[x_{2},x_{3}]]=-([x_{3},[x_{1},x_{2}]]+[x_{2},[x_{3},x_{1}]])=-s_{15}+s_{13},$
$\displaystyle
s_{6}=[x_{1},x_{7}]=[x_{1},[x_{2},x_{4}]]=-([x_{4},[x_{1},x_{2}]]+[x_{2},[x_{4},x_{1}]])=-s_{19}+\alpha_{3}s_{13},$
$\displaystyle
s_{15}=[x_{3},x_{5}]=[x_{3},[x_{1},x_{2}]]=-([x_{2},[x_{3},x_{1}]]+[x_{1},[x_{2},x_{3}]])=s_{13}-s_{5},$
$\displaystyle
s_{17}=[x_{3},x_{7}]=[x_{3},[x_{2},x_{4}]]=-([x_{4},[x_{3},x_{2}]]+[x_{2},[x_{4},x_{3}]])=s_{20}+\alpha_{6}s_{13},$
$\displaystyle
s_{19}=[x_{4},x_{5}]=[x_{4},[x_{1},x_{2}]]=-([x_{2},[x_{4},x_{1}]]+[x_{1},[x_{2},x_{4}]])=\alpha_{3}s_{13}-s_{6},$
$\displaystyle
s_{22}=[x_{4},x_{8}]=[x_{4},[x_{1},x_{3}]]=-([x_{3},[x_{4},x_{1}]]+[x_{1},[x_{3},x_{4}]])=\alpha_{3}s_{18}-\alpha_{6}s_{7},$
$\displaystyle
s_{23}=[x_{5},x_{6}]=[x_{5},[x_{2},x_{3}]]=-([x_{3},[x_{5},x_{2}]]+[x_{2},[x_{3},x_{5}]])=0,$
$\displaystyle
s_{24}=[x_{5},x_{7}]=[x_{5},[x_{2},x_{4}]]=-([x_{4},[x_{5},x_{2}]]+[x_{2},[x_{4},x_{5}]])=0,$
$\displaystyle
s_{25}=[x_{5},x_{8}]=-([x_{8},[x_{1},x_{2}]])=[x_{2},[x_{8},x_{1}]]+[x_{1},[x_{2},x_{8}]]=0,$
$\displaystyle
s_{26}=[x_{6},x_{7}]=[x_{6},[x_{2},x_{4}]]=-([x_{4},[x_{6},x_{2}]]+[x_{2},[x_{4},x_{6}]])=0,$
$\displaystyle
s_{27}=[x_{6},x_{8}]=-([x_{8},[x_{2},x_{3}]])=[x_{3},[x_{8},x_{2}]]+[x_{2},[x_{3},x_{8}]]=0,$
$\displaystyle
s_{28}=[x_{7},x_{8}]=-([x_{8},[x_{2},x_{4}]])=[x_{4},[x_{8},x_{2}]]+[x_{2},[x_{4},x_{8}]]=0,$

Put $x^{\prime}_{5}=x_{5}+s_{1},$ $x^{\prime}_{6}=x_{6}+s_{8},$
$x^{\prime}_{7}=x_{7}+s_{9}$ and $x^{\prime}_{8}=x_{8}+s_{2}.$ A change of
variables allows that $s_{1}=s_{2}=s_{8}=s_{9}=0.$ Thus

$\mathcal{M}(L)=\langle
s_{3},s_{4},s_{5},s_{6},s_{7},s_{10},s_{11},s_{12},s_{13},s_{14},s_{16},s_{17},s_{18},s_{21}\rangle$

and $\dim\mathcal{M}(L)=14$ and so $s(L)=8.$
In the case $\alpha_{3}$ or $\alpha_{12}$ is non-zero, we have $s(L)=8$ by
using a similar way. Therefore there is no such a nilpotent Lie algebra, and
the proof is completed. ∎

