Study on Poisson Algebra and Automorphism of a Special Class of Solvable Lie Algebras

Abstract

We define a four-dimensional Lie algebra g in this paper and then prove that this Lie algebra is solvable but not nilpotent. Due to the fact that g is a Lie algebra, $\forall x,y\in g$,$[x,y]=-[y,x]$, that is, the operation $[,]$ has anti symmetry. Symmetry is a very important law, and antisymmetry is also a very important law. We studied the structure of Poisson algebras on g using the matrix method. We studied the necessary and sufficient conditions for the automorphism of this class of Lie algebras, and give the decomposition of its automorphism group by $Aut\left(g\right)=G_3{G}_1{G}_2{G}_3{G}_4{G}_7{G}_8{G}_5$, or $Aut\left(g\right)=G_3{G}_1{G}_2{G}_3{G}_4{G}_7{G}_8{G}_5{G}_6$, or $Aut\left(g\right)=G_3{G}_1{G}_2{G}_3{G}_4{G}_7{G}_8{G}_5{G}_3$, where $G_i$ is a commutative subgroup of $Aut\left(g\right)$. We give some subgroups of g’s automorphism group and systematically studied the properties of these subgroups.

1. Introduction

In the past 20 years, Poisson algebra, which has a wide and profound application, has attracted the interests of many researchers, see [1,2,3,4,5,6] for details.

In [1,2,3], the authors studied DG Poisson algebras, Poisson Hopf algebras, Poisson ore extensions and their universal envelope algebras. Jie Tong and Quanqin Jin studied non-commutative Poisson algebra structures on the Lie algebra $s{o}_n\widetilde{\left({\mathbb{C}}_Q\right)}$ and $s{l}_n\widetilde{\left({\mathbb{C}}_Q\right)}$ in [4]. Poisson algebra structures on toroidal Lie algebras, Witt algebra, and Virasoro algebra were studied by researchers in [5,6]. In this paper, we studied the structure of Poisson algebras over four-dimensional Lie algebra g using the matrix method.

Scholars have obtained many profound results on the automorphism of Lie algebras. In [7,8,9], scholars studied the automorphisms of many kinds of Lie algebras, such as the Bianchi model Lie groups and matrix algebras over communicative rings. Automorphisms of some matrix algebras were discussed by scholars in [10,11,12]. Automorphisms of some triangular matrices over commutative rings were explored by researchers in [13,14,15]. In [16], Qiu Yu and Dengyin Wang and Shikun Ou studied the automorphism of standard Borel subalgebras of CM type Lie algebras over a co ring. In a word, many scholars have studied the automorphism of Lie algebras [17,18,19,20,21]. Determining the automorphism group $Aut\left(g\right)$ of a Lie algebra g is a basic problem in the study of the structure theory of Lie algebras. The structure problem of Lie algebras also occupies an irreplaceable position in the study of the structure theory of Lie algebras. The author of this paper has also studied the structure and representation of Lie algebras [22,23,24,25,26].

The set of all second-order square matrices on the complex field $\mathcal{C}$ is denoted as $gl(2,\mathcal{C})$. The definition of $gl(2,\mathcal{C})$ in the four-dimensional Lie algebra satisfies a very special lie operation different from the general one. This operation also satisfies the bilinear, antisymmetric condition and the square bracket product of Jacobi constant equation as follows:

$$A=\left(\begin{array}{cc}{a}_{11}\hfill & a_{12}\hfill \\ a_{21}\hfill & a_{22}\hfill \end{array}\right),B=\left(\begin{array}{cc}{b}_{11}\hfill & b_{12}\hfill \\ b_{21}\hfill & b_{22}\hfill \end{array}\right),$$

$$[A,B]=\left(\begin{array}{cc}0\hfill & a_{11}{b}_{12}+a_{12}{b}_{22}-b_{11}{a}_{12}-b_{12}{a}_{22}\hfill \\ 0\hfill & 0\hfill \end{array}\right).$$

A group of bases of $gl(2,\mathcal{C})$ is $e_{11},e_{12},e_{21},e_{22}$ and satisfies the following formula:

$$e_{11}=\left(\begin{array}{cc}1\hfill & 0\hfill \\ 0\hfill & 0\hfill \end{array}\right),e_{12}=\left(\begin{array}{cc}0\hfill & 1\hfill \\ 0\hfill & 0\hfill \end{array}\right),e_{21}=\left(\begin{array}{cc}0\hfill & 0\hfill \\ 1\hfill & 0\hfill \end{array}\right),e_{22}=\left(\begin{array}{cc}0\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right),$$

and

$$[e_{11},e_{12}]=e_{12},[e_{11},e_{21}]=0,[e_{11},e_{22}]=0,$$

$$[e_{12},e_{21}]=0,[e_{12},e_{22}]=e_{12},[e_{21},e_{22}]=0.$$

For convenience, Lie algebra $gl(2,\mathcal{C})$ is written as Lie algebra g, and represents its square bracket product [A,B] as function $F(A,B)$.

Let R be an elementary divisor ring or a local ring; [27] determined the automorphisms of the general Lie operation

$$[A,B]=AB-BA,$$

linear Lie algebra $sl(2,R)$ and the general linear Lie algebra $gl(2,R)$. However, in this paper, we discussed the automorphisms of the Lie operation of $gl(2,\mathcal{C})$ as:

$$A=\left(\begin{array}{cc}{a}_{11}\hfill & a_{12}\hfill \\ a_{21}\hfill & a_{22}\hfill \end{array}\right),B=\left(\begin{array}{cc}{b}_{11}\hfill & b_{12}\hfill \\ b_{21}\hfill & b_{22}\hfill \end{array}\right),$$

$$F(A,B)=\left(\begin{array}{cc}0\hfill & a_{11}{b}_{12}+a_{12}{b}_{22}-b_{11}{a}_{12}-b_{12}{a}_{22}\hfill \\ 0\hfill & 0\hfill \end{array}\right).$$

2. Main Results

After calculation, it can be verified that $g^{\left(1\right)}=[g,g]$ is a one-dimensional ideal generated by $e_{12}$. So, g is a four-dimensional solvable Lie algebra. The center $Z\left(g\right)$ of g is a two-dimensional subspace generated by $e_{21},e_{11}+e_{22}$. In order to save space, it is no longer verified.

Theorem 1.

g is not a nilpotent Lie algebra.

Proof. 

According to the operation law between the basis vectors of g, the following formula can be obtained.

$$F(e_{11},e_{12})=e_{12},$$

$$F(e_{11},F(e_{11},e_{12}))=e_{12},$$

$$F(e_{11},F(e_{11},F(e_{11},e_{12})))=e_{12},$$

$$F(e_{11},\cdots F(e_{11},F(e_{11},e_{12})))=e_{12}.$$

Thus, g is not a nilpotent Lie algebra. □

Poisson algebra is defined below.

