Isomorphism classes of four dimensional nilpotent associative algebras over a field

TL;DR

This paper classifies four-dimensional nilpotent associative algebras over various fields by analyzing isomorphism classes and regular subgroups of the affine group, providing explicit representatives for different types of fields.

Contribution

It offers a comprehensive classification of four-dimensional nilpotent associative algebras over finite, real, and algebraically closed fields, including explicit representatives.

Findings

Explicit classification of isomorphism classes over finite fields

Explicit classification over the real field R

Explicit classification over algebraically closed fields

Abstract

In this paper we classify the isomorphism classes of four dimensional nilpotent associative algebras over a field F, studying regular subgroups of the affine group AGL_4(F). In particular we provide explicit representatives for such classes when F is a finite field, the real field R or an algebraically closed field.

Tables

Table 1. Table 1. Four dimensional abelian nilpotent associative 𝔽 𝔽 \mathbb{F} -algebras and abelian regular subgroups of AGL 4 ( 𝔽 ) subscript AGL 4 𝔽 \mathop{\rm AGL}\nolimits_{4}(\mathbb{F}) .

Table 2. Table 2. Four dimensional nonabelian nilpotent associative 𝔽 𝔽 \mathbb{F} -algebras and nonabelian regular subgroups in Δ 4 ​ ( 𝔽 ) subscript Δ 4 𝔽 \Delta_{4}(\mathbb{F}) .

Table 3. Table 3. Representatives for conjugacy classes of nonabelian subgroups having 𝖽 ​ ( 𝐙 ​ ( R ) ) = 2 𝖽 𝐙 𝑅 2 \mathsf{d}(\mathbf{Z}(R))=2 and 𝗋 ​ ( 𝐙 ​ ( R ) ) = 1 𝗋 𝐙 𝑅 1 \mathsf{r}(\mathbf{Z}(R))=1 .

Equations

$$r=\left(\begin{array}{cc}1& v\\ 0& I_n+{\delta }_R(v)\end{array}\right)={\mu }_R(v),$$

$${\beta }_3=\frac{{\beta }_1{t}^2+{\beta }_{2}t-\rho }{t^2+{\beta }_{2}t-\rho {\beta }_1}\text{ and }{\beta }_4=\pm \frac{{\beta }_2{t}^2-2\rho ({\beta }_1+1)t-\rho {\beta }_2}{t^2+{\beta }_{2}t-\rho {\beta }_1}$$

$$\left\{\begin{array}[]{cl}9&\textrm{ if }\mathbb{F}=\overline{\mathbb{F}}\textrm{ is algebraically closed and }{\rm char\,}\mathbb{F}\neq 2,\\ 10&\textrm{ if }\mathbb{F}=\overline{\mathbb{F}}\textrm{ is algebraically closed and }{\rm char\,}\mathbb{F}=2,\\ 11&\textrm{ if }\mathbb{F}=\mathbb{F}_{q}\textrm{ is finite},\\ 12&\textrm{ if }\mathbb{F}=\mathbb{R},\\ \end{array}\right.$$

$${\delta }_R(v{\delta }_R(w))={\delta }_R(v){\delta }_R(w),\text{ for all }v,w\in {\mathbb{F}}^n,$$

$${\mu }_R(v){\mu }_R(w)={\mu }_R(w){\mu }_R(v)\text{ if and only if }v{\delta }_R(w)=w{\delta }_R(v).$$

$$\begin{array}[]{rcl}\mathsf{d}(H)&=&\max\{\deg\min_{\mathbb{F}}(h-I_{n+1}):h\in H\};\\ \mathsf{r}(H)&=&\max\{\mathrm{rk}(h-I_{n+1}):h\in H\};\\ \mathsf{k}(H)&=&\dim{\rm Ker\,}(\delta_{R}|_{H}).\end{array}$$

$$\tilde{R}=\mathcal{R}(R,D)=\left\{\left(\begin{array}{ccc}1& X& x_n\\ 0& I_{n-1}+{\delta }_R(X)& D{X}^T\\ 0& 0& 1\end{array}\right):X\in {\mathbb{F}}^{n-1},x_n\in \mathbb{F}\right\},$$

$$D{\delta }_R(e_i{)}^T{e}_j^T={\delta }_R(e_j)D{e}_i^T\text{for all }i,j=1,\dots ,n-1$$

$${\mathcal{R}}_D:=\mathcal{R}({\mathcal{T}}_{n-1},D)=\left\{\left(\begin{array}{ccc}1& X& x_n\\ 0& I_{n-1}& D{X}^T\\ 0& 0& 1\end{array}\right):X\in {\mathbb{F}}^{n-1},x_n\in \mathbb{F}\right\}.$$

$$\mathsf{d}({\mathcal{R}}_D)$$

$$\mathsf{r}({\mathcal{R}}_D)$$

$$\mathsf{k}({\mathcal{R}}_D)$$

$$m\mapsto \left(\begin{array}{cc}0& m_{\mathcal{B}}\\ 0& \delta (m)\end{array}\right),$$

$${\mathcal{L}}_R=\mathbb{F}{I}_{n+1}+R$$

$$N=\mathbf{J}({\mathcal{L}}_R)=R-I_{n+1}=\left\{\left(\begin{array}{cc}0& v\\ 0& {\delta }_R(v)\end{array}\right):v\in {\mathbb{F}}^n\right\}$$

$$R=\begin{pmatrix}1&\begin{array}[]{cccc}x_{1}&x_{2}&\ldots&x_{n}\end{array}\\ 0&I_{n}+\delta_{R}(x_{1},x_{2},\ldots,x_{n})\end{pmatrix}$$

$$R=\left\{\begin{pmatrix}1&\begin{array}[]{cccc}x_{1}&x_{2}&\ldots&x_{n}\end{array}\\ 0&I_{n}+\delta_{R}(x_{1},x_{2},\ldots,x_{n})\end{pmatrix}:x_{1},\ldots,x_{n}\in\mathbb{F}\right\}.$$

$$S_{(5)}=\left(\begin{array}{ccccc}1& x_1& x_2& x_3& x_4\\ 0& 1& x_1& x_2& x_3\\ 0& 0& 1& x_1& x_2\\ 0& 0& 0& 1& x_1\\ 0& 0& 0& 0& 1\end{array}\right).$$

$$S_{(4,1)}=\left(\begin{array}{ccccc}1& x_1& x_2& x_3& x_4\\ 0& 1& x_1& x_2& 0\\ 0& 0& 1& x_1& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right)\text{ and }{S}_{(3,2)}^{♯}=\left(\begin{array}{ccccc}1& x_1& x_2& x_3& x_4\\ 0& 1& x_1& 0& x_2\\ 0& 0& 1& 0& x_1\\ 0& 0& 0& 1& x_3\\ 0& 0& 0& 0& 1\end{array}\right).$$

$$U_1(\alpha ,\beta )=\left(\begin{array}{ccccc}1& x_1& x_2& x_3& x_4\\ 0& 1& 0& x_1& x_2\\ 0& 0& 1& \beta x_2& x_1+\alpha x_2\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right)$$

$$D=\left(\begin{array}{ccc}1& 0& 0\\ 0& {\beta }_2& {\beta }_3\\ 0& {\gamma }_2& {\gamma }_3\end{array}\right),$$

$$U_2({\alpha }_1,{\alpha }_3,{\gamma }_1,{\gamma }_3,\zeta )=\left(\begin{array}{ccccc}1& x_1& x_2& x_3& x_4\\ 0& 1& \zeta x_1& 0& {\alpha }_1{x}_1+x_2+{\alpha }_3{x}_3\\ 0& 0& 1& 0& x_1\\ 0& 0& 0& 1& {\gamma }_1{x}_1+{\gamma }_3{x}_3\\ 0& 0& 0& 0& 1\end{array}\right)$$

$$U_1(0,\rho )\text{ with }\rho \in {\mathbb{F}}^{\square }\cup \{0\}$$

$$U_1(1,ϵ)\text{ with }ϵ\in \mathsf{B}(\mathbb{F})\text{ or }{U}_1(0,\rho )\text{ with }\rho \in \mathsf{C}(\mathbb{F})$$

$$\Psi (t_1)=\omega \rho X_1,\Psi (t_2)=\alpha X_1-2X_2.$$

$$\Psi (t_1)=X_1,\Psi (t_2)=\alpha X_1-2X_2,$$

$$\Psi (t_1)=r{X}_1+X_2,\Psi (t_2)=\alpha X_1.$$

$$\Psi (t_1)=X_2,\Psi (t_2)=\omega X_1.$$

$$S_{(5)},S_{(4,1)},S_{(3,2)}^{♯},U_1(0,0),U_1(0,1),{\mathcal{R}}_{E_{1,1}},{\mathcal{R}}_{E_{1,1}+E_{2,2}},{\mathcal{R}}_{I_3},{\mathcal{T}}_4.$$

$$S_{(5)},S_{(4,1)},S_{(3,2)}^{♯},U_1(0,0),U_1(1,0),{\mathcal{R}}_{E_{1,1}},{\mathcal{R}}_{E_{1,1}+E_{2,2}},{\mathcal{R}}_{E_{2,3}+E_{3,2}},{\mathcal{R}}_{I_3},{\mathcal{T}}_4.$$

$$S_{(5)},S_{(4,1)},S_{(3,2)}^{♯},U_1(0,0),U_1(0,1),U_1(0,-1),$$

Full text

1. Introduction

The aim of this paper is the classification of the isomorphism classes of four dimensional nilpotent associative algebras over a field $\mathbb{F}$. This result will be achieved exploiting the properties of the regular subgroups of the affine group $\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$.

It is worth recalling that the affine group $\mathop{\rm AGL}\nolimits_{n}(\mathbb{F})$ can be identified with the subgroup of $\mathop{\rm GL}\nolimits_{n+1}(\mathbb{F})$ consisting of the invertible matrices having $(1,0,\ldots,0)^{\mathop{\rm T}\nolimits}$ as first column. It follows that $\mathop{\rm AGL}\nolimits_{n}(\mathbb{F})$ acts on the right on the set $\mathcal{A}=\left\{(1,v):v\in\mathbb{F}^{n}\right\}$ of affine points. A subgroup $R$ of $\mathop{\rm AGL}\nolimits_{n}(\mathbb{F})$ is called regular if it acts regularly on $\mathcal{A}$, namely if for every $v\in\mathbb{F}^{n}$ there exists a unique element in $R$ having $(1,v)$ as first row. Writing the elements $r$ of a regular subgroup $R\leq\mathop{\rm AGL}\nolimits_{n}(\mathbb{F})$ as

[TABLE]

we give rise to the functions $\delta_{R}:\mathbb{F}^{n}\rightarrow\mathop{\rm Mat}\nolimits_{n}(\mathbb{F})$ and $\mu_{R}:\mathbb{F}^{n}\rightarrow R$. For instance, the translation subgroup $\mathcal{T}_{n}$ of $\mathop{\rm AGL}\nolimits_{n}(\mathbb{F})$ is a regular subgroup, corresponding to the choice $\delta_{\mathcal{T}_{n}}$ equal to the zero function.

We focus our attention on the set $\Delta_{n}(\mathbb{F})$ of the regular subgroups $R$ of $\mathop{\rm AGL}\nolimits_{n}(\mathbb{F}$) with the property that $\delta_{R}$ is a linear function. Notice that if $R$ is abelian, then $R\in\Delta_{n}(\mathbb{F})$ (see [1, 8]); however, for $n\geq 3$, the set $\Delta_{n}(\mathbb{F})$ also contains nonabelian groups (see [5]). We also remark that if $R\in\Delta_{n}(\mathbb{F})$, then $R$ is unipotent and contains nontrivial translations (see [4, 6]). Our main motivation for studying the set $\Delta_{n}(\mathbb{F})$ is that there exists a bijection between the set of isomorphism classes of nilpotent associative algebras of dimension $n$ over a field $\mathbb{F}$ and the set of conjugacy classes of subgroups in $\Delta_{n}(\mathbb{F})$ (see [4, Proposition 2.2]). In fact, our classification of four dimensional nilpotent associative $\mathbb{F}$-algebras (Theorem 1.1) will follow from the classification of the conjugacy classes of regular subgroups in $\Delta_{4}(\mathbb{F})$.

