#  Invariant Gorenstein Rings On $3$-Dimensional Vector Spaces

By: Tamir Buqaie

Supervised by: Prof. Amiram Braun

Thesis submitted as partial fulfillment of the
requirements for the masters degree

University of Haifa
Faculty of Natural Sciences
Department of Mathematics

October, 2018

Approved by:  Date:
(Supervisor)
Approved by:  Date:
(Chairperson of M.sc. Committee)

Acknowledgments

I would like to thank my Master thesis advisor, Professor Amiram Braun for his
dedication, interest, professionality and human attitude.I appreciate his
constructive comments which helped me to learn and improve this paper.
I would like to thank my first mathematics teacher,Osima Shehady for her
creative,successful and outstanding education methods. I thank my family and
my friends,for their support, understanding and for giving me positive energy.

###### Contents

  1. 1 Introduction
  2. 2 Chapter $(A):V=Fv\oplus W,$ and $Fv,W$ are $G$-submodules.
  3. 3 Chapter($B$): $V$ is indecomposable and contains (at least) two $2-$ dimensional $G-$ submodules $W_{1},W_{2}.$
  4. 4 Chapter(C): Preliminaries to Chapters D,E,F and G.
  5. 5 Chapter($D$): $V$ is indecomposable,contains a unique $2-$ dimensional $G-$ submodule, but no $1-$ dimensional $G-$ submodule.
  6. 6 Chapter($E$): $V$ is indecomposable,contains a unique $1-$ dimensional $G-$ submodule $U,$ but no $2-$ dimensional $G-$ submodule.
  7. 7 Chapter($F$): $V$ is indecomposable and contains two $G-$ submodules $\mathbb{F}_{p}v_{0},W,$ where $\mathbb{F}_{p}v_{0}\subset W\subset V,$ and $dim_{\mathbb{F}_{p}}W=2.$
  8. 8 Chapter($G$): $V$ is indecomposable and contains (at least) two $1-$dimensional $G-$ submodules.
  9. 9 References

Invariant Gorenstein Rings On $3$-Dimensional Vector Spaces

Tamir Buqaie

Supervised by: Prof. Amiram Braun

Abstract

For a finite subgroup $G\subset SL(V),$ let $S(V)^{G}$ denote the subring of
$G-$invariants, where $V$ is a $3-$ dimensional $G-module$ over $F$ and $S(V)$
the symmetric algebra of $V.$ In this thesis we discuss the modular case of
invariant theory,that is,when $charF=p$ divides $|G|.$ The aim of this work is
to classify Gorenstein ring of the form $S(V)^{G}.\\\ \\\ $ A.Braun in [2] and
Fleischmann-Woodcock in [6] proved that if $S(V)^{G}$ is Cohen-Macaulay ring
and $S(V)^{W(G)}$ is a polynomial ring then $S(V)^{G}$ is Gorenstein if and
only if $G/W(G)\subseteq SL(m/m^{2}),$ where $W(G)$ is the $G-$subgroup
generated by all pseudo-reflections and $m$ is the maximal homogenous ideal of
$S(V)^{W(G)}.$ ( See [Theorem 1.2]).

Here, since $G\subset SL(V),$ we have one kind of pseudo-reflection ,called
transvection, in other words $W(G)=T(G),$ where $T(G)$ denote the $G-$
subgroup generated by all transvections of $G.$This work is divide into $7$
chapters, the classification of $S(V)^{G},$ in each chapter, depends on the
previous theorem which provides a necessary and suffiecient condition for $G$
such that $S(V)^{G}$ is Gorenstein.So to apply this theorem we might ask
ourselves, whether $S(V)^{T(G)}$ is polynomial ring and what are the
$T(G)-$invariants of this ring?Or at least, what are the degrees of the
$T(G)-$invariants ? Since $dim_{F}V=3,$ the Cohen-Macaulay property of
$S(V)^{G}$ holds [9, Proposition 5.6.10].These are some of the classical
questions that invariant theorists have studied in the modular case.
We prove the following theorem : Let $G\subset SL(3,\mathbb{F}_{p})$ be a
finite group and $V=\mathbb{F}_{p}^{3}$ is a reducible $G$-module,then
$S(V)^{G}$ is Gorenstein.(See [Theorem 1.3]).

We explain now how this work is organized.

As mentioned above, we divide our work into $7$ chapters.In chapters
$A,B,D,E,F$ and $G$ we fix a basis for $V,$ and represent each $g\in G$,by
matrices,with respect to this basis.Then we compute the ring of
$T(G)-$invariants. We are mainly concered with the case of $S(V)^{T(G)}$ being
a polynomial ring.Under the assumptions of each chapter we prove that
$S(V)^{T(G)}$ is a polynomial ring .Generally, we can not confirm in chapter
$D$ that the polynomial property of $S(V)^{T(G)}$ holds.We provide in chapters
$A,B,D,$ and $G,$ a sufficient and necessary condition on $G$ such that
$S(V)^{G}$ is Gorenstein ,when $G\subset SL(3,F)$.However, in chapter $E$ and
$F,$ we exclusively consider the case $G\subset SL(3,\mathbb{F}_{p}).$ The
technique in this work will be to exhibit the action of the group $G$ on
$m/m^{2}$ and using [Theorem 1.2].

In order to prove that $S(V)^{T(G)}$ is a polynomial ring ,in most cases, we
use Kemper’s theorem [4, Theorem 3.7.5].

In chapter $A,$ we consider $V,$ a reducible $G-module$ of the form
$V=Fv\oplus W$ and $Fv$ , $W$ are $G-$submodules with $dim_{F}W=2.$ This is a
special chapter, since by Proposition 2.3, the Gorenstein property of
$S(V)^{G}$ depends on $W.$ So we use the computation of [3],and proceed our
proof in steps by dividing it into $3$ cases, which are:

  1. 1.

$W$ is an irreducible and primitive $T(G)-module.$

  2. 2.

$W$ is a reducible $T(G)-module.$

  3. 3.

$W$ is an irreducible and imprimitive $T(G)-module.$

In fact,the results in chapter $A$ are obtained from the discussion in [3].The
difference between the situation here and the situation of [3], is that in
[3], $G\subset SL(2,F)$ and here $\bar{G}=G|W\subset GL(2,F).\\\ $
In chapter $B,$ we consider $V,$ a reducible, indecomposable $G-module$ which
has at least two $G$-submodules $W_{1},W_{2}$ where $dimW_{i}=2$ for $i=1,2.$
In addition to the main theorem, we prove,under specific condition, that
$S(V)^{G}$ is Gorenstein in the following cases (see [Theorem 3.5]) :

  1. (i)

$p=3.$

  2. (ii)

$|G|$ is prime to $3.$

  3. (iii)

$F=\mathbb{F}_{q},$ where $q=p^{s}$ and g.c.d$(3,s)=1.$

We define a $F-$vector space $X,$ where $dim_{F}X\in\\{1,2\\},$ and separate
this chapter into two cases. In the first case, we deal with
$dim(span_{F}X)=1.$ In the second case, we deal with $dim(span_{F}X)=2,$ where
$T(G)$ is a direct product of two transvection groups $T_{1}(G),T_{2}(G).$ In
Lemma 3.7,Lemma 3.8, and Proposition 3.9 we examine the normality of
$T_{1}(G),T_{2}(G).$ However this is not needed for Proposition 3.10
,Proposition 3.13, and Corollary 3.14.

Chapter $C$ is different, we provide in this chapter preliminaries to chapters
$D,E,F$ and $G.$

In chapter $D,$ we consider $V,$ a reducible and indecomposable $G-$module
which has a unique $G-$submodule $W$ with $dim_{F}W=2,$ and $V$ contains no
$1-$dimensional $G-$submodules.In this chapter,we can not conclude that
$S(V)^{T(G)}$ is always polynomial ring.So we make specific assumption (see
Assumption 5.2 )to prove [Theorem 1.3].This is applicable in case
$F=\mathbb{F}_{p}.$

In chapter $E,$ we consider $V,$ a reducible and indecomposable $G-module$
which has a unique $1-$dimensional $G-$submodule $U=Fv_{0}$ and $V$ contains
no $2-$dimensional $G-$submodules.Hence $W=V/U$ is an irreducible $G-$ module.
We exclusively consider here the case $F=\mathbb{F}_{p}.\\\ $ There are two
cases of $W$ that Lemma 6.2 have provided .The first case is when $W$ is an
unfaithful irreducible $T(G)-$module, and the second one is when $W$ is a
faithful irreducible $T(G)-$module.

In chapter $F,$ we consider $V,$ a reducible and indecomposable $G-$module
which has a unique $G-$submodule $W$ with $dim_{F}W=2,$ and a unique
$1-$dimensional $G-$submodule $Fv_{0},$ and $Fv_{0}\subset W\subset V.$ We
exclusively consider here the case $F=\mathbb{F}_{p}.$

Finally,in chapter $G,$ we consider $V,$ a reducible and indecomposable
$G-module$ and contains (at least) two $1-$dimensional $G-$submodules.By using
Proposition 4.3 , we provide a sufficient and necessary condition on $G$ for
the Gorenstein property of $S(V)^{G},$ and present the main result in
Corollary 8.1.

Index of notations

$\mathbb{F}_{p}-$ The prime field of order $p.$

$\indent|G|-$ The order of group $G.$

$\indent F-$ A finite field with $charF=p.$

$\indent\bar{F}-$ The algebraic closure of $F.$

$\indent S(V)-$ The symmetric algebra of $V.$

$\indent S(V)^{G}-$ The subring of $G-$ invariants.

$\indent T(G)-$ The $G-$subgroup generated by all transvections.

$\indent W(G)-$ The $G-$subgroup generated by all pseudoreflections.

$\indent m-$ The unique homogenous maximal ideal of $S(V)^{T(G)}.$

$\indent V^{*}-$ The hom-dual $hom_{F}(V,F).$

$\indent GL(V)-$ The general linear group of $V.$

$\indent GL(n,F)-$ The linear automorphisms of $F^{n}.$

$\indent SL(V)-$ The special linear group of $V.$

$\indent SL(n,F)-$ The linear automorphisms of $F^{n}$ of determinant $1.$

$\indent G|W-$ The restriction of $G$ to $W.$

$\indent<\sigma_{1},...,\sigma_{s}>-$ The subgroup generated by
$\\{\sigma_{1},...,\sigma_{s}\\}.$

$\indent Fix_{G}(W)-$ The elements of $G,$ that acts trivially on subspace
$W.$

##  1 Introduction

Let $G\subset SL(V)$ be a finite subgroup of SL(V). Let $F$ be a field with
$charF=p>0$ , $V$ a $3-$ dimensional $F-$ vector space and $S(V)$ the
symmetric algebra of $V$ .We denote by $S(V)^{G}$ the subring of
$G-invariants$. One of the main purposes of this thesis is to classify
Gorenstein rings of the form ${S(V)}^{G},$ when $p$ divides $|G|.$

The following are results of K.Watanabe ,Fleischmann-Woodcock and A.Braun .

######  Theorem 1.1.

(K.Watanabe [10],[11])
Suppose that $(|G|,p)=1$ and $G$ contains no pseudo-reflection.Then the
following are equivalent:

  1. 1.

$G\subset SL(V).$

  2. 2.

$S(V)^{G}$ is a Gorenstein ring.

######  Theorem 1.2.

(A.Braun [2],Fleischmann-Woodcock [6])
Suppose that $S(V)^{G}$ is Cohen-Macaulay and $S(V)^{W(G)}$ is a polynomial
ring. Then $S(V)^{G}$ is Gorenstein iff $G/W(G)\subseteq SL(m/m^{2}),$ where
$m$ is the unique homogenous maximal ideal of $S(V)^{W(G)}$ and $W(G)$ is the
$G-subgroup$ generated by all pseudo-reflections(of all types).

Here an element $g\in G$ is called pseudoreflection if $rank(g-I)=1$. If
$(|G|,p)=1$ then every pseudoreflection is diagonalizable with one’s on all
but a single diagonal entry. In the modular case, that is if $p$ divides $|G|$
we have a new type of pseudoreflection, called transvection, having the
additional property $(g-I)^{2}=0,$ so it is not diagonalizable.Denote by
$T(G)$ the $G-subgroup$ generated by all transvections .

Eagon and Hochster proved in [5] that $S(V)^{G}$ is Cohen-Macaulay for all
nonmodular groups $G,$ i.e when $(|G|,p)=1.$ But there are finite $p-groups,$
such that $S(V)^{G}$ is not Cohen-Macaulay.

######  Example 1.1.

Let $F$ be a field of characteristic $p$ and $V$ is a faithful representation
of a non-trivial $p$-group $P$ and $mV$ denote the faithful representation of
$G$ formed by taking the direct sum of $m$ copies of $V.$ Then $S(mV)^{P}$ is
not Cohen-Maculay when $m\geq 3.$[12, Theorem 1.2]

However in this thesis we assume that $dim_{F}V=3$ and by [9, Proposition
5.6.10] $S(V)^{G}$ is always Cohen-Macaulay ring.

It is shown in [3, Section 2] that $S(V)^{G}$ can fail to be Gorenstein even
if $G\subset SL(V)$ and $dimV=2.$ Moreover, [3] provides necessary and
sufficient conditions to ensure the Gorenstein property of $S(V)^{G}$ in case
$dim_{F}V=2$ and $G\subset SL(V).$ We mostly examine here the case of
$dim_{F}V=3,G\subset SL(V)$ and $V$ is a reducible $G$-module. The analysis is
carried out by separation into different cases. A noteworthy consequence is
perhaps the following:

######  Theorem 1.3.

Let $G\subset SL(3,\mathbb{F}_{p})$ be a finite group and
$V=\mathbb{F}_{p}^{3}$ is a reducible $G$-module. Then $S(V)^{G}$ is
Gorenstein.

This extends a similar result in [3, Corollary (C)].The separation into cases
is summed up in the following table.

Table
---
Chapter |  $V$ is a reducible $G$-module.
$A$ |  $V=Fv\oplus W$ and $Fv,W$ are $G$-submodules.
$B$ |  $V$ is indecomposable and contains (at least) two $2-$ dimensional $G-$ submodules $W_{1},W_{2}.$
$D$ |  $V$ is indecomposable,contains a unique $2-$ dimensional $G-$ submodule, but no $1-$ dimensional $G-$ submodule.
$E$ |  $V$ is indecomposable,contains a unique $1-$ dimensional $G-$ submodule $U,$ but no $2-$ dimensional $G-$ submodule.
$F$ |  $V$ is indecomposable and $\mathbb{F}_{p}v_{0}\subset W\subset V,$where $\mathbb{F}_{p}v_{0},W$ are $G-$ submodules and $dim_{\mathbb{F}_{p}}W=2.$
$G$ |  $V$ is indecomposable and contains (at least) two $1-$ dimensional $G-$ submodules.

Convention: Given a basis $\\{v_{1},...,v_{n}\\}$ of $V,$ and $g\in G.$ Then
$g(v_{i})=\sum_{j=1}^{n}a_{ij}v_{j},a_{ij}\in F$ and the matrix $A:=(a_{ij}),$
representing $g,$ acts on coordinates (=row vectors) from the right .

##  2 Chapter $(A):V=Fv\oplus W,$ and $Fv,W$ are $G$-submodules.

In this chapter we consider $V,$ a reducible $G-module$ of the form
$V=Fv\oplus W$ and $Fv$ , $W$ are $G-modules$ with $dim_{F}W=2$ . Proposition
2.3 provides a sufficient and necessary condition for $G$ as above,to ensure
the Gorenstein property of $S(V)^{G}.$ Our proof will proceed in steps by
dividing it into $3$ cases:

  1. 1.

$W$ is an irreducible and primitive $T(G)-module.$

  2. 2.

$W$ is a reducible $T(G)-module.$

  3. 3.

$W$ is an irreducible and imprimitive $T(G)-module.$

The next two results are usefull in this chapter.

######  Lemma 2.1.

$W$ is a faithful $G$-module.

###### Proof.

Let $1_{G}\neq g\in G$ , with $g|W=Id$ .Then

$g=\left(\begin{array}[]{ccc}\lambda(g)&0&0\\\ 0&1&0\\\
0&0&1\end{array}\right).$

Where $g(v)=\lambda(g)v$ , for all $g\in G$ .Since $G\subset SL(V)$ ,
$\lambda(g)=1$ and therefore

$g=\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\ 0&0&1\end{array}\right).$

A contradiction.

∎

The next result does not require the assumption $V=Fv\oplus W.$

######  Lemma 2.2.

Suppose that $G\subset SL(V)$ and $U$ a $1$-dimensional $G-submodule.$ Then
$U\subseteq ker(\sigma-I)$ for all transvection $\sigma$ in $G.$ Equivalently,
$T(G)$ acts trivially on $U.$

###### Proof.

Let $\sigma\in T(G)$ be a transvection ,$\sigma\neq I,$ and $M=ker(\sigma-I).$
Suppose that $Fu=U\not\subset M.$ Then with respect to a basis
$\\{u,m_{1},m_{2}\\},m_{1},m_{2}\in M,$ we have :
$\sigma=\left(\begin{matrix}\lambda(\sigma)&0&0\\\ 0&1&0\\\
0&0&1\end{matrix}\right),$ where $\sigma(u)=\lambda(\sigma)u$ and
$\sigma|M=Id.$ But $G\subset SL(V),$ therefore $\lambda(\sigma)=1,$ implying
that $\sigma=I,$ a contardiction.

∎

######  Proposition 2.3.

Let $G\subset SL(V)$ be a finite group and $V=Fv\oplus W$ a decomposition of
$V$ into $G-submodules$ with $dim_{F}W=2$.Then the following are equivalent:

  1. 1.

$S(V)^{G}$ is Gorenstein.

  2. 2.

$det(g|W)=det(g|{m/m^{2}}),$ for each $g\in G$ where $m$ is the unique
homogenous maximal ideal of $S(W)^{T(G)}$.

###### Proof.

By Lemma 2.2 ${S(V)}^{T(G)}={S(W)}^{T(G)}[v]$.Now by [8],
$S(W)^{T(G)}=F[a_{1},a_{2}]$ is a polynomial ring,where $m=(a_{1},a_{2})$. Let
$P$ be the unique homogenous maximal ideal of $S(V)^{T(G)},$ then
$P=(a_{1},a_{2},v).$ By [Theorem 1.2] $S(V)^{G}$ is Gorenstein if and only if
$1=det(g|P/P^{2})=\lambda(g)det(g|m/m^{2})$ , where $g(v)=\lambda(g)v$ , for
all $g\in G.$ Since $G\subset SL(V),det(g|W)={\lambda(g)}^{-1},$ hence
$S(V)^{G}$ is Gorenstein if and only if $det(g|W)=det(g|{m/m^{2}}),$ for each
$g\in G.$

∎

The transvections group $T(G)|W$ had been classified by Kantor whenever
$T(G)|W$ is an irreducible primitive linear group [7, Theorem 1.5]. So with
the above notation one of the following holds:

  1. (i)

$T(G)|W=SL(2,\mathbb{F}_{q}),$ where $p|q.$

  2. (ii)

$T(G)|W\cong SL(2,\mathbb{F}_{5}),T(G)|W\subset SL(2,\mathbb{F}_{9})$ and
$\mathbb{F}_{9}\subseteq F.\\\ $

We now consider the above case (i).