Definition 1.

Define the Poisson algebra $(g,\ast ,[-,-\left]\right)$ on the base field $\mathcal{C}$, which is a vector space g on $\mathcal{C}$, and has bilinear product ∗ and Lie algebra structure $[-,-]$, and the following Leibniz rule holds:

$$F(z\ast x,y)=F(z,y)\ast x+z\ast F(x,y).\forall z,x,y\in g.$$

For any $z,x,y$ in g, ∗ does not necessarily satisfy the associative law and commutative law.

Since ∗ is a bilinear binary operation,

$$x\ast (k_{1}y+k_{2}z)=k_1(x\ast y)+k_2(x\ast z),\forall x,y,z\in g,\forall k_1,k_2\in \mathcal{C},$$

Thus, ∗ induces a left multiply linear transformation $L_x$. Since $\forall x\in g,x$ can be linearly represented by base $e_{11},e_{12},e_{21},e_{22}$. We only need to calculate the matrices of the linear transformation

$$L_{e_{11}},L_{e_{12}},L_{e_{21}},L_{e_{22}}$$

under the basis of

$$e_{11},e_{12},e_{21},e_{22}.$$

When studying the Poisson algebra structure of Lie algebra g, we marke $e_{11}$ as $e_1$, $e_{12}$ as $e_2$, $e_{21}$ as $e_3$, and $e_{22}$ as $e_4$. Note that we only simplify the sign in this way when we study the Poisson algebra structure of the Lie algebra g. When we study the automorphism of the Lie algebra g, we still use the original sign, because

$$F(e_{11},e_{12})=e_{12},F(e_{11},e_{21})=0,F(e_{11},e_{22})=0,$$

$$F(e_{12},e_{21})=0,F(e_{12},e_{22})=e_{12},F(e_{21},e_{22})=0.$$

thus,

$$F(e_1,e_2)=e_2,F(e_1,e_3)=0,F(e_1,e_4)=0,$$

$$F(e_2,e_3)=0,F(e_2,e_4)=e_2,F(e_3,e_4)=0.$$

Theorem 2.

$(g,\ast ,[-,-\left]\right)$ is a Poisson algebra on $(g,[-,-\left]\right)$, then:

$$L_{e_1}(e_1,e_2,e_3,e_4)=(e_1\ast e_1,e_1\ast e_2,e_1\ast e_3,e_1\ast e_4)=(e_1,e_2,e_3,e_4)A,$$

$$A=\left(\begin{array}{cccc}{a}_{11}& a_{21}& a_{31}& a_{41}\\ a_{12}& a_{22}& a_{32}& a_{42}\\ a_{13}& a_{23}& a_{33}& a_{43}\\ a_{14}& a_{24}& a_{34}& a_{44}\end{array}\right)=\left(\begin{array}{cccc}{a}_{11}& 0& a_{31}& a_{41}\\ 0& a_{22}& 0& 0\\ a_{13}& 0& a_{33}& a_{43}\\ a_{14}& 0& a_{34}& a_{44}\end{array}\right)$$

$$L_{e_2}(e_1,e_2,e_3,e_4)=(e_2\ast e_1,e_2\ast e_2,e_2\ast e_3,e_2\ast e_4)=(e_1,e_2,e_3,e_4)B,$$

$$B=\left(\begin{array}{cccc}{b}_{11}& b_{21}& b_{31}& b_{41}\\ b_{12}& b_{22}& b_{32}& b_{42}\\ b_{13}& b_{23}& b_{33}& b_{43}\\ b_{14}& b_{24}& b_{34}& b_{44}\end{array}\right)=\left(\begin{array}{cccc}0& 0& 0& 0\\ b_{12}& 0& b_{32}& b_{42}\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$$

$$L_{e_3}(e_1,e_2,e_3,e_4)=(e_3\ast e_1,e_3\ast e_2,e_3\ast e_3,e_3\ast e_4)=(e_1,e_2,e_3,e_4)C,$$

$$C=\left(\begin{array}{cccc}{c}_{11}& c_{21}& c_{31}& c_{41}\\ c_{12}& c_{22}& c_{32}& c_{42}\\ c_{13}& c_{23}& c_{33}& c_{43}\\ c_{14}& c_{24}& c_{34}& c_{44}\end{array}\right)=\left(\begin{array}{cccc}0& 0& c_{31}& c_{41}\\ 0& 0& 0& 0\\ 0& 0& c_{33}& c_{43}\\ 0& 0& c_{34}& c_{44}\end{array}\right)$$

$$L_{e_4}(e_1,e_2,e_3,e_4)=(e_4\ast e_1,e_4\ast e_2,e_4\ast e_3,e_4\ast e_4)=(e_1,e_2,e_3,e_4)D,$$

$$D=\left(\begin{array}{cccc}{d}_{11}& d_{21}& d_{31}& d_{41}\\ d_{12}& d_{22}& d_{32}& d_{42}\\ d_{13}& d_{23}& d_{33}& d_{43}\\ d_{14}& d_{24}& d_{34}& d_{44}\end{array}\right)=\left(\begin{array}{cccc}{d}_{11}& 0& d_{31}& d_{41}\\ 0& d_{22}& 0& 0\\ d_{13}& 0& d_{33}& d_{43}\\ d_{14}& 0& d_{34}& d_{44}\end{array}\right).$$

Proof. 

Since the Leibniz law is established, there are

$$F(z\ast x,y)=F(z,y)\ast x+z\ast F(x,y),\forall z,x,y\in g.$$

(1)

Since ∗ and $[,]$ are bilinear operations. If $Z,x,y$ can only select $e_1,e_2,e_3,e_4$, so that there are (1) cases of $4^3=64$, one by one can be verified. When

$$z=e_1,x=e_1,y=e_1,$$

the following equation can be obtained from (1):

$$F(e_1\ast e_1,e_1)=F(e_1,e_1)\ast e_1+e_1\ast F(e_1,e_1),$$

(2a)

left side of (2a)

$$=F(a_{11}{e}_1+a_{12}{e}_2+a_{13}{e}_3+a_{14}{e}_4,e_1)=F(a_{12}{e}_2,e_1)=-a_{12}{e}_2,$$

right side of (2a)

$$=F(e_1,e_1)\ast e_1+e_1\ast F(e_1,e_1)=0,$$

thus

$$a_{12}=0.$$

Because

$$F(e_1\ast e_1,e_2)=F(e_1,e_2)\ast e_1+e_1\ast F(e_1,e_2),$$

(2b)

left side of (2b)

$$=F(a_{11}{e}_1+a_{12}{e}_2+a_{13}{e}_3+a_{14}{e}_4,e_2)=(a_{11}-a_{14})e_2,$$

right side of (2b)

$$=F(e_1,e_2)\ast e_1+e_1\ast F(e_1,e_2)=e_2\ast e_1+e_1\ast e_2=b_{11}{e}_1+b_{12}{e}_2+b_{13}{e}_3+b_{14}{e}_4+a_{21}{e}_1+a_{22}{e}_2+a_{23}{e}_3+a_{24}{e}_4,$$

thus

$$b_{11}+a_{21}=0,b_{13}+a_{23}=0,b_{14}+a_{24}=0,b_{12}+a_{22}=(a_{11}-a_{14}).$$

Because

$$F(e_1\ast e_1,e_3)=F(e_1,e_3)\ast e_1+e_1\ast F(e_1,e_3),$$

(2c)

left side of (2c) = right side of (2c), so Equation (2c) holds.