The paper is organized as follows. In Section 2 we recall some useful results concerning $\Delta_{n}(\mathbb{F})$ and in the following sections we classify, up to conjugation, the regular subgroups in $\Delta_{4}(\mathbb{F})$, see Theorems 4.2 and 5.12. More in detail, in Section 4 we deal with abelian regular subgroups and in Section 5 we consider nonabelian subgroups. Corollaries 4.3 and 4.5 and Propositions 5.15 and 5.23 give the explicit classification for algebraically closed fields or finite fields. The classification for the real field is given in Corollary 4.4 and Proposition 5.20.

Before to describe the results we obtained, we need to fix some notation. The sets $\Pi_{S}(\mathbb{F})$ and $\Pi_{A}(\mathbb{F})$ are complete sets of representatives of projective congruent classes for, respectively, symmetric and asymmetric matrices in $\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$, see Definition 2.2. Given a field $\mathbb{F}$, we denote by $\mathbb{F}^{\square}$ a transversal for $(\mathbb{F}^{\ast})^{2}$ in $\mathbb{F}^{\ast}$. Also, for any $\rho\in\mathbb{F}^{\square}$, we denote by $\mathsf{A}_{\rho}(\mathbb{F})$ a fixed set of representatives for the equivalence classes with respect to the following relation $\divideontimes_{\rho}$ defined on $\overline{\mathsf{A}}(\mathbb{F})=\left(\mathbb{F}\setminus\{1\}\right)\times\mathbb{F}$: given $(\beta_{1},\beta_{2}),(\beta_{3},\beta_{4})\in\overline{\mathsf{A}}(\mathbb{F})$, we write $(\beta_{1},\beta_{2})\divideontimes_{\rho}(\beta_{3},\beta_{4})$ if and only if either $(\beta_{3},\beta_{4})=(\beta_{1},\pm\beta_{2})$ or

[TABLE]

for some $t\in\mathbb{F}$ such that $(t^{2}+\rho)(t^{2}+\beta_{2}t-\rho\beta_{1})\neq 0$. If ${\rm char\,}\mathbb{F}=2$, we fix two additional sets $\mathsf{B}(\mathbb{F})$ and $\mathsf{C}(\mathbb{F})$. The set $\mathsf{B}(\mathbb{F})$ is a transversal for the subgroup $\overline{\mathsf{B}}(\mathbb{F})=\{\lambda^{2}+\lambda:\lambda\in\mathbb{F}\}$ in $\mathbb{F}$. The set $\mathsf{C}(\mathbb{F})$ is a set of representatives for the equivalence classes with respect to the following relation $\star$ defined on $\mathbb{F}^{\square}$: given $\rho_{1},\rho_{2}\in\mathbb{F}^{\square}$, we write $\rho_{1}\star\rho_{2}$ if and only if $\rho_{2}=\frac{x_{1}^{2}+x_{2}^{2}\rho_{1}}{x_{3}^{2}+x_{4}^{2}\rho_{1}}$ for some $x_{1},x_{2},x_{3},x_{4}\in\mathbb{F}$ such that $x_{1}x_{4}\neq x_{2}x_{3}$. For convenience, we assume $1\in\mathbb{F}^{\square}$ and $0\in\mathsf{B}(\mathbb{F})$.

We can now state our main result.

Theorem 1.1**.**

The distinct isomorphism classes of four dimensional nilpotent associative algebras over a field $\mathbb{F}$ can be represented by the algebras described in the second column of Table 1 (abelian case) and Table 2 (nonabelian case).

In particular, there are exactly

[TABLE]

nonisomorphic abelian nilpotent associative $\mathbb{F}$-algebras of dimension four. We point out that the classification given in Theorem 1.1 for the abelian case was already obtained, using other techniques, by De Graaf [2] (if $\mathbb{F}=\mathbb{F}_{q}$ or $\mathbb{R}$) and by Poonen [7] (for algebraically closed fields). It is also interesting to observe that if $\mathbb{F}=\mathbb{F}_{q}$ is a finite field, then there are exactly $5q+9$ (if $q$ is odd) or $5q+6$ (if $q$ is even) nonisomorphic nonabelian nilpotent associative $\mathbb{F}_{q}$-algebras of dimension four (see Proposition 5.23).

2. Preliminary results

Our classification of regular subgroups in $\Delta_{4}(\mathbb{F})$ basically relies on the key results of [4, 5] that we recall in this section.

Proposition 2.1**.**

[5, Theorems 2.4 and 4.4]** Let $R\in\Delta_{n}(\mathbb{F})$. Then,

• (a)

$R$* is a unipotent subgroup;*

• (b)

up to conjugation under $\mathop{\rm AGL}\nolimits_{n}(\mathbb{F})$, the center $\mathbf{Z}(R)$ of $R$ contains a nontrivial element $z$ that coincides with its Jordan form $\mathop{\rm diag}\nolimits\left(J_{n_{1}},\dots,J_{n_{k}}\right)$ having upper unitriangular Jordan blocks $J_{n_{i}}$ of respective sizes $n_{i}\geq n_{i+1}$ for all $i\geq 1$.

If $\delta_{R}$ is a linear function then a regular subset $R$ of $\mathop{\rm AGL}\nolimits_{n}(\mathbb{F})$ is a subgroup if and only if

[TABLE]

see [5]. Also, given $v,w\in\mathbb{F}^{n}$,

[TABLE]

We also recall that two subgroups $R_{1},R_{2}\in\Delta_{n}(\mathbb{F})$ are conjugate in $\mathop{\rm AGL}\nolimits_{n}(\mathbb{F})$ if and only if there exists a matrix $g=\mathop{\rm diag}\nolimits(1,\bar{g})$ with $\bar{g}\in\mathop{\rm GL}\nolimits_{n}(\mathbb{F})$ such that $R_{1}^{g}=R_{2}$ (see [5, Proposition 3.4]). On the other hand it is useful, mainly to prove that two regular subgroups are not conjugate, to introduce the following three parameters. Let $H\in\{R,\mathbf{Z}(R)\}$, where $R\in\Delta_{n}(\mathbb{F})$. Then we may define:

[TABLE]

The last parameter is justified since the set $\{v\in\mathbb{F}^{n}:\mu_{R}(v)\in\mathbf{Z}(R)\}$ is a subspace of $\mathbb{F}^{n}$ (see [4, Proposition 2.1]).

We will proceed considering the possible values of $(\mathsf{d}(\mathbf{Z}(R)),\mathsf{r}(\mathbf{Z}(R))$ for $R\in\Delta_{4}(\mathbb{F})$ and working in the centralizer of a Jordan form as described in Proposition 2.1(b). Unfortunately, the case $\mathsf{r}(\mathbf{Z}(R))=2$ will require a different approach, based on the classification of the regular subgroups in $\Delta_{3}(\mathbb{F})$ obtained in [5]. In particular, we will make use of the following observation.

Any regular subgroup $\widetilde{R}\in\Delta_{n}(\mathbb{F})$, $n\geq 2$, can be written as

[TABLE]

for some $R\in\Delta_{n-1}(\mathbb{F})$ and some square matrix $D\in\mathop{\rm Mat}\nolimits_{n-1}(\mathbb{F})$ such that

[TABLE]

($\{e_{1},\ldots,e_{n-1}\}$ denotes the canonical basis of $\mathbb{F}^{n-1}$), see [4]. In particular, we can take $R=\mathcal{T}_{n-1}$: in this case any square matrix $D\in\mathop{\rm Mat}\nolimits_{n-1}$ satisfies (2.2). We set

[TABLE]

The study of these subgroups $\mathcal{R}_{D}$ is justified mainly because of their connection with projectively congruent matrices.

Definition 2.2**.**

[9] Two matrices $A,B\in\mathop{\rm Mat}\nolimits_{n}(\mathbb{F})$ are said to be projectively congruent if $PAP^{\mathop{\rm T}\nolimits}=\lambda B$, for some non-zero element $\lambda\in\mathbb{F}^{\ast}$ and some invertible matrix $P\in\mathop{\rm GL}\nolimits_{n}(\mathbb{F})$.

Lemma 2.3**.**

[4, Lemma 3.2]** Given two matrices $A,B\in\mathop{\rm Mat}\nolimits_{n-1}(\mathbb{F})$, the subgroups $\mathcal{R}_{A}$ and $\mathcal{R}_{B}$ are conjugate in $\mathop{\rm AGL}\nolimits_{n}(\mathbb{F})$ if and only if $A$ and $B$ are projectively congruent.

Furthermore, it is easy to see that

[TABLE]

A given nilpotent associative algebra $N$ of dimension $n$ can be embedded into $\mathop{\rm Mat}\nolimits_{n+1}(\mathbb{F})$ via

[TABLE]

where, for any $m\in N$, $m_{\mathcal{B}}$ and $\delta(m)$ denote, respectively, the coordinate row vector of $m$ and the matrix of the right multiplication by $m$ with respect to a fixed basis $\mathcal{B}$ of $N$ over $\mathbb{F}$. Identifying $N$ with its image, $\mathcal{L}=\mathbb{F}I_{n+1}+N$ is a split local subalgebra of $\mathop{\rm Mat}\nolimits_{n+1}(\mathbb{F})$ with Jacobson radical $\mathbf{J}(\mathcal{L})=N$. The subset $R=\{I_{n+1}+m:m\in N\}\subseteq\mathcal{L}$ consists of invertible matrices and is closed under multiplication, since $\delta(m_{1}\delta(m_{2}))=\delta(m_{1})\delta(m_{2})$ for all $m_{1},m_{2}\in N$. Hence, $R$ is a regular subgroup lying in $\Delta_{n}(\mathbb{F})$, by [5, Lemma 2.1]. On the other hand, given a regular subgroup $R\in\Delta_{n}(\mathbb{F})$, we have that

[TABLE]

is a split local $\mathbb{F}$-algebra of dimension $n$, by [5, Theorem 3.3]. Notice that set

[TABLE]

is a nilpotent associative $\mathbb{F}$-algebra of dimension $n$.

Proposition 2.4**.**

[4, 5]** Let $R_{1},R_{2}\in\Delta_{n}(\mathbb{F})$ and $\mathcal{L}_{R_{1}},\mathcal{L}_{R_{2}}$ be as in (2.6). Then the following conditions are equivalent:

• (a)

the subgroups $R_{1}$ and $R_{2}$ are conjugate in $\mathop{\rm AGL}\nolimits_{n}(\mathbb{F})$;

• (b)

the split local algebras $\mathcal{L}_{R_{1}}$ and $\mathcal{L}_{R_{2}}$ are isomorphic;

• (c)

the nilpotent associative algebras $\mathbf{J}(\mathcal{L}_{R_{1}})$ and $\mathbf{J}(\mathcal{L}_{R_{2}})$ are isomorphic.

For sake of brevity, we write

[TABLE]

to indicate the regular subgroup

[TABLE]

For all $i\leq n$, we denote by $X_{i}$ the matrix of $R-I_{n+1}$ obtained taking $x_{i}=1$ and $x_{j}=0$ for all $j\neq i$. Finally, $E_{i,j}$ is the elementary matrix having $1$ at position $(i,j)$, [math] elsewhere.

3. The regular subgroups in $\Delta_{4}(\mathbb{F})$

For classifying the regular subgroups $R\in\Delta_{4}(\mathbb{F})$, where $\mathbb{F}$ is any field, we consider the possible values of $(\mathsf{d}(\mathbf{Z}(R)),\mathsf{r}(\mathbf{Z}(R)))$. By Proposition 2.1 the subgroup $R$ is unipotent and its center contains an element conjugate in $\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$ to one of the following Jordan forms $z$:

[TABLE]

Some of these cases have been already studied in [5]: we recall here the results obtained in Lemmas 5.2, 5.4 and 7.4. First of all, if $\mathsf{d}(\mathbf{Z}(R))\geq 4$, then $R$ is abelian. In particular, if $\mathsf{d}(R)=5$, then $R$ is conjugate to

[TABLE]

If $\mathsf{d}(R)=4$, we have two conjugacy classes of regular subgroups: when $\mathsf{k}(R)=2$ the subgroup $R$ is conjugate to $S_{(4,1)}$ and when $\mathsf{k}(R)=1$ it is conjugate to $S_{(3,2)}^{\sharp}$, where

[TABLE]

Now, if $\mathsf{d}(\mathbf{Z}(R))=\mathsf{r}(\mathbf{Z}(R))=3$, then again $R$ is abelian. Conjugating the subgroups obtained in [5, Lemma 7.4] by $g=\mathop{\rm diag}\nolimits(I_{2},E_{1,2}+E_{2,1},1)$, we obtain that $R$ is conjugate to

[TABLE]

for some $\alpha,\beta\in\mathbb{F}$.