######  Proposition 2.4.

Suppose $G\subset SL(V)$ is a finite group with $T(G)|W=SL(2,\mathbb{F}_{q})$
,where $q=p^{s}$ and $\mathbb{F}_{q}\subseteq F.$ Let $V=F^{3}=Fv\oplus W$ be
the natural $SL(3,F)-module.$ Then $S(V)^{G}$ is Gorenstein if and only if
$G|W\subset GL(2,\mathbb{F}_{q^{2}}).$

###### Proof.

Recall as in [3, Proposition 3.5] that
$S(W)^{T(G)}=S(W)^{SL(2,\mathbb{F}_{q})}=F[u,c_{21}],deg(u)=q+1,deg(c_{21})=q^{2}-q.$
Let $g\in G$ be arbitrary and set $g|W=\bigl{(}\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\bigr{)}$ with respect to the basis $\\{x,y\\}.$ So
$g^{-1}|W=\frac{1}{det(g|W)}\bigl{(}\begin{smallmatrix}d&-b\\\
-c&a\end{smallmatrix}\bigr{)}.$Since $gT(G)g^{-1}=T(G),$ we get after
restriction to $W$ that $(g|W)\bigl{(}\begin{smallmatrix}1&1\\\
0&1\end{smallmatrix}\bigr{)}(g^{-1}|W)=\frac{1}{det(g|W)}\bigl{(}\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\bigr{)}\bigl{(}\begin{smallmatrix}1&1\\\
0&1\end{smallmatrix}\bigr{)}\bigl{(}\begin{smallmatrix}d&-b\\\
-c&a\end{smallmatrix}\bigr{)}=\frac{1}{det(g|W)}\bigl{(}\begin{smallmatrix}ad-
bc-ac&a^{2}\\\ -c^{2}&ad-bc+ac\end{smallmatrix}\bigr{)}$ is in
$SL(2,\mathbb{F}_{q}).$

Similarly, $(g|W)\bigl{(}\begin{smallmatrix}1&0\\\
1&1\end{smallmatrix}\bigr{)}(g^{-1}|W)=\frac{1}{det(g|W)}\bigl{(}\begin{smallmatrix}ad-
bc+bd&-b^{2}\\\ d^{2}&ad-bc-bd\end{smallmatrix}\bigr{)}\in
SL(2,\mathbb{F}_{q}).$Set $e=det(g|W)=ad-bc.$ Then
$\frac{ac}{e},\frac{a^{2}}{e},\frac{c^{2}}{e},\frac{b^{2}}{e},\frac{d^{2}}{e},\frac{bd}{e}\in\mathbb{F}_{q}.$
Assume firstly that $a\neq 0.$Then
$\gamma:=\frac{c}{a}=\frac{(\frac{c^{2}}{e})}{(\frac{ac}{e})}\in\mathbb{F}_{q}.$
If $b\neq 0$ then
$\mu:=\frac{d}{b}=\frac{(\frac{bd}{e})}{(\frac{b^{2}}{e})}\in\mathbb{F}_{q}.$Hence
$e=ad-bc=ab\mu-ba\gamma=ab(\mu-\gamma)$ and therefore
$\beta:=\frac{b}{a}=\frac{ab(\mu-\gamma)}{a^{2}(\mu-\gamma)}=\frac{e}{a^{2}(\mu-\gamma)}\in\mathbb{F}_{q},$
as well as $\delta:=\frac{d}{a}=\frac{\mu b}{a}=\mu\beta\in\mathbb{F}_{q}.$If
$b=0$ then take $\beta=0$ and
$\delta:=\frac{d}{a}=\frac{ad}{a^{2}}=\frac{e}{a^{2}}\in\mathbb{F}_{q}.$So in
both cases $b=a\beta,c=a\gamma,d=a\delta,$
$\beta,\gamma,\delta\in\mathbb{F}_{q}.$ Therefore
$g|W=\bigl{(}\begin{smallmatrix}a&0\\\
0&a\end{smallmatrix}\bigr{)}\bigl{(}\begin{smallmatrix}1&\beta\\\
\gamma&\delta\end{smallmatrix}\bigr{)},$ with
$h:=\bigl{(}\begin{smallmatrix}1&\beta\\\
\gamma&\delta\end{smallmatrix}\bigr{)}\in GL(2,\mathbb{F}_{q}).$

If $a=0$ then $e=det(g|W)=-bc$ implies $b\neq 0\neq c.$ So we get equality
$g|W=\bigl{(}\begin{smallmatrix}c&0\\\
0&c\end{smallmatrix}\bigr{)}\bigl{(}\begin{smallmatrix}0&\beta\\\
1&\delta\end{smallmatrix}\bigr{)}$ with $c$ replacing $a$ and
$\beta,\delta\in\mathbb{F}_{q}.$Consequently if $a\neq 0$ we get:
$g(c_{21})=(aI_{W})(h(c_{21}))=a^{q^{2}-q}c_{21},$ and as in the proof of [3,
Proposition 3.5] we get that
$g(u)=(aI_{W})(h(u))=det(h)a^{degu}u=det(h)a^{q+1}u.$ But
$det(g|W)=a^{2}det(h),$ hence $g(u)=det(g|W)a^{q-1}u.$

Let $M=Fu+Fc_{21}$ be the $2-$dimensional subspace of the polynomial ring
$S(W)^{T(G)}=F[u,c_{21}],m=(u,c_{21}),$ so $m/m^{2}\cong M.$ Then the matrix
representing $g|M$ with respect to the basis $\\{u,c_{21}\\}$ is
$\bigl{(}\begin{smallmatrix}a^{q-1}det(g|W)&0\\\
0&a^{q^{2}-q}\end{smallmatrix}\bigr{)}.$ Therefore the condition of
Proposition 2.3 : $det(g|W)=det(g|m/m^{2})=det(g|M)$ is equivalent to:
$det(g|W)=a^{q-1}det(g|W)a^{q^{2}-q}=a^{q^{2}-1}det(g|W).$ Hence it is
equivalent to $a^{q^{2}-1}=1,$ namely $a\in\mathbb{F}_{q^{2}}.$ Using $h\in
GL(2,\mathbb{F}_{q}),$ this is also equivalent to
$g|W=\bigl{(}\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix}\bigr{)}\in
GL(2,\mathbb{F}_{q^{2}}).$ A similar conclusion is obtained if $a=0.$

∎

We next consider the above case (ii).

######  Proposition 2.5.

Suppose $G\subset SL(3,F)$ is a finite group with $T(G)|W\cong
SL(2,\mathbb{F}_{5}),T(G)|W\subseteq SL(2,\mathbb{F}_{9})$ and
$\mathbb{F}_{9}\subseteq F$ (here $p=3$).Let $V=F^{3}=Fv\oplus W$ be the
natural $SL(3,F)-module.$ Then $S(V)^{G}$ is Gorenstein iff
$G|W\subseteq<T(G)|W,\eta I_{W}>,$ where $\eta$ is a $20^{th}$-primitive root
of unity.

###### Proof.

Let $g\in G,$ choose $\xi\in\bar{F}$ such that $\xi^{2}det(g|W)=1.$ Hence
$\hat{g}:=\xi I_{W}g|W\in SL(2,\bar{F})$ where $\bar{F}$ is the algebraic
closure of $F$.Clearly $\hat{g}(T(G)|W)\hat{g}^{-1}=T(G)|W.$ Therefore the
computation in [3, Proposition 3.7], with $\hat{g}$ instead of $g|W,$ shows
that $\hat{g}\in T(G)|W.$ Hence $\bigl{(}\begin{smallmatrix}\xi&0\\\
0&\xi\end{smallmatrix}\bigr{)}=\hat{g}({g}^{-1}|W)\in(T(G)|W)(G|W)=G|W.$
Consequently $G|W\subseteq<T(G)|W,(\begin{smallmatrix}\lambda&0\\\
0&\lambda\end{smallmatrix}\bigr{)}|\lambda\in\bar{F}>.\\\ $

Now $\bigl{(}\begin{smallmatrix}\xi&0\\\
0&\xi\end{smallmatrix}\bigr{)}g|W=\hat{g}|W\in T(G)|W.$ Recall from [3,
Proposition 6.6.] that $S(W)^{SL(2,\mathbb{F}_{5})}=F[f_{10},f_{12}],$ where
$f_{10}=x_{1}^{9}x_{2}-x_{1}x_{2}^{9}$ and
$f_{12}=x_{1}^{12}+x_{1}^{10}x_{2}^{2}-x_{1}^{6}x_{2}^{6}+x_{1}^{2}x_{2}^{10}-x_{2}^{12}.$
Hence $\hat{g}(f_{10})=f_{10},\hat{g}(f_{12})=f_{12}$ implying that
$g(f_{10})=\xi^{-10}f_{10},g(f_{12})=\xi^{-12}f_{12}.\\\ $ Let
$U=Ff_{10}+Ff_{12}$ be the $2-$dimensional subspace of the polynomial ring
$S(W)^{SL(2,\mathbb{F}_{5})}=F[f_{10},f_{12}],m=(f_{10},f_{12}),$ so
$m/m^{2}\cong U.$ Then the matrix representing $g|U$ with respect to the basis
$\\{f_{10},f_{12}\\}$ is $\bigl{(}\begin{smallmatrix}\xi^{-10}&0\\\
0&\xi^{-12}\end{smallmatrix}\bigr{)}.$ Therefore, by Proposition 2.3
$det(g|W)=det(g|m/m^{2})=det(g|U)$ is translated into $\xi^{-2}=\xi^{-22}$ or
$\xi^{20}=1.$ Therefore $S(V)^{G}$ is Gorenstein iff $G|W\subseteq<T(G)|W,\eta
I_{W}>$ where $\eta\in\bar{F}$ is a primitive $20-th$ root of unity.

∎

We now consider case $2,$ namely the possibility of $W$ being a reducible
$T(G)-module.$

######  Proposition 2.6.

Let $G\subset SL(3,F)$ be a finite group.Let $V=F^{3}=Fv\oplus W$ be the
natural $SL(3,F)-module$ and suppose $W$ is a reducible $T(G)$-submodule.Then
$S(V)^{G}$ is Gorenstein if and only if
$G|W\subseteq\\{\bigl{(}\begin{smallmatrix}a&b\\\
0&d\end{smallmatrix}\bigr{)}|a^{p^{n}-1}=1,a,b,d\in F\\}.$

###### Proof.

Set $\hat{g}:=\xi I_{W}g|W,$ with $det(\hat{g})=\xi^{2}det(g|W)=1.$ Since $W$
is a reducible $T(G)|W-module,$there exists a basis $\\{x_{1},x_{2}\\}$ of $W$
with respect to which $T(G)|W\subseteq\bigl{(}\begin{smallmatrix}F&F\\\
0&F\end{smallmatrix}\bigr{)}.$ Every transvection acts trivially on every one
dimensional submodule of $W$ by Lemma 2.2. Hence it must stabilize $Fx_{2}.$
Therefore since $T(G)|W\subset SL(W),$
$T(G)|W\subseteq\bigl{(}\begin{smallmatrix}1&F\\\
0&1\end{smallmatrix}\bigr{)}.$ Consequently $T(G)|W$ is elementary abelian
group.So $T(G)|W=<\sigma_{1}>\times\cdot\cdot\cdot\times<\sigma_{n}>,$ where
$\sigma_{i}=\bigl{(}\begin{smallmatrix}1&\alpha_{i}\\\
0&1\end{smallmatrix}\bigr{)},$ for $i=1,...,n,$ and
$\\{\alpha_{1},...,\alpha_{n}\\}$ are linearly independent over
$\mathbb{F}_{p}.$Let $g|W=\bigl{(}\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\bigr{)}.$ Now the computation in the proof of [3, Theorem
3.1] with $\hat{g}$ instead of $g|W$ shows that $\xi c=0,$ hence $c=0,$
$\hat{g}=\bigl{(}\begin{smallmatrix}\lambda&\pi\\\
0&\lambda^{-1}\end{smallmatrix}\bigr{)}.\\\ $ Hence
$\hat{g}|m/m^{2}=\bigl{(}\begin{smallmatrix}\lambda^{p^{n}}&0\\\
0&\lambda^{-1}\end{smallmatrix}\bigr{)},$ where $\xi a=\lambda,$ $\xi
d=\lambda^{-1},\pi\in F$ and $m=(b,x_{2}),$ where $b$ which is of degree
$p^{n}$ is defined as in the proof of [3, Theorem 3.1].Hence
$\hat{g}(b)=\lambda^{p^{n}}b=(\xi a)^{p^{n}}b=(\xi)^{p^{n}}g(b)$ and
$\hat{g}(x_{2})=\lambda^{-1}x_{2}=(\xi d)x_{2}=\xi g(x_{2}).$ Therefore
$g|m/m^{2}=\bigl{(}\begin{smallmatrix}a^{p^{n}}&0\\\
0&d\end{smallmatrix}\bigr{)}.$ Consequently
$det(g|{m/m^{2}})=a^{p^{n}}d=a^{p^{n}-1}(ad)=a^{p^{n}-1}det(g|W)$.Hence, by
Proposition 2.3 $S(V)^{G}$ is Gorenstein if and only if $a^{p^{n}-1}=1$.In
other words $S(V)^{G}$ is Gorenstein if and only if
$G|W\subseteq\\{\bigl{(}\begin{smallmatrix}a&b\\\
0&d\end{smallmatrix}\bigr{)}|a^{p^{n}-1}=1,b,d\in F\\}.$

∎

Our next objective is to handle case $3,$ i.e when $T(G)|W$ acts irreducibly
and imprimitively on $W.$ It is well known(e.g.[13]) that $T(G)|W$ is a
monomial subgroup.Recall that $H\subset GL(W)$ is called monomial if $W$ has a
basis with respect to which the matrix of each element of $H$ has exactly one
non-zero entry in each row and column. If $charF\neq 2$ , then by [3, Lemma
3.9.] $T(G)|W={1}.$This contradicts the assumption that $W$ is an irreducible
$T(G)|W-$module.Hence we only deal with $p=2.$

######  Proposition 2.7.

Assume that $charF=2.$ Let $G\subset SL(3,F)$ be a finite group with $T(G)|W$
acting imprimitively and irreducibly on the submodule $W$ and $V=Fv\oplus W.$
Then $S(V)^{G}$ is Gorenstein iff $G|W\subseteq<T(G)|W,\delta I_{W}>,$ where
$\delta$ is a primitive $d-th$ root of unity and $d$ defined in the proof.

###### Proof.

Let $\\{x_{1},x_{2}\\}$ be the basis of $W$ with respect to which $T(G)|W$ is
irreducible and monomial.Let $\\{g_{1},...,g_{n}\\}$ be the set of
transvections generating $T(G)|W.$The $T(G)|W-$ irreducibility of $W$ forces
$n\geq 2.$ So either $g_{i}=\bigl{(}\begin{smallmatrix}*&0\\\
0&*\end{smallmatrix}\bigr{)}$ or $\bigl{(}\begin{smallmatrix}0&*\\\
*&0\end{smallmatrix}\bigr{)}.$ The first possibility leads , as in [3, Lemma
3.9.] to $g_{i}|W=I_{W}.$ So $g_{i}=\bigl{(}\begin{smallmatrix}0&\alpha_{i}\\\
\alpha_{i}^{-1}&0\end{smallmatrix}\bigr{)},$ for $i=1,...,n,$ with respect to
the basis $\\{x_{1},x_{2}\\}$ of $W.$ Let $n_{ij}$ be the minimal number such
that $(g_{i}g_{j})^{n_{ij}}=1,$where $i\neq j,$ clearly $g_{i}^{-1}=g_{i}$ for
$i=1,...,n.$ Also $n_{ij}=n_{ji}>1$ and since $charF=2,$ $n_{ij}$ is odd.
Recall that $S(W)^{T(G)}$ is a polynomial ring [8, Theorem 2.4].We next
compute the actual generators of $S(W)^{T(G)}.$

Let $[(T(G)|W),(T(G)|W)]=<g_{i}g_{j}g_{i}g_{j}>$ be the commutator subgroup of
$T(G)|W.$ We have $(g_{1}g_{2})^{n_{12}}=1.$ Therefore
$g_{2}=(g_{1}g_{2})^{n_{12}}g_{2}=(g_{1}g_{2})^{n_{12}-1}(g_{1}g_{2})g_{2}=(g_{1}g_{2})^{n_{21}-1}g_{1}.$
But $n_{12}-1$ is even, since $n_{12}$ is odd .Hence
$g_{2}=(g_{1}g_{2}g_{1}g_{2})^{\frac{n_{12}-1}{2}}g_{1}\in[T(G)|W,T(G)|W]g_{1}.$
This similarly holds for $g_{i},i\geq 3,$ hence
$\frac{|(T(G)|W)|}{|([T(G)|W,T(G)|W])|}=2.$ Let $d=l.c.m\\{n_{ij}|i<j\\}.$
Since
$g_{i}g_{j}g_{i}g_{j}=\bigl{(}\begin{smallmatrix}\alpha_{i}\alpha_{j}^{-1}\alpha_{i}\alpha_{j}^{-1}&0\\\
0&\alpha_{i}^{-1}\alpha_{j}\alpha_{i}^{-1}\alpha_{j}\end{smallmatrix}\bigr{)},$
it follows that $[(T(G)|W),(T(G)|W)]=\\{\bigl{(}\begin{smallmatrix}\zeta&0\\\
0&\zeta^{-1}\end{smallmatrix}\bigr{)}|\zeta^{d}=1\\}.$ Since
$(\alpha_{i}\alpha_{j}^{-1})^{n_{ij}}=1$ it follows that
$\alpha_{i}^{n_{ij}}=\alpha_{j}^{n_{ij}}.$ Consequently
$\alpha_{i}^{d}:=\beta$ for $i=\\{1,2,3,..,n\\}.$

Now we have:

  1. 1.

$g_{i}(x_{1}^{d}+\alpha_{i}^{d}x_{2}^{d})=g_{i}(x_{1})^{d}+\alpha_{i}^{d}g_{i}(x_{2})^{d}=\alpha_{i}^{d}x_{2}^{d}+\alpha_{i}^{d}(\alpha_{i}^{-1}x_{1})^{d}=x_{1}^{d}+\alpha_{i}^{d}x_{2}^{d}$.

  2. 2.

$g_{i}(x_{1}x_{2})=g_{i}(x_{1})g_{i}(x_{2})=\alpha_{i}x_{2}\cdot\alpha_{i}^{-1}x_{1}=x_{1}x_{2}$.

Hence $\\{x_{1}x_{2},x_{1}^{d}+\beta x_{2}^{d}\\}\in S(W)^{T(G)}$ and
$deg(x_{1}x_{2})deg(x_{1}^{d}+\beta x_{2}^{d})=2d=|T(G)|W|.$ So by [4, Theorem
3.7.5] we get that $S(W)^{T(G)}=F[x_{1}x_{2},x_{1}^{d}+\beta x_{2}^{d}].$ Let
$g|W=\bigl{(}\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix}\bigr{)}.$ Set
$\hat{g}=\xi I_{W}g|W,$ where $\xi^{2}det(g|W)=1.$ Now the computation of [3,
Theorem 3.10],with $\hat{g}$ instead of $g|W,$ shows that $\hat{g}\in T(G)|W.$

We have:

  1. (i)

$g(x_{1}x_{2})=\bigl{(}\begin{smallmatrix}\xi^{-1}&0\\\
0&\xi^{-1}\end{smallmatrix}\bigr{)}\hat{g}(x_{1}x_{2})=\xi^{-2}x_{1}x_{2}.$

  2. (ii)

$g(x_{1}^{d}+\beta x_{2}^{d})=\bigl{(}\begin{smallmatrix}\xi^{-1}&0\\\
0&\xi^{-1}\end{smallmatrix}\bigr{)}\hat{g}(x_{1}^{d}+\beta
x_{2}^{d})=\bigl{(}\begin{smallmatrix}\xi^{-1}&0\\\
0&\xi^{-1}\end{smallmatrix}\bigr{)}(x_{1}^{d}+\beta
x_{2}^{d})=\xi^{-d}(x_{1}^{d}+\beta x_{2}^{d}).$

Consequently, $g|{m/m^{2}}=\bigl{(}\begin{smallmatrix}\xi^{-2}&0\\\
0&\xi^{-d}\end{smallmatrix}\bigr{)}$ ,where
$m/m^{2}=U=F(x_{1}x_{2})+F(x_{1}^{d}+\beta x_{2}^{d}).$ So
$det(g|{m/m^{2}})=\xi^{-2}\xi^{-d}=det(g|W)\xi^{-d},$ and
$det(g|W)=det(g|{m/m^{2}})$ if and only if $\xi^{-d}=1.$ Consequently, by
Proposition 2.3 $S(V)^{G}$ is Gorenstein if and only if
$G|W\subseteq<T(G)|W,\delta I_{W}>$ where $\delta$ is a primitive $d-th$ root
of unity.

∎

######  Corollary 2.8.

Suppose $G\subset SL(3,\mathbb{F}_{p}),$ and $V=\mathbb{F}_{p}v\oplus W$ a
decomposition of $V=\mathbb{F}_{p}^{3}$ into $G-$submodules.Then $S(V)^{G}$ is
Gorenstein.

###### Proof.

This is a consequence of Proposition 2.4 , Proposition 2.6 and Proposition 2.7
. ∎

##  3 Chapter($B$): $V$ is indecomposable and contains (at least) two $2-$
dimensional $G-$ submodules $W_{1},W_{2}.$

In this chapter we consider $V,$ a reducible, indecomposable $G-module$ which
has at least two $G$-submodules $W_{1},W_{2}$ where $dimW_{i}=2$ for $i=1,2.$
Fix a basis $\\{w_{2},w_{1},v\\}$ such that $W_{i}=Fw_{i}+Fv.$ Then $W_{1}\cap
W_{2}=Fv$ is $1$-dimensional $G$-submodule of $V.$ Therefore for every $g\in
G$ we have a matrix representation with respect to this basis.

$g=\left(\begin{array}[]{ccc}\lambda_{2}(g)&0&\delta_{13}(g)\\\
0&\lambda_{1}(g)&\delta_{23}(g)\\\ 0&0&\lambda(g)\end{array}\right).$

Let $\sigma\in T(G)$ be a transvection . $\sigma:W_{i}\to W_{i}$ for $i=1,2.$
So $\sigma(w_{i})=\alpha_{i}w_{i}+\beta_{i}v$ where $\alpha_{i},\beta_{i}\in
F.$ Since $|<\sigma>|=p$ and $\sigma$ acts trivially on $Fv$ by Lemma 2.2,
$w_{i}=\sigma^{p}(w_{i})=\alpha^{p}_{i}w_{i}+\gamma_{i}v,$ hence
$\gamma_{i}=0$ and $\alpha_{i}^{p}=1,$ then $\alpha_{i}=1,$ for $i=1,2.$ As a
result, $\sigma$ is represented by the matrix :

$\left(\begin{array}[]{ccc}1&0&a\\\ 0&1&b\\\ 0&0&1\end{array}\right)$

with respect the basis $\\{w_{2},w_{1},v\\}.$

######  Corollary 3.1.

$T(G)$ is an elementary abelian group.

###### Proof.

Suppose that $\tau$ is another transvection where

$\tau=\left(\begin{array}[]{ccc}1&0&a_{1}\\\ 0&1&b_{1}\\\
0&0&1\end{array}\right).$

Then it easy to see that $\tau\sigma=\sigma\tau.$ Therefore $T(G)$ is an
elementary abelian group.Say it is minimaly generated by
$\\{\sigma_{1},...,\sigma_{n}\\},$ where

$\sigma_{i}=\left(\begin{array}[]{ccc}1&0&a_{i}\\\ 0&1&b_{i}\\\
0&0&1\end{array}\right)$

for $i=1,...,n.$ So $|T(G)|=p^{n}.$

∎

Let $X:=\\{\bigl{(}\begin{smallmatrix}a_{i}\\\
b_{i}\end{smallmatrix}\bigr{)}\in(\begin{smallmatrix}F\\\
F\end{smallmatrix}\bigr{)}|i=1,...,n\\},$ then the elements of $X$ are
linearly independent over $\mathbb{F}_{p}.$ Clearly
$dim(span_{F}X)\in\\{1,2\\}.$

### Case 1: $dim(span_{F}X)=1$

Assume that $dim(span_{F}X)=1.$So $\bigl{(}\begin{smallmatrix}a_{i}\\\
b_{i}\end{smallmatrix}\bigr{)}=\mu_{i}\bigl{(}\begin{smallmatrix}a\\\
b\end{smallmatrix}\bigr{)}$ for $i=1,...,n.$ Where
$\bigl{(}\begin{smallmatrix}a_{1}\\\
b_{1}\end{smallmatrix}\bigr{)}=\bigl{(}\begin{smallmatrix}a\\\
b\end{smallmatrix}\bigr{)},\mu_{1}=1$ and $\mu_{i}\in F$ for $i=2,...,n.$
Since the elements of $X$ are linearly independent over $\mathbb{F}_{p},$ this
is equivalent to $\\{1,\mu_{2},...,\mu_{n}\\}$ being of linearly independent
over $\mathbb{F}_{p}.$

######  Lemma 3.2.

$S(V)^{T(G)}$ is a polynomial ring.