Because

$$F(e_1\ast e_1,e_4)=F(e_1,e_4)\ast e_1+e_1\ast F(e_1,e_4),$$

(2d)

left side of (2d)

$$=F(a_{11}{e}_1+a_{12}{e}_2+a_{13}{e}_3+a_{14}{e}_4,e_4)=a_{12}{e}_2,$$

right side of (2d)

$$=F(e_1,e_4)\ast e_1+e_1\ast F(e_1,e_4)=0\ast e_1+e_1\ast 0=0,$$

thus,

$$a_{12}=0.$$

Because

$$F(e_1\ast e_2,e_1)=F(e_1,e_1)\ast e_2+e_1\ast F(e_2,e_1),$$

(2e)

left side of (2e)

$$=F(a_{21}{e}_1+a_{22}{e}_2+a_{23}{e}_3+a_{24}{e}_4,e_1)=F(a_{22}{e}_2,e_1)=-a_{22}{e}_2,$$

right side of (2e)

$$=F(e_1,e_1]\ast e_2+e_1\ast F(e_2,e_1)=0-e_1\ast e_2=-(a_{21}{e}_1+a_{22}{e}_2+a_{23}{e}_3+a_{24}{e}_4),$$

thus,

$$a_{21}=0,a_{23}=0,a_{24}=0.$$

Because

$$F(e_1\ast e_2,e_2)=F(e_1,e_2)\ast e_2+e_1\ast F(e_2,e_2),$$

(2f)

left side of (2f)

$$=F(a_{21}{e}_1+a_{22}{e}_2+a_{23}{e}_3+a_{24}{e}_4,e_2)=a_{21}{e}_2-a_{24}{e}_2,$$

right side of (2f)

$$=e_2\ast e_2=b_{21}{e}_1+b_{22}{e}_2+b_{23}{e}_3+b_{24}{e}_4,$$

thus,

$$b_{21}=0,b_{23}=0,b_{24}=0,(a_{21}-a_{24})=b_{22}.$$

Because

$$F(e_1\ast e_2,e_3)=F(e_1,e_2)\ast e_3+e_1\ast F(e_2,e_3),$$

(2g)

left side of (2g) = right side of (2g) so Equation (2g) holds.

Because

$$F(e_1\ast e_2,e_4)=F(e_1,e_4)\ast e_2+e_1\ast F(e_2,e_4),$$

(2h)

and it can be known from (2e):

$$a_{21}=0,a_{23}=0,a_{24}=0.$$

left side of (2h)

$$F(a_{22}{e}_2,e_4)=a_{22}{e}_2,$$

right side of (2h)

$$F(e_1,e_4)\ast e_2+e_1\ast F(e_2,e_4)=0+e_1\ast e_2=a_{22}{e}_2,$$

So the equation holds.

By analogy from the remaining 56 cases:

$$a_{32}=0,b_{31}=0,b_{33}=0,b_{34}=0,(a_{31}-a_{34})=b_{32},a_{42}=0,$$

$$b_{21}=b_{23}=b_{22}=b_{24}=0,b_{41}=a_{21},b_{42}-a_{22}=a_{41}-a_{44},b_{43}=a_{23},b_{44}=a_{24}.$$

$$b_{11}=b_{13}=b_{14}=0,b_{31}=b_{33}=b_{34}=0,b_{22}=b_{11}-b_{14},b_{41}=0,b_{43}=0,b_{44}=0,$$

$$c_{12}=0,c_{21}=c_{23}=c_{24}=0,c_{22}=c_{11}-c_{14}=0,c_{22}=0,c_{11}=0,c_{13}=0,c_{14}=0,c_{24}=0.$$

$$c_{32}=0,c_{31}=c_{34},c_{42}=0,c_{41}-c_{44}=0,$$

$$d_{12}=0,d_{32}=0,d_{21}=d_{23}=d_{24}=0,d_{31}-d_{34}=-b_{32},d_{42}=0,d_{41}-d_{44}=-b_{42}-d_{22}.$$

Since ∗ is a bilinear binary operation,

$$(k_{1}y+k_{2}z)\ast x=k_1(y\ast x)+k_2(z\ast x),\forall x,y,z\in g,\forall k_1,k_2\in \mathcal{C}.$$

Thus, ∗ induces a right multiply linear transformation $R_x$. Since any x in g can be linearly represented by base $e_{11},e_{12},e_{21},e_{22}$, we only need to calculate the matrices of linear transformation

$$R_{e_{11}},R_{e_{12}},R_{e_{21}},R_{e_{22}}$$

under base

$$e_{11},e_{12},e_{21},e_{22}.$$

□

Since it is similar to Theorem 2, in order to save space, it will not be described again.

In addition, if any $z,x,y$ in g, ∗ satisfies the associative law or the commutative law, there will be more strict requirements for the matrix $A,B,C,D$. In order to save space, we will not repeat it.

Definition 2.

Let $g_1,g_2$ be a Lie algebra over field F. If the linear mapping φ of $g_1$ to $g_2$ satisfies

$$F(\phi \left(x\right),\phi \left(y\right))=\phi \left(F(x,y)\right),\forall x,yϵg_1,$$

then 6φ is said to be a homomorphic mapping or homomorphism of $g_1$ to $g_2$.

Definition 3.

The homomorphism of a Lie algebra g to itself is called the endomorphism of g, and all endomorphisms of g are denoted as $End\left(g\right)$. The isomorphism from g to itself is called automorphism and all automorphisms of g form a group, which is called the automorphism group of g and is called $Aut\left(g\right)$.

Theorem 3.