If $\mathsf{d}(\mathbf{Z}(R))=3$ and $\mathsf{r}(\mathbf{Z}(R))=2$, we may assume that $z=\mathop{\rm diag}\nolimits(J_{3},I_{2})^{g}\in\mathbf{Z}(R)$, where $g=\mathop{\rm diag}\nolimits(I_{2},E_{1,3}+E_{2,1}+E_{3,2})$. So, working in $\mathbf{C}_{\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})}(z)$ and using the linearity of $\delta_{R}$ we obtain that $R$ is conjugate to a subgroup $\mathcal{R}_{D}$, where

[TABLE]

for some $\beta_{2},\beta_{3},\gamma_{2},\gamma_{3}\in\mathbb{F}$.

If $\mathsf{d}(\mathbf{Z}(R))=\mathsf{r}(\mathbf{Z}(R))=2$, we may assume that the center of $R$ contains the element $z=\mathop{\rm diag}\nolimits(J_{2},J_{2},I_{1})^{g}$, where $g=\mathop{\rm diag}\nolimits(1,E_{1,2}+E_{2,1},E_{1,2}+E_{2,1})$. Again, working in $\mathbf{C}_{\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})}(z)$ and using the linearity of $\delta_{R}$ we obtain that $R$ is conjugate to the subgroup

[TABLE]

for some $\alpha_{1},\alpha_{3},\gamma_{1},\gamma_{3},\zeta\in\mathbb{F}$.

Finally, if $R$ is abelian and $(\mathsf{d}(R),\mathsf{r}(R))=(2,1)$, then $R=\mathcal{T}_{4}$ by [5, Lemma 5.3].

4. The abelian case

In this section we assume that $R\in\Delta_{4}(\mathbb{F})$ is abelian. In view of the results recalled in Section 3, we are left to determine the conjugacy classes of subgroups of type $U_{1}(\alpha,\beta)$, $U_{2}(\alpha_{1},\alpha_{3},\gamma_{1},\gamma_{3},\zeta)$ and $\mathcal{R}_{D}$ where $D\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$ is a symmetric matrix. We start with the subgroups $U_{1}(\alpha,\beta)$.

Lemma 4.1**.**

Let $R$ be an abelian regular subgroup of $\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$ such that $\mathsf{d}(R)=\mathsf{r}(R)=3$. Then $R$ is conjugate to exactly one of the following subgroups:

[TABLE]

if ${\rm char\,}\mathbb{F}\neq 2$;

[TABLE]

if ${\rm char\,}\mathbb{F}=2$.

Proof.

By the previous argument, we can assume that $R=U_{1}(\alpha,\beta)$, as in (3.1), for some $\alpha,\beta\in\mathbb{F}$. First, suppose that ${\rm char\,}\mathbb{F}\neq 2$ and set $\Delta=\alpha^{2}+4\beta$. If $\Delta\neq 0$, write $\Delta=\omega^{2}\rho$ for some $\omega\in\mathbb{F}^{\ast}$ and a unique $\rho\in\mathbb{F}^{\square}$. We can define an epimorphism (of algebras) $\Psi:\mathbb{F}[t_{1},t_{2}]\rightarrow\mathcal{L}_{U_{1}(\alpha,\beta)}$ by setting

[TABLE]

We have ${\rm Ker\,}(\Psi)=\langle t_{1}^{3},\;t_{2}^{3},\;t_{1}^{2}-\rho t_{2}^{2}\rangle$. The subgroup $U_{1}(\alpha,\beta)$ is conjugate to $U_{1}(0,\rho)$ by Proposition 2.4. Now, it is quite easy to see that two subgroups $U_{1}(0,\rho_{1})$ and $U_{1}(0,\rho_{2})$ ($\rho_{1},\rho_{2}\in\mathbb{F}^{\square}$) are conjugate in $\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$ if and only if $\rho_{1}=\rho_{2}$.

If $\alpha^{2}+4\beta=0$, we consider the epimorphism $\Psi:\mathbb{F}[t_{1},t_{2}]\rightarrow\mathcal{L}_{U_{1}(\alpha,\beta)}$ defined by

[TABLE]

whose kernel is ${\rm Ker\,}(\Psi)=\langle t_{1}^{3},\;t_{1}^{2}t_{2},\;t_{2}^{2}\rangle$. Again by Proposition 2.4, $U_{1}(\alpha,\beta)$ is conjugate to $U_{1}(0,0)$ (which is not conjugate to $U_{1}(0,\rho)$, $\rho\in\mathbb{F}^{\square}$).

Next, suppose that ${\rm char\,}\mathbb{F}=2$. If $\alpha\neq 0$, take $\epsilon\in\mathsf{B}(\mathbb{F})$ such that $\epsilon+\frac{\beta}{\alpha^{2}}\in\overline{\mathsf{B}}(\mathbb{F})$. Then, there exists $r\in\mathbb{F}$ such that $r^{2}+\alpha r+\beta+\alpha^{2}\epsilon=0$ (a polynomial $t^{2}+t+\lambda$ is reducible in $\mathbb{F}[t]$ if and only if $\lambda\in\overline{\mathsf{B}}(\mathbb{F})$). In this case, we can define an epimorphism $\Psi:\mathbb{F}[t_{1},t_{2}]\rightarrow\mathcal{L}_{U_{1}(\alpha,\beta)}$ by setting

[TABLE]

We obtain ${\rm Ker\,}(\Psi)=\langle t_{1}^{3},\;t_{2}^{3},\;t_{1}^{2}+t_{1}t_{2}+\epsilon t_{2}^{2}\rangle$, proving that $U_{1}(\alpha,\beta)$ is conjugate to $U_{1}(1,\epsilon)$. Now, $U_{1}(1,\epsilon_{1})$ and $U_{1}(1,\epsilon_{2})$ are conjugate in $\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$ if and only if $\epsilon_{1}+\epsilon_{2}\in\overline{\mathsf{B}}(\mathbb{F})$. It follows that $U(\alpha,\beta)$ is conjugate to exactly one subgroup $U_{1}(1,\epsilon)$ with $\epsilon\in\mathsf{B}(\mathbb{F})$.

Finally, assume $\alpha=0$. For $\beta=0$ observe that $U_{1}(0,0)^{g}=U_{1}(0,1)$, where $g=I_{5}+E_{3,2}+E_{5,4}$. If $\beta\neq 0$, write $\beta=\omega^{2}\rho$ for some $\omega\in\mathbb{F}^{\ast}$ and a unique $\rho\in\mathbb{F}^{\square}$. An epimorphism $\Psi:\mathbb{F}[t_{1},t_{2}]\rightarrow\mathcal{L}_{U_{1}(0,\beta)}$ can be defined by taking

[TABLE]

We have ${\rm Ker\,}(\Psi)=\langle t_{1}^{3},\;t_{2}^{3},\;t_{1}^{2}+\rho t_{2}^{2}\rangle$ and so $U_{1}(0,\beta)$ is conjugate to $U_{1}(0,\rho)$. Furthermore, given $\rho_{1},\rho_{2}\in\mathbb{F}^{\square}$, the subgroup $U_{1}(0,\rho_{1})$ is conjugate to $U_{1}(0,\rho_{2})$ if and only if $\rho_{1}\star\rho_{2}$, that is if and only if $\rho_{2}=\frac{x_{1}^{2}+x_{2}^{2}\rho_{1}}{x_{3}^{2}+x_{4}^{2}\rho_{1}}$ for some $x_{1},x_{2},x_{3},x_{4}\in\mathbb{F}$ such that $x_{1}x_{4}\neq x_{2}x_{3}$ (see the notation described in the Introduction).

To conclude we observe that subgroups $U_{1}(0,\rho)$, $\rho\in\mathsf{C}(\mathbb{F})$, are not conjugate to subgroups $U_{1}(1,\epsilon)$, $\epsilon\in\mathsf{B}(\mathbb{F})$. ∎

We can now classify the abelian regular subgroups of $\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$.

Theorem 4.2**.**

Let $\mathbb{F}$ be a field. The distinct conjugacy classes of abelian regular subgroups of $\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$ can be represented by the subgroups described in the first column of Table 1.

Proof.

Let $R\leq\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$ be an abelian regular subgroup. Clearly, $2\leq\mathsf{d}(R)\leq 5$. As seen in Section 3, if $\mathsf{d}(R)=5$ then $R$ is conjugate to $S_{(5)}$, and if $\mathsf{d}(R)=4$ then $R$ is conjugate either to $S_{(4,1)}$ or to $S_{(3,2)}^{\sharp}$.

Suppose that $\mathsf{d}(R)=3$. Then $2\leq\mathsf{r}(R)\leq 3$. If $\mathsf{r}(R)=3$, we apply Lemma 4.1: if ${\rm char\,}\mathbb{F}\neq 2$, $R$ is conjugate to $U_{1}(0,\rho)$ for a unique $\rho\in\mathbb{F}^{\square}\cup\{0\}$; if ${\rm char\,}\mathbb{F}=2$, it is conjugate either to $U_{1}(1,\epsilon)$ for a unique $\epsilon\in\mathsf{B}(\mathbb{F})$ or to $U_{1}(0,\rho)$ for a unique $\rho\in\mathsf{C}(\mathbb{F})$. If $\mathsf{r}(R)=2$, then $R$ is conjugate to a subgroup $\mathcal{R}_{D}$ for some $D\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$.

Suppose $\mathsf{d}(R)=2$. Then $1\leq\mathsf{r}(R)\leq 2$. If $\mathsf{r}(R)=2$ we apply [5, Lemma 7.7], obtaining that ${\rm char\,}\mathbb{F}=2$ and $R$ is conjugate to $U_{2}(0,0,0,0,0)=\mathcal{R}_{D}$, where $D=E_{1,2}+E_{2,1}$. If $\mathsf{r}(R)=1$, then $U=\mathcal{T}_{4}=\mathcal{R}_{0}$, where $0\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$ is the zero matrix.

Notice that in three cases, $R$ is conjugate to a subgroup $\mathcal{R}_{D}$. By Lemma 2.3 the statement of the theorem is proved considering a set $\Pi_{S}(\mathbb{F})$ of representatives for projective congruent classes of symmetric matrices in $\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$. ∎

We now provide explicit representatives for algebraically closed fields, for $\mathbb{R}$ and for finite fields.

Corollary 4.3**.**

Let $\mathbb{F}=\overline{\mathbb{F}}$ be an algebraically closed field. If ${\rm char\,}\mathbb{F}\neq 2$, there are exactly $9$ distinct conjugacy classes of abelian regular subgroups of $\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$, that can be represented by

[TABLE]

If ${\rm char\,}\mathbb{F}=2$, there are exactly $10$ distinct conjugacy classes of abelian regular subgroups of $\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$, that can be represented by

[TABLE]

Proof.

Since $\mathbb{F}$ is algebraically closed, $\mathbb{F}^{\square}=\{1\}$ and, when ${\rm char\,}\mathbb{F}=2$, $\mathsf{B}(\mathbb{F})=\{0\}$ and $\mathsf{C}(\mathbb{F})=\{1\}$. The set $\Pi_{S}(\mathbb{F})$ can be obtained applying, for instance, [3, Theorem 2.1]. Hence, the statement is proved noticing that, as observed in the proof of Lemma 4.1, when ${\rm char\,}\mathbb{F}=2$ the subgroups $U_{1}(0,0)$ and $U_{1}(0,1)$ are conjugate. ∎

Corollary 4.4**.**

There are exactly $12$ distinct conjugacy classes of abelian regular subgroups of $\mathop{\rm AGL}\nolimits_{4}(\mathbb{R})$, that can be represented by

[TABLE]

[TABLE]

Proof.

First of all, we can take $\mathbb{R}^{\square}=\{1,-1\}$. The set $\Pi_{S}(\mathbb{F})$ can be obtained again applying [3, Theorem 2.1], noticing that two matrices $A,B\in\mathop{\rm Mat}\nolimits_{n}(\mathbb{F})$ are projectively congruent if and only if $A$ is congruent to $\pm B$. ∎

Corollary 4.5**.**

Let $\mathbb{F}=\mathbb{F}_{q}$ be a finite field. There are exactly $11$ distinct conjugacy classes of abelian regular subgroups of $\mathop{\rm AGL}\nolimits_{4}(q)$, that can be represented by:

• (a)

for $q$ odd,

[TABLE]

where $x^{2}-\xi$ is a fixed irreducible polynomial of $\mathbb{F}_{q}[x]$ and

[TABLE]

• (b)

for $q$ even,

[TABLE]

where $x^{2}+x+\xi$ is a fixed irreducible polynomial of $\mathbb{F}_{q}[x]$ and

[TABLE]

Proof.