###### Proof.

We shall firstly change the basis of $V.$ We have $\sigma_{1}(w_{1})=w_{1}+bv$
and $\sigma_{1}(w_{2})=w_{2}+av,$ where $a=a_{1}$ and $b=b_{1}.$Hence
$\sigma_{1}(aw_{1}-bw_{2})=a\sigma(w_{1})-b\sigma(w_{2})=a(w_{1}+bv)-b(w_{2}+av)=aw_{1}-bw_{2}.\\\
$ So $0\neq(aw_{1}-bw_{2})\in S(V)^{<\sigma_{1}>}.$We therefore change the
basis of $V$ into $\\{w_{2},aw_{1}-bw_{2},v\\},$ and then $\sigma_{1}$ is
represented by the matrix:

$\left(\begin{array}[]{ccc}1&0&a\\\ 0&1&0\\\ 0&0&1\end{array}\right).$

We also have:

$\sigma_{i}(aw_{1}-bw_{2})=a\sigma_{i}(w_{1})-b\sigma_{i}(w_{2})=a(w_{1}+b_{i}v)-b(w_{2}+av_{i})=aw_{1}-bw_{2}+(ab_{i}-ba_{i})v.$
But $\bigl{(}\begin{smallmatrix}a_{i}\\\
b_{i}\end{smallmatrix}\bigr{)}=\mu_{i}\bigl{(}\begin{smallmatrix}a\\\
b\end{smallmatrix}\bigr{)},$ and therefore $(ab_{i}-ba_{i})=0.$ Hence
$0\neq(aw_{1}-bw_{2})\in S(V)^{<\sigma_{i}>},$ for $i=1,...,n.$ So each
$\sigma_{i}$ is represented by the matrix:

$\left(\begin{array}[]{ccc}1&0&a_{i}\\\ 0&1&0\\\ 0&0&1\end{array}\right)$

where $a_{i}=\mu_{i}a.$ So we are in the same situation as in the modular
binary case (see [3]).Consequently, by [3, Lemma 3.13]
$S(V)^{T(G)}=F[x,aw_{1}-bw_{2},v]$ is a polynomial ring, and $x$ is an
homogenous polynomial in $\\{v,aw_{1}-bw_{2}\\}$ of degree $p^{n}$ having
$w_{2}^{p^{n}}$ as one of its monomial.

∎

We now need to shift to the action of $G$ on $S(V)^{T(G)}.$

######  Proposition 3.3.

$S(V)^{G}$ is Gorenstein if and only if every element of $G/T(G)$ has order
dividing $|T(G)|-1=p^{n}-1.$

###### Proof.

If $a=0$ or $b=0$ we argue exactly as in the binary form paper (see[3]).So we
may assume that $a\neq 0$ and $b\neq 0.$ Since $T(G)\lhd
G,g\sigma_{i}g^{-1}\in T(G),\forall i=1,...,n.$ Hence

$g\sigma_{i}g^{-1}=\varphi_{i}=\left(\begin{array}[]{ccc}1&0&\epsilon_{i}a\\\
0&1&\epsilon_{i}b\\\
0&0&1\end{array}\right),\epsilon_{i}\in\mathbb{F}_{p}1+\mathbb{F}_{p}\mu_{2}...+\mathbb{F}_{p}\mu_{n}.$

So $g\sigma_{i}=\varphi_{i}g$ is written explicitly by:

$\left(\begin{array}[]{ccc}\lambda_{2}(g)&0&\delta_{13}(g)\\\
0&\lambda_{1}(g)&\delta_{23}(g)\\\
0&0&\lambda(g)\end{array}\right)\left(\begin{array}[]{ccc}1&0&\mu_{i}a\\\
0&1&\mu_{i}b\\\
0&0&1\end{array}\right)=\left(\begin{array}[]{ccc}1&0&\epsilon_{i}a\\\
0&1&\epsilon_{i}b\\\
0&0&1\end{array}\right)\left(\begin{array}[]{ccc}\lambda_{2}(g)&0&\delta_{13}(g)\\\
0&\lambda_{1}(g)&\delta_{23}(g)\\\ 0&0&\lambda(g)\end{array}\right).$

By equating the $(1,3)$ and $(2,3)$ entries in both sides , we get :

$\left\\{\begin{array}[]{ll}\lambda_{2}(g)\mu_{i}a=\epsilon_{i}a\lambda(g)\\\
\lambda_{1}(g)\mu_{i}b=\epsilon_{i}b\lambda(g)\par\end{array}\right.$

Since $a\neq 0\neq b$ we get in particular that
$\lambda_{1}(g)=\lambda_{2}(g)$ for each $g\in G.$ Recall that $G\subset
SL(V),$hence $\lambda_{1}(g)\cdot\lambda_{2}(g)\cdot\lambda(g)=1.$ Let
$m=<x,aw_{1}-bw_{2},v>$ be the unique graded maximal ideal of
$S(V)^{T(G)}=F[x,aw_{1}-bw_{2},v].$ Then,

$g|m/m^{2}=\left(\begin{array}[]{ccc}\lambda_{2}(g)^{p^{n}}&0&0\\\
0&\lambda_{1}(g)&a\delta_{23}(g)-b\delta_{13}(g)\\\
0&0&\lambda(g)\end{array}\right).$

So by [Theorem 1.2] $S(V)^{G}$ is Gorenstein iff
$\lambda_{2}(g)^{p^{n}}\cdot\lambda_{1}(g)\cdot\lambda(g)=1.$ Since
$\lambda_{2}(g)=\lambda_{1}(g)$ and
$\lambda_{1}(g)\cdot\lambda_{2}(g)\cdot\lambda(g)=1,$ then $S(V)^{G}$ is
Gorenstein if and only if $\lambda_{1}(g)^{p^{n}-1}=1.$ Since
$\lambda_{1}(g)=\lambda_{2}(g),|T(G)|=p^{n},$ then
$|gT(G)|=|g|m/m^{2}|=|\lambda_{1}(g)|,$ for $g\in G.$ So $S(V)^{G}$ is
Gorenstein if and only if every $gT(G)\in G/T(G)$ has order dividing
$|T(G)|-1=p^{n}-1.$

∎

######  Remark 3.4.

The difference between the previous result and the results of [3], is that in
[3], $G/T(G)$ was a cyclic group and here it is not necessarily so.

### Case 2:$dim(span_{F}X)=2$

We also assume that
$T(G)=<\sigma_{1},...,\sigma_{n_{1}},\tau_{1},...,\tau_{n_{2}}>,$ where

$\sigma_{i}=\left(\begin{array}[]{ccc}1&0&a\mu_{i}\\\ 0&1&b\mu_{i}\\\
0&0&1\end{array}\right)$

and

$\tau_{i}=\left(\begin{array}[]{ccc}1&0&c\alpha_{i}\\\ 0&1&d\alpha_{i}\\\
0&0&1\end{array}\right),$

where $\bigl{(}\begin{smallmatrix}a\\\
b\end{smallmatrix}\bigr{)},\bigl{(}\begin{smallmatrix}c\\\
d\end{smallmatrix}\bigr{)}\in\bigl{(}\begin{smallmatrix}F\\\
F\end{smallmatrix}\bigr{)}$ are $F$-linearly independent.Also
$\\{\mu_{1},...,\mu_{n_{1}}\\}\subset F$ are $\mathbb{F}_{p}$-linearly
independent and $\\{\alpha_{1},...,\alpha_{n_{2}}\\}\subset F$ are
$\mathbb{F}_{p}$-linearly independent. Let
$T_{1}(G)=<\sigma_{1},...,\sigma_{n_{1}}>$ and
$T_{2}(G)=<\tau_{1},...,\tau_{n_{2}}>.$

We shall prove the following result.

######  Theorem 3.5.

Let $F$ be a field with $charF=p>0.$ Let $G\subset SL(3,F)$ be a finite
subgroup and $V=F^{3},$the natural $SL(3,F)-$module with the above notation,
assumption and $\lambda_{1}(g)=\lambda_{2}(g)$ $\forall g\in G.$ Then
$S(V)^{G}$ is Gorenstein in the following cases:

  1. (i)

$p=3.$

  2. (ii)

$|G|$ is prime to $3.$

  3. (iii)

$F=\mathbb{F}_{q},$ where $q=p^{s}$ and g.c.d$(3,s)=1.$

######  Lemma 3.6.

$S(V)^{T(G)}$ is a polynomial ring.

###### Proof.

$\sigma_{i}(w_{1})=w_{1}+b\mu_{i}v$ and $\sigma_{i}(w_{2})=w_{2}+a\mu_{i}v,$
implying that $:\sigma_{i}(aw_{1}-bw_{2})=aw_{1}-bw_{2},$ for $i=1,...,n_{1}.$
Similarly, $\tau_{i}(cw_{1}-dw_{2})=cw_{1}-dw_{2},$ for $i=1,...,n_{2}.$ Since
$\\{aw_{1}-bw_{2},cw_{1}-dw_{2}\\}$ are linearly independent (otherwise
$\bigl{(}\begin{smallmatrix}a\\\
b\end{smallmatrix}\bigr{)}=\lambda\bigl{(}\begin{smallmatrix}c\\\
d\end{smallmatrix}\bigr{)},$ in contradiction to their $F$-linear
independence), $\\{aw_{1}-bw_{2},cw_{1}-dw_{2},v\\}$ is a basis of $V$ and
they are $F-$algebraically independent.Now
$\tau_{i}(aw_{1}-bw_{2})=a\tau_{i}(w_{1})-b\tau_{i}(w_{2})=a(w_{1}+d\alpha_{i}v)-b(w_{2}+c\alpha_{i}v)=(aw_{1}-bw_{2})+(ad-
bc)\alpha_{i}v.$ Hence by [3, Lemma 3.13]
$S(F(aw_{1}-bw_{2})+Fv)^{T(G)}=F[x,v]$ is a polynomial ring, where $x$ is
homogenous polynomial (in $(aw_{1}-bw_{2}),v$) of degree $p^{n_{2}},$ having
$(aw_{1}-bw_{2})^{p^{n_{2}}}$ as one of its monomials.Similarly
$\sigma_{i}(cw_{1}-dw_{2})=(cw_{1}-dw_{2})+(cb-da)\mu_{i}v,$ and again
$S(F(cw_{1}-dw_{2})+Fv)^{T(G)}=F[y,v],$ is a polynomial ring, where $y$ is
homogenous polynomial (in $(cw_{1}-dw_{2}),v$) of degree $p^{n_{1}},$ having
$(cw_{1}-dw_{2})^{p^{n_{1}}}$ as one of its monomials. Since
$\\{aw_{1}-bw_{2},cw_{1}-dw_{2},v\\}$ are algebraic independent , then
$\\{x,y,v\\}$ are algebraically independent.Since
$(degx)\cdot(degy)\cdot(degv)=p^{n_{2}}\cdot p^{n_{1}}\cdot 1=|T(G)|,$ we have
by [4, Theorem 3.9.4] that $S(V)^{T(G)}=F[x,y,v]$ is a polynomial ring.

∎

In the next three results we examine the normality of $T_{1}(G),T_{2}(G).$
This is not needed for Proposition 3.10 ,Proposition 3.13, and Corollary 3.14.

######  Lemma 3.7.

Let $G\subset GL(V)$ be a finite subgroup,$T(G)=T_{1}(G)\times T_{2}(G)$ and
suppose that $T_{k}(G)\lhd G$ for $k=1,2.$ Then one of the following holds:

  1. (a)

$\lambda_{2}(g)=\lambda_{1}(g)$ $\forall g\in G.$

  2. (b)

$\lambda_{2}(g)=\delta(g)\lambda_{1}(g),$ where $\delta(g)\in\mathbb{F}_{p}$
$\forall g\in G.$

###### Proof.

Since $g$ normalizes $T_{1}(G)$ $\forall g\in G,$ we have for each
$i=1,...,n_{1}:$

$\gamma_{i}=g\sigma_{i}g^{-1}=\left(\begin{array}[]{ccc}1&0&\sum_{j=1}^{n_{1}}a\mu_{j}r_{ji}\\\
0&1&\sum_{j=1}^{n_{1}}b\mu_{j}r_{ji}\par\\\ 0&0&1\end{array}\right),$

where $r_{ji}\in\mathbb{F}_{p}.$ Since $g\sigma_{i}=\gamma_{i}g$ we get by
equating the $(1,3)$ and $(2,3)$ entries in both sides:

  1. (1)

$\left\\{\begin{array}[]{ll}\lambda_{2}(g)a\mu_{i}=\lambda(g)(\sum_{j=1}^{n_{1}}\mu_{j}r_{ji})a\\\
\lambda_{1}(g)b\mu_{i}=\lambda(g)(\sum_{j=1}^{n_{1}}\mu_{j}r_{ji})b\end{array}\right.$

Similarly,since $g$ normalizes $T_{2}(G)$ $\forall g\in G,$ we have for each
$i=1,...,n_{2}:$

  2. (2)

$\left\\{\begin{array}[]{ll}\lambda_{2}(g)c\alpha_{i}=\lambda(g)(\sum_{j=1}^{n_{2}}\alpha_{j}r_{ji})c\\\
\lambda_{1}(g)d\alpha_{i}=\lambda(g)(\sum_{j=1}^{n_{2}}\alpha_{j}r_{ji})d\par\end{array}\right.$

Recall that $\bigl{(}\begin{smallmatrix}a\\\
b\end{smallmatrix}\bigr{)},\bigl{(}\begin{smallmatrix}c\\\
d\end{smallmatrix}\bigr{)}\in\bigl{(}\begin{smallmatrix}F\\\
F\end{smallmatrix}\bigr{)}$ are $F$-linearly independent. If $a\neq 0\neq b$
(or $c\neq 0\neq d$ ) then we get by (1)(or (2)) that :
$\lambda_{1}(g)=\lambda_{2}(g)$ $\forall g\in G.$ On the other hand,if
$\bigl{(}\begin{smallmatrix}a&c\\\
b&d\end{smallmatrix}\bigr{)}=\bigl{(}\begin{smallmatrix}a&0\\\
0&d\end{smallmatrix}\bigr{)},$ then from (1) and (2) we get that
$\delta(g)=\frac{\lambda_{2}(g)}{\lambda_{1}(g)}=\frac{\alpha_{i}\sum_{j=1}^{n_{1}}\mu_{j}r_{ji}}{\mu_{i}\sum_{j=1}^{n_{2}}\alpha_{j}r_{ji}},$
and if $\bigl{(}\begin{smallmatrix}a&c\\\
b&d\end{smallmatrix}\bigr{)}=\bigl{(}\begin{smallmatrix}0&c\\\
b&0\end{smallmatrix}\bigr{)},$ then from (1) and (2) we get that
$\delta(g)=\frac{\lambda_{2}(g)}{\lambda_{1}(g)}=\frac{\mu_{i}\sum_{j=1}^{n_{2}}\alpha_{j}r_{ji}}{\alpha_{i}\sum_{j=1}^{n_{1}}\mu_{j}r_{ji}}.$
Since $\mu_{i},\alpha_{i},r_{ji}\in\mathbb{F}_{p}$ then
$\delta(g)\in\mathbb{F}_{p}.$

∎

######  Lemma 3.8.

$T_{k}(G)\lhd G$ for $k=1,2$ in the two following cases:

  1. (a)

$\lambda_{2}(g)=\lambda_{1}(g)$ $\forall g\in G.$

  2. (b)

$\bigl{(}\begin{smallmatrix}a&c\\\
b&d\end{smallmatrix}\bigr{)}\in\\{\bigl{(}\begin{smallmatrix}0&c\\\
b&0\end{smallmatrix}\bigr{)},\bigl{(}\begin{smallmatrix}a&0\\\
0&d\end{smallmatrix}\bigr{)}\\}.$

###### Proof.

Suppose to the contrary that $T_{1}(G)\ntriangleleft G.$ Then
$gT(G)g^{-1}=T(G)$ implies that
$g\sigma_{i}g^{-1}=\sigma_{1}^{e_{1}}\cdot...\cdot\sigma_{n_{1}}^{e_{n_{1}}}\cdot\tau_{1}^{f_{1}}\cdot...\cdot\tau_{n_{2}}^{f_{n_{2}}},$
where not all of the $f_{i}^{\prime}s$ are $0.$ By a simple calculation we
have:

$g\sigma_{i}g^{-1}=\left(\begin{array}[]{ccc}1&0&\lambda(g)^{-1}\mu_{i}\lambda_{2}(g)a\\\
0&1&\lambda(g)^{-1}\mu_{i}\lambda_{1}(g)b\\\ 0&0&1\end{array}\right).$

On the other hand we have :

$\sigma_{1}^{e_{1}}\cdot...\cdot\sigma_{n_{1}}^{e_{n_{1}}}=\left(\begin{array}[]{ccc}1&0&$$\sum_{i=1}^{n_{1}}a\mu_{i}e_{i}$$\\\
0&1&$$\sum_{i=1}^{n_{1}}b\mu_{i}e_{i}$$\\\ 0&0&1\end{array}\right).$

as well as

$\tau_{1}^{f_{1}}\cdot...\cdot\tau_{n_{2}}^{f_{n_{2}}}=\left(\begin{array}[]{ccc}1&0&$$\sum_{i=1}^{n_{2}}c\alpha_{i}f_{i}$$\\\
0&1&$$\sum_{i=1}^{n_{2}}d\alpha_{i}f_{i}$$\\\ 0&0&1\end{array}\right).$

Hence by equating the $(1,3)$ and $(2,3)$ entries in both sides :

$\left\\{\begin{array}[]{ll}\lambda(g)^{-1}\mu_{i}\lambda_{2}(g)a=\sum_{j=1}^{n_{1}}a\mu_{j}e_{j}+\sum_{j=1}^{n_{2}}c\alpha_{j}f_{j}\\\
\lambda(g)^{-1}\mu_{i}\lambda_{1}(g)b=\sum_{j=1}^{n_{1}}b\mu_{j}e_{j}+\sum_{j=1}^{n_{2}}d\alpha_{j}f_{j}\par\end{array}\right.$

We firstly assume that $\lambda_{2}(g)=\lambda_{1}(g).$ Then
$\lambda_{2}(g)\lambda(g)^{-1}\mu_{i}(\begin{smallmatrix}a\\\
b\end{smallmatrix}\bigr{)}=(\sum_{j=1}^{n_{1}}\mu_{j}e_{j})(\begin{smallmatrix}a\\\
b\end{smallmatrix}\bigr{)}+(\sum_{j=1}^{n_{2}}\alpha_{j}f_{j})(\begin{smallmatrix}c\\\
d\end{smallmatrix}\bigr{)}.$ So since $(\begin{smallmatrix}a\\\
b\end{smallmatrix}\bigr{)},(\begin{smallmatrix}c\\\
d\end{smallmatrix}\bigr{)}$ are $F$-linearly independent we get that
$\sum_{j=1}^{n_{2}}\alpha_{j}f_{j}=0.$ Since
$\\{\alpha_{1},...,\alpha_{n_{2}}\\}$ are $\mathbb{F}_{p}$-linearly
independent , $f_{j}=0$ for $j=1,...,n_{2}.$ A contradiction, hence
$T_{1}(G)\lhd G.$

Assume now that item (b) holds.Say $\bigl{(}\begin{smallmatrix}a&c\\\
b&d\end{smallmatrix}\bigr{)}=\bigl{(}\begin{smallmatrix}a&0\\\
0&d\end{smallmatrix}\bigr{)},$ hence
$(\begin{smallmatrix}\lambda(g)^{-1}\mu_{i}\lambda_{2}(g)a\\\
0\end{smallmatrix}\bigr{)}=(\sum_{j=1}^{n_{1}}\mu_{j}e_{j})(\begin{smallmatrix}a\\\
0\end{smallmatrix}\bigr{)}+(\sum_{j=1}^{n_{2}}\alpha_{j}f_{j})(\begin{smallmatrix}0\\\
d\end{smallmatrix}\bigr{)},$ so $(\sum_{j=1}^{n_{2}}\alpha_{j}f_{j})=0.$ Since
$\\{\alpha_{1},...,\alpha_{n_{2}}\\}$ are $\mathbb{F}_{p}$-linearly
independent, $f_{j}=0$ for $j=1,...,n_{2},$ a contradiction. If
$\bigl{(}\begin{smallmatrix}a&c\\\
b&d\end{smallmatrix}\bigr{)}=\bigl{(}\begin{smallmatrix}0&c\\\
b&0\end{smallmatrix}\bigr{)},$then by the same steps above we get that
$(\sum_{j=1}^{n_{2}}\alpha_{j}f_{j})=0,$ again $f_{j}=0$ for $j=1,...,n_{2},$
again a contradiction.By a similar discussion we get that $T_{2}(G)\lhd G.$

∎

The combination of Lemma 3.7 and Lemma 3.8 yields.

######  Proposition 3.9.

The following conditions are equivalent .

  1. (a)

$T_{k}(G)\lhd G,$ for $k=1,2.$

  2. (b)

one of the following holds:

  1. 1.

$\bigl{(}\begin{smallmatrix}a&c\\\
b&d\end{smallmatrix}\bigr{)}\in\\{\bigl{(}\begin{smallmatrix}0&c\\\
b&0\end{smallmatrix}\bigr{)},\bigl{(}\begin{smallmatrix}a&0\\\
0&d\end{smallmatrix}\bigr{)}\\}$

  2. 2.