The linear mapping in the four-dimensional Lie algebra g is established as follows:

$$\phi (e_{11},e_{12},e_{21},e_{22})=(e_{11},e_{12},e_{21},e_{22})\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}.\end{array}\right).$$

If φ is an automorphism of Lie algebra g, then there must be

$$a_{12}=a_{32}=a_{42}=0,$$

$$a_{23}=0,$$

$$a_{44}-a_{14}=1,$$

$$a_{11}-a_{41}=1,$$

$$a_{43}-a_{13}=0,$$

$$a_{24}+a_{21}=0.$$

Proof. 

Let

$$\overline{A}=\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right).$$

Let $\phi$ be an automorphism on g, then $\phi$ must be a linear transformation on g.

$\phi (e_{11},e_{12},e_{21},e_{22})=(e_{11},e_{12},e_{21},e_{22})\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right)=(e_{11},e_{12},e_{21},e_{22})\overline{A},$

thus,

$$\phi \left(e_{11}\right)=a_{11}{e}_{11}+a_{21}{e}_{12}+a_{31}{e}_{21}+a_{41}{e}_{22},$$

$$\phi \left(e_{12}\right)=a_{12}{e}_{11}+a_{22}{e}_{12}+a_{32}{e}_{21}+a_{42}{e}_{22},$$

$$\phi \left(e_{21}\right)=a_{13}{e}_{11}+a_{23}{e}_{12}+a_{33}{e}_{21}+a_{43}{e}_{22},$$

$$\phi \left(e_{22}\right)=a_{14}{e}_{11}+a_{24}{e}_{12}+a_{34}{e}_{21}+a_{44}{e}_{22}.$$

According to the definition of isomorphism, $F(\phi \left(e_{ij}\right),\phi \left(e_{mn}\right))=\phi \left(F(e_{ij},e_{mn})\right),$$(i,j,m,n=1,2),$

(1) Because

$$F(e_{11},e_{12})=e_{12},$$

so

$$F(\phi \left(e_{11}\right),\phi \left(e_{12}\right))=\phi \left(F(e_{11},e_{12})\right)=\phi \left(e_{12}\right),$$

$$F(a_{11}{e}_{11}+a_{21}{e}_{12}+a_{31}{e}_{21}+a_{41}{e}_{22},a_{12}{e}_{11}+a_{22}{e}_{12}+a_{32}{e}_{21}+a_{42}{e}_{22})=a_{12}{e}_{11}+a_{22}{e}_{12}+a_{32}{e}_{21}+a_{42}{e}_{22},$$

$$(a_{11}{a}_{22}-a_{12}{a}_{21}+a_{21}{a}_{42}-a_{41}{a}_{22})e_{12}=a_{12}{e}_{11}+a_{22}{e}_{12}+a_{32}{e}_{21}+a_{42}{e}_{22};$$

thus,

$$a_{12}=a_{32}=a_{42}=0,a_{11}{a}_{22}-a_{12}{a}_{21}+a_{21}{a}_{42}-a_{41}{a}_{22}=a_{22}.$$

Therefore, there are

$$a_{11}{a}_{22}-a_{41}{a}_{22}=a_{22}.$$

(2) Because

$$F(e_{11},e_{21})=0,$$

so

$$F(\phi \left(e_{11}\right),\phi \left(e_{21}\right))=\phi \left(F(e_{11},e_{21})\right)=0,$$

$$F(a_{11}{e}_{11}+a_{21}{e}_{12}+a_{31}{e}_{21}+a_{41}{e}_{22},a_{13}{e}_{11}+a_{23}{e}_{12}+a_{33}{e}_{21}+a_{43}{e}_{22})=0,$$

$$(a_{11}{a}_{23}-a_{21}{a}_{13}+a_{21}{a}_{43}-a_{41}{a}_{23})e_{12}=0,$$

thus

$$a_{11}{a}_{23}-a_{21}{a}_{13}+a_{21}{a}_{43}-a_{41}{a}_{23}=0.$$

(3) Because

$$F(e_{11},e_{22})=0,$$

so

$$F(\phi \left(e_{11}\right),\phi \left(e_{22}\right))=\phi \left(F(e_{11},e_{22})\right)=0,$$

$$F(a_{11}{e}_{11}+a_{21}{e}_{12}+a_{31}{e}_{21}+a_{41}{e}_{22},a_{14}{e}_{11}+a_{24}{e}_{12}+a_{34}{e}_{21}+a_{44}{e}_{22})=0,$$

$$(a_{11}{a}_{24}-a_{21}{a}_{14}+a_{21}{a}_{44}-a_{41}{a}_{24})e_{12}=0,$$

thus

$$a_{11}{a}_{24}-a_{21}{a}_{14}+a_{21}{a}_{44}-a_{41}{a}_{24}=0.$$

(4) Because

$$F(e_{12},e_{21})=0,$$

so

$$F(\phi \left(e_{12}\right),\phi \left(e_{21}\right))=\phi \left(F(e_{12},e_{21})\right)=0,$$

$$F(a_{12}{e}_{11}+a_{22}{e}_{12}+a_{32}{e}_{21}+a_{42}{e}_{22},a_{13}{e}_{11}+a_{23}{e}_{12}+a_{33}{e}_{21}+a_{43}{e}_{22})=0,$$

$$(a_{12}{a}_{23}-a_{22}{a}_{13}+a_{22}{a}_{43}-a_{42}{a}_{23})e_{12}=0,$$

thus

$$a_{12}{a}_{23}-a_{22}{a}_{13}+a_{22}{a}_{43}-a_{42}{a}_{23}=0.$$

(5) Because

$$F(e_{12},e_{22})=e_{12},$$

so

$$F(\phi \left(e_{12}\right),\phi \left(e_{22}\right))]=\phi \left(F(e_{12},e_{22})\right)=\phi \left(e_{12}\right),$$

$$F(a_{12}{e}_{11}+a_{22}{e}_{12}+a_{32}{e}_{21}+a_{42}{e}_{22},a_{14}{e}_{11}+a_{24}{e}_{12}+a_{34}{e}_{21}+a_{44}{e}_{22})=a_{12}{e}_{11}+a_{22}{e}_{12}+a_{32}{e}_{21}+a_{42}{e}_{22},$$

$$(a_{12}{a}_{24}-a_{22}{a}_{14}+a_{22}{a}_{44}-a_{42}{a}_{24})e_{12}=a_{12}{e}_{11}+a_{22}{e}_{12}+a_{32}{e}_{21}+a_{42}{e}_{22};$$

thus,

$$a_{12}=a_{32}=a_{42}=0,a_{12}{a}_{24}-a_{22}{a}_{14}+a_{22}{a}_{44}-a_{42}{a}_{24}=a_{22}.$$