For $q$ odd we have $\mathbb{F}^{\square}=\{1,\xi\}$ and for $q$ even we have $\mathbb{F}^{\square}=\mathsf{C}(\mathbb{F})=\{1\}$ and $\mathsf{B}(\mathbb{F})=\{0,\xi\}$. The set $\Pi_{S}(\mathbb{F}_{q})$ has been determined in [9, Theorem 4]. ∎

5. The nonabelian case

In this section we assume that $R\in\Delta_{4}(\mathbb{F})$ is nonabelian. We are reduced to study the cases when $(\mathsf{d}(\mathbf{Z}(R)),\mathsf{r}(\mathbf{Z}(R)))=(3,2)$, $(2,2)$ or $(2,1)$. We recall that if $(\mathsf{d}(\mathbf{Z}(R)),\mathsf{r}(\mathbf{Z}(R)))=(3,2)$ then $R$ is conjugate to a subgroup of shape $\mathcal{R}_{D}$.

If $\mathsf{d}(\mathbf{Z}(R))=\mathsf{r}(\mathbf{Z}(R))=2$, then $R$ is conjugate to $U_{2}(\alpha_{1},\alpha_{3},\gamma_{1},\gamma_{3},\zeta)$ as in (3.2). Clearly, $U_{2}(\alpha_{1},\alpha_{3},\gamma_{1},\gamma_{3},0)=\mathcal{R}_{D}$, where

[TABLE]

So, we have to consider now the case $\zeta\neq 0$. Notice that since $R$ is nonabelian, we must have $\gamma_{1}\neq\alpha_{3}$. Furthermore, we have $\mathsf{d}(R)=4$ and $\mathsf{r}(R)=3$.

Lemma 5.1**.**

There are exactly $2$ conjugacy classes of nonabelian regular subgroups $R\in\Delta_{4}(\mathbb{F})$ such that $\mathsf{d}(\mathbf{Z}(R))=\mathsf{r}(\mathbf{Z}(R))=2$, not conjugate to $\mathcal{R}_{A}$ for any $A\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$. Such classes can be represented by the subgroups $U_{2}(0,1,0,1,1)$ and $U_{2}(0,1,0,0,1)$.

Proof.

By the previous considerations, we may assume that $R=U_{2}(\alpha_{1},\alpha_{3},\gamma_{1},\gamma_{3},\zeta)$ where $\zeta\neq 0$ and $\gamma_{1}\neq\alpha_{3}$.

First, suppose $\gamma_{3}\neq 0$: in this case $\mathsf{k}(R)=1$. Consider the algebra $\mathcal{L}=\mathcal{L}_{U_{2}(0,1,0,1,1)}$ defined in (2.6):

[TABLE]

The function $\Psi:\mathcal{L}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism: by Proposition 2.4, $R$ is conjugate to $U_{2}(0,1,0,1,1)$.

Suppose now that $\gamma_{3}=0$: in this case $\mathsf{k}(R)=2$ and so $R$ is not conjugate to $U_{2}(0,1,0,1,1)$. Consider the algebra $\mathcal{L}=\mathcal{L}_{U_{2}(0,1,0,0,1)}$:

[TABLE]

The function $\Psi:\mathcal{L}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism and so $R$ is conjugate to $U_{2}(0,1,0,0,1)$. ∎

We now consider the case when $\mathsf{d}(\mathbf{Z}(R))=2$ and $\mathsf{r}(\mathbf{Z}(R))=1$. Instead of working in $\mathbf{C}_{\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})}(\mathop{\rm diag}\nolimits(J_{2},I_{3}))$, it is more convenient to write $R$ as $\mathcal{R}(\tilde{R},D)$, for some $\tilde{R}\in\Delta_{3}(\mathbb{F})$ and $D\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$, see (2.1). The regular subgroups in $\Delta_{3}(\mathbb{F})$ have been classified in [5].

Theorem 5.2**.**

[5, Section 7.2]** Let $\mathbb{F}$ be a field. The distinct conjugacy classes of regular subgroups in $\Delta_{3}(\mathbb{F})$ are represented by

[TABLE]

[TABLE]

Suppose that $R=\mathcal{R}(S_{(4)},D)$ for some $D\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$. By (2.2), $D=\alpha_{1}E_{1,1}+\alpha_{2}(E_{1,2}E_{2,1})+\alpha_{3}(E_{1,3}+E_{2,2}+E_{3,1})$ for some $\alpha_{1},\alpha_{2},\alpha_{3}\in\mathbb{F}$. It follows that $R$ is abelian. Furthermore, $\mathcal{R}(\mathcal{T}_{3},D)=\mathcal{R}_{D}$. Hence, we have to consider the six subcases corresponding to $\tilde{R}=S_{(3,1)},R_{\rho},U_{1}^{3},N_{1},N_{2},N_{3,\lambda}$.

Lemma 5.3**.**

There are exactly $|\mathbb{F}|+1$ distinct conjugacy classes of nonabelian subgroups $R=\mathcal{R}(S_{(3,1)},D)\in\Delta_{4}(\mathbb{F})$ such that $\mathsf{d}(\mathbf{Z}(R))=2$ and $\mathsf{r}(\mathbf{Z}(R))=1$. Such classes can be represented by the subgroups corresponding to

[TABLE]

These subgroups are not conjugate to $\mathcal{R}_{A}$ for any $A\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$.

Proof.

Let $R=\mathcal{R}(S_{(3,1)},D)$ for some $D\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$ such that $\mathsf{d}(\mathbf{Z}(R))=2$ and $\mathsf{r}(\mathbf{Z}(R))=1$. By (2.2) and the hypothesis that $R$ is nonabelian, we obtain $D=\begin{pmatrix}\bar{\alpha}_{1}&\bar{\alpha}_{2}&\bar{\alpha}_{3}\\ \bar{\alpha}_{2}&0&0\\ \bar{\gamma}_{1}&0&\bar{\gamma}_{3}\end{pmatrix}$ with $\bar{\gamma}_{1}\neq\bar{\alpha}_{3}$. Moreover, the condition $\mathsf{r}(\mathbf{Z}(R))=1$ implies $\bar{\alpha}_{2}=0$. Then, $R$ is conjugate via $g=I_{5}-\bar{\alpha}_{1}E_{3,5}$ to

[TABLE]

for some $\alpha_{3},\gamma_{1},\gamma_{3}\in\mathbb{F}$ such that $\gamma_{1}\neq\alpha_{3}$.

Suppose that $\gamma_{3}\neq 0$. Then $\mathsf{d}(R)=\mathsf{r}(R)=3$ and $\mathsf{k}(R)=2$. Consider the algebra $\mathcal{L}=\mathcal{L}_{U_{3}(0,1,1)}$:

[TABLE]

The function $\Psi:\mathcal{L}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism and so $R$ is conjugate to $U_{3}(0,1,1)$.

Suppose now that $\gamma_{3}=\alpha_{3}=0$ (and so $\gamma_{1}\neq 0$). Then $\mathsf{d}(R)=\mathsf{r}(R)=\mathsf{k}(R)=3$. In this case, consider the algebra $\mathcal{L}=\mathcal{L}_{U_{3}(0,1,0)}$:

[TABLE]

The function $\Psi:\mathcal{L}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism: it follows that $R$ is conjugate to $U_{3}(0,1,0)$.

Finally, suppose that $\gamma_{3}=0$ and $\alpha_{3}\neq 0$. For any $\lambda\in\mathbb{F}\setminus\{1\}$, consider the algebra $\mathcal{L}_{\lambda}=\mathcal{L}_{U_{3}(1,\lambda,0)}$:

[TABLE]

The function $\Psi:\mathcal{L}_{\lambda}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism and hence $R$ is conjugate to $U_{3}(1,\lambda,0)$ for some $\lambda\neq 1$. Now, notice that if $\lambda=0$ then $\mathsf{d}(U_{3}(1,0,0))=3$ and $\mathsf{r}(U_{3}(1,0,0))=\mathsf{k}(U_{3}(1,0,0))=2$. Otherwise, $\mathsf{d}(U_{3}(1,\lambda,0))=\mathsf{r}(U_{3}(1,\lambda,0))=3$ and $\mathsf{k}(U_{3}(1,\lambda,0))=2$. Furthermore, $U_{3}(1,0,0)$ is not conjugate to any $\mathcal{R}_{A}$ (direct computations) and the same holds for the other subgroups, since $\mathsf{r}(\mathcal{R}_{A})\leq 2$. The subgroups $U_{3}(1,\lambda_{1},0)$ and $U_{3}(1,\lambda_{2},0)$ are conjugate if and only if $\lambda_{1}=\lambda_{2}$. To conclude, we observe that $U_{3}(1,\lambda,0)$ is not conjugate to $U_{3}(0,1,1)$. ∎

Before to consider the next subcase, we recall the following notation given in the Introduction. For any $\rho\in\mathbb{F}^{\square}$, we denote by $\mathsf{A}_{\rho}(\mathbb{F})$ a fixed set of representatives for the equivalence classes with respect to the following relation $\divideontimes_{\rho}$ defined on $\overline{\mathsf{A}}(\mathbb{F})=\left(\mathbb{F}\setminus\{1\}\right)\times\mathbb{F}$: given $(\beta_{1},\beta_{2}),(\beta_{3},\beta_{4})\in\overline{\mathsf{A}}(\mathbb{F})$, we write $(\beta_{1},\beta_{2})\divideontimes_{\rho}(\beta_{3},\beta_{4})$ if and only if either $(\beta_{3},\beta_{4})=(\beta_{1},\pm\beta_{2})$ or

[TABLE]

for some $t\in\mathbb{F}$ such that $(t^{2}+\rho)(t^{2}+\beta_{2}t-\rho\beta_{1})\neq 0$.

Lemma 5.4**.**

The distinct conjugacy classes of nonabelian subgroups $R=\mathcal{R}(R_{\rho},D)$ $\in\Delta_{4}(\mathbb{F})$ such that $\mathsf{d}(\mathbf{Z}(R))=2$ and $\mathsf{r}(\mathbf{Z}(R))=1$ can be represented by

[TABLE]

with $\rho\in\mathbb{F}^{\square}$ and $(\beta_{1},\beta_{2})\in\mathsf{A}_{\rho}(\mathbb{F})$. Furthermore, $\mathsf{d}(R)=\mathsf{r}(R)=3$ and $\mathsf{k}(R)=2$.

Proof.

Let $R=\mathcal{R}(R_{\rho},D)\in\Delta_{4}(\mathbb{F})$, with $\rho\in\mathbb{F}^{\square}$ and $D\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$, be a nonabelian subgroup such that $\mathsf{d}(\mathbf{Z}(R))=2$ and $\mathsf{r}(\mathbf{Z}(R))=1$. By (2.2) and the hypothesis that $R$ is nonabelian, we have $D=\begin{pmatrix}\bar{\alpha}_{1}&\bar{\alpha}_{2}&0\\ \bar{\beta}_{1}&\bar{\beta}_{2}&0\\ 0&0&0\end{pmatrix}$, with $\bar{\beta}_{1}\neq\bar{\alpha}_{2}$. Taking $g=\mathop{\rm diag}\nolimits\left(I_{3},\begin{pmatrix}1&-\bar{\alpha}_{1}\bar{\alpha}_{2}^{-1}\\ 0&\bar{\alpha}_{2}^{-1}\end{pmatrix}\right)$ if $\bar{\alpha}_{2}\neq 0$ and $g=\mathop{\rm diag}\nolimits\left(1,\begin{pmatrix}0&1\\ \rho&0\end{pmatrix},\begin{pmatrix}\rho&-\bar{\beta}_{2}\bar{\beta}_{1}^{-1}\\ 0&\rho\bar{\beta}_{1}^{-1}\end{pmatrix}\right)$ if $\bar{\alpha}_{2}=0$, we obtain that $R$ is conjugate via $g$ to the subgroup

[TABLE]