######  Theorem 3.9.

Let $L$ an $n$-dimensional nilpotent Lie algebra with $s(L)=7$ and $\dim
L^{2}=4.$ Then $L$ is isomorphic to $L_{6,21}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F}^{*}/(\stackrel{{\scriptstyle*}}{{\sim}}).$

###### Proof.

If $Z(L)\not\subseteq L^{2},$ then there is a central one dimensional ideal
$I$ of $L$ such that $I\cap L^{2}=0.$ Similar to the proof of Theorem 2.7, we
have $s(L/I)\leq 4.$ Since $\dim(L/I)^{2}=4,$ we have there is no such a
nilpotent Lie algebra $L/I$ with $s(L/I)\leq 4$ by using Theorem 1.6.
Let $Z(L)\subseteq L^{2}.$ Then there exist a one dimension ideal $I$ of $L$
such that $I\subseteq Z(L)\cap L^{2}.$ By using Theorem 1.2, we have

$\displaystyle\frac{1}{2}(n-1)(n-2)-5\leq\frac{1}{2}(n-2)(n-3)+1-s(L/I)+n-4$

and so $s(L/I)\leq 4.$ Since $s(L/I)\leq 4$ and $\dim(L/I)^{2}=3,$ we have
$L/I$ is isomorphic to $L_{6,26,}$ $L_{5,6},$ $L_{5,7},$ $L_{5,9}$ or $37A.$
If $L/I$ is isomorphic to $L_{6,26,}$ $L_{5,6},$ $L_{5,7}$ or $L_{5,9},$ then
$n=6$ or $7.$ If $n=6,$ then $L\cong L_{6,21}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F}^{*}/(\stackrel{{\scriptstyle*}}{{\sim}})$ by using
Table LABEL:ta81. Let $L$ a $7$-dimensional Lie algebra. Then in this case
there is no nilpotent Lie algebra with $s(L)=7$ and $\dim L^{2}=4$ by using
Lemma 2.6.
Let $L/I\cong 37A.$ If $L$ is of nilpotency class two, then by using
Proposition 3.8 there is no such a Lie algebra with $\dim L^{2}=4$ and
$s(L)=7.$
If $L$ is of nilpotency class $3$, then by using (2.1) and (2.2), we have

$\displaystyle 15=\dim\mathcal{M}(L)$
$\displaystyle\leq\dim\mathcal{M}(L/\gamma_{3}(L))-\dim\gamma_{3}(L)+\dim(L/\gamma_{2}(L))\otimes\gamma_{3}(L)-2$
$\displaystyle\leq 14,$

which is contradiction. The result follows. ∎

We are ready to have the Main Theorem of this section.

######  Theorem 3.10.

Let $L$ be an $n$-dimensional nilpotent Lie algebra with $s(L)=7.$ Then $L$ is
isomorphic to one of the nilpotent Lie algebras $L_{5,8}\oplus A(6)$
$L_{4,3}\oplus A(5),$ $L_{5,5}\oplus A(4),$ $L_{6,22}(\varepsilon)\oplus A(4)$
for all $\varepsilon\in\mathbb{F}/(\stackrel{{\scriptstyle*}}{{\sim}}),$
$27B,$ $L_{6,10}\oplus A(2),$ $27A\oplus A(1),$ $157\oplus A(1),$
$L_{6,10}\dotplus H(1),$ $H(1)\oplus H(2),$ $S_{1},$ $L_{6,23}\oplus A(1),$
$L_{6,25}\oplus A(1),$ $37B\oplus A(1),$ $37C\oplus A(1),$ $37D\oplus A(1),$
$L_{6,26}\oplus A(2),$ $L_{6,13},$ $257A,$ $257C,$ $257F$ or
$L_{6,21}(\varepsilon)$ for all
$\varepsilon\in\mathbb{F}^{*}/(\stackrel{{\scriptstyle*}}{{\sim}}).$

###### Proof.

By using Proposition 3.1, Theorems 3.3, 3.7 and 3.9, we can obtain the result.
∎

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