Therefore, there are

$$-a_{22}{a}_{14}+a_{22}{a}_{44}=a_{22}.$$

(6) Because

$$F(e_{21},e_{22})=0,$$

so

$$F(\phi \left(e_{21}\right),\phi \left(e_{22}\right))=\phi \left([e_{21},e_{22})\right)=0,$$

$$F(a_{13}{e}_{11}+a_{23}{e}_{12}+a_{33}{e}_{21}+a_{43}{e}_{22},a_{14}{e}_{11}+a_{24}{e}_{12}+a_{34}{e}_{21}+a_{44}{e}_{22})=0,$$

$$(a_{13}{a}_{24}-a_{23}{a}_{14}+a_{23}{a}_{44}-a_{43}{a}_{24})e_{12}=0;$$

thus,

$$a_{13}{a}_{24}-a_{23}{a}_{14}+a_{23}{a}_{44}-a_{43}{a}_{24}=0.$$

Based on the above six cases, the following equation holds:

$$a_{12}=a_{32}=a_{42}=0,$$

$$(a_{43}-a_{13})a_{22}=0,$$

$$a_{22}(1+a_{14}-a_{44})=0,$$

$$a_{22}(1+a_{41}-a_{11})=0,$$

$$(a_{11}-a_{41})a_{23}+(a_{43}-a_{13})a_{21}=0,$$

$$(a_{11}-a_{41})a_{24}+(a_{44}-a_{14})a_{21}=0,$$

$$(a_{13}-a_{43})a_{24}+(a_{44}-a_{14})a_{23}=0.$$

Since $\phi$ is isomorphic, $a_{12}=a_{32}=a_{42}=0$ is known from the previous reasoning, so there must be $a_{22}\ne 0$, otherwise:

$$\left|\begin{array}{cccc}{a}_{11}& 0& a_{13}& a_{14}\\ a_{21}& 0& 0& -a_{21}\\ a_{31}& 0& a_{33}& a_{34}\\ a_{11}-1& 0& a_{13}& a_{14}+1\end{array}\right|=0;$$

it is an isomorphic contradiction with $\phi$. Thus, $a_{22}\ne 0$, and the following equation holds:

$$a_{23}=0,$$

$$a_{44}-a_{14}=1,$$

$$a_{11}-a_{41}=1,$$

$$a_{43}-a_{13}=0,$$

$$a_{24}+a_{21}=0;$$

so:

$$\overline{A}=\left(\begin{array}{cccc}{a}_{11}& 0& a_{13}& a_{14}\\ a_{21}& a_{22}& 0& -a_{21}\\ a_{31}& 0& a_{33}& a_{34}\\ a_{11}-1& 0& a_{13}& a_{14}+1\end{array}\right),$$

$|\overline{A}|=a_{22}(a_{13}(a_{34}+a_{31})-a_{33}(a_{14}+a_{11}))\ne 0.$□

Theorem 4.

Let

$$G_1=\left\{\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1,\end{array}\right)\right|\forall a_{21}\in \mathcal{C}\},$$

then $g_1$ is a commutative subgroup of $Aut\left(g\right)$.

Proof. 

Obviously, $G_1\subseteq Aut\left(g\right)$ holds. For any ${\gamma }_1,{\gamma }_2$ in $G_1$, there are

$${\gamma }_1{\gamma }_2=\left(\begin{array}{cccc}1& 0& 0& 0\\ m_1& 1& 0& -m_1\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ m_2& 1& 0& -m_2\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ m_1+m_2& 1& 0& -(m_1+m_2)\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),$$

so ${\gamma }_1{\gamma }_2$ belongs to $G_1$, easy-to-know ${\gamma }_1{\gamma }_2$ is equal to ${\gamma }_2{\gamma }_1$. Let

$${\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ m_1& 1& 0& -m_1\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\in G_1,$$

$${\gamma }_1^{-1}=\left(\begin{array}{cccc}1& 0& 0& 0\\ -m_1& 1& 0& m_1\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\in G_1.$$

So, $G_1$ is a commutative subgroup of $Aut\left(g\right)$. □

Theorem 5.

Let

$$G_2=\left\{\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& a_{34}\\ 0& 0& 0& 1\end{array}\right)\right|\forall a_{34}\in \mathcal{C}\};$$

then, $G_2$ is a commutative subgroup of $Aut\left(g\right)$.

Proof. 

Obviously, $G_2\subseteq Aut\left(g\right)$ holds. For any ${\gamma }_1,{\gamma }_2$ in $G_2$, there are

$${\gamma }_1{\gamma }_2=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& m_1\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& m_2\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& m_1+m_2\\ 0& 0& 0& 1\end{array}\right);$$

so ${\gamma }_1{\gamma }_2$ belongs to $G_2$, easy-to-know ${\gamma }_1{\gamma }_2$ is equal to ${\gamma }_2{\gamma }_1$.

$$\forall {\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& m_1\\ 0& 0& 0& 1\end{array}\right)\in G_2,\forall m_1\in \mathcal{C},$$

$${\gamma }_1^{-1}=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& -m_1\\ 0& 0& 0& 1\end{array}\right)\in G_2;$$

so, $G_2$ is a commutative subgroup of $Aut\left(g\right)$. □

Theorem 6.

Let

$$G_3=\left\{\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ a_{31}& 0& 1& 0\\ 0& 0& 0& 1,\end{array}\right)\right|\forall a_{31}\in \mathcal{C}\},$$

then $G_3$ is a commutative subgroup of $Aut\left(g\right)$.

Proof. 

Obviously, $G_3\subseteq Aut\left(g\right)$ holds. For any ${\gamma }_1,{\gamma }_2$ in $G_3$, there are

$${\gamma }_1{\gamma }_2=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ m_1& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ m_2& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ m_1+m_2& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right);$$

so, ${\gamma }_1{\gamma }_2$ belongs to $G_3$, easy-to-know ${\gamma }_1{\gamma }_2$ is equal to ${\gamma }_2{\gamma }_1$.

$$\forall {\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ m_1& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\in G_1,\forall m_1\in \mathcal{C},$$

$${\gamma }_1^{-1}=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -m_1& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\in G_3;$$

so, $G_3$ is a commutative subgroup of $Aut\left(g\right)$. □

Theorem 7.

Let

$$G_4=\left\{\left(\begin{array}{cccc}1& 0& a_{13}& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& a_{13}& 1\end{array}\right)\right|\forall a_{13}\in \mathcal{C}\},$$

then $G_4$ is a commutative subgroup of $Aut\left(g\right)$.

Proof. 