$\lambda_{2}(g)=\lambda_{1}(g),$ for all $g\in G.$

###### Proof.

That (b) implies (a) is a consequence of Lemma 3.8. That (a) implies (b)
follows from the proof of Lemma 3.7.

∎

We can now give a necessary and sufficient condition for a subgroup $G$ such
that $S(V)^{G}$ is Gorenstein.

######  Proposition 3.10.

Suppose $n_{1}<n_{2}.$ Then $S(V)^{G}$ is Gorenstein if and only if :

  1. 1.

$\lambda_{1}(g)^{p^{n_{1}}+p^{n_{2}}}\lambda(g)=1,$ in case $c\neq 0\neq d.$

  2. 2.

$\lambda_{1}(g)^{p^{n_{2}}}\lambda_{2}(g)^{p^{n_{1}}}\lambda(g)=1,$ in case
$c=0\neq d.$

  3. 3.

$\lambda_{2}(g)^{p^{n_{2}}}\lambda_{1}(g)^{p^{n_{1}}}\lambda(g)=1,$ in case
$c\neq 0=d.$

The case $n_{2}<n_{1}$ is similar.Note that we do not assume any normality
assumption on $T_{k}(G).$

###### Proof.

By Lemma 3.6 , we have that
$S(V)^{T(G)}=F[x,y,v],deg(y)=p^{n_{1}},deg(x)=p^{n_{2}}.$ Since
$g^{-1}T(G)g=T(G)$ we have $g(y)\in S(V)^{T(G)}.$ Therefore,since $g$ preserve
degrees,$g(y)=\alpha y+\beta v^{p^{n_{1}}}.$ Recall that
$(cw_{1}-dw_{2})^{p^{n_{1}}}$ appears as a monomial in $y$ and all other
monomials involve $v$ in addition.

We have :

$g(cw_{1}-dw_{2})=cg(w_{1})-dg(w_{2})=c(\lambda_{1}(g)w_{1}+\delta_{23}(g)v)-d(\lambda_{2}(g)w_{2}+\delta_{13}(g)v)=(c\lambda_{1}(g)w_{1}-d\lambda_{2}(g)w_{2})+\chi
v,$ where $\chi=c\delta_{23}(g)-d\delta_{13}(g).$ Hence
$g((cw_{1}-dw_{2})^{p^{n_{1}}})=c^{p^{n_{1}}}\lambda_{1}(g)^{p^{n_{1}}}w_{1}^{p^{n_{1}}}-d^{p^{n_{1}}}\lambda_{2}(g)^{p^{n_{1}}}w_{2}^{p^{n_{1}}}+\chi^{p^{n_{1}}}v^{p^{n_{1}}}.$
So by comparing coefficients of $y$ in $g(y)$ we get:

$c^{p^{n_{1}}}\lambda_{1}(g)^{p^{n_{1}}}w_{1}^{p^{n_{1}}}-d^{p^{n_{1}}}\lambda_{2}(g)^{p^{n_{1}}}w_{2}^{p^{n_{1}}}=\alpha(cw_{1}-dw_{2})^{p^{n_{1}}}.$
Consequently

  1. (1)

$c^{p^{n_{1}}}\lambda_{1}(g)^{p^{n_{1}}}=\alpha
c^{p^{n_{1}}},d^{p^{n_{1}}}\lambda_{2}(g)^{p^{n_{1}}}=\alpha d^{p^{n_{1}}}.$

Similarly $g(x)=\delta x+\varepsilon y^{p^{n_{2}-n_{1}}}+$ mixed terms (in
$y,v$)+$\xi v^{p^{n_{2}}}.$ Recall that $(aw_{1}-bw_{2})^{p^{n_{2}}}$ appears
as a monomial in $x$ and all other monomials involve $v$ in addition.So the
contribution to the terms with $w_{1}^{p^{n_{2}}}$ in the right hand side of
the last equation are
$\delta(aw_{1}-bw_{2})^{p^{n_{2}}}+\varepsilon((cw_{1}-dw_{2})^{p^{n_{1}}})^{p^{n_{2}-n_{1}}}=\delta(aw_{1}-bw_{2})^{p^{n_{2}}}+\varepsilon(cw_{1}-dw_{2})^{p^{n_{2}}}.\\\
$ The contribution of the left hand side are from
$g((aw_{1}-bw_{2})^{p^{n_{2}}})=a^{p^{n_{2}}}(\lambda_{1}(g)w_{1}+\delta_{23}(g)v)^{p^{n_{2}}}-b^{p^{n_{2}}}(\lambda_{2}(g)w_{2}+\delta_{13}(g)v)^{p^{n_{2}}}.$
Consequently we get :

  1. (2)

$a^{p^{n_{2}}}\lambda_{1}(g)^{p^{n_{2}}}=\delta a^{p^{n_{2}}}+\varepsilon
c^{p^{n_{2}}},b^{p^{n_{2}}}\lambda_{2}(g)^{p^{n_{2}}}=\delta
b^{p^{n_{2}}}+\varepsilon d^{p^{n_{2}}}.$

If $c\neq 0\neq d,$ then (1) shows that
$\lambda_{1}(g)^{p^{n_{1}}}=\alpha=\lambda_{2}(g)^{p^{n_{1}}},$ so
$\lambda_{1}(g)=\lambda_{2}(g).$ Now (2) shows that
$a^{p^{n_{2}}}(\lambda_{1}(g)^{p^{n_{2}}}-\delta)=\varepsilon
c^{p^{n_{2}}},b^{p^{n_{2}}}(\lambda_{2}(g)^{p^{n_{2}}}-\delta)=\varepsilon
d^{p^{n_{2}}}.\\\ $ So if $\varepsilon\neq 0$ then
$\frac{a^{p^{n_{2}}}}{\varepsilon
c^{p^{n_{2}}}}=\frac{b^{p^{n_{2}}}}{\varepsilon d^{p^{n_{2}}}}.$ Hence
$\frac{a}{c}=\frac{b}{d}$ or $\frac{a}{b}=\frac{c}{d},$ which is a
contradiction,so $\varepsilon=0$ in case $c\neq 0\neq d.$ Therefore
$\lambda_{1}(g)^{p^{n_{2}}}=\delta=\lambda_{2}(g)^{p^{n_{2}}}.$ Let
$m=<x,y,v>$ be the unique homogenous maximal ideal of $S(V)^{T(G)}=F[x,y,v].$
The matrix representing $g$ on $m/m^{2}$ whenever $c\neq 0\neq d$ is

$\left(\begin{array}[]{ccc}\lambda_{1}(g)^{p^{n_{2}}}&0&0\\\
0&\lambda_{1}(g)^{p^{n_{1}}}&0\\\ 0&0&\lambda(g)\end{array}\right),$

so the condition for $S(V)^{G}$ to be Gorenstein is
$\lambda_{1}(g)^{p^{n_{2}}+p^{n_{1}}}\lambda(g)=1.$ Since $G\subset
SL(V),\lambda_{1}(g)=\lambda(g)^{-2}.$ Hence $S(V)^{G}$ is Gorenstein if and
only if $\lambda_{1}(g)^{p^{n_{2}}+p^{n_{1}}-2}=1.$ Moreover, by Lemma 3.8,
$T_{k}(G)\lhd G$ for $k=1,2.$
Assume now that $c=0\neq d,$ hence from (1) we get
$\lambda_{2}(g)^{p^{n_{1}}}=\alpha$ and from (2), since $a\neq 0$ in this
case, $\lambda_{1}(g)^{p^{n_{2}}}=\delta.$ So the matrix representing $g$ on
$m/m^{2}$ is

$\left(\begin{array}[]{ccc}\lambda_{1}(g)^{p^{n_{2}}}&0&0\\\
0&\lambda_{2}(g)^{p^{n_{1}}}&0\\\ 0&0&\lambda(g)\end{array}\right),$

and $S(V)^{G}$ is Gorenstein if and only if
$\lambda_{1}(g)^{p^{n_{2}}}\lambda_{2}(g)^{p^{n_{1}}}\lambda(g)=1.$

Assume now that $c\neq 0=d.$ Then from (1) we get
$\lambda_{1}(g)^{p^{n_{1}}}=\alpha$ and from (2), since $b\neq 0$ in this
case, $\lambda_{2}(g)^{p^{n_{2}}}=\delta.$ So the matrix representing $g$ on
$m/m^{2}$ is

$\left(\begin{array}[]{ccc}\lambda_{2}(g)^{p^{n_{2}}}&0&0\\\
0&\lambda_{1}(g)^{p^{n_{1}}}&0\\\ 0&0&\lambda(g)\end{array}\right),$

and $S(V)^{G}$ is Gorenstein if and only if
$\lambda_{2}(g)^{p^{n_{2}}}\lambda_{1}(g)^{p^{n_{1}}}\lambda(g)=1.$ This
settles the case $n_{1}<n_{2}.$ If $n_{1}>n_{2}$ then we do the same analysis
but with $\\{a,b\\}$ instead of $\\{c,d\\}.$

∎

Our next result will be used in the proof of Theorem 3.5.

######  Lemma 3.11.

Suppose $T_{k}(G)\lhd G$.Then
$(\frac{\lambda_{2}(g)}{\lambda(g)})^{p^{n_{k}}-1}=(\frac{\lambda_{1}(g)}{\lambda(g)})^{p^{n_{k}}-1}=1,$
for $k=1,2.$

###### Proof.

We have by Lemma 3.6 that $S(F(aw_{1}-bw_{2})+Fv)^{T_{2}(G)}=F[x,v],$ where
$x$ is homogenous polynomial (in $(aw_{1}-bw_{2}),v$) of degree $p^{n_{2}},$
having $(aw_{1}-bw_{2})^{p^{n_{2}}}$ as one of its monomials and all other
monomials involve $v.$ In other words
$x=(aw_{1}-bw_{2})^{p^{n_{2}}}+v^{p^{n_{2}}-1}(aw_{1}-bw_{2})+$ monomials
involve $v.$ By a simple calculating we get that :

$\left\\{\begin{array}[]{ll}g(aw_{1}-bw_{2})^{p^{n_{2}}}=a^{p^{n_{2}}}\lambda_{1}(g)^{p^{n_{2}}}w_{1}^{p^{n_{2}}}-b^{p^{n_{2}}}\lambda_{2}(g)^{p^{n_{2}}}w_{2}^{p^{{n_{2}}}}+(a\delta_{23}(g)-b\delta_{13}(g))^{p^{n_{2}}}v^{p^{n_{2}}}.\\\
g(v^{p^{n_{2}}-1}(aw_{1}-bw_{2}))=a\lambda(g)^{p^{n_{2}}-1}\lambda_{1}(g)w_{1}v^{p^{n_{2}}-1}-b\lambda(g)^{p^{n_{2}}-1}\lambda_{2}(g)w_{2}v^{p^{n_{2}}-1}+\lambda(g)^{p^{n_{2}}-1}v^{p^{n_{2}}}(a\delta_{23}(g)-b\delta_{13}(g)).\par\end{array}\right.$

Now $T_{2}(G)\lhd G$ imply that $F[x,v]$ is $g-$stable.Therfeore , since $g$
preserve degrees , we get $g(x)=\rho x+\kappa v^{p^{n_{2}}},\rho,\kappa\in F.$
Assume firstly,using Lemma 3.7, that $\lambda_{1}(g)=\lambda_{2}(g).$ Hence ,
if $a\neq 0$ then by comparing coefficients of $w_{1}^{p^{n_{2}}}$ and
$v^{p^{n_{2}}-1}w_{1}$ in $g(x)$ we get
$\rho=\lambda_{1}(g)^{p^{n_{2}}}=\lambda(g)^{p^{n_{2}}-1}\lambda_{1}(g).$ So
$(\frac{\lambda_{1}(g)}{\lambda(g)})^{p^{n_{2}}-1}=1,$ and if $b\neq 0$ then
by comparing coefficients of $w_{2}^{p^{n_{2}}}$ and $v^{p^{n_{2}}-1}w_{2}$ in
$g(x)$ we get
$\rho=\lambda_{2}(g)^{p^{n_{2}}}=\lambda(g)^{p^{n_{2}}-1}\lambda_{2}(g).$ So
$(\frac{\lambda_{2}(g)}{\lambda(g)})^{p^{n_{2}}-1}=1.\\\ $

Assume now, using Lemma 3.7, that $\lambda_{2}(g)=\delta(g)\lambda_{1}(g),$
where $\delta(g)\in\mathbb{F}_{p}.$ Consequently, by Proposition 3.9 ,
$\bigl{(}\begin{smallmatrix}a&c\\\
b&d\end{smallmatrix}\bigr{)}\in\\{\bigl{(}\begin{smallmatrix}0&c\\\
b&0\end{smallmatrix}\bigr{)},\bigl{(}\begin{smallmatrix}a&0\\\
0&d\end{smallmatrix}\bigr{)}\\}.$ Hence if $a\neq 0=b,$ then by comparing
coefficient of $w_{1}^{p^{n_{2}}}$ and $v^{p^{n_{2}}-1}w_{1}$ in $g(x)$ we get
$\rho=\lambda_{1}(g)^{p^{n_{2}}}=\lambda(g)^{p^{n_{2}}-1}\lambda_{1}(g).$ So
$(\frac{\lambda_{1}(g)}{\lambda(g)})^{p^{n_{2}}-1}=1,$ hence
$(\frac{\delta(g)^{-1}\lambda_{2}(g)}{\lambda(g)})^{p^{n_{2}}-1}=1.$ By little
Fermat’s theorem we get that $\delta(g)^{p^{n_{2}}-1}=1,$ then
$(\frac{\lambda_{2}(g)}{\lambda(g)})^{p^{n_{2}}-1}=1.$ If $b\neq 0=a,$ then by
comparing coefficient of $w_{2}^{p^{n_{2}}}$ and $v^{p^{n_{2}}-1}w_{2}$ in
$g(x)$ we get
$\rho=\lambda_{2}(g)^{p^{n_{2}}}=\lambda(g)^{p^{n_{2}}-1}\lambda_{2}(g).$ So
$(\frac{\lambda_{2}(g)}{\lambda(g)})^{p^{n_{2}}-1}=1,$ hence
$(\frac{\delta(g)\lambda_{1}(g)}{\lambda(g)})^{p^{n_{2}}-1}=1.$ By little
Fermat’s theorem we get that $\delta(g)^{p^{n_{2}}-1}=1,$ then
$(\frac{\lambda_{2}(g)}{\lambda(g)})^{p^{n_{2}}-1}=1.\\\ $

Similarly, if we look at $S(F(cw_{1}-dw_{2})+Fv)^{T_{1}(G)}=F[y,v]$ from Lemma
3.6 , then the above reasoning shows again that
$(\frac{\lambda_{k}(g)}{\lambda(g)})^{p^{n_{1}}-1}=1,$ for $k=1,2.$

∎

######  Lemma 3.12.

$(|G|/|T(G)|,p)=1,$ and if $\lambda_{1}(g)=\lambda_{2}(g)$ $\forall g\in G$
then $G/T(G)$ is a cyclic group.

###### Proof.

Since $G\subset SL(V),\lambda(g)=\lambda_{1}(g)^{-1}\lambda_{2}(g)^{-1}.$ We
have:

$g^{p}=\left(\begin{array}[]{ccc}\lambda_{2}(g)^{p}&0&*\\\
0&\lambda_{1}(g)^{p}&*\\\
0&0&\lambda_{1}(g)^{-p}\lambda_{2}(g)^{-p}\end{array}\right)$

If $g^{p}\in T(G)$ then $\lambda_{1}(g)^{p}=\lambda_{2}(g)^{p}=1,$ and then
$\lambda_{1}(g)=\lambda_{2}(g)=1,$ so :

$g=\left(\begin{array}[]{ccc}1&0&*\\\ 0&1&*\\\ 0&0&1\end{array}\right)\in
T(G).$

This shows that $(|G|/|T(G)|,p)=1.$ Assume now that
$\lambda_{1}(g)=\lambda_{2}(g)$ $\forall g\in G.$ Set
$U=\mathbb{F}_{p}\mu_{1}+\cdot\cdot\cdot\mathbb{F}_{p}\mu_{n_{1}}.$ By Lemma
3.8, $T_{1}(G)\lhd G$ then $g\sigma_{i}g^{-1}\in T_{1}(G).$ Now since

$g\sigma_{i}g^{-1}=\left(\begin{array}[]{ccc}1&0&\lambda(g)^{-1}\mu_{i}\lambda_{2}(g)a\\\
0&1&\lambda(g)^{-1}\mu_{i}\lambda_{1}(g)b\\\ 0&0&1\end{array}\right)$

for $i=1,...,n_{1},$ we get that $\lambda(g)^{-1}\lambda_{1}(g)\mu_{i}\in U,$
hence $\lambda(g)^{-1}\lambda_{1}(g)U\subseteq U$ Since $G\subset SL(V)$ and
$\lambda_{1}(g)=\lambda_{2}(g)$ then
$\lambda(g)^{-1}\lambda_{1}(g)=\lambda_{1}(g)^{3}.$Consequently by [3, Lemma
3.12.] $|\lambda_{1}(g)^{3}|$ divides $(p^{n_{1}}-1).$ By a similar discussion
on $T_{2}(G)$ we get that $|\lambda_{1}(g)^{3}|$ divides $(p^{n_{2}}-1).$ Let
$H=\\{\epsilon\in F|\epsilon^{3(p^{n_{1}}-1)}=1\\}.$ Define
$\Phi:G/T(G)\longrightarrow H,$ by $\Phi(gT(G))=\lambda_{1}(g),$ for each
$g\in G.$ Since

$gT(G)=g|m/m^{2}=\left(\begin{array}[]{ccc}\lambda_{1}(g)^{p^{n_{2}}}&0&0\\\
0&\lambda_{1}(g)^{p^{n_{1}}}&0\\\ 0&0&\lambda_{1}(g)^{-2}\end{array}\right),$

it follows that $\Phi$ is an injective homomorphism with its image in $H$
since $|gT(G)|=|\lambda_{1}(g)|$ divides $3(p^{n_{1}}-1).$ Since $H$ is cyclic
group it follows that $G/T(G)$ is cyclic.

∎

Recall that $n_{1}$ and $n_{2}$ are the numbers of the generators of
$T_{1}(G)$ and $T_{2}(G)$ respectively.
The next result considers the case $n_{1}=n_{2}$ which was omitted from
Proposition 3.10.

######  Proposition 3.13.

Suppose that $n_{1}=n_{2}.$ Then $S(V)^{G}$ is Gorenstein iff
$\lambda_{2}(g)^{p^{n_{1}}}\lambda_{1}(g)^{p^{n_{1}}}\lambda(g)=1.$

###### Proof.

We have by Lemma 3.6 that
$S(V)^{T(G)}=F[x,y,v],deg(y)=p^{n_{1}},deg(x)=p^{n_{2}}.$ Since
$g^{-1}T(G)g=T(G)$ we have $g(y),g(x)\in S(V)^{T(G)}.$ Hence, since $g$
preserve degrees, we get :

  1. (1)

$g(y)=\alpha y+\gamma x+\beta v^{p^{n_{1}}},\alpha,\gamma,\beta\in F.$

  2. (2)

$g(x)=\delta x+\varepsilon y+\chi v^{p^{n_{1}}},\delta,\varepsilon,\chi\in F.$

Therefore, $S(V)^{G}=[S(V)^{T(G)}]^{G/T(G)}.$ Let $m=(x,y,v)$ be the unique
graded maximal ideal of $S(V)^{T(G)}=F[y,x,v],$ then the matrix of $\bar{g}\in
G/T(G)$ on $m/m^{2}=F\bar{y}+F\bar{x}+F\bar{v}$ is :

$\left(\begin{array}[]{ccc}\alpha&\gamma&0\\\ \varepsilon&\delta&0\\\
0&0&\lambda(g)\end{array}\right).$

So by [Theorem 1.2] $S(V)^{G}$ is Gorenstein if and only if
$(\alpha\delta-\varepsilon\gamma)\lambda(g)=1$ $\forall g\in G.$

The computation of $\alpha\delta-\varepsilon\gamma:$

We shall assume that $c\neq 0\neq d$ and $a\neq 0\neq b.$ We know that
$(cw_{1}-dw_{2})^{p^{n_{1}}}$ appears as a monomial in $y,$
$(aw_{1}-bw_{2})^{p^{n_{1}}}$ appears as a monomial in $x,$ and all other
monomials involve also $v.$ Therefore the coefficients of $w_{1}^{p^{n_{1}}}$
in (1) are :
$c^{p^{n_{1}}}\lambda_{1}(g)^{p^{n_{1}}}$ in the left hand side and $\alpha
c^{p^{n_{1}}}+\gamma a^{p^{n_{1}}}$ in the right hand side.Hence
$c^{p^{n_{1}}}\lambda_{1}(g)^{p^{n_{1}}}=\alpha c^{p^{n_{1}}}+\gamma
a^{p^{n_{1}}}.$ Similarly, for the coefficients of $w_{2}^{p^{n_{1}}},$ we
have $d^{p^{n_{1}}}\lambda_{2}(g)^{p^{n_{1}}}=\alpha d^{p^{n_{1}}}+\gamma
b^{p^{n_{1}}}$.From (2) we similarly have:

$\left\\{\begin{array}[]{ll}a^{p^{n_{1}}}\lambda_{1}(g)^{p^{n_{1}}}=\delta
a^{p^{n_{1}}}+\varepsilon c^{p^{n_{1}}}\\\
b^{p^{n_{1}}}\lambda_{2}(g)^{p^{n_{1}}}=\delta b^{p^{n_{1}}}+\varepsilon
d^{p^{n_{1}}}\par\end{array}\right.$

Hence from (1) and (2) we have :