Let $\rho_{1},\rho_{2}\in\mathbb{F}^{\square}$, $\beta_{1},\beta_{2},\beta_{3},\beta_{4}\in\mathbb{F}$ with $\beta_{1},\beta_{3}\neq 1$. It is quite easy to verify that the subgroups $U_{4}(\rho_{1},\beta_{1},\beta_{2})$ and $U_{4}(\rho_{2},\beta_{3},\beta_{4})$ are conjugate in $\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$ if and only if there exists an element $g=\mathop{\rm diag}\nolimits(1,A,B)\in\mathop{\rm GL}\nolimits_{5}(\mathbb{F})$, with $A=\begin{pmatrix}a_{1}&a_{2}\\ a_{3}&a_{4}\end{pmatrix}$, $B=\begin{pmatrix}b_{1}&b_{2}\\ b_{3}&b_{4}\end{pmatrix}$, such that $U_{4}(\rho_{1},\beta_{1},\beta_{2})^{g}=U_{4}(\rho_{2},\beta_{3},\beta_{4})$. This holds if and only if $b_{1}=\rho_{2}a_{2}^{2}+a_{1}^{2}$, $b_{2}=(\beta_{3}+1)a_{1}a_{2}+\beta_{4}a_{2}^{2}$, $b_{3}=\rho_{2}a_{2}a_{4}+a_{1}a_{3}$, $b_{4}=\beta_{3}a_{2}a_{3}+\beta_{4}a_{2}a_{4}+a_{1}a_{4}$ and

[TABLE]

[TABLE]

provided that

[TABLE]

Now, if $a_{2}=0$, from (5.2) and (5.4) we get $a_{3}=0$ and in this case (5.3) gives $\beta_{3}=\beta_{1}$, $\rho_{2}=\rho_{1}$ and $\beta_{4}=\pm\beta_{2}$ (notice that, taking $g=\mathop{\rm diag}\nolimits(1,1,-1,1,-1)$, we have $U_{4}(\rho_{1},\beta_{1},\beta_{2})^{g}=U_{4}(\rho_{1},\beta_{1},-\beta_{2})$). Hence, we may suppose $a_{2}\neq 0$ and $a_{4}=-a_{1}a_{3}(a_{2}\rho_{2})^{-1}$ by (5.2). In this case (5.3) gives $\rho_{2}=\rho_{1}$, $a_{3}=\pm a_{2}\rho_{1}$, $f_{0}=0$ and $f_{+}=0$ or $f_{-}=0$, where

[TABLE]

provided that

[TABLE]

Taking $a_{2}=\pm 1$, it follows that $U_{4}(\rho_{1},\beta_{1},\beta_{2})$ is conjugate to $U_{4}(\rho_{2},\beta_{3},\beta_{4})$ if and only if $\rho_{2}=\rho_{1}$ and either $(\beta_{3},\beta_{4})=(\beta_{1},\pm\beta_{2})$ or

[TABLE]

for some $t\in\mathbb{F}$ such that $(t^{2}+\rho)(t^{2}+\beta_{2}t-\rho_{1}\beta_{1})\neq 0$. In other words, this holds if and only if $\rho_{2}=\rho_{1}$ and $(\beta_{1},\beta_{2})\divideontimes_{\rho_{1}}(\beta_{3},\beta_{4})$. ∎

Remark 5.5*.*

Let $\rho\in\mathbb{F}^{\square}$, $\beta_{1},\beta_{3}\in\mathbb{F}\setminus\{1\}$ and $\beta_{2},\beta_{4}\in\mathbb{F}$. Then:

• (a)

$U_{4}(\rho,\beta_{1},\beta_{2})$ and $U_{4}(\rho,-1,0)$ are conjugate if and only if $(\beta_{1},\beta_{2})=(-1,0)$;

• (b)

$U_{4}(\rho,\beta_{1},\beta_{2})$ and $U_{4}(\rho,\beta_{1},\beta_{4})$ are conjugate if and only if $\beta_{4}=\pm\beta_{2}$;

• (c)

$U_{4}(\rho,\beta_{1},0)$ and $U_{4}(\rho,\beta_{3},0)$ are conjugate if and only if either $\beta_{3}=\beta_{1}$ or $\beta_{1}\neq 0$ and $\beta_{3}=\beta_{1}^{-1}$;

• (d)

$U_{4}(\rho,\beta_{1},0)$ and $U_{4}(\rho,0,\beta_{4})$ are conjugate if and only if $\beta_{1}=\rho\omega^{2}$ and $\beta_{4}=\frac{2\rho\omega}{\rho\omega^{2}-1}$ for some $\omega\in\mathbb{F}$ such that $\rho\omega^{2}\neq\pm 1$.

Remark 5.6*.*

It will be useful to give the presentation of the following algebras:

• (a)

$\mathcal{L}_{U_{4}(\rho,0,\lambda)}=\mathrm{Span}(t_{1},t_{2})\;\textrm{ such that: }\;t_{1}^{3}=t_{2}t_{1}=0;\quad t_{2}^{2}=\rho t_{1}^{2}+\lambda t_{1}t_{2}$;

• (b)

$\mathcal{L}_{U_{4}(\rho,-1,0)}=\mathrm{Span}(t_{1},t_{2})\;\textrm{ such that: }\;t_{1}^{3}=t_{1}t_{2}+t_{2}t_{1}=0;\quad t_{2}^{2}=\rho t_{1}^{2}$.

Lemma 5.7**.**

Assume ${\rm char\,}\mathbb{F}=2$. There are exactly $|\mathsf{B}(\mathbb{F})|$ distinct conjugacy classes of nonabelian subgroups $R=\mathcal{R}(U_{1}^{3},D)\in\Delta_{4}(\mathbb{F})$ such that $\mathsf{d}(\mathbf{Z}(R))=2$ and $\mathsf{r}(\mathbf{Z}(R))=1$. Such classes can be represented by the subgroups corresponding to

[TABLE]

Furthermore, $\mathsf{d}(R)=\mathsf{r}(R)=3$ and $\mathsf{k}(R)=2$.

Proof.

Let $R=\mathcal{R}(U_{1}^{3},D)$ for some $D\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$. By (2.2) and the hypothesis that $R$ is nonabelian, we have $D=\begin{pmatrix}\bar{\alpha}_{1}&\bar{\alpha}_{2}&0\\ \bar{\beta}_{1}&\bar{\beta}_{2}&0\\ 0&0&0\end{pmatrix}$ with $\bar{\beta}_{1}\neq\bar{\alpha}_{2}$. Then, $R$ is conjugate via $g=I_{5}+\bar{\beta}_{1}E_{4,5}$ to

[TABLE]

where $\alpha_{2}\neq 0$. Then $\mathsf{d}(R)=\mathsf{r}(R)=3$ and $\mathsf{k}(U)=2$.

For any $\lambda\in\mathbb{F}$, consider the algebra $\mathcal{L}_{\lambda}=\mathcal{L}_{U_{5}(1,1,\lambda)}$:

[TABLE]

(clearly, when $\lambda=0$, the condition $t_{2}^{3}=0$ can be omitted). We now define an isomorphism of split local algebras $\Psi:\mathcal{L}_{\lambda}\rightarrow\mathcal{L}_{R}$ in the following way:

[TABLE]

where $\lambda=\frac{\alpha_{1}\beta_{2}}{\alpha_{2}^{2}}$. Hence, $R$ is conjugate to $U_{5}(1,1,\lambda)$ for some $\lambda\in\mathbb{F}$. Now, $U_{5}(1,1,\lambda_{1})$ and $U_{5}(1,1,\lambda_{2})$ are conjugate if and only if $\lambda_{1}+\lambda_{2}\in\overline{\mathsf{B}}(\mathbb{F})$. We conclude that are exactly $|\mathsf{B}(\mathbb{F})|$ conjugacy classes. ∎

Lemma 5.8**.**

Let $R=\mathcal{R}(N_{1},D)\in\Delta_{4}(\mathbb{F})$ be a nonabelian subgroup such that $\mathsf{d}(\mathbf{Z}(R))=2$ and $\mathsf{r}(\mathbf{Z}(R))=1$. Then $R$ is conjugate to one of the following subgroups: $\mathcal{R}_{E_{2,1}}$, $U_{3}(1,0,0)$, $U_{3}(0,1,0)$, $U_{3}(0,1,1)$, $U_{4}(\rho,-1,0)$, $U_{4}(\rho,0,\lambda)$ or $U_{5}(1,1,0)$, for some $\rho\in\mathbb{F}^{\square}$ and $\lambda\in\mathbb{F}$.

Proof.

Let $R=\mathcal{R}(N_{1},D)$ for some $D\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$. By (2.2), $D=\begin{pmatrix}\bar{\alpha}_{1}&\bar{\alpha}_{2}&0\\ \bar{\beta}_{1}&\bar{\beta}_{2}&0\\ 0&0&0\end{pmatrix}$. Then, $R$ is conjugate via $g=I_{5}-\bar{\beta}_{1}E_{4,5}$ to

[TABLE]

for some $\alpha_{1},\alpha_{2},\beta_{2}\in\mathbb{F}$. It is easy to see that $U_{6}(\alpha_{1},\alpha_{2},\beta_{2})$ is conjugate to some $\mathcal{R}_{A}$ if and only if $\alpha_{1}=\alpha_{2}=\beta_{2}=0$. More in detail, $U_{6}(0,0,0)^{g}=\mathcal{R}_{E_{2,1}}$, where $g=\mathop{\rm diag}\nolimits(I_{3},E_{1,2}+E_{2,1})$.

If $\alpha_{1}=\alpha_{2}=0$ and $\beta_{2}\neq 0$, then $\mathsf{d}(R)=3$ and $\mathsf{r}(R)=\mathsf{k}(R)=2$. The function $\Psi:\mathcal{L}_{U_{3}(1,0,0)}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism. By Proposition 2.4, $R$ is conjugate to the subgroup $U_{3}(1,0,0)$.

If $\alpha_{1}\neq 0$ and $\alpha_{2}=\beta_{2}=0$, then $\mathsf{d}(R)=\mathsf{r}(R)=\mathsf{k}(R)=3$. The function $\Psi:\mathcal{L}_{U_{3}(0,1,0)}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism and so $R$ is conjugate to $U_{3}(0,1,0)$.

For the other values of $\alpha_{1},\alpha_{2},\beta_{2}$ we have $\mathsf{d}(R)=\mathsf{r}(R)=3$ and $\mathsf{k}(R)=2$. If $\alpha_{1},\beta_{2}\neq 0$ and $\alpha_{2}=0$, write $\frac{\alpha_{1}}{\beta_{2}}=\omega^{2}\rho$ for some $\omega\in\mathbb{F}^{\ast}$ and a unique $\rho\in\mathbb{F}^{\square}$: the function $\Psi:\mathcal{L}_{U_{4}(\rho,0,0)}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism. Hence, $R$ is conjugate to $U_{4}(\rho,0,0)$.

We are left to consider the case where $\alpha_{2}\neq 0$. Suppose $\alpha_{1}=\beta_{2}=0$. If ${\rm char\,}\mathbb{F}\neq 2$, write $-1=\omega^{2}\rho$ for some $\omega\in\mathbb{F}^{\ast}$ and a unique $\rho\in\mathbb{F}^{\square}$: the function $\Psi:\mathcal{L}_{U_{4}(\rho,-1,0)}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism and hence $R$ is conjugate to $U_{4}(\rho,-1,0)$. If ${\rm char\,}\mathbb{F}=2$, the function $\Psi:\mathcal{L}_{U_{5}(1,1,0)}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism and hence $R$ is conjugate to $U_{5}(1,1,0)$.

If $(\alpha_{1},\beta_{2})\neq(0,0)$, set $\Delta=\alpha_{1}\beta_{2}-\alpha_{2}^{2}$. If $\Delta=0$ (and so $\beta_{2}\neq 0$) then the function $\Psi:\mathcal{L}_{U_{3}(0,1,1)}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism. Hence, $U$ is conjugate to $U_{3}(0,1,1)$. If $\Delta\neq 0$, write $\frac{\Delta}{\alpha_{2}^{2}}=\omega^{2}\rho$ for some $\omega\in\mathbb{F}^{\ast}$ and a unique $\rho\in\mathbb{F}^{\square}$. The function $\Psi:\mathcal{L}_{U_{4}(\rho,0,\omega^{-1})}\rightarrow\mathcal{L}_{R}$ defined by

[TABLE]

is an isomorphism. Hence, $R$ is conjugate to $U_{4}(\rho,0,\omega^{-1})$. ∎

Lemma 5.9**.**

Assume ${\rm char\,}\mathbb{F}\neq 2$. Let $R=\mathcal{R}(N_{2},D)\in\Delta_{4}(\mathbb{F})$ be a nonabelian subgroup such that $\mathsf{d}(\mathbf{Z}(R))=2$ and $\mathsf{r}(\mathbf{Z}(R))=1$. Then $R$ is conjugate to one of the following subgroups: $\mathcal{R}_{(E_{1,2}-E_{2,1})}$, $U_{3}(1,-1,0)$ or $U_{4}(\rho,-1,0)$ for some $\rho\in\mathbb{F}^{\square}$.