Obviously, $G_4\subseteq Aut\left(g\right)$ holds. For any ${\gamma }_1,{\gamma }_2$ in $G_4$, there are

$${\gamma }_1{\gamma }_2=\left(\begin{array}{cccc}1& 0& m_1& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& m_1& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& m_2& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& m_2& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& m_1+m_2& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& -(m_1+m_2)& 1\end{array}\right);$$

so, ${\gamma }_1{\gamma }_2$ belongs to $G_4$, easy-to-know ${\gamma }_1{\gamma }_2$ is equal to ${\gamma }_2{\gamma }_1$.

$$\forall {\gamma }_1=\left(\begin{array}{cccc}1& 0& m_1& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& m_1& 1\end{array}\right)\in G_4,\forall m_1\in \mathcal{C},$$

$${\gamma }_1^{-1}=\left(\begin{array}{cccc}1& 0& -m_1& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& -m_1& 1\end{array}\right)\in G_4;$$

so, $G_4$ is a commutative subgroup of $Aut\left(g\right)$. □

Theorem 8.

Let

$$G_5=\left\{\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& 0& 0\\ 0& 0& a_{33}& 0\\ 0& 0& 0& 1\end{array}\right)\right|\forall a_{22}\ne 0,a_{33}\ne 0\in \mathcal{C}\},$$

then $G_5$ is a commutative subgroup of $Aut\left(g\right)$.

Proof. 

Obviously, $G_5\subseteq Aut\left(g\right)$ holds. For any ${\gamma }_1,{\gamma }_2$ in $G_5$, $\forall a_{22}\ne 0,a_{33}\ne 0\in \mathcal{C},$$b_{22}\ne 0,b_{33}\ne 0\in \mathcal{C}$, there are

$${\gamma }_1{\gamma }_2=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& 0& 0\\ 0& 0& a_{33}& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& b_{22}& 0& 0\\ 0& 0& b_{33}& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}{b}_{22}& 0& 0\\ 0& 0& a_{33}{b}_{33}& 0\\ 0& 0& 0& 1\end{array}\right);$$

so, ${\gamma }_1{\gamma }_2$ belongs to $G_5$, easy-to-know ${\gamma }_1{\gamma }_2$ is equal to ${\gamma }_2{\gamma }_1$.

$$\forall {\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& 0& 0\\ 0& 0& a_{33}& 0\\ 0& 0& 0& 1\end{array}\right)\in G_5,\forall a_{22}\ne 0,a_{33}\ne 0,\in \mathcal{C},$$

$${\gamma }_1^{-1}=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}^{-1}& 0& 0\\ 0& 0& a_{33}^{-1}& 0\\ 0& 0& 0& 1\end{array}\right)\in G_5;$$

so, $G_5$ is a commutative subgroup of $Aut\left(g\right)$. □

Theorem 9.

Let

$$G_6=\{\left(\begin{array}{cccc}0& 0& 0& -1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right),\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\};$$

then $G_6$ is a second order cyclic subgroup of $Aut\left(g\right)$.

Proof. 

Obviously, $G_6\subseteq Aut\left(g\right)$ holds.

$$\left(\begin{array}{cccc}0& 0& 0& -1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right)\left(\begin{array}{cccc}0& 0& 0& -1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),$$

$$\forall {\gamma }_1=\left(\begin{array}{cccc}0& 0& 0& -1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right)\in G_6,$$

$${\gamma }_1^{-1}=\left(\begin{array}{cccc}0& 0& 0& -1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right)\in G_6;$$

so, $G_6$ is a second order cyclic subgroup of $Aut\left(g\right)$. □

Theorem 10.

Let

$$G_7=\left\{\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ a_{11}-1& 0& 0& 1\end{array}\right)\right|\forall a_{11}\in \mathcal{C}\},$$

then $G_7$ is a commutative subgroup of $Aut\left(g\right)$.

Proof. 

Obviously, $G_7\subseteq Aut\left(g\right)$ holds. For any ${\gamma }_1,{\gamma }_2$ in $G_7$, $\forall a_{11}\ne 0,b_{11}\ne 0\in \mathcal{C},$ there are

$${\gamma }_1{\gamma }_2=\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ a_{11}-1& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}{b}_{11}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ b_{11}-1& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}{a}_{11}{b}_{11}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ a_{11}{b}_{11}-1& 0& 0& 1\end{array}\right);$$

so, ${\gamma }_1{\gamma }_2$ belongs to $G_7$, easy-to-know ${\gamma }_1{\gamma }_2$ is equal to ${\gamma }_2{\gamma }_1$. $\forall a_{11}\ne 0\in \mathcal{C}$,

$${\gamma }_1=\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ a_{11}-1& 0& 0& 1\end{array}\right)\in G_7,$$

$${\gamma }_1^{-1}=\left(\begin{array}{cccc}{a}_{11}^{-1}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ a_{11}^{-1}-1& 0& 0& 1\end{array}\right)\in G_7;$$

so, $G_7$ is a commutative subgroup of $Aut\left(g\right)$. □

Theorem 11.

Let

$$G_8=\left\{\left(\begin{array}{cccc}1& 0& 0& a_{14}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1+a_{14}\end{array}\right)\right|\forall a_{14}\in \mathcal{C}\};$$

then $G_8$ is a commutative subgroup of $Aut\left(g\right)$.

Proof. 

Obviously, $G_8\subseteq Aut\left(g\right)$ holds. For any ${\gamma }_1,{\gamma }_2$ in $G_8$, $\forall a_{14}\ne -1,b_{14}\ne -1\in \mathcal{C},$

$${\gamma }_1{\gamma }_2=\left(\begin{array}{cccc}1& 0& 0& a_{14}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1+a_{14}\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& b_{14}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1+b_{14}\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& a_{14}{b}_{14}+a_{14}+b_{14}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1+a_{14}{b}_{14}+a_{14}+b_{14}\end{array}\right);$$

so, ${\gamma }_1{\gamma }_2$ belongs to $G_8$, easy-to-know ${\gamma }_1{\gamma }_2$ is equal to ${\gamma }_2{\gamma }_1$. $\forall a_{14}\ne -1\in \mathcal{C}$,

$${\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& a_{14}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1+a_{14}\end{array}\right)\in G_8,$$

$${\gamma }_1^{-1}=\left(\begin{array}{cccc}1& 0& 0& -\frac{a_{14}}{1+a_{14}}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1-\frac{a_{14}}{1+a_{14}}\end{array}\right)\in G_8;$$

so, $G_8$ is a commutative subgroup of $Aut\left(g\right)$. □

Theorem 12.

$G_1$ and $G_2$ are interchangeable.

Proof. 