$\left\\{\begin{array}[]{llll}\lambda_{1}(g)^{p^{n_{1}}}=\alpha+\gamma(\frac{a}{c})^{p^{n_{1}}}\\\
\lambda_{2}(g)^{p^{n_{1}}}=\alpha+\gamma(\frac{b}{d})^{p^{n_{1}}}\\\
\lambda_{1}(g)^{p^{n_{1}}}=\delta+\varepsilon(\frac{c}{a})^{p^{n_{1}}}\\\
\lambda_{2}(g)^{p^{n_{1}}}=\delta+\varepsilon(\frac{d}{b})^{p^{n_{1}}}\par\par\end{array}\right.$

Hence,

$\left\\{\begin{array}[]{ll}(\lambda_{1}(g)-\lambda_{2}(g))^{p^{n_{1}}}=\gamma(\frac{a}{c}-\frac{b}{d})^{p^{n_{1}}}\\\
(\lambda_{1}(g)-\lambda_{2}(g))^{p^{n_{1}}}=\varepsilon(\frac{c}{a}-\frac{d}{b})^{p^{n_{1}}}\end{array}\right.$

It follows that:

  1. (A)

:=
$(\lambda_{1}(g)-\lambda_{2}(g))^{2p^{n_{1}}}=\gamma\varepsilon((\frac{a}{c}-\frac{b}{d})(\frac{c}{a}-\frac{d}{b}))^{p^{n_{1}}}.$

Also we have :

$\left\\{\begin{array}[]{ll}(\frac{c}{a}\lambda_{1}(g)-\frac{d}{b}\lambda_{2}(g))^{p^{n_{1}}}=\alpha(\frac{c}{a}-\frac{d}{b})^{p^{n_{1}}}\\\
(\frac{a}{c}\lambda_{1}(g)-\frac{b}{d}\lambda_{2}(g))^{p^{n_{1}}}=\delta(\frac{a}{c}-\frac{b}{d})^{p^{n_{1}}}\end{array}\right.$

And it follows that :

  2. (B)

:=
$(\frac{c}{a}\lambda_{1}(g)-\frac{d}{b}\lambda_{2}(g))^{p^{n_{1}}}(\frac{a}{c}\lambda_{1}(g)-\frac{b}{d}\lambda_{2}(g))^{p^{n_{1}}}=\alpha\delta[(\frac{c}{a}-\frac{d}{b})(\frac{a}{c}-\frac{b}{d})]^{p^{n_{1}}}.$

So
$\textbf{(B)-(A)}=(\alpha\delta-\varepsilon\gamma)[(\frac{c}{a}-\frac{d}{b})(\frac{a}{c}-\frac{b}{d})]^{p^{n_{1}}}.$
Observe that
$(\frac{c}{a}-\frac{d}{b})(\frac{a}{c}-\frac{b}{d})=2-(\frac{cb}{ad}+\frac{ad}{bc}).$
Also
$\textbf{(B)-(A)}=(\lambda_{1}(g)^{2}+\lambda_{2}(g)^{2}-(\frac{cb}{ad}+\frac{ad}{bc})\lambda_{1}(g)\lambda_{2}(g))^{p^{n_{1}}}-(\lambda_{1}(g)^{2}+\lambda_{2}(g)^{2}-2\lambda_{1}(g)\lambda_{2}(g))^{p^{n_{1}}}=(\lambda_{1}(g))^{p^{n_{1}}}(\lambda_{2}(g))^{p^{n_{1}}}[2-(\frac{cb}{ad}+\frac{ad}{bc})]^{p^{n_{1}}}.$

Hence
$\frac{(B)-(A)}{((\frac{c}{a}-\frac{d}{b})(\frac{a}{c}-\frac{b}{a}))^{p^{n_{1}}}}=(\alpha\delta-\varepsilon\gamma).$
Therefore
$\alpha\delta-\varepsilon\gamma=\lambda_{1}(g)^{p^{n_{1}}}\lambda_{2}(g)^{p^{n_{1}}},$
hence $S(V)^{G}$ is Gorenstein if and only if
$\lambda_{2}(g)^{p^{n_{1}}}\lambda_{1}(g)^{p^{n_{1}}}\lambda(g)=1.\\\ $

The cases where one of $\\{a,b,c,d\\}$ is $0$ are easier.For example, if
$a=0,$ hence since $\bigl{(}\begin{smallmatrix}a\\\
b\end{smallmatrix}\bigr{)},\bigl{(}\begin{smallmatrix}c\\\
d\end{smallmatrix}\bigr{)}$ are $F$-linearly independent , $c\neq 0,$ .Hence
$\lambda_{1}(g)^{p^{n_{1}}}=\alpha$ and $\varepsilon=0,$ therefore
$\delta=\lambda_{2}(g)^{p^{n_{1}}}.$ Hence
$\alpha\delta-\varepsilon\gamma=\alpha\delta=\lambda_{1}(g)^{p^{n_{1}}}\lambda_{2}(g)^{p^{n_{1}}}.$So
again $S(V)^{G}$ Gorenstein if and only if
$\lambda_{2}(g)^{p^{n_{1}}}\lambda_{1}(g)^{p^{n_{1}}}\lambda(g)=1.$
Equivalently $\lambda(g)^{p^{n_{1}}-1}=1.$

∎

We finally arrive at: The proof of Theorem 3.5.

###### Proof.

We have by Lemma 3.12 that $(|G|/|T(G)|,p)=1$ and
$G/T(G)\cong<\lambda_{1}(g)>$ is a cyclic group. Let
$\frac{|G|}{|T(G)|}=|\lambda_{1}(g)|\equiv k,$ since $G\subset
SL(V),\frac{\lambda_{1}(g)}{\lambda(g)}=\frac{\lambda_{1}(g)}{\lambda_{1}(g)^{-1}\lambda_{2}(g)^{-1}}=\frac{\lambda_{1}(g)}{\lambda_{1}(g)^{-1}\lambda_{1}(g)^{-1}}=\lambda_{1}(g)^{3}.$
Hence,by Lemma 3.11 ,
$(\lambda_{1}(g)^{3})^{p^{n_{1}}-1}=(\lambda_{1}(g)^{3})^{p^{n_{2}}-1}=1.$So
$(\lambda_{1}(g)^{3})^{p^{n_{1}}+p^{n_{2}}-2}=1,$ which means that $k$ divides
$[3\cdot({p^{n_{1}}+p^{n_{2}}-2})].$

If $p=3$ then since $(3,k)=1,k$ divides $(p^{n_{1}}+p^{n_{2}}-2),$
consequently by Proposition 3.10 and Proposition 3.13, $S(V)^{G}$ is
Gorenstein.This settles item $(i).$

Assume now that $3\nmid|G|.$The action of $g^{3}$ on $m/m^{2}$ where $m$ is
the unique maximal ideal of $S(V)^{T(G)}$ is :

$\left(\begin{array}[]{ccc}\lambda_{1}(g)^{3\cdot{p^{n_{2}}}}&0&0\\\
0&\lambda_{1}(g)^{3\cdot{p^{n_{1}}}}&0\\\
0&0&\lambda_{1}(g)^{-6}\end{array}\right).$

Since $|\lambda_{1}(g)^{3}|$ divides $p^{n_{1}}+p^{n_{2}}-2,$ the above
reasoning shows that $S(V)^{<g^{3},T(G)>}=S(m/m^{2})^{<g^{3}|m/m^{2}>}$ is
Gorenstein.Consequently if $3\nmid|G|,$ then $G=<g^{3},T(G)>$ and $S(V)^{G}$
is therefore Gorenstein,so item $(ii)$ is verified.

We have by Lemma 3.8 that $T_{k}(G)\lhd G$ for $k=1,2,$ hence by using the
proof of Lemma 3.7 we get :

  1. (1)

$\left\\{\begin{array}[]{ll}\lambda_{2}(g)a\mu_{i}=\lambda(g)($$\sum_{j=1}^{n_{1}}\mu_{j}r_{ji}$$)a\\\
\lambda_{1}(g)b\mu_{i}=\lambda(g)($$\sum_{j=1}^{n_{1}}\mu_{j}r_{ji}$$)b\par\end{array}\right.$

  2. (2)

$\left\\{\begin{array}[]{ll}\lambda_{2}(g)c\alpha_{i}=\lambda(g)($$\sum_{j=1}^{n_{2}}\alpha_{j}r_{ji}$$)c\\\
\lambda_{1}(g)d\alpha_{i}=\lambda(g)($$\sum_{j=1}^{n_{2}}\alpha_{j}r_{ji}$$)d\par\end{array}\right.$

Given the assumption of item $(iii)$.Since $\lambda_{1}(g)=\lambda_{2}(g),$ it
follows that $(a\neq 0\neq b)$ or $(c\neq 0\neq d).$ Say $a,c,d\neq 0,$ the
equations above imply that
$\frac{\lambda(g)}{\lambda_{1}(g)}($$\sum_{j=1}^{n_{1}}\mu_{j}r_{ji}$$)=\mu_{i}$
and
$\frac{\lambda(g)}{\lambda_{1}(g)}($$\sum_{j=1}^{n_{2}}\alpha_{j}r_{ji}$$)=\alpha_{i}.$
Set $M_{1}=\mathbb{F}_{p}\mu_{1}+...+\mathbb{F}_{p}\mu_{n_{1}}$ and
$M_{2}=\mathbb{F}_{p}\alpha_{1}+...+\mathbb{F}_{p}\alpha_{n_{2}},$ so
$dim_{\mathbb{F}_{p}}M_{1}=n_{1}$ and $dim_{\mathbb{F}_{p}}M_{2}=n_{2}.$
Thus $\frac{\lambda_{1}(g)}{\lambda(g)}M_{1}=\lambda_{1}(g)^{3}M_{1}\subseteq
M_{1}$ and
$\frac{\lambda_{1}(g)}{\lambda(g)}M_{2}=\lambda_{1}(g)^{3}M_{2}\subseteq
M_{2}.$
Suppose to the contrary that
$\lambda_{1}(g)\not\in\mathbb{F}_{p}(\lambda_{1}(g)^{3}),$ then
$[\mathbb{F}_{p}(\lambda_{1}(g)):\mathbb{F}_{p}(\lambda_{1}(g)^{3})]=3.$ But
$[F:\mathbb{F}_{p}]=s$ is relatively prime to $3,$ a contradiction.Thus
$\lambda_{1}(g)\in\mathbb{F}_{p}(\lambda_{1}(g)^{3}),$ hence
$\lambda_{1}(g)M_{1}\subseteq M_{1},$ similarly,we get that
$\lambda_{2}(g)M_{2}\subseteq M_{2}.$ Therefore, by [3, Lemma 3.12.]
$k|(p^{n_{1}}-1)$ and $k|(p^{n_{2}}-1),$ hence $k|(p^{n_{1}}+p^{n_{2}}-2).$
Consequently,by Proposition 3.10 item (1) , $S(V)^{G}$ is Gorenstein and item
$(iii)$ holds.

∎

We also have here the following.

######  Corollary 3.14.

Suppose $G\subset SL(3,\mathbb{F}_{p})$ and $V=\mathbb{F}_{p}^{3}$ contains
two $2-$dimensional $G-$submodules. Then $S(V)^{G}$ is Gorenstein.

###### Proof.

Since $F=\mathbb{F}_{p},dim_{{F}_{p}}(spanX)\in\\{1,2\\},$ and either
$T(G)=T_{1}(G)\times T_{2}(G)$ or $T(G)=T_{1}(G).$ If
$dim_{\mathbb{F}_{p}}(spanX)=2,$ then $n_{1}=n_{2}=1.$ So by Proposition 3.13
the condition for being Gorenstein is
$\lambda_{2}(g)^{p}\lambda_{1}(g)^{p}\lambda(g)=1,$ for each $g\in G.$ Since
$\lambda(g)=\lambda_{1}(g)^{-1}\lambda_{2}(g)^{-1},$ it follows that it is
equivalent to $\lambda_{2}(g)^{p-1}\lambda_{1}(g)^{p-1}=1.$ Now since
$\lambda_{1}(g),\lambda_{2}(g)\in\mathbb{F}_{p}\setminus\\{0\\}$ it follows
that $\lambda_{1}(g)^{p-1}=1=\lambda_{2}(g)^{p-1},$ and the condition is
fullfilled. If $dim_{\mathbb{F}_{p}}(spanX)=1,$ then $n=1,$ that is
$|T(G)|=p,$ so since $|gT(G)|=|\lambda_{1}(g)|,$ it follows from
$\lambda_{1}(g)\in\mathbb{F}_{p}\setminus\\{0\\}$ that $|gT(G)|$ divides
$|T(G)|-1,$ and again the Gorenstein property holds.

∎

##  4 Chapter(C): Preliminaries to Chapters D,E,F and G.

Let $G\subset GL(V)$ be a group and $W\subset V$ a $G-$submodule with
$dim_{F}W=dim_{F}V-1.$ With respesct to a fixed basis, each $g\in G$ is
represented by the matrix :

$g=\left(\begin{matrix}g_{11}&g_{12}&\ldots&g_{1n}\\\
0&g_{22}&\ldots&g_{2n}\\\ \vdots&\vdots&\ddots&\vdots\\\
0&g_{n2}&\ldots&g_{nn}\end{matrix}\right)$

######  Lemma 4.1.

Suppose $g\in G\cap SL(V).$ Then

$g^{-1}hg=\left(\begin{matrix}1&\indent\indent\indent
G_{11}(\alpha_{1},...,\alpha_{n-1})g|W\\\ 0&1&0&\ldots&\ldots&\ldots&&0\\\
0&0&1&0&\ldots&\ldots&0&0&\\\ \vdots&\vdots&0&\vdots&&&\vdots&\vdots\\\
0&\ldots&\vdots&0&\ldots&0&1&0\\\
0&0&0&\ldots&\ldots&\ldots&0&1\end{matrix}\right)=\\\ $

$=\left(\begin{matrix}1&&G_{11}(\alpha_{1},...,\alpha_{n-1})g|W&\\\ 0&\\\
\vdots&\indent\indent&I_{n-1}\\\ 0&\end{matrix}\right),where\\\ $

$\indent g|W=\left(\begin{matrix}g_{22}&g_{23}&\ldots&g_{2n}\\\
g_{32}&\ldots&&g_{3n}\\\ \vdots&\vdots&\ddots&\vdots\\\
g_{n2}&\ldots&&g_{nn}\end{matrix}\right),G_{11}=det(g|W),and$
$h=\left(\begin{matrix}1&\alpha_{1}&\ldots&\ldots&&&\ldots&\alpha_{n-1}\\\
0&1&0&\ldots&\ldots&\ldots&&0\\\ 0&0&1&0&\ldots&\ldots&0&0&\\\
\vdots&\vdots&0&&&&\vdots&\vdots\\\ 0&\ldots&\vdots&0&\ldots&0&1&0\\\
0&0&0&\ldots&\ldots&\ldots&0&1\end{matrix}\right)=\left(\begin{matrix}1&\alpha_{1}&\ldots&\alpha_{n-1}\\\
0&\\\ \vdots&\indent\indent&I_{n-1}\\\ 0&\end{matrix}\right).$

###### Proof.

$g^{-1}=\left(\begin{matrix}G_{11}&-G_{21}&\ldots&(-1)^{n+1}G_{n1}\\\ \\\
\indent\indent(-1)^{i+j}G_{ji}\end{matrix}\right)=\\\ $

$=\left(\begin{matrix}G_{11}&-G_{21}&\ldots&(-1)^{n+1}G_{n1}\\\ 0&\\\
\vdots&\indent\indent g^{-1}|W\\\ 0&\end{matrix}\right).$

Hence,

$g^{-1}\left(\begin{matrix}1&\alpha_{1}&\ldots&\ldots&&&\ldots&\alpha_{n-1}\\\
0&1&0&\ldots&\ldots&\ldots&&0\\\ 0&0&1&0&\ldots&\ldots&0&0&\\\
\vdots&\vdots&0&&&&\vdots&\vdots\\\ 0&\ldots&\vdots&0&\ldots&0&1&0\\\
0&0&0&\ldots&\ldots&\ldots&0&1\end{matrix}\right)g=g^{-1}hg=\\\ $
$\left(\begin{matrix}G_{11}&-G_{21}&\ldots&(-1)^{n+1}G_{n1}\\\ 0&\\\
\vdots&\indent\indent g^{-1}|W\\\ 0&\end{matrix}\right){\Large
h}\left(\begin{matrix}g_{11}&g_{12}&\ldots&g_{1n}\\\ 0&\\\
\vdots&\indent\indent g|W\\\ 0&\end{matrix}\right)=\\\ $
$\left(\begin{matrix}G_{11}&-G_{21}&\ldots&(-1)^{n+1}G_{n1}\\\ 0&\\\
\vdots&\indent\indent g^{-1}|W\\\
0&\end{matrix}\right)\left(\begin{matrix}g_{11},&(g_{12},...,g_{1n})+(\alpha_{1},...,\alpha_{n-1})g|W\\\
0&\\\ \vdots&\indent g|W\\\ 0&\end{matrix}\right)=\\\ $
$\left(\begin{matrix}G_{11}&-G_{21}&\ldots&(-1)^{n+1}G_{n1}\\\ 0&\\\
\vdots&\indent\indent g^{-1}|W\\\
0&\end{matrix}\right)\left(\begin{matrix}g_{11}&g_{12}&\ldots&g_{1n}\\\ 0&\\\
\vdots&\indent\indent g|W\\\ 0&\end{matrix}\right)+\\\ $

$\left(\begin{matrix}G_{11}&-G_{21}&\ldots&(-1)^{n+1}G_{n1}\\\ 0&\\\
\vdots&\indent\indent g^{-1}|W\\\
0&\end{matrix}\right)\left(\begin{matrix}0&\indent\indent(\alpha_{1},...,\alpha_{n-1})g|W\\\
0&0&\ldots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
0&0&\ldots&0\end{matrix}\right)=\\\ $ $I+\left(\begin{matrix}0&\indent\indent
G_{11}(\alpha_{1},...,\alpha_{n-1})g|W\\\ 0&0&\ldots&0\\\
\vdots&\vdots&\ddots&\vdots\\\ 0&0&\ldots&0\end{matrix}\right),$

as claimed.

∎

######  Corollary 4.2.

Suppose $g\in G$ is a transvection.Then

$g^{-1}hg=\left(\begin{matrix}1&\indent\indent\indent(\alpha_{1},...,\alpha_{n-1})g|W\\\
0&1&0&\ldots&\ldots&\ldots&&0\\\ 0&0&1&0&\ldots&\ldots&0&0&\\\
\vdots&\vdots&0&\vdots&&&\vdots&\vdots\\\ 0&\ldots&\vdots&0&\ldots&0&1&0\\\
0&0&0&\ldots&\ldots&\ldots&0&1\end{matrix}\right)=\left(\begin{matrix}1&&(\alpha_{1},...,\alpha_{n-1})g|W&\\\
0&\\\ \vdots&\indent\indent&I_{n-1}\\\ 0&\end{matrix}\right)$

###### Proof.

Since $g$ is transvection we have $g\in SL(V).$ Now $g|W$ is either the
identity or a transvection on $W.$ Hence in the $1^{st}$ case there is nothing
to prove and in the $2^{nd}$ case $det(g|W)=1$ and $g_{11}=1.$ Since
$G_{11}=det(g|W),$ the result follows from Lemma 4.1. ∎

Let $V$ be a $G-$module and $W$ a $G-$submodule. Recall that
$Fix_{G}(W):=\\{g\in G|g|W=id\\}.$

######  Proposition 4.3.

Suppose $G$ is generated by transvections and $dim_{F}W=dim_{F}V-1.$ Then
$S(V)^{Fix_{G}(W)}=S(W)[z],$ where
$z=v^{p^{t}}+q_{1}v^{p^{t-1}}+...+q_{t}v,q_{i}\in S(W)^{G|W}$ is homogenous of
degree $p^{t}-p^{t-i}$ and $|Fix_{G}(W)|=p^{t}.$

###### Proof.

We fix a basis $\\{v,w_{n-1},...,w_{1}\\}$ of $V$ and
$\\{w_{n-1},...,w_{1}\\}$ a basis of $W.$ The action of all elements of $G$
are given by matrices with respect to this basis. So
$Fix_{G}(W)=<\sigma_{1},...,\sigma_{t}>,$ where
$\\{\sigma_{1},...,\sigma_{t}\\}$ is a minimal generating set and

$\sigma_{i}=\left(\begin{matrix}1&\alpha_{1}^{(i)}&\ldots&\ldots&&&\ldots&\alpha_{n-1}^{(i)}\\\
0&1&0&\ldots&\ldots&\ldots&&0\\\ 0&0&1&0&\ldots&\ldots&0&0&\\\
\vdots&\vdots&0&&&&\vdots&\vdots\\\ 0&\ldots&\vdots&0&\ldots&0&1&0\\\
0&0&0&\ldots&\ldots&\ldots&0&1\end{matrix}\right)=\left(\begin{matrix}1&\alpha_{1}^{(i)}&\ldots&\alpha_{n-1}^{(i)}\\\
0&\\\ \vdots&\indent\indent&I_{n-1}\\\ 0&\end{matrix}\right),i=1,...,t.$

Hence $\\{(\alpha_{1}^{(i)},...,\alpha_{n-1}^{(i)})|i=1,...,t\\}$ is a
$\mathbb{F}_{p}-$linearly independent set. Since, by Corollary 4.2 ,
$Fix_{G}(W)$ is a normal subgroup of $G,$ then for a transvection $g\in
G,\\{g^{-1}\sigma_{1}g,...,g^{-1}\sigma_{t}g\\}$ is also a minimal generating
set of $Fix_{G}(W).$ Consequently,
$span_{\mathbb{F}_{p}}\\{(\alpha_{1}^{(i)},...,\alpha_{n-1}^{(i)})|i=1,...,t\\}=span_{\mathbb{F}_{p}}\\{(\alpha_{1}^{(i)},...,\alpha_{n-1}^{(i)})(g|W)|i=1,...,t\\}.\\\
$