Proof.

Let $R=\mathcal{R}(N_{2},D)$ for some $D\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$. By (2.2) we must have $D=\begin{pmatrix}\bar{\alpha}_{1}&\bar{\alpha}_{2}&0\\ \bar{\beta}_{1}&\bar{\beta}_{2}&0\\ 0&0&0\end{pmatrix}$. Then, $R$ is conjugate via $g=I_{5}-\bar{\beta}_{1}E_{4,5}$ to

[TABLE]

for some $\alpha_{1},\alpha_{2},\beta_{2}\in\mathbb{F}$. It is easy to see that $U_{7}(\alpha_{1},\alpha_{2},\beta_{2})$ is conjugate to some $\mathcal{R}_{A}$ if and only if $\alpha_{1}=\alpha_{2}=\beta_{2}=0$. In this case $\mathsf{d}(R)=\mathsf{r}(R)=\mathsf{k}(R)=2$ and $U_{7}(0,0,0)^{g}=\mathcal{R}_{(E_{1,2}-E_{2,1})}$ where $\mathop{\rm diag}\nolimits(I_{3},(E_{2,1}-E_{1,2}))$.

Suppose now $(\alpha_{1},\alpha_{2},\beta_{2})\neq(0,0,0)$ and let $\Delta=4\alpha_{1}\beta_{2}-\alpha_{2}^{2}$. Notice that in this case $\mathsf{d}(R)=\mathsf{r}(R)=3$ and $\mathsf{k}(R)=2$. Assume $\Delta=0$ and consider the function $\Psi:\mathcal{L}_{U_{3}(1,-1,0)}\rightarrow\mathcal{L}_{R}$ given by

[TABLE]

Since $\Psi$ is an isomorphism, we obtain that $U_{7}(\alpha_{1},\alpha_{2},\beta_{2})$ is conjugate to $U_{3}(1,-1,0)$. Now, assume $\Delta\neq 0$ and write $\Delta=\omega^{2}\rho$ for some $\omega\in\mathbb{F}^{\ast}$ and a unique $\rho\in\mathbb{F}^{\square}$. The function $\Psi:\mathcal{L}_{U_{4}(\rho,-1,0)}\rightarrow\mathcal{L}_{R}$ given by

[TABLE]

is an isomorphism: $R$ is conjugate to $U_{4}(\rho,-1,0)$. ∎

Lemma 5.10**.**

Let $R=\mathcal{R}(N_{3,\lambda},D)\in\Delta_{4}(\mathbb{F})$, $\lambda\in\mathbb{F}^{\ast}$, be a nonabelian subgroup such that $\mathsf{d}(\mathbf{Z}(R))=2$ and $\mathsf{r}(\mathbf{Z}(R))=1$. Then, $R$ is conjugate to one of the following subgroups: $\mathcal{R}_{A}$, where $A=E_{1,1}+E_{2,1}+\lambda E_{2,2}$, $U_{3}(0,1,1)$, $U_{3}(1,\mu,0)$ for some $\mu\in\mathbb{F}\setminus\{0,1\}$, $U_{4}(\rho,\beta_{1},\beta_{2})$ for some $\rho\in\mathbb{F}^{\square}$ and $\beta_{1},\beta_{2}\in\mathbb{F}$, or $U_{5}(1,1,\omega)$ where $\omega\in\mathbb{F}$ is such that $\omega+\lambda\in\overline{\mathsf{B}}(\mathbb{F})$.

Proof.

Let $R=\mathcal{R}(N_{3,\lambda},D)$ for some $\lambda\in\mathbb{F}^{\ast}$ and some $D\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$. By (2.2) $D=\begin{pmatrix}\bar{\alpha}_{1}&\bar{\alpha}_{2}&0\\ \bar{\beta}_{1}&\bar{\beta}_{2}&0\\ 0&0&0\end{pmatrix}$. Then, $R$ is conjugate via $g=I_{5}-\frac{\bar{\beta}_{2}}{\lambda}E_{4,5}$ to

[TABLE]

for some $\alpha_{1},\alpha_{2},\beta_{1}\in\mathbb{F}$. It is easy to see that $\tilde{U}_{8}(\lambda,\alpha_{1},\alpha_{2},\beta_{1})$ is conjugate to some $\mathcal{R}_{A}$ if and only if $\alpha_{1}=\alpha_{2}=\beta_{1}=0$. In this case $\mathsf{d}(R)=3$ and $\mathsf{r}(R)=\mathsf{k}(R)=2$ and $\tilde{U}_{8}(\lambda,0,0,0)^{g}=\mathcal{R}_{A}$, where $g=E_{1,1}+E_{2,3}+E_{5,4}-\lambda E_{2,2}+\lambda E_{3,3}+\lambda^{2}E_{4,5}$ and $A=E_{1,1}+E_{2,1}+\lambda E_{2,2}$.

So, suppose $(\alpha_{1},\alpha_{2},\beta_{1})\neq(0,0,0)$: we have $\mathsf{d}(R)=\mathsf{r}(R)=3$ and $\mathsf{k}(R)=2$. We may also assume $\beta_{1}=1$. Namely, if $\beta_{1}\neq 0$ it suffices to conjugate $\tilde{U}_{8}(\lambda,\alpha_{1},\alpha_{2},\beta_{1})$ by $g=\mathop{\rm diag}\nolimits(1,\beta_{1},\beta_{1},\beta_{1}^{2},\beta_{1})$. If $\beta_{1}=0$ and $\alpha_{2}\neq 0$, then we conjugate by the matrix $\mathop{\rm diag}\nolimits\left(1,\begin{pmatrix}1&-1\\ \lambda&0\end{pmatrix},\begin{pmatrix}\lambda&(\lambda\alpha_{1}-\alpha_{2})\alpha_{2}^{-1}\\ 0&-\lambda\alpha_{2}^{-1}\end{pmatrix}\right)$ and if $\alpha_{2}=\beta_{1}=0$, we conjugate by the matrix $\mathop{\rm diag}\nolimits\left(1,\begin{pmatrix}0&\alpha_{1}\\ -\lambda\alpha_{1}&\alpha_{1}\end{pmatrix},\begin{pmatrix}\lambda\alpha_{1}^{2}&-\lambda\alpha_{1}^{2}\\ 0&\lambda\alpha_{1}\end{pmatrix}\right)$.

Now, set $U_{8}(\lambda,\alpha_{1},\alpha_{2})=\tilde{U}_{8}(\lambda,\alpha_{1},\alpha_{2},1)$ and $\Delta=-\lambda(\alpha_{2}-1)(\alpha_{1}-\alpha_{2}+1)-1$. Suppose $\alpha_{2}=1$. In this case $\Delta=-1$ that we write as $-1=\omega^{2}\rho$ for some $\omega\in\mathbb{F}^{\ast}$ and a unique $\rho\in\mathbb{F}^{\square}$. If $\alpha_{1}\neq 0,1$, then $U_{8}(\lambda,\alpha_{1},1)^{g}=U_{4}\left(\rho,\frac{1}{1-\alpha_{1}},\frac{\alpha_{1}^{2}\lambda-\alpha_{1}+2}{\omega(\alpha_{1}-1)}\right)$, where

[TABLE]

If $\alpha_{1}=0$ and ${\rm char\,}\mathbb{F}\neq 2$, then $U_{8}(\lambda,0,1)^{g}=U_{4}\left(\rho,1-\frac{2}{\lambda},-\frac{2(\lambda+1)}{\omega\lambda}\right)$, where $g=\mathop{\rm diag}\nolimits\left(1,\begin{pmatrix}\omega^{-1}&1\\ \omega^{-1}&-1\end{pmatrix},\begin{pmatrix}0&-\frac{4}{\omega\lambda}\\ 2\omega^{-2}&\frac{2(\lambda+2)}{\omega\lambda}\end{pmatrix}\right)$. If $\alpha_{1}=0$ and ${\rm char\,}\mathbb{F}=2$, then $U_{8}(\lambda,0,1)^{g}=U_{5}(1,1,\lambda+\mu+\mu^{2})$, where $g=\mathop{\rm diag}\nolimits\left(1,\begin{pmatrix}1&0\\ \mu&1\end{pmatrix},\begin{pmatrix}0&1\\ 1&\mu\end{pmatrix}\right)$. If $\alpha_{1}=1$, then $U_{8}(\lambda,1,1)^{g}=U_{4}\left(\rho,0,-(\lambda+1)\omega^{-1}\right)$, where

[TABLE]

Hence, we may now suppose $\alpha_{2}\neq 1$. If $\Delta\neq 0$, we write $\Delta=\rho\omega^{2}$ for some $\omega\in\mathbb{F}^{\ast}$ and a unique $\rho\in\mathbb{F}^{\square}$. Then $U_{8}(\lambda,\alpha_{1},\alpha_{2})^{g}=U_{4}(\rho,\alpha_{2},-(\alpha_{1}\alpha_{2}\lambda-\alpha_{1}\lambda+\alpha_{2}+1)\omega^{-1})$ where

[TABLE]

If $\Delta=0$, i.e. $\alpha_{1}=\frac{\lambda(\alpha_{2}-1)^{2}-1}{\lambda(\alpha_{2}-1)}$, we need to distinguish two cases. If $\lambda(\alpha_{2}-1)^{2}+\alpha_{2}\neq 0$, then $U_{8}(\lambda,\alpha_{1},\alpha_{2})^{g}=U_{3}(0,1,1)$, where

[TABLE]

and if $\lambda=\frac{-\alpha_{2}}{(\alpha_{2}-1)^{2}}$, then $U_{8}(\lambda,\alpha_{1},\alpha_{2})^{g}=U_{3}(1,\alpha_{2},0)$, where

[TABLE]

∎

Corollary 5.11**.**

Let $R\in\Delta_{4}(\mathbb{F})$ be a nonabelian subgroup such that $\mathsf{d}(\mathbf{Z}(R))=2$ and $\mathsf{r}(\mathbf{Z}(R))=1$. Suppose that $R$ is not conjugate to any $\mathcal{R}_{A}$, $A\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$. Then $R$ is conjugate to exactly one of the subgroups listed in Table 3.

Proof.

In view of Lemmas 5.3, 5.4, 5.7, 5.8, 5.9 and 5.10, $R$ is conjugate to one of the subgroups listed in Table 3. Since $\mathsf{d}(\mathcal{R}_{A})\leq 2$, it follows from Lemma 5.3 that these subgroups are not conjugate to $\mathcal{R}_{A}$ for any $A\in\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$. Direct computations show that no pair of subgroups $U_{3}(0,1,1)$, $U_{3}(1,\lambda,0)$, $U_{4}(\rho,\beta_{1},\beta_{2})$ and $U_{5}(1,1,\epsilon)$ is conjugate in $\mathop{\rm AGL}\nolimits_{4}(\mathbb{F})$. ∎

We can now give the classification of the nonabelian subgroups in $\Delta_{4}(\mathbb{F})$.

Theorem 5.12**.**

Let $\mathbb{F}$ be a field. The distinct conjugacy classes of nonabelian regular subgroups in $\Delta_{4}(\mathbb{F})$ can be represented by the subgroups described in the first column of Table 2.

Proof.

Let $R\in\Delta_{4}(\mathbb{F})$ be a nonabelian regular subgroup. By the considerations given in Section 3 we have the following possibilities: $(\mathsf{d}(\mathbf{Z}(R)),\mathsf{r}(\mathbf{Z}(R)))=(3,2)$, $(2,2)$ or $(2,1)$. If $(\mathsf{d}(\mathbf{Z}(R)),\mathsf{r}(\mathbf{Z}(R)))=(2,2)$ then $R$ is conjugate either to $U_{2}(0,1,0,1,1)$ or to $U_{2}(0,1,0,0,1)$, by Lemma 5.1. If $(\mathsf{d}(\mathbf{Z}(R)),\mathsf{r}(\mathbf{Z}(R)))=(2,1)$, then $R$ is conjugate to either one of the subgroups of Table 3 or to a subgroup $\mathcal{R}_{A}$, by Corollary 5.11. If $(\mathsf{d}(\mathbf{Z}(R)),\mathsf{r}(\mathbf{Z}(R)))=(3,2)$, then $R$ is conjugate to a subgroup $\mathcal{R}_{A}$. The statement of the theorem is proved considering a set $\Pi_{A}(\mathbb{F})$ of representatives for projective congruent classes of asymmetric matrices in $\mathop{\rm Mat}\nolimits_{3}(\mathbb{F})$. ∎

We would like to give explicit sets $\mathsf{A}_{\rho}$, at least when $\mathbb{F}$ is algebraically closed, the real field $\mathbb{R}$ or a finite field. Observe that, for every field $\mathbb{F}$ of characteristic $\neq 2$, $(-1,0)\in\mathsf{A}_{\rho}$ for all $\rho\in\mathbb{F}^{\square}$, by Remark 5.5(a).