$\forall {\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)ϵG_1,{\gamma }_2=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& a_{34}\\ 0& 0& 0& 1\end{array}\right)ϵG_2$, because

$${\gamma }_1{\gamma }_2=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& a_{34}\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& a_{34}\\ 0& 0& 0& 1\end{array}\right),$$

$${\gamma }_2{\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& a_{34}\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& a_{34}\\ 0& 0& 0& 1\end{array}\right);$$

so,

$${\gamma }_1{\gamma }_2={\gamma }_2{\gamma }_1.$$

Thus, $G_1$ and $G_2$ are interchangeable. □

Definition 4.

Given that $G_1$ and $G_2$ are two subgroups of $Aut\left(g\right)$, let $G_1\circ G_2=\left\{xy\right|\forall x\in G_1,\forall y\in G_2\}.$

Theorem 13.

$G_1\circ G_2=\left\{xy\right|\forall x\in G_1,\forall y\in G_2\}$ is a subgroup of $Autg$.

Proof. 

Because $e\in G_1$ and $e\in G_2$, so $ee=e\in G_1\circ G_2$.

$\forall x_1,x_2\in G_1;y_1,y_2\in G_2$, due to $G_1$ and $G_2$ is exchangeable, so

$$x_1{y}_1{x}_2{y}_2=x_1{x}_2{y}_1{y}_2\in G_1\circ G_2,$$

and

$${\left(x_1{y}_1\right)}^{-1}=y_1^{-1}{x}_1^{-1}=x_1^{-1}{y}_1^{-1}\in G_1\circ G_2.$$

□

Theorem 14.

$G_1$ and $G_3$ are interchangeable.

Proof. 

$\forall {\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)ϵG_1,{\gamma }_3=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ a_{31}& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)ϵG_3$, because

$${\gamma }_1{\gamma }_3=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ a_{31}& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ a_{31}& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),$$

$${\gamma }_3{\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ a_{31}& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ a_{31}& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),$$

so

$${\gamma }_1{\gamma }_3={\gamma }_3{\gamma }_1.$$

Thus, $G_1$ and $G_3$ are interchangeable. □

Theorem 15.

$G_1$ and $G_4$ are interchangeable.

Proof. 

$\forall {\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)ϵG_1,{\gamma }_4=\left(\begin{array}{cccc}1& 0& a_{13}& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& a_{13}& 1\end{array}\right)ϵG_4$, because

$${\gamma }_1{\gamma }_4=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& a_{13}& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& a_{13}& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& a_{13}& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& a_{13}& 1\end{array}\right),$$

$${\gamma }_4{\gamma }_1=\left(\begin{array}{cccc}1& 0& a_{13}& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& a_{13}& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& a_{13}& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& a_{13}& 1\end{array}\right);$$

so,

$${\gamma }_1{\gamma }_4={\gamma }_4{\gamma }_1.$$

Thus, $G_1$ and $G_4$ are interchangeable. □

Theorem 16.

$G_1$ and $G_5$ are not necessarily interchangeable.

Proof. 

$\forall {\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)ϵG_1,{\gamma }_5=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& 0& 0\\ 0& 0& a_{33}& 0\\ 0& 0& 0& 1\end{array}\right)ϵG_5$, because

$${\gamma }_1{\gamma }_5=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& 0& 0\\ 0& 0& a_{33}& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& a_{22}& 0& -a_{21}\\ 0& 0& a_{33}& 0\\ 0& 0& 0& 1\end{array}\right),$$

$${\gamma }_5{\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& 0& 0\\ 0& 0& a_{33}& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}{a}_{22}& a_{22}& 0& -a_{21}{a}_{22}\\ 0& 0& a_{33}& 0\\ 0& 0& 0& 1\end{array}\right);$$

so, ${\gamma }_1{\gamma }_5$ is not necessarily equal to ${\gamma }_5{\gamma }_1.$

Thus, $G_1$ and $G_5$ are not necessarily interchangeable. □

Theorem 17.

$G_1$ and $G_6$ are interchangeable.

Proof. 

$\forall {\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)ϵG_1,{\gamma }_{61}=\left(\begin{array}{cccc}0& 0& 0& -1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right),{\gamma }_{62}=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),$ because

$${\gamma }_1{\gamma }_{61}=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}0& 0& 0& -1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right)=\left(\begin{array}{cccc}0& 0& 0& -1\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right),$$

$${\gamma }_{61}{\gamma }_1=\left(\begin{array}{cccc}0& 0& 0& -1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}0& 0& 0& -1\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right);$$

so,

$${\gamma }_1{\gamma }_{61}={\gamma }_{61}{\gamma }_1,$$

obviously,

$${\gamma }_1{\gamma }_{62}={\gamma }_{62}{\gamma }_1.$$

Thus, $G_1$ and $G_4$ are interchangeable. □

Theorem 18.

$G_1$ and $G_7$ are interchangeable.

Proof. 

$\forall {\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)ϵG_1,{\gamma }_7=\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ a_{11}-1& 0& 0& 1\end{array}\right)ϵG_7$, bcause

$${\gamma }_1{\gamma }_7=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ a_{11}-1& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ a_{11}-1& 0& 0& 1\end{array}\right),$$

$${\gamma }_7{\gamma }_1=\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ a_{11}-1& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ a_{11}-1& 0& 0& 1\end{array}\right);$$

so,

$${\gamma }_1{\gamma }_7={\gamma }_7{\gamma }_1.$$

Thus, $G_1$ and $G_7$ are interchangeable. □

Theorem 19.

$G_1$ and $G_8$ are interchangeable.

Proof. 

$\forall {\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)ϵG_1,{\gamma }_8=\left(\begin{array}{cccc}1& 0& 0& a_{14}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1+a_{14}\end{array}\right)ϵG_8$, because

$${\gamma }_1{\gamma }_8=\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& a_{14}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1+a_{14}\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& a_{14}\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1+a_{14}\end{array}\right),$$

$${\gamma }_8{\gamma }_1=\left(\begin{array}{cccc}1& 0& 0& a_{14}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1+a_{14}\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& a_{14}\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1+a_{14}\end{array}\right);$$

so,

$${\gamma }_1{\gamma }_8={\gamma }_8{\gamma }_1.$$

Thus, $G_1$ and $G_8$ are interchangeable. □

Similarly, we can study the commutativity between $G_2,G_3,G_4,G_5,G_6,G_7$ and $G_8$. For example, we can prove that $G_2$ and $G_3$ can be exchanged, $G_5$ and $G_7$ can be exchanged, and $G_5$ and $G_8$ can be exchanged. Studying this commutativity is certainly helpful for the subsequent study of whether the decomposition of automorphism groups is unique.

Theorem 20.