Recall that $S(V)^{Fix_{G}(W)}=S(W)[z],$where $z$ is obtained via the
following procedure:

$\left\\{\begin{array}[]{lll}z_{1}=v^{p}-(\sigma_{1}-I)(v)^{p-1}v=v^{p}-(\alpha_{1}^{(1)}w_{n-1}+...+\alpha_{n-1}^{(1)}w_{1})^{p-1}v.\\\
z_{i}=z_{i-1}^{p}-(\sigma_{i}-I)(z_{i-1})^{p-1}z_{i-1},i=2,...,t.\\\
z=z_{t}.\par\par\par\end{array}\right.$

Similarly, set

$\sigma_{i,g}=\left(\begin{matrix}1&\indent\indent\indent(\alpha_{1}^{(i)},...,\alpha_{n-1}^{(i)})g|W\\\
0&1&0&\ldots&\ldots&\ldots&&0\\\ 0&0&1&0&\ldots&\ldots&0&0&\\\
\vdots&\vdots&0&\vdots&&&\vdots&\vdots\\\ 0&\ldots&\vdots&0&\ldots&0&1&0\\\
0&0&0&\ldots&\ldots&\ldots&0&1\end{matrix}\right)=\left(\begin{matrix}1&&(\alpha_{1}^{(i)},...,\alpha_{n-1}^{(i)})g|W&\\\
0&\\\ \vdots&\indent\indent&I_{n-1}\\\ 0&\end{matrix}\right),$

for $i=1,...,t.$ Hence
$z_{1}^{{}^{\prime}}=v^{p}-(\sigma_{1,g}-I)(v)^{p-1}v=v^{p}-(\beta_{1}^{(1)}w_{n-1}+...+\beta_{n-1}^{(1)}w_{1})^{p-1}v,$
where
$(\beta_{1}^{(i)},...,\beta_{n-1}^{(i)})=(\alpha_{1}^{(i)},...,\alpha_{n-1}^{(i)})(g|W).$
Hence
$g((\sigma_{i}-I)(v))=g(\alpha_{1}^{(i)}w_{n-1}+...+\alpha_{n-1}^{(i)}w_{1})=\beta_{1}^{(i)}w_{n-1}+...+\beta_{n-1}^{(i)}w_{1},$
for $i=1,...,t.$ If
$z_{i}^{{}^{\prime}}=(z_{i-1}^{{}^{\prime}})^{p}-[(\sigma_{i,g}-I)(z_{i-1}^{{}^{\prime}})]^{p-1}z_{i-1}^{{}^{\prime}}$
and $z^{{}^{\prime}}=z_{t}^{{}^{\prime}},$ then, by induction, starting with
$i=1,$ we have :

$\left\\{\begin{array}[]{lll}z_{i}=v^{p^{i}}+\gamma_{1i}v^{p^{i-1}}+...+\gamma_{ii}v,\gamma_{ji}\in
S(W),1\leq j\leq i.\\\
z_{i}^{{}^{\prime}}=v^{p^{i}}+\delta_{1i}v^{p^{i-1}}+...+\delta_{ii}v,\delta_{ji}\in
S(W),1\leq j\leq i.\\\ g(\gamma_{ji})=\delta_{ji},\forall
j=1,...,i.\par\par\end{array}\right.$

Since $S(V)^{Fix_{G}(W)}=S(W)[z]=S(W)[z^{{}^{\prime}}]$ and
$z,z^{{}^{\prime}}$ are monic polynomials in $v$ with $0$ as the free
coefficient, it follows that $z^{{}^{\prime}}=z$ and consequently
$\delta_{jt}=\gamma_{jt},$ for $j=1,..,t.$ Therefore
$g(\gamma_{jt})=\gamma_{jt}$ for $1\leq j\leq t,$ and all transvections $g\in
G.$ Hence $q_{j}:=\gamma_{jt}\in S(W)^{G|W}$ for $j=1,...,t.$ ∎

Let $V$ be a $F-$finite dimensional $G$-module,then $V^{*}=Hom_{F}(V,F)$ is a
$G-$module by the action $(g\cdot f)(v)=f(g^{-1}v),$ for $f\in V^{*},g\in G$
and $v\in V.$ Let $W\subset V$ be a $G-$submodule. Set $W^{\bot}=\\{f\in
V^{*}|f(W)=0\\}.$ Clearly $W^{\bot}$ is a $G-$submodule of $V^{*}$ and
$dim_{F}W^{\bot}=dim_{F}V-dim_{F}W.$

######  Lemma 4.4.

$V^{*}/W^{\bot}\cong W^{*}$ as $G-$modules.

###### Proof.

Consider the map $\pi:V^{*}\longrightarrow W^{*}$ given by $\pi(f)=f|W,$ that
is the restriction of $f$ to $W.Ker(\pi)=\\{f\in V^{*}|f|W=0\\}=W^{\bot}.$
Hence the induced map $\bar{\pi}:V^{*}/W^{\bot}\longrightarrow W^{*}$ is
injective. Since
$dim_{F}V^{*}-dim_{F}W^{\bot}=dim_{F}V-dim_{F}W^{\bot}=dim_{F}W=dim_{F}W^{*}$
we get that $\bar{\pi}$ is an isomorphism. It is also easy to cheak that $\pi$
and hence $\bar{\pi}$ are $G-$module maps. ∎

We also have the following results.

######  Corollary 4.5.

$V/W\cong(W^{\bot})^{*}.$

###### Proof.

$(W^{\bot})^{\bot}=\\{v\in V|g(v)=0,\forall g\in W^{\bot}\\}.$ Hence
$W\subset(W^{\bot})^{\bot}.$ Since
$dim_{F}(W^{\bot})^{\bot}=dim_{F}V^{*}-dim_{F}W^{\bot}=dim_{F}V^{*}-(dim_{F}V-dim_{F}W)=dim_{F}W$
we get that $(W^{\bot})^{\bot}=W.$ We now apply Lemma 4.4 to the pair
$W^{\bot}\subset V^{*}.$ Hence
$V/W=(V^{*})^{*}/(W^{\bot})^{\bot}\cong(W^{\bot})^{*}.$

∎

######  Corollary 4.6.

$\\{g\in G|g(V)\subseteq W\\}=\\{g\in G|(g|W^{\bot})=0\\}.$

###### Proof.

This is a consequence of Corollary 4.5, since $\\{g\in G|g(V)\subseteq
W\\}=Ann_{G}V/W=Ann_{G}(W^{\bot})^{*}=Ann_{G}W^{\bot}=\\{g\in
G|g|W^{\bot}=0\\}.$ ∎

######  Corollary 4.7.

Let $W\subset V$ be a $G-$submodule. Then

  1. 1.

$W$ is faithful $G-$module iff $W^{*}$ is a faithful $G-$module iff
$V^{*}/W^{\bot}$ is a faithful $G-$module.

  2. 2.

$W$ is an irreducible $G-$module iff $W^{*}$ is an irreducible $G-$module iff
$V^{*}/W^{\bot}$ is an irreducible $G-$module.

###### Proof.

This follows from Lemma 4.4 . ∎

##  5 Chapter($D$): $V$ is indecomposable,contains a unique $2-$ dimensional
$G-$ submodule, but no $1-$ dimensional $G-$ submodule.

In this chapter we consider $V,$ a reducible and indecomposable $G-module$
which has a unique $G$-submodule $W$ with $dim_{F}W=2,$ and $V$ contains no
$1-$dimensional $G-$submodules, hence $W$ is an irreducible $G-module$.

######  Lemma 5.1.

One of the following holds:

  1. 1.

$W$ is a reducible $T(G)-$ module and $\sigma|W=id,\forall\sigma\in T(G).$

  2. 2.

$W$ is an irreducible $T(G)-$ module.

###### Proof.

Assume that $W$ is a reducible $T(G)-$ module.Then there exists $w\in W$ such
that $Fw$ is a $T(G)-$ module, and it follows,using Lemma 2.2, that
$\sigma(w)=w$ $\forall\sigma\in T(G).$ Let $g\in G,$ since $T(G)\lhd G,$
$(g^{-1}\sigma g)=\tau$ for some $\tau\in T(G),$ hence
$\sigma(g(w))=g(\tau(w))=g(w).$ If $g(w)\not\in Fw$ then $\\{w,g(w)\\}$ is a
basis for $W$ and $\sigma$ acts trivially on $W$ $\forall\sigma\in T(G)$ so
(1) is established. On the other hand if $g(w)\in Fw$ $\forall g\in G,$ then
$Fw$ is $1-$dimensional $G-$module.This contradicts the assumption that $V$
contains no $1-$dimensional $G-$modules.

∎

For $W$ an irreducible $T(G)-$ module, the transvection group $T(G)|W$ had
been classified by Kantor whenever $T(G)|W$ is an irreducible primitive linear
group, see e.g [7, Theorem 1.5]. So one of the following holds in this case :

  1. (i)

$T(G)|W=SL(2,\mathbb{F}_{q}),$ where $p|q.$

  2. (ii)

$T(G)|W\cong SL(2,\mathbb{F}_{5}),T(G)|W\subset SL(2,\mathbb{F}_{9})$ and
$\mathbb{F}_{9}\subseteq F.$

Let $\\{v,w_{2},w_{1}\\}$ be a basis of $V=F^{3}$ where $\\{w_{1},w_{2}\\}$ is
a basis of $W.$ So every $g\in G$ is represented with respect to this basis,
by the matrix:

$(5.a)\left(\begin{array}[]{ccc}\lambda(g)&\delta_{12}(g)&\delta_{13}(g)\\\
0&\delta_{22}(g)&\delta_{23}(g)\\\
0&\delta_{32}(g)&\delta_{33}(g)\end{array}\right).$

We shall make the following assumption:

######  Assumption 5.2.

Let $\alpha:T(G)\longrightarrow(T(G)|W)$ be the map defined by
$\alpha(\tau)=(\tau|W)$ and let $N:=ker(\alpha).$ Then $S(V)^{T(G)}$ is a
polynomial ring, $S(V)^{T(G)}=S(W)^{T(G)}[n]=F[n,m_{1},m_{2}]$ where
$m=(m_{1},m_{2})$ is the homogenous maximal ideal of
$S(W)^{T(G)}=F[m_{1},m_{2}],$ and $S(V)^{N}=S(W)[n].$

The next result provides a necessary and sufficient condition on $G$ to make
$S(V)^{G}$ into a Gorenstein ring .

######  Proposition 5.3.

We keep the above notation and Assumption 5.2. Let $G\subset SL(V)$ be a
finite group.Then the following are equivalent:

  1. 1.

$S(V)^{G}$ is Gorenstein.

  2. 2.

$det(g|m/m^{2})=\lambda(g)^{1-d}det(g|W),$ where
$d=deg(n)=\frac{|T(G)|}{deg(m_{1})\cdot deg(m_{2})}$ and $m=(m_{1},m_{2}).$

###### Proof.

Suppose firstly that $S(V)^{G}$ is Gorenstein.Then by [Theorem 1.2]
$det(g|p/p^{2})=1,$ where $p=(n,m_{1},m_{2})$ is the homogenous maximal ideal
of $S(V)^{T(G)}.$ Clearly $g$ acts linearly on $p/p^{2}$ and is represented by
the matrix:

$\left[\begin{array}[]{c|c}\lambda(g)^{d}&\,*\quad*\\\
\hline\cr\begin{matrix}0\\\ 0\end{matrix}&g|m/m^{2}\end{array}\right]$

So $det(g|p/p^{2})=\lambda(g)^{d}det(g|m/m^{2})=1.$ Since $G\subset
SL(V),\lambda(g)det(g|W)=1.$ Consequently,
$\lambda(g)^{d}det(g|m/m^{2})=\lambda(g)det(g|W),$ hence
$det(g|m/m^{2})=\lambda(g)^{1-d}det(g|W).$ Suppose now that
$det(g|m/m^{2})=\lambda(g)^{1-d}det(g|W).$ Hence,using $G\subset SL(V),$
$\lambda(g)^{d}det(g|m/m^{2})=\lambda(g)det(g|W)=1.$ So $det(g|p/p^{2})=1,$
then by [Theorem 1.2] $S(V)^{G}$ is Gorenstein .

∎

######  Theorem 5.4.

Suppose $W\subset V$ is an irreducible $G-$submodule and $\sigma|W=Id,$ for
all $\sigma\in T(G)$ (as in Lemma 5.1 (1)). Then $S(V)^{G}$ is Gorenstein iff
$\lambda(g)^{1-d}=1.$ Consequently if $V=\mathbb{F}_{p}^{3}$ then $S(V)^{G}$
is always Gorenstein.

###### Proof.

By assumption $T(G)=ker(\alpha),$ where $\alpha:T(G)\longrightarrow(T(G)|W)$
is the restriction map. Therefore by Proposition 4.3 $S(V)^{T(G)}=S(W)[n],$
where $d=deg(n)=p^{t}.$ Hence Assumption 5.2 holds here with
$m_{1}=w_{1},m_{2}=w_{2},m=(w_{1},w_{2}),p=(n,w_{1},w_{2}).$ Since
$m/m^{2}=Fw_{1}+Fw_{2}=W$ we have that $det(g|m/m^{2})=det(g|W),$ implying by
Proposition 5.3 that $S(V)^{G}$ is Gorenstein if and pnly if
$\lambda(g)^{1-d}=1.$ Consequently, if $V=\mathbb{F}_{p}^{3}$ then , since
$\lambda(g)\in\mathbb{F}_{p},\lambda(g)^{1-d}=1.$

∎

The next result applies in case (i), where $W$ is an irreducible primitive
$T(G)-$module.

######  Proposition 5.5.

Suppose $G\subset SL(V)$ is a finite group with $T(G)|W=SL(2,\mathbb{F}_{q})$
,where $q=p^{s},$ $\mathbb{F}_{q}\subseteq F$ and satisfies Assumption 5.2.
Let $V=F^{3}$ be the natural $SL(3,F)-module.$Let $g\in G$ be arbitrary and
set $g|W=\bigl{(}\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix}\bigr{)}$ with
respect to the $\mathbb{F}_{q}-$ basis $\\{w_{1},w_{2}\\}.$Then $S(V)^{G}$ is
Gorenstein if and only if $a^{q^{2}-1}=\lambda(g)^{1-d}.$

###### Proof.

The computation in Proposition 2.4 shows that the matrix representing
$g|m/m^{2}$ is $\bigl{(}\begin{smallmatrix}a^{q-1}det(g|W)&0\\\
0&a^{q^{2}-q}\end{smallmatrix}\bigr{)},$ hence
$det(g|m/m^{2})=a^{q^{2}-1}det(g|W).$ Hence,by Proposition 5.3, $S(V)^{G}$ is
Gorenstein if and only if $a^{q^{2}-1}det(g|W)=det(g|W)\lambda(g)^{1-d}.$ Thus
$S(V)^{G}$ is Gorenstein if and only if $a^{q^{2}-1}=\lambda(g)^{1-d}.$

∎

######  Remark 5.6.

For $W$ an irreducible and imprimitive $T(G)-module,T(G)|W$ is a monomial
subgroup (e.g.[13]).In this case, if $charF\neq 2$ , then by [3, Lemma 3.9.]
$T(G)|W={1},$ which contradicts the assumption that $W$ is an irreducible
$T(G)-$module. On the other hand, if $p=2,$ then $T(G)|W=<\sigma>$,where
$\sigma=\bigl{(}\begin{smallmatrix}0&1\\\ 1&0\end{smallmatrix}\bigr{)}.$ This
implies that $(w_{2}+w_{1})\in S(V)^{G}.$ Hence $\mathbb{F}_{2}(w_{2}+w_{1})$
is $G-$module,which contradicts the assumption that $W$ is an irreducible
$G-$module.Hence we deal only with the case when $W$ is an irreducible
primitive $T(G)-$module.

######  Proposition 5.7.

Suppose $F=\mathbb{F}_{p}$ and $W$ is an irreducible $T(G)-$module. Then
Assumption 5.2 holds for $G.$

###### Proof.

Clearly

$N=ker(\alpha)\subseteq\left(\begin{array}[]{ccc}1&\mathbb{F}_{p}&\mathbb{F}_{p}\\\
0&1&0\\\ 0&0&1\end{array}\right).$

Since $F=\mathbb{F}_{p},$ we have by Remark 5.6 that $W$ is a primitive
$T(G)-$module, and therefore $T(G)|W=SL(2,\mathbb{F}_{p}).$ If $N\neq 1$ then
there exists

$\sigma:=\left(\begin{array}[]{ccc}1&\alpha&\beta\\\ 0&1&0\\\
0&0&1\end{array}\right),(\alpha,\beta)\neq(0,0),\sigma\in N.$

Consequently by Corollary 4.2 we have :

$\left(\begin{array}[]{ccc}1&\indent&(\alpha,\beta)SL(2,\mathbb{F}_{p})\\\
0&1&0\\\ 0&0&1\end{array}\right)\subseteq N,$

and therefore since
$(\alpha,\beta)SL(2,\mathbb{F}_{p})=(\mathbb{F}_{p},\mathbb{F}_{p})$ we
conclude that

$N=\left(\begin{array}[]{ccc}1&\mathbb{F}_{p}&\mathbb{F}_{p}\\\ 0&1&0\\\
0&0&1\end{array}\right).$

Let

$g=\left(\begin{array}[]{ccc}1&\delta_{12}(g)&\delta_{13}(g)\\\
0&\delta_{22}(g)&\delta_{23}(g)\\\
0&\delta_{32}(g)&\delta_{33}(g)\end{array}\right)\in T(G),$

be an arbitrary element.Then

$\left(\begin{array}[]{ccc}1&\delta_{12}(g)&\delta_{13}(g)\\\
0&\delta_{22}(g)&\delta_{23}(g)\\\
0&\delta_{32}(g)&\delta_{33}(g)\end{array}\right)\left(\begin{array}[]{ccc}1&-\delta_{12}(g)&-\delta_{13}(g)\\\
0&1&0\\\ 0&0&1\end{array}\right)=\left(\begin{array}[]{ccc}1&0&0\\\
0&\delta_{22}(g)&\delta_{23}(g)\\\
0&\delta_{32}(g)&\delta_{33}(g)\end{array}\right)\in T(G)N=T(G).$

Consequently,since
$T(G)|W=\\{g|W=\bigl{(}\begin{smallmatrix}\delta_{22}(g)&\delta_{23}(g)\\\
\delta_{32}(g)&\delta_{33}(g)\end{smallmatrix}\bigr{)}|g\in T(G)\\}$ and
$SL(2,\mathbb{F}_{p})=T(G)|W,$ then

$H:=\left(\begin{array}[]{c|c}1&\,0\quad 0\\\ \hline\cr\begin{matrix}0\\\
0\end{matrix}&SL(2,\mathbb{F}_{p})\end{array}\right)\subseteq T(G).$

Moreover $H\cap N=\\{1\\}.$Hence
$|HN|=|H|\cdot|N|=(p^{2}-p)\cdot(p+1)\cdot|N|=|T(G)|W|\cdot|N|=\frac{|T(G)|}{|N|}\cdot|N|=|T(G)|,$
implying that $T(G)=HN.$ Let $\\{v,w_{2},w_{1}\\}$ be the basis representing
the matrices above,where $W=span_{\mathbb{F}_{p}}\\{w_{1},w_{2}\\}.$ Now by
Proposition 4.3 , there exists $n=v^{p^{2}}+q_{1}v^{p}+q_{2}v,q_{i}\in
S(W)^{T(G)|W},$ with $S(V)^{N}=S(W)[n].$ Since $h(v)=v$ for each $h\in H,$ it
follows that $n\in S(V)^{H},$ and therefore
$S(V)^{T(G)}=S(W)[n]^{T(G)/N}=S(W)^{H}[n]=S(W)^{SL(2,\mathbb{F}_{p})}[n]=F[d_{1},d_{2},n],$
where $\\{d_{1},d_{2}\\}$ are the Dickson invariants of degrees
$\\{p+1,p^{2}-p\\}$ respectively.
Suppose $N=\\{1\\}.$ Hence $T(G)\cong T(G)|W=SL(2,\mathbb{F}_{p}).$ Now
$T(G)=<\phi_{1},...,\phi_{k}>,$ where $\phi_{i},$ is a transvection, for
$i=1,...,k.$ Moreover $k\geq 2$ (since $T(G)\cong SL(2,\mathbb{F}_{p})$), and
we may assume that at least two out of $\\{\phi_{1}|W,...,\phi_{k}|W\\}$ do
not commute, say $\\{\phi_{1}|W,\phi_{2}|W\\}.$ Therefore
$<\phi_{1}|W,\phi_{2}|W>\cong SL(2,\mathbb{F}_{p}),$ implying that
$\phi_{i}|W\in<\phi_{1}|W,\phi_{2}|W>,$ for each $i\geq 3.$ Therefore, since
$\alpha$ is injective, we have: $\phi_{i}\in<\phi_{1},\phi_{2}>,$ for $i\geq
3,$ that is $T(G)=<\phi_{1},\phi_{2}>.$ Hence $ker(\phi_{1}-I)\cap
ker(\phi_{2}-I):=Fv,$ is a one dimensional trivial $T(G)-$module. Moreover,
since $W$ is an irrreducible $T(G)-$module, $v\not\in W,$ hence $V=Fv\oplus W$
as $T(G)-$modules. Consequently with respect to the basis
$\\{v,w_{1},w_{2}\\}$ (where $\\{w_{1},w_{2}\\}$ is any basis of $W$), $T(G)$
is represent by matrices of the form :

$H:=\left(\begin{array}[]{c|c}1&\,0\quad 0\\\ \hline\cr\begin{matrix}0\\\
0\end{matrix}&SL(2,\mathbb{F}_{p})\end{array}\right),$

and since $|H|=|SL(2,\mathbb{F}_{p})|=|T(G)|W|=|T(G)|$ we have $H=T(G).$ Hence
$S(V)^{T(G)}=S(W)^{SL(2,\mathbb{F}_{p})}[v]=F[d_{1},d_{2},v]$ and we take
$n=v.$

∎

######  Theorem 5.8.

Suppose $G\subset SL(3,\mathbb{F}_{p})$ and $W$ is an irreducible
$T(G)-$submodule of $V=\mathbb{F}_{p}^{3}.$ Then $S(V)^{G}$ is Gorenstein.