Fixing $\rho\in\mathbb{F}^{\square}$, $\beta_{1}\in\mathbb{F}\setminus\{1\}$ and $\beta_{2}\in\mathbb{F}$ such that $(\beta_{1},\beta_{2})\neq(-1,0)$, define the following polynomials in $\mathbb{F}[t]$:

[TABLE]

Lemma 5.13**.**

Suppose that $\beta_{1}\neq 1$ and $(\beta_{1},\beta_{2})\neq(-1,0)$. If either $\beta_{1}=0$ or $f_{1}(t)$ is reducible, then $U_{4}(\rho,\beta_{1},\beta_{2})$ is conjugate to $U_{4}(\rho,0,\lambda)$ for some $\lambda\in\mathbb{F}$.

Proof.

Clearly, the statement holds if $\beta_{1}=0$. So, assume $\beta_{1}\neq 0$. Since $f_{1}(t)$ is reducible, we can take $a\in\mathbb{F}$ such that $f_{1}(a)=0$ and from (5.1) we obtain that $U_{4}(\rho,\beta_{1},\beta_{2})$ is conjugate to $U_{4}\left(\rho,0,\frac{f_{2}(a)}{g_{1}(a)}\right)$ provided that $g_{1}(a)g_{2}(a)\neq 0$. Suppose that $f_{1}(a)=g_{1}(a)g_{2}(a)=0$. Observe that $f_{1}(a)=g_{1}(a)=0$ implies $g_{2}(a)=0$, since $f_{1}(t)=g_{1}(t)+(\beta_{1}-1)g_{2}(t)$. So, we may suppose that any root of $f_{1}(t)$ is also a root of $g_{2}(t)$. Clearly, this implies that $\rho=-\omega^{2}$ for some $\omega\in\mathbb{F}$. Take $a=a_{i}=(-1)^{i}\omega$, $i=0,1$. If $f_{1}(a_{i})=0$ for some $i$, then $\beta_{2}=(-1)^{i+1}\omega(\beta_{1}+1)$. In this case, $f_{1}(t)=(t-a_{i})(\beta_{1}t-(-1)^{i}\omega)$ implies that $\beta_{1}(-1)^{j}\omega-(-1)^{i}\omega=0$ for some $j=0,1$, whence $\beta_{1}=-1$ and $(i,j)\in\{(0,1),(1,0)\}$. It follows that $\beta_{2}=0$, contradicting our assumption $(\beta_{1},\beta_{2})\neq(-1,0)$. ∎

In the following, given a subset $S$ of $\mathbb{F}^{\ast}$, we denote by $-S$ and $S^{-1}$ the sets $\{-s:s\in S\}$ and $\{s^{-1}:s\in S\}$, respectively.

Corollary 5.14**.**

Let $\mathbb{F}$ be a field with no quadratic extensions.

• (a)

If ${\rm char\,}\mathbb{F}\neq 2$, then $\mathsf{A}_{1}=\{(-1,0),(0,0)\}\cup\{(0,\lambda):\lambda\in\mathsf{N}(\mathbb{F})\}$, where $\mathsf{N}(\mathbb{F})$ is a fixed subset of $\mathbb{F}^{\ast}$ such that $\mathbb{F}^{\ast}=\mathsf{N}(\mathbb{F})\sqcup-\mathsf{N}(\mathbb{F})$.

• (b)

If ${\rm char\,}\mathbb{F}=2$, then $\mathsf{A}_{1}=\{(0,\lambda):\lambda\in\mathbb{F}\}$.

Proof.

It follows from Lemma 5.13 and Remark 5.5. ∎

Proposition 5.15**.**

Let $\mathbb{F}=\overline{\mathbb{F}}$ be an algebraically closed field. The distinct conjugacy classes of nonabelian regular subgroups in $\Delta_{4}(\mathbb{F})$ can be represented by

[TABLE]

[TABLE]

and

[TABLE]

where, if ${\rm char\,}\mathbb{F}\neq 2$,

[TABLE]

and $\mathsf{M}(\mathbb{F})$ and $\mathsf{N}(\mathbb{F})$ are subsets of $\mathbb{F}^{\ast}$ such that $\mathbb{F}\setminus\{0,\pm 1\}=\mathsf{M}(\mathbb{F})\sqcup\mathsf{M}(\mathbb{F})^{-1}$ and $\mathbb{F}^{\ast}=\mathsf{N}(\mathbb{F})\sqcup-\mathsf{N}(\mathbb{F})$; if ${\rm char\,}\mathbb{F}=2$,

[TABLE]

and $\mathsf{M}(\mathbb{F})$ is a subset of $\mathbb{F}^{\ast}$ such that $\mathbb{F}\setminus\{0,1\}=\mathsf{M}(\mathbb{F})\sqcup\mathsf{M}(\mathbb{F})^{-1}$.

Proof.

By Theorem 5.12 and Corollary 5.14 we are left to determine the set $\Pi_{A}(\mathbb{F})$. Applying [3, Theorem 2.1], the elements of $\Pi_{A}(\mathbb{F})$ are the asymmetric matrices obtained as diagonal sum of the following matrices:

[TABLE]

where $\mu\in\mathsf{M}(\mathbb{F})$. If ${\rm char\,}\mathbb{F}\neq 2$ we also allow $\mu=-1$; if ${\rm char\,}\mathbb{F}=2$ we must omit the matrices

[TABLE]

∎

Lemma 5.16**.**

Let $\rho=1$ and ${\rm char\,}\mathbb{F}\neq 2$. Suppose that there exists an element $\iota\in\mathbb{F}$ such that $\iota^{2}=-1$. Then

[TABLE]

Proof.

We show that every subgroup $U_{4}(1,\beta_{1},\beta_{2})$, $\beta_{1}\neq 1$, is conjugate to a unique subgroup $U_{4}(1,\beta_{3},\beta_{4})$ with $(\beta_{3},\beta_{4})\in\mathsf{A}_{1}$ as in the statement. This clearly holds if $\beta_{4}=\pm 2\iota$, so suppose $\beta_{4}\neq\pm 2\iota$. Keeping the notation of (5.6), let $h_{1}(t)=f_{2}(t)-2\iota g_{1}(t)\in\mathbb{F}[t]$. Then, $a=\frac{2\beta_{1}+\iota\beta_{2}}{\beta_{2}-2\iota}$ is a root of $h_{1}(t)$. It follows that $R$ is conjugate to $U_{4}(1,\lambda+1,2\iota)$ for some $\lambda\in\mathbb{F}^{\ast}$, provided that $g_{1}(a)g_{2}(a)\neq 0$. Notice that $g_{1}(a)g_{2}(a)=0$ if and only if $(\beta_{1}+\iota\beta_{2}+1)(\beta_{2}^{2}+4\beta_{1})=0$.

Suppose that $\beta_{2}^{2}+4\beta_{1}=0$. If $\beta_{2}\neq 0$, take $b=2\beta_{2}^{-1}$. In this case, $f_{1}(b)=f_{2}(b)=0$, $g_{1}(b)=\frac{(\beta_{2}^{2}+4)^{2}}{4\beta_{2}^{2}}\neq 0$ and $g_{2}(b)=\frac{\beta_{2}^{2}+4}{\beta_{2}^{2}}\neq 0$. It follows that $R$ is conjugate to $U_{4}(1,0,0)$ (this clearly holds even if $\beta_{2}=0$). Now, suppose that $\beta_{1}+\iota\beta_{2}+1=0$ and let $h_{2}(t)=f_{2}(t)+2\iota g_{1}(t)$. Take $c=-\frac{3\iota\beta_{2}+2}{\beta_{2}+2\iota}$. Then $h_{2}(c)=0$, $g_{1}(c)=\frac{-2\iota\beta_{2}(\beta_{2}-2\iota)^{2})}{(\beta_{2}+2\iota)^{2}}\neq 0$ and $g_{2}(c)=\frac{-8\beta_{2}(\beta_{2}-2\iota)}{(\beta_{2}+2\iota)^{2}}\neq 0$. It follows that $R$ is conjugate to $U_{4}(1,-3,-2\iota)$ and then to $U_{4}(1,-3,2\iota)$, by Remark 5.5.

Finally, it is an easy computation to verify that $U_{4}(1,\lambda_{1}+1,2\iota)$ is conjugate to $U_{4}(1,\lambda_{2}+1,2\iota)$ if and only if $\lambda_{1}=\lambda_{2}$. ∎

Clearly the previous lemma gives an alternative set $\mathsf{A}_{1}$ for algebraically closed field of characteristic $\neq 2$. In a very similar way we can prove the following.

Lemma 5.17**.**

Let $\rho=-1$, ${\rm char\,}\mathbb{F}\neq 2$ and suppose that $-1\not\in(\mathbb{F})^{2}$. Then

[TABLE]

Lemma 5.18**.**

Suppose that ${\rm char\,}\mathbb{F}\neq 2$, $\beta_{1}\neq 0,1$, $\beta_{2}\neq 0$ and $(\beta_{1},\beta_{2})\neq(-1,0)$. If $f_{1}(t)$ is irreducible and $f_{2}(t)$ is reducible, then $U_{4}(\rho,\beta_{1},\beta_{2})$ is conjugate to $U_{4}(\rho,\lambda,0)$ for some $\lambda\in\mathbb{F}\setminus\{0,1\}$.

Proof.

Since $f_{2}(t)$ is reducible, we can take $a\in\mathbb{F}$ such that $f_{2}(a)=0$ and from (5.1) we obtain that $U_{4}(\rho,\beta_{1},\beta_{2})$ is conjugate to $U_{4}\left(\rho,\frac{f_{1}(a)}{g_{1}(a)},0\right)$ provided that $g_{1}(a)g_{2}(a)\neq 0$. So, first, suppose that $f_{2}(a)=g_{2}(a)=0$ for some $a\in\mathbb{F}$. As in Lemma 5.13, this implies that $\rho=-\omega^{2}$ for some $\omega\in\mathbb{F}^{\ast}$ and $a=a_{i}=(-1)^{i}\omega$ for some $i=0,1$. In particular, we get $\beta_{2}=(-1)^{i+1}\omega(\beta_{1}+1)$ (whence $\beta_{1}\neq-1$). In this case, we obtain that $f_{1}(a_{i})=0$, an absurd, since $f_{1}(t)$ is irreducible. Next, suppose that any root of $f_{2}(t)$ is a root of $g_{1}(t)=0$ (but not a root of $g_{2}(t)$). Let $r$ be the resultant between $f_{2}(t)$ and $g_{1}(t)$ with respect to $t$. From $r=0$ we get $(\beta_{2}^{2}+4\rho\beta_{1})(\rho(\beta_{1}+1)^{2}+\beta_{2}^{2})=0$. Since $f_{1}(t)$ is irreducible, this implies $\rho(\beta_{1}+1)^{2}+\beta_{2}^{2}=0$ and so $\rho=-\omega^{2}$ for some $\omega\in\mathbb{F}^{\ast}$ and $\beta_{2}=\pm\omega(\beta_{1}+1)$. As before, this condition implies that $f_{1}(t)$ is reducible, a contradiction. ∎

Corollary 5.19**.**

Let $\mathbb{F}=\mathbb{R}$. Then

[TABLE]

Proof.

The result follows from Lemmas 5.13 and 5.18 and Remark 5.5, observing that $f_{2}(t)\in\mathbb{R}[t]$ is reducible provided that $\beta_{2}\neq 0$. ∎

From Theorem 5.12 and [3, Theorem 2.1] we obtain the following classification.