The automorphism group $Aut\left(g\right)=g$ of Lie algebra g can be decomposed into the following form:

(1) When $b_{11}\ne 0$, there are

$$G=G_3{G}_1{G}_2{G}_3{G}_4{G}_7{G}_8{G}_5;$$

(2) When $b_{11}=0,b_{14}\ne 0$, there are

$$G=G_3{G}_1{G}_2{G}_3{G}_4{G}_7{G}_8{G}_5{G}_6;$$

(3) When $b_{11}=0,b_{14}=0,b_{13}\ne 0$, there are

$$G=G_3{G}_1{G}_2{G}_3{G}_4{G}_7{G}_8{G}_5{G}_3.$$

Proof. 

Take any B in G at $Aut\left(g\right)$; let

$$B=\left(\begin{array}{cccc}{b}_{11}& 0& b_{13}& b_{14}\\ b_{21}& b_{22}& 0& -b_{21}\\ b_{31}& 0& b_{33}& b_{34}\\ b_{11}-1& 0& b_{13}& b_{14}+1\end{array}\right).$$

Using the undetermined coefficient method, for

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -a_{34}& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\in G_3,\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\in G_1,\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& a_{34}\\ 0& 0& 0& 1\end{array}\right)\in G_2,$$

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ a_{31}& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\in G_3\left(\begin{array}{cccc}1& 0& a_{13}& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& a_{13}& 1\end{array}\right)\in G_4,$$

$$\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ a_{11}-1& 0& 0& 1\end{array}\right)\in G_7,\left(\begin{array}{cccc}1& 0& 0& a_{14}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1+a_{14}\end{array}\right)\in G_8,\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\in G_5,$$

it is advisable to set

$$B=\left(\begin{array}{cccc}{b}_{11}& 0& b_{13}& b_{14}\\ b_{21}& b_{22}& 0& -b_{21}\\ b_{31}& 0& b_{33}& b_{34}\\ b_{11}-1& 0& b_{13}& b_{14}+1\end{array}\right)=$$

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -a_{34}& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ a_{21}& 1& 0& -a_{21}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& a_{34}\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ a_{31}& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& a_{13}& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& a_{13}& 1\end{array}\right)$$

$$\left(\begin{array}{cccc}{a}_{11}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ a_{11}-1& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& a_{14}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1+a_{14}\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& a_{22}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)$$

$$=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -a_{34}& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}{a}_{11}& 0& a_{13}& a_{11}{a}_{14}\\ a_{21}& a_{22}& 0& -a_{21}\\ a_{11}{a}_{31}+a_{11}{a}_{34}-a_{34}& 0& a_{13}{a}_{31}+a_{13}{a}_{34}+1& a_{11}{a}_{14}{a}_{31}+a_{11}{a}_{14}{a}_{34}+a_{34}\\ a_{11}-1& 0& a_{13}& a_{11}{a}_{14}+1\end{array}\right)$$

$$=\left(\begin{array}{cccc}{a}_{11}& 0& a_{13}& a_{11}{a}_{14}\\ a_{21}& a_{22}& 0& -a_{21}\\ a_{11}{a}_{31}-a_{34}& 0& a_{13}{a}_{31}+1& a_{11}{a}_{14}{a}_{31}+a_{34}\\ a_{11}-1& 0& a_{13}& a_{11}{a}_{14}+1\end{array}\right).$$

Case 1:

$$b_{11}\ne 0,$$

because

$$B=\left(\begin{array}{cccc}{b}_{11}& 0& b_{13}& b_{14}\\ b_{21}& b_{22}& 0& -b_{21}\\ b_{31}& 0& b_{33}& b_{34}\\ b_{11}-1& 0& b_{13}& b_{14}+1\end{array}\right)=\left(\begin{array}{cccc}{a}_{11}& 0& a_{13}& a_{11}{a}_{14}\\ a_{21}& a_{22}& 0& -a_{21}\\ a_{11}{a}_{31}-a_{34}& 0& a_{13}{a}_{31}+1& a_{11}{a}_{14}{a}_{31}+a_{34}\\ a_{11}-1& 0& a_{13}& a_{11}{a}_{14}+1\end{array}\right),$$

so

$$a_{11}=b_{11},$$

$$a_{13}=b_{13},$$

$$a_{11}{a}_{14}=b_{14}\Rightarrow a_{14}=\frac{b_{14}}{b_{11}},$$

$$a_{21}=b_{21},$$

$$a_{22}=b_{22},$$

$$a_{11}{a}_{31}+1=b_{33}\Rightarrow a_{31}=\frac{b_{33}-1}{b_{11}},$$

$$a_{11}{a}_{14}{a}_{31}+a_{34}=b_{34}\Rightarrow a_{34}=b_{34}-b_{14}\frac{b_{33}-1}{b_{11}}.$$

Thus, the original proposition holds.

Case 2:

$$b_{11}=0,b_{14}\ne 0,$$

Take

$$\left(\begin{array}{cccc}0& 0& 0& -1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right)\in G_6,$$

$$\left(\begin{array}{cccc}0& 0& b_{13}& b_{14}\\ b_{21}& b_{22}& 0& -b_{21}\\ b_{31}& 0& b_{33}& b_{34}\\ -1& 0& b_{13}& b_{14}+1\end{array}\right)\left(\begin{array}{cccc}0& 0& 0& -1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1& 0& 0& 0\end{array}\right)$$

$$=\left(\begin{array}{cccc}-b_{14}& 0& b_{13}& 0\\ b_{21}& b_{22}& 0& -b_{21}\\ -b_{34}& 0& b_{33}& -b_{31}\\ -b_{14}-1& 0& b_{13}& 1\end{array}\right),$$

At this time, because $b_{14}\ne 0$, it is converted to case 1.

Case 3:

$$b_{11}=0,b_{14}=0,b_{13}\ne 0,$$

Take

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 1& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\in G_3,$$

because

$$\left(\begin{array}{cccc}0& 0& b_{13}& 0\\ b_{21}& b_{22}& 0& -b_{21}\\ b_{31}& 0& b_{33}& b_{34}\\ -1& 0& b_{13}& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 1& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)$$

$$=\left(\begin{array}{cccc}{b}_{13}& 0& b_{13}& 0\\ b_{21}& b_{22}& 0& -b_{21}\\ b_{31}+b_{33}& 0& b_{33}& b_{34}\\ b_{13}-1& 0& b_{13}& 1\end{array}\right),$$

At this time, when $b_{13}\ne 0$, it is changed to case 1. □

3. Conclusions

This article cleverly utilized the elementary transformation of partitioned matrices to study the subgroups of a four-dimensional solvable Lie algebra g and obtain the necessary and sufficient conditions for its automorphism. It also characterized three decomposition scenarios of the automorphism group of g, presenting the structure of its automorphism group more clearly. This article added new methods to the study of low-dimensional Lie algebra automorphism, which can provide assistance for the structure of research of general Lie algebra automorphism groups.