###### Proof.

By Proposition 5.5 and Proposition 5.7, we have to check that
$a^{p^{2}-1}=\lambda(g)^{1-d},$ where $d=deg(n)\in\\{1,p^{2}\\}$ and
$g|W=\bigl{(}\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix}\bigr{)}.$ If $N=1$
then $d=1,$ and if $N\neq 1,$ then $d=p^{2}.$ Since
$a,\lambda(g)\in\mathbb{F}_{p}$ we have $a^{p-1}=\lambda(g)^{p-1}=1.$ Hence
$a^{p^{2}-1}=1=\lambda(g)^{1-d}$ for $d\in\\{1,p^{2}\\}.$ ∎

######  Corollary 5.9.

Suppose $G\subset SL(3,\mathbb{F}_{p}),W\subset V=\mathbb{F}_{p}^{3}$ is an
irreducible $G-$submodule and $dim_{\mathbb{F}_{p}}W=2.$ Then $S(V)^{G}$ is
Gorenstein.

###### Proof.

This is a combination of Lemma 5.1, Theorem 5.4 and Theorem 5.8 ∎

##  6 Chapter($E$): $V$ is indecomposable,contains a unique $1-$ dimensional
$G-$ submodule $U,$ but no $2-$ dimensional $G-$ submodule.

In this chapter we consider $V,$ a $3-$dimensional reducible and
indecomposable $G-module$ which has a unique $1-$dimensional $G$-submodule
$U=Fv_{0}$ and $V$ contains no $2-$dimensional $G-$submodules.Hence $W=V/U$ is
an irreducible $G-$ module. We shall exclusively consider here the case
$F=\mathbb{F}_{p}.\\\ $ Let $\\{w_{1},w_{2},v_{0}\\}$ be any basis of
$V=F^{3}.$ So every $g\in G$ is represented with respect to this basis, by the
matrix:

$(6.a)\indent
g=\left(\begin{array}[]{ccc}\delta_{11}(g)&\delta_{12}(g)&\delta_{13}(g)\\\
\delta_{21}(g)&\delta_{22}(g)&\delta_{23}(g)\\\
0&0&\lambda(g)\end{array}\right).$

Let $\alpha:T(G)\longrightarrow(T(G)|W)$ be the map which is defined by
$\alpha(\tau)=\tau|W,$ hence $ker(\alpha)=\\{\sigma\in T(G)|\sigma(x)-x\in
Fv_{0},\forall x\in V\\}.$Consequently,

$ker(\alpha)\subseteq\left(\begin{array}[]{ccc}1&0&*\\\ 0&1&*\\\
0&0&1\end{array}\right).$

We shall firstly consider the case that $W=V/\mathbb{F}_{p}v_{0}$ is an
irreducible $T(G)-module$ (on top of being $G-$irreducible).Since
$F=\mathbb{F}_{p},$it follows from Remark 5.6, that
$T(G)|W=SL(2,\mathbb{F}_{p}).$

######  Proposition 6.1.

Suppose $V=\mathbb{F}_{p}^{3},$ and $V/\mathbb{F}_{p}v_{0}$ is an unfaithful,
irreducible $T(G)-$module.Then $S(V)^{T(G)}$ is a polynomial ring.

###### Proof.

Clearly $T(G)\subset SL(V)$.Let $N:=ker(\alpha),$ then by assumption $N\neq
1.$ Set $U=\mathbb{F}_{p}v_{0}.$ Then by Corollary 4.6 ,
$N=Ann_{T(G)}V/U=Ann_{T(G)}U^{\bot}.\\\ V/U$ is an irreducible $T(G)-$module
hence by Corollary 4.7 , $U^{\bot}$ is a $2-$dimensional irreducible $T(G)-$
submodule of $V^{*}.$ Therefore by the proof of the early part of Proposition
5.7 , $|N|=p^{2}.$

Hence,

$N=\left(\begin{array}[]{ccc}1&0&\mathbb{F}_{p}\\\ 0&1&\mathbb{F}_{p}\\\
0&0&1\end{array}\right),$

with generators

$\\{\left(\begin{array}[]{ccc}1&0&1\\\ 0&1&0\\\
0&0&1\end{array}\right),\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&1\\\
0&0&1\end{array}\right)\\}.$

Let

$g=\left(\begin{array}[]{ccc}\delta_{11}(g)&\delta_{12}(g)&\delta_{13}(g)\\\
\delta_{21}(g)&\delta_{22}(g)&\delta_{23}(g)\\\ 0&0&1\end{array}\right),$

be a transvection in $T(G).$Then

$\left(\begin{array}[]{ccc}1&0&-\delta_{13}(g)\\\ 0&1&-\delta_{23}(g)\\\
0&0&1\end{array}\right)\left(\begin{array}[]{ccc}\delta_{11}(g)&\delta_{12}(g)&\delta_{13}(g)\\\
\delta_{21}(g)&\delta_{22}(g)&\delta_{23}(g)\\\
0&0&1\end{array}\right)=\left(\begin{array}[]{ccc}\delta_{11}(g)&\delta_{12}(g)&0\\\
\delta_{21}(g)&\delta_{22}(g)&0\\\ 0&0&1\end{array}\right)\in NT(G)=T(G).$

Since $\\{\bigl{(}\begin{smallmatrix}\delta_{11}(g)&\delta_{12}(g)\\\
\delta_{21}(g)&\delta_{22}(g)\end{smallmatrix}\bigr{)}|g\in
T(G)\\}=SL(2,\mathbb{F}_{p})=T(G)|W$ then

$H:=\left(\begin{array}[]{c|c}SL(2,\mathbb{F}_{p})&\begin{matrix}0\\\
0\end{matrix}\\\
\hline\cr\begin{matrix}0&0\end{matrix}&1\end{array}\right)\subseteq T(G).$

Since $|T(G)|=|SL(2,\mathbb{F}_{p})|\cdot|N|=(p+1)(p^{2}-p)p^{2},$ then
$T(G)=<H,N>.$ We have $S(V)^{N}=\mathbb{F}_{p}[v_{0},a,b]$ where
$a=w_{1}^{p}-v_{0}^{p-1}w_{1}$ and $b=w_{2}^{p}-v_{0}^{p-1}w_{2}.$ Consider
$\\{\sigma,\tau\\},$ generators of $H,$ where

$\tau=\left(\begin{array}[]{ccc}1&1&0\\\ 0&1&0\\\
0&0&1\end{array}\right),\sigma=\left(\begin{array}[]{ccc}1&0&0\\\ 1&1&0\\\
0&0&1\end{array}\right).$

Then $\tau(b)=b,\tau(a)=a+b,\sigma(a)=a,\sigma(b)=a+b.$ Hence
$A=a^{p}b-b^{p}a$ and $B=\frac{a^{p^{2}}b-b^{p^{2}}a}{A},$ are Dickson
invariants of $({S(V)^{N}})^{H}=S(V)^{T(G)}.$So $deg(A)\cdot deg(B)\cdot
deg(v_{0})=p(p+1)\cdot p(p^{2}-p)=|T(G)|.$ Hence by [4, Theorem 3.7.5]
$S(V)^{T(G)}=\mathbb{F}_{p}[v_{0},A,B]$ is a polynomial ring.

∎

######  Lemma 6.2.

Suppose $V=\mathbb{F}_{p}^{3},$ and $W=V/\mathbb{F}_{p}v_{0}$ is an
irreducible $T(G)-$module.
Then one of the following holds.

  1. 1.

$W$ is an unfaithful irreducible $T(G)-$module and $T(G)=<H,N>.$

  2. 2.

$W$ is a faithful irreducible $T(G)-$module and $T(G)=H,$ where $H$ is defined
in Proposition 6.1.

###### Proof.

If item (1) holds then by Proposition 6.1 the result follows.So we may assume
that $W$ is a faithful irreducible $T(G)-$module. Let
$T(G)=<\phi_{1},...,\phi_{k}>,$ so $(T(G)|W)=<\phi_{1}|W,...,\phi_{k}|W>.$ If
$\phi_{i}|W=Id$ then $\phi_{i}\in N=1.$ Hence $\phi_{i}|W$ are non trivial
transvections for $i=1,...,k.$ Since $W$ is an irreducible $T(G)-$module then
there exist at least two transvections that do not commute ,say $(\phi_{1}|W)$
and $(\phi_{2}|W).$ But then $W$ is an irreducible
$<\phi_{1}|W,\phi_{2}|W>-$module.Hence $<\phi_{1}|W,\phi_{2}|W>\cong
SL(2,\mathbb{F}_{p}).$Since $|(T(G)|W)|=|SL(2,\mathbb{F}_{p})|$ then
$<\phi_{1}|W,\phi_{2}|W>=(T(G)|W).$ Therefore, $\forall\phi_{i}\in T(G)$ with
$i\geq 3,(\phi_{i}|W)$ is generated by $(\phi_{1}|W)$ and $(\phi_{2}|W).$ So
$\phi_{i}=h\cdot Z$ where $h\in<\phi_{1},\phi_{2}>$ and $Z\in N=ker(\alpha).$
Since $N=1,\phi_{i}=h.$ Therefore $T(G)=<\phi_{1},\phi_{2}>.\\\ \\\ $ Consider
$U^{\bot}\subset V^{*}.$ Since $W=V/U$ is a faithful and irreducible
$T(G)-$module it follows from Corollary 4.5,Corollary 4.7 that $U^{\bot}$ is a
faithful and irreducible $T(G)-$submodule of $V^{*}.\\\ $ Consider
$ker(\phi_{1}-I)\cap ker(\phi_{2}-I)\subset V^{*}.$ It is a $1-$dimensional
subspace (since $\phi_{1},\phi_{2}$ are transvections on $V^{*}$). Let
$Fg=ker(\phi_{1}-I)\cap ker(\phi_{2}-I)\subset V^{*}.$ Since
$T(G)=<\phi_{1},\phi_{2}>$ it follows that $Fg$ is a trivial $T(G)-$submodule
of $V^{*}.$ Since $U^{\bot}$ is irreducible it follows that $Fg\oplus
U^{\bot}=V^{*},$ a direct sum decomposition of $V^{*}$ as $T(G)-$modules.Since
$U^{\bot}$ is irreducible and faithful we have , as in Proposition 5.7, a
representation of $T(G)$ as matrices of the form :

$\left(\begin{array}[]{c|c}1&\,0\quad 0\\\ \hline\cr\begin{matrix}0\\\
0\end{matrix}&SL(2,\mathbb{F}_{p})\end{array}\right),$

with respect to the basis $\\{g,h_{1},h_{2}\\},$ where $\\{h_{1},h_{2}\\}$ is
any basis of $U^{\bot}.$ Let $\\{x,w_{1},w_{2}\\}$ be the dual basis of
$\\{g,h_{1},h_{2}\\}.$ So
$g(x)=1,h_{1}(x)=h_{2}(x)=0,h_{i}(w_{j})=\delta_{ij}$ and
$g(w_{1})=g(w_{2})=0.$ Hence $x\in(Fh_{1}+Fh_{2})^{\bot}=(U^{\bot})^{\bot}=U$
and $\mathbb{F}_{p}x=\mathbb{F}_{p}v_{0}.\\\ $ $(Fg)^{\bot}=Fw_{1}+Fw_{2}$ is
a $2-$dimensional $T(G)-$submodule of $V$ and
$(Fg)^{\bot}\oplus(U^{\bot})^{\bot}=V.$ Hence the matrices representing $T(G)$
with respect to the basis $\\{w_{1},w_{2},v_{0}\\}$ are

$H:=\left(\begin{array}[]{c|c}SL(2,\mathbb{F}_{p})&\begin{matrix}0\\\
0\end{matrix}\\\ \hline\cr\begin{matrix}0&0\end{matrix}&1\end{array}\right).$

∎

The following theorem deals with the special case,when
$W=V/\mathbb{F}_{p}v_{0}$ is an irreducible primitive $T(G)-$module and
$F=\mathbb{F}_{p}.$ Since $F=\mathbb{F}_{p},$ by [7, Theorem 1.5]
$(T(G)|W)=SL(2,\mathbb{F}_{p})$ and $p\neq 2.$

######  Theorem 6.3.

Suppose that $G\subset SL(3,\mathbb{F}_{p}),W=V/\mathbb{F}_{p}v_{0}$ is an
irreducible $T(G)-$module. Then $S(V)^{G}$ is Gorenstein.

###### Proof.

Assume firstly that $W$ is a faithful irreducible $T(G)$-module.By Lemma 6.2
(2), $S(V)^{T(G)}=S(V)^{H}=\mathbb{F}_{p}[u,y,v_{0}]$ is a polynomial ring
where
$u=w_{1}w_{2}^{p}-w_{2}w_{1}^{p},y=\frac{w_{1}w_{2}^{p^{2}}-w_{2}w_{1}^{p^{2}}}{u}$
are the Dickson’s invariants.Now for $g\in G$ since
$gT(G)g^{-1}=T(G),\mathbb{F}_{p}[u,y,v_{0}]$ is $g$-stable and since $g$
preserve degrees we get $g(u)=\alpha u+\beta v_{0}^{p+1}$ with
$\alpha,\beta\in\mathbb{F}_{p}.$ By Fermat’ little theorem and formula (6.a)
we get:

$g(u)=(\delta_{11}(g)w_{1}+\delta_{12}(g)w_{2}+\delta_{13}(g)v_{0})(\delta_{21}(g)w_{1}^{p}+\delta_{22}(g)w_{2}^{p}+\delta_{23}(g)v_{0}^{p})-(\delta_{21}(g)w_{1}+\delta_{22}(g)w_{2}+\delta_{23}(g)v_{0})(\delta_{11}(g)w_{1}^{p}+\delta_{12}(g)w_{2}^{p}+\delta_{13}(g)v_{0}^{p})=\delta_{11}(g)\delta_{22}(g)(w_{1}w_{2}^{p}-w_{2}w_{1}^{p})+\delta_{11}(g)\delta_{23}(g)(w_{1}v_{0}^{p}-v_{0}w_{1}^{p})+\delta_{12}(g)\delta_{23}(g)(w_{2}v_{0}^{p}-v_{0}w_{2}^{p})+\delta_{13}(g)\delta_{22}(g)(w_{2}^{p}v_{0}-w_{2}v_{0}^{p})+\delta_{12}(g)\delta_{21}(g)(w_{2}w_{1}^{p}-w_{1}w_{2}^{p})+\delta_{13}(g)\delta_{21}(g)(w_{1}^{p}v_{0}-w_{1}v_{0}^{p})=(\delta_{11}(g)\delta_{22}(g)-\delta_{12}(g)\delta_{21}(g))(w_{1}w_{2}^{p}-w_{2}w_{1}^{p})+(\delta_{11}(g)\delta_{23}(g)-\delta_{13}(g)\delta_{21}(g))(w_{1}v_{0}^{p}-v_{0}w_{1}^{p})+(\delta_{12}(g)\delta_{23}(g)-\delta_{13}(g)\delta_{22}(g))(w_{2}v_{0}^{p}-v_{0}^{p}w_{2}).$

Now by comparing coefficients of $g(u)$ we get:

$\textbf{(6.b)}\indent\left\\{\begin{array}[]{llll}\alpha=\delta_{11}(g)\delta_{22}(g)-\delta_{12}(g)\delta_{21}(g)\\\
\beta=0\\\ \delta_{11}(g)\delta_{23}(g)-\delta_{13}(g)\delta_{21}(g)=0\\\
\delta_{12}(g)\delta_{23}(g)-\delta_{13}(g)\delta_{22}(g)=0\par\par\end{array}\right.$

Since $G\subset
SL(V),\lambda(g)^{-1}=\delta_{11}(g)\delta_{22}(g)-\delta_{12}(g)\delta_{21}(g),$then
$g(u)=\lambda(g)^{-1}u.$ Similarly,$g(y)=\gamma y+q(u,v_{0}),$ where
$\gamma\in\mathbb{F}_{p}$ and $q(u,v_{0})$ is a homogenous polynomial (in
$u,v_{0}$) with $deg(q)=deg(y)=p^{2}-p.$

By Fermat’ little theorem , formula (6.a) and (6.b) we get:

$g(w_{1}w_{2}^{p^{2}}-w_{2}w_{1}^{p^{2}})=(\delta_{11}(g)\delta_{22}(g)-\delta_{12}(g)\delta_{21}(g))(w_{1}w_{2}^{p^{2}}-w_{2}w_{1}^{p^{2}})+(\delta_{11}(g)\delta_{23}(g)-\delta_{13}(g)\delta_{21}(g))(w_{1}v_{0}^{p^{2}}-v_{0}w_{1}^{p^{2}})+(\delta_{12}(g)\delta_{23}(g)-\delta_{13}(g)\delta_{22}(g))(w_{2}v_{0}^{p^{2}}-v_{0}^{p^{2}}w_{2})=\lambda(g)^{-1}(w_{1}w_{2}^{p^{2}}-w_{2}w_{1}^{p^{2}})+0+0,$

using
$\delta_{11}(g)\delta_{22}(g)-\delta_{12}(g)\delta_{21}(g)=\lambda(g)^{-1}.$
Since $uy=w_{1}w_{2}^{p^{2}}-w_{2}w_{1}^{p^{2}}$ we get
$g(y)=\frac{g(w_{1}w_{2}^{p^{2}}-w_{2}w_{1}^{p^{2}})}{g(u)}=\frac{\lambda(g)^{-1}(w_{1}w_{2}^{p^{2}}-w_{2}w_{1}^{p^{2}})}{\lambda(g)^{-1}u}=\frac{w_{1}w_{2}^{p^{2}}-w_{2}w_{1}^{p^{2}}}{u}=y.$
So by comparing coefficients we get that $\gamma=1$ and $q(u,v_{0})=0.$

Clearly,$g$ acts linearly on
$\mathbb{F}_{p}u+\mathbb{F}_{p}y+\mathbb{F}_{p}v_{0}$ and is represented by
the matrix :

$\left(\begin{array}[]{ccc}\lambda(g)^{-1}&0&0\\\ 0&1&0\\\
0&0&\lambda(g)\end{array}\right).$

Therefore, by [Theorem 1.2] $S(V)^{G}$ is Gorenstein.

Assume now that $W$ is an unfaithful irreducible $T(G)-$module.By Proposition
6.1 $S(V)^{T(G)}=\mathbb{F}_{p}[A,B,v_{0}]$ is a polynomial ring where
$A=a^{p}b-b^{p}a,B=\frac{a^{p^{2}}b-b^{p^{2}}a}{A}$ and
$deg(A)=p^{2}+p,deg(B)=p(p^{2}-p).$ We have :

$A=a^{p}b-b^{p}a=(w_{1}^{p^{2}}-v_{0}^{p(p-1}w_{1}^{p})(w_{2}^{p}-v_{0}^{p(p-1)}w_{2})-(w_{2}^{p^{2}}-v_{0}^{p(p-1)})w_{2}^{p}(w_{1}^{p}-v_{0}^{p-1}w_{1})=(w_{1}^{p^{2}}w_{2}^{p}-w_{2}^{p^{2}}w_{1}^{p})-v_{0}^{p-1}(w_{1}^{p^{2}}w_{2}-w_{2}^{p^{2}}w_{1})+v_{0}^{p^{2}-1}(w_{1}^{p}w_{2}-w_{2}^{p}w_{1})=(-u)^{p}-v_{0}^{p-1}(w_{1}^{p^{2}}w_{2}-w_{2}^{p^{2}}w_{1})+v_{0}^{p^{2}-1}(-u),$
where $u=w_{1}w_{2}^{p}-w_{2}w_{1}^{p}.$ We get in the previous calculation
that :

$\left\\{\begin{array}[]{ll}g(u)=\lambda(g)^{-1}u.\\\
g(w_{1}^{p^{2}}w_{2}-w_{2}^{p^{2}}w_{1})=\lambda(g)^{-1}(w_{1}^{p^{2}}w_{2}-w_{2}^{p^{2}}w_{1}).\\\
\end{array}\right.$

Since
$g(v_{0}^{p^{2}-1})=(\lambda(g)^{-1})^{p^{2}-1}v_{0}^{p^{2}-1}=v_{0}^{p^{2}-1},$
then $g(A)=\lambda(g)^{-1}A.$

We also have :

$a^{p^{2}}b-b^{p^{2}}a=(w_{1}^{p^{3}}-v_{0}^{p^{2}(p-1)}w_{1}^{p^{2}})(w_{2}^{p}-v_{0}^{p-1}w_{2})-(w_{2}^{p^{3}}v_{0}^{p^{2}(p-1)}w_{2}^{p^{2}})(w_{1}-v_{0}^{p-1}w_{1})=(w_{1}^{p^{3}}w_{2}^{p^{2}}-w_{2}^{p^{3}}w_{1}^{p})-v_{0}^{(p-1)}(w_{1}^{p^{3}}w_{2}-w_{2}^{p^{3}}w_{1})+v_{0}^{p^{2}(p-1)}(w_{2}^{p^{2}}w_{1}^{p}-w_{2}^{p^{2}}w_{2}^{p})+v_{0}^{(p^{2}+1)(p-1)}(w_{1}^{p^{2}}w_{2}-w_{2}^{p^{2}}w_{1})=(-u)^{p^{2}}-v_{0}^{(p-1)}(w_{1}^{p^{3}}w_{2}-w_{2}^{p^{3}}w_{1})+v_{0}^{p^{2}(p-1)}(u)^{p}+v_{0}^{(p^{2}+1)(p-1)}(-uy).\\\
\\\ $ We get in the previous calculation that $g(u)=\lambda(g)u$ and $g(y)=1.$
On the other hand we have :

$g(v_{0}^{(p-1)}(w_{1}^{p^{3}}w_{2}-w_{2}^{p^{3}}w_{1}))=(\delta_{11}(g)\delta_{22}(g)-\delta_{12}(g)\delta_{21}(g))(w_{1}^{p^{3}}w_{2}-w_{2}^{p^{3}}w_{1})+(\delta_{23}(g)\delta_{11}(g)-\delta_{13}(g)\delta_{21}(g))(v_{0}^{p^{3}}w_{2}-w_{2}^{p^{3}}v_{0})+(\delta_{22}(g)\delta_{13}(g)-\delta_{12}(g)\delta_{23}(g))(w_{1}^{p^{3}}v_{0}-w_{2}^{p^{3}}v_{0}).$
Hence by
$\textbf{(6.b)},g(v_{0}^{(p-1)}(w_{1}^{p^{3}}w_{2}-w_{2}^{p^{3}}w_{1}))=\lambda(g)^{-1}(v_{0}^{(p-1)}(w_{1}^{p^{3}}w_{2}-w_{2}^{p^{3}}w_{1}).\\\
$ Consequently
$g(a^{p^{2}}b-b^{p^{2}}a)=\lambda(g)^{-1}(a^{p^{2}}b-b^{p^{2}}a),$ hence
$g(B)=\frac{g(a^{p^{2}}b-b^{p^{2}}a)}{g(A)}=\frac{\lambda(g)^{-1}(a^{p^{2}}b-b^{p^{2}}a)}{\lambda(g)^{-1}A}=B.$
So $g$ acts linearly on $\mathbb{F}_{p}A+\mathbb{F}_{p}B+\mathbb{F}_{p}v_{0}$
and is represented by the matrix :

$\left(\begin{array}[]{ccc}\lambda(g)^{-1}&0&0\\\ 0&1&0\\\
0&0&\lambda(g)\end{array}\right).$

Again,by [Theorem 1.2] $S(V)^{G}$ is Gorenstein .