Proposition 5.20**.**

The distinct conjugacy classes of nonabelian regular subgroups in $\Delta_{4}(\mathbb{R})$ can be represented by:

[TABLE]

[TABLE]

[TABLE]

where

[TABLE]

Lemma 5.21**.**

Suppose that ${\rm char\,}\mathbb{F}\neq 2$, $\beta_{1}\neq 0,1$ and $(\beta_{1},\beta_{2})\neq(-1,0)$. Let $\mathbb{F}^{\square}=\{1,\xi\}$ and let $\vartheta\in\mathbb{F}$ such that $\vartheta^{2}+1=\sigma^{2}\xi$ for some $\sigma\in\mathbb{F}^{\ast}$. If $f_{1}(t)$ and $f_{2}(t)$ are both irreducible, then $U_{4}(1,\beta_{1},\beta_{2})$ and $U_{4}(\xi,\beta_{1},\beta_{2})$ are conjugate, respective, to $U_{4}(1,\lambda_{1},\vartheta(\lambda_{1}+1))$ and $U_{4}(\xi,-1,\lambda_{2})$ for some $\lambda_{1}\in\mathbb{F}\setminus\{\pm 1\}$ and $\lambda_{2}\in\mathbb{F}^{\ast}$.

Proof.

Since $f_{1}(t)$ and $f_{2}(t)$ are both irreducible polynomials, then $\beta_{2}^{2}+4\rho\beta_{1}=\omega^{2}\xi$ and $(\beta_{1}+1)^{2}\rho^{2}+\rho\beta_{2}^{2}=\nu^{2}\xi$ for some $\omega,\nu\in\mathbb{F}^{\ast}$. First, consider $\rho=1$. The statement is obvious if $\beta_{2}=\vartheta(\beta_{1}+1)$. So, assume $\beta_{2}\neq\vartheta(\beta_{1}+1)$ and let $h(t)=f_{2}(t)-\vartheta(f_{1}(t)+g_{1}(t))$. Since the discriminant of $h(t)$ is $4(\vartheta^{2}+1)\nu^{2}\xi=(2\sigma\nu\xi)^{2}$, there exists $a\in\mathbb{F}$ such that $h(a)=0$. Now, denote by $r_{i}$ the resultant between $h(t)$ and $g_{i}(t)$ with respect to $t$: we obtain $r_{1}=\nu^{2}\xi((\beta_{1}-1)^{2}\vartheta^{2}-\omega^{2}\xi)$ and $r_{2}=4(\vartheta^{2}+1)\nu^{2}\xi$. From $r_{1}=0$ we get $\xi=((\beta_{1}-1)\vartheta\omega^{-1})^{2}$, an absurd; from $r_{2}=0$ we obtain $\vartheta^{2}+1=0$, which is in contradiction with the initial assumption on $\vartheta$. Hence, $g_{1}(a)g_{2}(a)\neq 0$ and so $U_{4}(1,\beta_{1},\beta_{2})$ is conjugate to $U_{4}\left(1,\frac{f_{1}(a)}{g_{1}(a)},\frac{\vartheta(f_{1}(a)+g_{1}(a))}{g_{1}(a)}\right)$.

Next, consider $\rho=\xi$. The statement is obvious if $\beta_{1}=-1$. So, assume $\beta_{1}\neq-1$ and let $h(t)=f_{1}(t)+g_{1}(t)$. Since the discriminant of $h(t)$ is $4\nu^{2}$, there exists $a\in\mathbb{F}$ such that $h(a)=0$. Clearly, if $h(a)=g_{1}(a)=0$, then $f_{1}(a)=0$, contradicting the assumption that $f_{1}(t)$ is irreducible. So, suppose that $h(a)=g_{2}(a)=0$ and denote by $r$ the resultant between $h(t)$ and $g_{2}(t)$ with respect to $t$. Since $r=4\nu^{2}\xi\neq 0$, we obtain that $g_{2}(a)\neq 0$ and so $U_{4}(\xi,\beta_{1},\beta_{2})$ is conjugate to $U_{4}\left(\xi,-1,\frac{f_{2}(a)}{g_{1}(a)}\right)$. ∎

Lemma 5.22**.**

Let $\mathbb{F}=\mathbb{F}_{q}$ be a finite field and keep the previous notation.

• (a)

Suppose $q$ even. Then we can take $\mathbb{F}^{\square}=\{1\}$ and

[TABLE]

where, for every $a\in\mathbb{F}\setminus\{0,1\}$, $\kappa_{a}\in\mathbb{F}^{\ast}$ is a fixed element such that $\kappa_{a}^{2}+\kappa_{a}\not\in a^{2}\overline{\mathsf{B}}(\mathbb{F})$. In particular, $|\mathsf{A}_{1}|=2q-2$.

• (b)

Suppose $q\equiv 1\pmod{4}$. Then we can take $\mathbb{F}^{\square}=\{1,\xi\}$ and

[TABLE]

where $\iota^{2}=-1$ and

[TABLE]

and

[TABLE]

In particular, $|\mathsf{A}_{1}|+|\mathsf{A}_{\xi}|=2q+1$.

• (c)

Suppose $q\equiv 3\pmod{4}$. Then we can take $\mathbb{F}^{\square}=\{1,-1\}$ and

[TABLE]

where $\vartheta$ is such that $\vartheta^{2}+\sigma^{2}+1=0$ for some $\sigma\in\mathbb{F}^{\ast}$ and

[TABLE]

and

[TABLE]

In particular, $|\mathsf{A}_{1}|+|\mathsf{A}_{-1}|=2q+1$.

Proof.

Let $f_{1}(t),f_{2}(t),g_{1}(t),g_{2}(t)$ as in (5.6). Suppose $q$ even. If either $\beta_{1}\neq 0$ or $f_{1}(t)$ is reducible, then $U_{4}(1,\beta_{1},\beta_{2})$ is conjugate to $U_{4}(1,0,\lambda)$ for a unique $\lambda\in\mathbb{F}$, by Lemma 5.13 and Remark 5.5. Now, suppose $\beta_{1}\neq 0,1$ and $f_{1}(t)$ irreducible. Then $\beta_{2}\neq 0$, $\beta_{1}\beta_{2}^{-2}\not\in\overline{\mathsf{B}}(\mathbb{F})$ and $g_{1}(t)$ is also irreducible. We obtain that $U_{4}(1,\beta_{1},\beta_{2})$ is conjugate to $U_{4}(1,\beta_{3},\beta_{4})$ ($\beta_{3}\neq 0,1$, $\beta_{4}\neq 0$, $\beta_{3}\beta_{4}^{-2}\not\in\overline{\mathsf{B}}(\mathbb{F})$) if and only if $\beta_{2}(\beta_{3}+1)=\beta_{4}(\beta_{1}+1)$. So, $U_{4}(1,\beta_{1},\beta_{2})$ is conjugate to $U_{1}(1,1+\kappa_{a}^{-1},a\kappa_{a}^{-1})$, where $a=\beta_{2}(\beta_{1}+1)^{-1}\neq 0,1$ and $\kappa_{a}\in\mathbb{F}^{\ast}$ is chosen in such a way that $\kappa_{a}^{2}+\kappa_{a}\not\in a^{2}\overline{\mathsf{B}}(\mathbb{F})$.

Suppose $q\equiv 1\pmod{4}$. The set $\mathsf{A}_{1}$ is given in Lemma 5.16. So, consider the case $\rho=\xi$ with $(\beta_{1},\beta_{2})\neq(-1,0)$. If either $\beta_{1}=0$ or $f_{1}(t)$ is reducible, then $R=U_{4}(\xi,\beta_{1},\beta_{2})$ is conjugate to $U_{4}(\xi,0,\lambda)$ for a unique $\lambda\in\mathsf{N}(\mathbb{F}_{q})$, by Lemma 5.13 and Remark 5.5. So, assume $\beta_{1}\neq 0$ and $f_{1}(t)$ irreducible. If $f_{2}(t)$ is reducible, then $R$ is conjugate to $U_{4}(\xi,\lambda,0)$ for some $\lambda\neq 0,1$, by Lemma 5.18. Recall that, by Remark 5.5, $U_{4}(\xi,\lambda_{1},0)$ is conjugate to $U_{4}(\xi,\lambda_{2},0)$ if and only if either $\lambda_{2}=\lambda_{1}$ or $\lambda_{2}=\lambda_{1}^{-1}$; $U_{4}(\xi,\lambda_{1},0)$ is conjugate to $U_{4}(\xi,0,\lambda_{2})$ if and only if $\lambda_{1}=\xi\omega^{2}$ and $\lambda_{2}=2\xi\omega(\xi\omega^{2}-1)^{-1}$ for some $\omega\in\mathbb{F}$; $U_{4}(\xi,-1,\lambda_{1})$ is conjugate to $U_{4}(\xi,-1,\lambda_{2})$ if and only if $\lambda_{2}=\pm\lambda_{1}$. Finally, if $\beta_{1}\neq 0,1$ and both $f_{1}(t)$ and $f_{2}(t)$ are reducible, then $R$ is conjugate to $U_{4}(\xi,-1,\lambda)$ for some $\lambda\in\mathbb{F}$, by Lemma 5.21. Now, $U_{4}(\xi,-1,\lambda_{1})$ is not conjugate to $U_{4}(\xi,\lambda_{2},0)$ if $\lambda_{1}\neq 0$. Furthermore, $U_{4}(\xi,-1,\lambda_{1})$ is conjugate to $U_{4}(\xi,0,\lambda_{2})$ if and only if $\lambda_{1}=a+\xi a^{-1}$ and $\lambda_{2}=\frac{\pm a^{2}-\xi}{2a}$ for some $a\in\mathbb{F}^{\ast}$.

Suppose $q\equiv 3\pmod{4}$. The set $\mathsf{A}_{-1}$ is given in Lemma 5.17. So, consider the case $\rho=1$ with $(\beta_{1},\beta_{2})\neq(-1,0)$. Proceeding in a way similar to what we did for the case $q\equiv 1\pmod{4}$, $R=U_{4}(1,\beta_{1},\beta_{2})$ is conjugate to a subgroup having one of the following shape $U_{4}(1,0,\lambda)$ for a unique $\lambda\in\mathsf{N}(\mathbb{F}_{q})$, $U_{4}(1,\lambda,0)$ for some $\lambda\neq 0,1$, or $U_{4}(1,\lambda,\vartheta(\lambda+1))$ for some $\lambda\neq 1$. Now, $U_{4}(1,\lambda_{1},0)$ is conjugate to $U_{4}(1,\lambda_{2},0)$ if and only if either $\lambda_{2}=\lambda_{1}$ or $\lambda_{2}=\lambda_{1}^{-1}$; $U_{4}(1,\lambda_{1},0)$ is conjugate to $U_{4}(1,0,\lambda_{2})$ if and only if $\lambda_{1}=\omega^{2}$ and $\lambda_{2}=2\omega(\omega^{2}-1)^{-1}$ for some $\omega\in\mathbb{F}\setminus\{\pm 1\}$; $U_{4}(1,\lambda_{1},\vartheta(\lambda_{1}+1))$ is conjugate to $U_{4}(1,\lambda_{2},\vartheta(\lambda_{2}+1))$ if and only if $\lambda_{2}=\lambda_{1}^{\pm 1}$.

Furthermore, $U_{4}(1,\lambda_{1},\vartheta(\lambda_{1}+1))$ is conjugate to $U_{4}(1,0,\lambda_{2})$ if and only if $\lambda_{1}=\left(\frac{1-a\vartheta}{a(a+\vartheta)}\right)^{\pm 1}$, $\lambda_{2}=\mp\frac{a^{2}\vartheta-2a-\vartheta}{a^{2}+2a\vartheta-1}$, $a\neq 0,-\vartheta,\vartheta^{-1}$; $U_{4}(1,\lambda,\vartheta(\lambda_{1}+1))$ is not conjugate to $U_{4}(1,\lambda_{2},0)$. ∎

Proposition 5.23**.**

Let $\mathbb{F}=\mathbb{F}_{q}$ be a finite field. If $q$ is odd, there are exactly $5q+9$ distinct conjugacy classes of nonabelian regular subgroups in $\Delta_{4}(\mathbb{F}_{q})$, that can be represented by:

[TABLE]

[TABLE]

where

[TABLE]

and $x^{2}-\xi$ is a fixed irreducible polynomial of $\mathbb{F}_{q}[x]$ (take $\xi=-1$ if $q\equiv 3\pmod{4}$).

If $q$ is even, there are exactly $5q+6$ distinct conjugacy classes of nonabelian regular subgroups in $\Delta_{4}(\mathbb{F}_{q})$, that can be represented by:

[TABLE]

[TABLE]

where

[TABLE]

and $x^{2}+x+\xi$ is a fixed irreducible polynomial of $\mathbb{F}_{q}[x]$.

Proof.

The set $\Pi_{S}(\mathbb{F}_{q})$ have been determined in [9, Theorem 4], so the result follows from Theorem 5.12 and Lemma 5.22. ∎

We conclude recalling that Theorem 1.1 follows immediately from Theorems 4.2 and 5.12, applying Proposition 2.4.