∎

We shall now consider the possibility that $W$ is a reducible $T(G)-$module.

######  Lemma 6.4.

Suppose $W=V/Fv$ is a reducible $T(G)-$module.Then $W$ is a trivial
$T(G)-$module, that is $ker(\alpha)=T(G).$

###### Proof.

Let $F\bar{w_{1}}$ be a $T(G)-$ submodule of $W.$ Hence
$\phi(\bar{w_{1}})=\bar{w_{1}},\forall\phi\in T(G).$ Since $W$ is irreducible
$G-$submodule,$F\bar{w_{1}}$ is not a $G-$submodule.Consequently,there exists
$g\in G$ such that $g(\bar{w_{1}})=\bar{w_{2}}\not\in F\bar{w_{1}}.$ Since
$T(G)\lhd G,\forall\sigma\in T(G),g^{-1}\sigma g=\tau,$ for some $\tau\in
T(G).$Hence $(g^{-1}\sigma g)(\bar{w_{1}})=\tau(\bar{w_{1}})=\bar{w_{1}},$ and
therefore $\sigma(g(\bar{w_{1}}))=g(\bar{w_{1}}),$ hence
$\sigma(\bar{w_{2}})=\bar{w_{2}}.$ But $\\{\bar{w_{1}},\bar{w_{2}}\\}$ is a
basis of $W.$Consequently $W$ is a trivial $T(G)-$module.

∎

######  Corollary 6.5.

Suppose $W=V/\mathbb{F}_{p}v_{0}$ is a reducible $T(G)-$module, then
$S(V)^{T(G)}$ is a polynomial ring.

###### Proof.

$ker(\alpha)=T(G)\subseteq\left(\begin{array}[]{ccc}1&0&\mathbb{F}_{p}\\\
0&1&\mathbb{F}_{p}\\\ 0&0&1\end{array}\right).$

Hence $|T(G)|\in\\{p,p^{2}\\}.$ If $|T(G)|=p^{2},$ then

$T(G)=\left(\begin{array}[]{ccc}1&0&\mathbb{F}_{p}\\\ 0&1&\mathbb{F}_{p}\\\
0&0&1\end{array}\right),$

with generators :

$\left(\begin{array}[]{ccc}1&0&1\\\ 0&1&0\\\
0&0&1\end{array}\right),\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&1\\\
0&0&1\end{array}\right).$

Hence $S(V)^{T(G)}=\mathbb{F}_{p}[v_{0},x,y],$ where
$x=w_{1}^{p}-v_{0}^{p-1}w_{1},y=w_{2}^{p}-v_{0}^{p-1}w_{2}.$ If $|T(G)|=p,$
then $T(G)$ is generated by :

$\left(\begin{array}[]{ccc}1&0&\alpha\\\ 0&1&\beta\\\
0&0&1\end{array}\right),\alpha,\beta\in\mathbb{F}_{p}.$

Then as in Lemma 3.2 $S(V)^{T(G)}=\mathbb{F}_{p}[v_{0},\alpha w_{2}-\beta
w_{1},z]$ is a polynomial ring and $z=w_{1}^{p}-(\alpha v_{0})^{p-1}w_{1}.$ ∎

######  Corollary 6.6.

Suppose $W=V/\mathbb{F}_{p}v_{0}$ is a reducible $T(G)-$ module and an
irreducible $G-$module. Then $|T(G)|=p^{2}.$

###### Proof.

Suppose as in Corollary 6.5, that $|T(G)|=p,$ so

$T(G)=<\left(\begin{array}[]{ccc}1&0&\alpha\\\ 0&1&\beta\\\
0&0&1\end{array}\right)>.$

Since $\mathbb{F}_{p}v_{0},\mathbb{F}_{p}(\alpha w_{1}-\beta w_{2})$ are
trivial $T(G)-$ modules we have that
$A=\mathbb{F}_{p}v_{0}+\mathbb{F}_{p}(\alpha w_{1}-\beta w_{2})=ker(\sigma-I)$
for each transvection $\sigma\in T(G).$ Let $g\in G$ and $\sigma$ a
transvection in $G.\\\ $Then $g^{-1}\sigma g=\tau$ for some transvection
$\tau.$ Then for each $a\in A$ we have $(g^{-1}\sigma g)(a)=\tau(a)=a$ and
hence $\sigma(g(a))=g(a),$ implying that $g(a)\in A$ for each $g\in G.$ Hence
$A$ is a $2-$dimensional $G-$submodule of $V.$ This contradicts the assumption
on $V.$

∎

######  Theorem 6.7.

Suppose $W=V/\mathbb{F}_{p}v_{0}$ is a reducible $T(G)-$ module and an
irreducible $G-$module.Then $S(V)^{G}$ is Gorenstein.

###### Proof.

We follow here the argument in Proposition 3.13 with
$\bigl{(}\begin{smallmatrix}a\\\
b\end{smallmatrix}\bigr{)}=\bigl{(}\begin{smallmatrix}1\\\
0\end{smallmatrix}\bigr{)},\bigl{(}\begin{smallmatrix}c\\\
d\end{smallmatrix}\bigr{)}=\bigl{(}\begin{smallmatrix}0\\\
1\end{smallmatrix}\bigr{)}.$ Recall that by Corollary 6.5 and Corollary 6.6,

$T(G)=\left(\begin{array}[]{ccc}1&0&\mathbb{F}_{p}\\\ 0&1&\mathbb{F}_{p}\\\
0&0&1\end{array}\right),$

with generators :

$\left(\begin{array}[]{ccc}1&0&1\\\ 0&1&0\\\
0&0&1\end{array}\right),\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&1\\\
0&0&1\end{array}\right).$

Hence $S(V)^{T(G)}=\mathbb{F}_{p}[v_{0},x,y],$ where
$x=w_{1}^{p}-v_{0}^{p-1}w_{1},y=w_{2}^{p}-v_{0}^{p-1}w_{2}.$ So if
$m=(v_{0},x,y)$ then $\bar{g}\in G/T(G)$ is represented on
$m/m^{2}=\mathbb{F}_{p}\bar{y}+\mathbb{F}_{p}\bar{x}+\mathbb{F}_{p}v_{0}$ by
the matrix :

$\left(\begin{array}[]{ccc}\alpha&\gamma&0\\\ \varepsilon&\delta&0\\\
0&0&\lambda(g)\end{array}\right),$

and $g\in G$ is represented on
$V=\mathbb{F}_{p}w_{2}+\mathbb{F}_{p}w_{1}+\mathbb{F}_{p}v_{0},$ by the matrix
:

$\left(\begin{array}[]{ccc}\lambda_{2}(g)&\delta_{12}(g)&\delta_{13}(g)\\\
\delta_{21}(g)&\lambda_{1}(g)&\delta_{23}(g)\\\
0&0&\lambda(g)\end{array}\right).$

We shall show that $det(\bar{g})=1,$ for each $\bar{g}\in G/T(G),$ implying by
[Theorem 1.2] that $S(V)^{G}$ is Gorenstein. Now $g(y)=\alpha y+\gamma x+\beta
v_{0}^{p}$ and $g(x)=\delta x+\varepsilon y+\chi v_{0}^{p}$ (keeping the
notation in Lemma 3.13). Since
$g(w_{2})=\lambda_{2}(g)w_{2}+\delta_{12}(g)w_{1}+\delta_{13}(g)v_{0},$ we
have
$g(y)=g(w_{2})^{p}-(\lambda(g)v_{0})^{p-1}g(w_{2})=(\lambda_{2}(g)w_{2}+\delta_{12}(g)w_{1}+\delta_{13}(g)v_{0})^{p}-(\lambda(g)v_{0})^{p-1}(\lambda_{2}(g)w_{2}+\delta_{12}(g)w_{1}+\delta_{13}(g)v_{0})=(\lambda_{2}(g)^{p}w_{2}^{p}-\lambda(g)^{p-1}\lambda_{2}(g)v_{0}^{p-1}w_{2})+(\delta_{12}(g)^{p}w_{1}^{p}-\lambda(g)^{p-1}\delta_{12}(g)v_{0}^{p-1}w_{1})+(\delta_{13}(g)^{p}-\lambda(g)^{p-1}\delta_{13}(g))v_{0}^{p}.$

Consequently
$\alpha=\lambda_{2}(g)^{p}=\lambda(g)^{p-1}\lambda_{2}(g),\gamma=\delta_{12}(g)^{p}=\lambda(g)^{p-1}\delta_{12}(g).$
Consequently,since $\lambda_{2}(g),\delta_{12}(g)\in\mathbb{F}_{p},$ it
follows that $\alpha=\lambda_{2}(g)$ and $\gamma=\delta_{12}(g).\\\ \\\ $
Similarly using
$g(w_{1})=\delta_{21}(g)w_{2}+\lambda_{1}(g)w_{1}+\delta_{23}(g)v_{0},$ we
have
$g(x)=g(w_{1})^{p}-(\lambda(g)v_{0})^{p-1}g(w_{1})=(\delta_{21}(g)w_{2}+\lambda_{1}(g)w_{1}+\delta_{23}(g)v_{0})^{p}-(\lambda(g)v_{0})^{p-1}(\delta_{21}(g)w_{2}+\lambda_{1}(g)w_{1}+\delta_{23}(g)v_{0})=(\delta_{21}(g)^{p}w_{2}^{p}-\lambda(g)^{p-1}\delta_{21}(g)v_{0}^{p-1}w_{2})+(\lambda_{1}(g)^{p}w_{1}^{p}-\lambda(g)^{p-1}\lambda_{1}(g)v_{0}^{p-1}w_{1})+(\delta_{23}(g)^{p}-\lambda(g)^{p-1}\delta_{23}(g))v_{0}^{p}.\\\
$ Therefore $\varepsilon=\delta_{21}(g)^{p},\delta=\lambda_{1}(g)^{p},$ and
since $\lambda_{1}(g),\delta_{21}(g)\in\mathbb{F}_{p},$ we have
$\varepsilon=\delta_{21}(g),\delta=\lambda_{1}(g).$ Therefore
$det(\bar{g})=(\alpha\delta-\varepsilon\gamma)\lambda(g)=(\lambda_{2}(g)\lambda_{1}(g)-\delta_{21}(g)\delta_{12}(g))\lambda(g)=det(g)=1$
(since $G\subset SL(3,\mathbb{F}_{p})$).

∎

As a consequence of Theorem 6.3 and Theorem 6.7 we have :

######  Corollary 6.8.

Suppose $W=V/\mathbb{F}_{p}v_{0}$ is an irreducible $G-$module. Then
$S(V)^{G}$ is Gorenstein.

##  7 Chapter($F$): $V$ is indecomposable and contains two $G-$ submodules
$\mathbb{F}_{p}v_{0},W,$ where $\mathbb{F}_{p}v_{0}\subset W\subset V,$ and
$dim_{\mathbb{F}_{p}}W=2.$

In this chapter we conside $V,$ a reducible indecomposable $G-$module,which
has a unique $1-$dimensional $G-$ submodule $Fv_{0},$ and a unique
$2-$dimensional $G-$ submodule $W,$ and $Fv_{0}\subset W\subset V.$ We shall
exclusively consider the case $F=\mathbb{F}_{p}.\\\ $ Let
$\\{w_{2},w_{1},v_{0}\\}$ be a basis of $V=\mathbb{F}_{p}^{3},$ where
$W=\mathbb{F}_{p}w_{1}+\mathbb{F}_{p}v_{0}.$ So every $g\in G$ is represented
with respect to this basis , by the matrix :

$g=\left(\begin{array}[]{ccc}\lambda_{2}(g)&\delta_{12}(g)&\delta_{13}(g)\\\
0&\lambda_{1}(g)&\delta_{23}(g)\\\ 0&0&\lambda(g)\end{array}\right).$

By Lemma 2.2, $T(G)$ acts trivially on $\mathbb{F}_{p}v_{0},$ and also on
$W/\mathbb{F}_{p}v_{0},$ hence

$T(G)\subseteq\left(\begin{array}[]{ccc}1&\mathbb{F}_{p}&\mathbb{F}_{p}\\\
0&1&\mathbb{F}_{p}\\\ 0&0&1\end{array}\right).$

So one of the following three cases holds:

  1. (i)

$|T(G)|=p,$

  2. (ii)

$|T(G)|=p^{2},$

  3. (iii)

$|T(G)|=p^{3}.$

######  Corollary 7.1.

If case (i) holds. Then $S(V)^{G}$ is Gorenstein.

###### Proof.

Since $|T(G)|=p,T(G):=<\sigma>$ or $<\tau>,$ where

$\sigma=\left(\begin{array}[]{ccc}1&0&a\\\ 0&1&b\\\
0&0&1\end{array}\right),\tau=\left(\begin{array}[]{ccc}1&c&d\\\ 0&1&0\\\
0&0&1\end{array}\right),a,b,c,d\in\mathbb{F}_{p}.$

Assume firstly that $T(G)=<\sigma>.$ We shall change the basis of $V$ into
$\\{w_{2},aw_{1}-bw_{2},v_{0}\\}.$We have,
$\sigma(aw_{1}-bw_{2})=a(w_{1}+bv_{0})-b(w_{2}+av_{0})=aw_{1}-bw_{2}.$ Suppose
$a\neq 0,$ so $0\neq w_{1}^{\prime}:=(aw_{1}-bw_{2})\in S(V)^{T(G)}.$
Therefore $\sigma$ is represented with respect to
$\\{w_{2},w_{1}^{\prime},v_{0}\\}$ by the matrix:

$\sigma=\left(\begin{array}[]{ccc}1&0&a\\\ 0&1&0\\\ 0&0&1\end{array}\right).$

Consequently, by [3, Lemma 3.13]
$S(V)^{T(G)}=\mathbb{F}_{p}[x,w_{1}^{\prime},v_{0}]$ is a polynomial ring, and
$x=w_{2}^{p}-(av_{0})^{p-1}w_{2}.$ Let $m=(x,w_{1}^{\prime},v_{0})$ be the
graded maximal homogenous ideal of $S(V)^{T(G)}.$
Since $T(G)\lhd G,\mathbb{F}_{p}[x,w_{1}^{\prime},v_{0}]$ is $g-$stable and
since $g$ preserve degrees we get that
$g(\mathbb{F}_{p}w_{1}^{\prime}+\mathbb{F}_{p}v_{0})\in(\mathbb{F}_{p}w_{1}^{\prime}+\mathbb{F}_{p}v_{0})$
for every $g\in G.$This implies that
$(\mathbb{F}_{p}w_{1}^{\prime}+\mathbb{F}_{p}v_{0})$ is $G-$module.Therefore
$G$ has a triangular presentation with respect to the new basis
$\\{w_{2},w_{1}^{\prime},v_{0}\\}.$Hence we still use the same
$\lambda_{2}(g),\delta_{12}(g),\delta_{13}(g),\lambda_{1}(g),\delta_{23}(g),$
for the new presentation of $g,$ with respect to the basis
$\\{w_{2},w_{1}^{\prime},v_{0}\\}.$

We have:

$(7.a)\left\\{\begin{array}[]{llll}g(x)=\lambda_{2}(g)^{p}w_{2}^{p}+\delta_{12}(g)^{p}(w_{1}^{\prime})^{p}+\delta_{13}(g)^{p}v_{0}^{p}-(a\lambda(g)v_{0})^{p-1}(\lambda_{2}(g)w_{2}+\delta_{12}(g)w_{1}^{\prime}+\delta_{13}(g)v_{0}).\\\
g(w_{2})=\lambda_{2}(g)w_{2}+\delta_{12}(g)w_{1}^{\prime}+\delta_{13}(g)v_{0}.\\\
g(w_{1}^{\prime})=\lambda_{1}(g)w_{1}^{\prime}+\delta_{23}(g)v_{0}.\\\
g(v_{0})=\lambda(g)v_{0}.\end{array}\right.$

Now since $gT(G)g^{-1}=T(G),\mathbb{F}_{p}[x,w_{1}^{\prime},v_{0}]$ is
$g-$stable and since $g$ preserve degrees we get:

$(7.b)\left\\{\begin{array}[]{l}g(x)=\alpha_{1}(w_{2}^{p}-(av_{0})^{p-1}w_{2})+\alpha_{2}(w_{1}^{\prime})^{p}+\alpha_{3}v_{0}^{p}+$mixed
terms in$\\{w_{1}^{\prime},v_{0}\\}.\\\
\alpha_{1},\alpha_{2},\alpha_{3}\in\mathbb{F}_{p}.\par\end{array}\right.$

Hence by comparing coefficient of $w_{2}^{p}$ in $g(x),$ we get that
$\alpha_{1}=\lambda_{2}(g)^{p}.$ So $g$ acts linearly on $m/m^{2}$ and is
represented by the matrix:

$\left(\begin{array}[]{ccc}\lambda_{2}(g)^{p}&0&0\\\
0&\lambda_{1}(g)&\delta_{23}(g)\\\
0&0&\lambda(g)\end{array}\right)=\left(\begin{array}[]{ccc}\lambda_{2}(g)&0&0\\\
0&\lambda_{1}(g)&\delta_{23}(g)\\\ 0&0&\lambda(g)\end{array}\right).$

Hence by [Theorem 1.2] $S(V)^{G}$ is Gorenstein.If $a=0,$ then
$S(V)^{T(G)}=\mathbb{F}_{p}[w_{2},x,v_{0}]$ where
$x=w_{1}^{p}-(bv_{0})^{p-1}w_{1}.$So we have:

$(7.c)\left\\{\begin{array}[]{ll}g(x)=(\lambda_{1}(g)w_{1}+\delta_{23}(g)v_{0})^{p}-(bv_{0})^{p-1}(\lambda_{1}(g)w_{1}+\delta_{23}(g)v_{0}).\\\
g(x)=\gamma_{1}x+\gamma_{2}w_{2}^{p}+\gamma_{3}v_{0}^{p}+$mixed terms
in$\\{w_{2},v_{0}\\}.\indent\gamma_{1},\gamma_{2},\gamma_{3}\in\mathbb{F}_{p}.\\\
\end{array}\right.$

Hence by comparing coefficient of $w_{1}^{p}$ in both expressions,we get that
$\gamma_{1}=\lambda_{1}(g)^{p}=\lambda_{1}(g).$ Hence by taking
$m=(w_{2},x,v_{0}),$ the matrix represinting $g|m/m^{2}$ with respect to
$\bar{w_{2}},\bar{x},\bar{v_{0}}$ is

$\left(\begin{array}[]{ccc}\lambda_{2}(g)&\delta_{12}(g)&\delta_{13}(g)\\\
0&\lambda_{1}(g)^{p}&0\\\ 0&0&\lambda(g)\end{array}\right),$

so its determinant is $\lambda_{2}(g)\cdot\lambda_{1}(g)\cdot\lambda(g)=1,$
for each $g\in G,$ and $S(V)^{G}$ is Gorenstein.

Assume now that $T(G)=<\tau>.$ If $c=0,$ then we are exactly in the first
case.So we may assume that $c\neq 0.$ Consider
$w_{1}^{\prime}:=cw_{1}+dv_{0},$ then $\\{w_{2},w_{1}^{\prime},v_{0}\\}$ is a
basis of $V,$with $\\{w_{1}^{\prime},v_{0}\\}$ a basis of $W.$ Then the matrix
of $\tau$ with respect to this new basis is

$\tau=\left(\begin{array}[]{ccc}1&1&0\\\ 0&1&0\\\ 0&0&1\end{array}\right).$

Hence $S(V)^{<\tau>}=\mathbb{F}_{p}[z,w_{1}^{\prime},v_{0}],$where
$z=w_{2}^{p}-(w_{1}^{\prime})^{p-1}w_{2}.$

By the same discussion as above,we have:

$(7.d)\left\\{\begin{array}[]{llll}g(z)=\lambda_{2}(g)^{p}w_{2}^{p}+\delta_{12}(g)^{p}(w_{1}^{\prime})^{p}+\delta_{13}(g)^{p}v_{0}^{p}+g(-(w_{1}^{\prime})^{p-1})(\lambda_{2}(g)w_{2}+\delta_{12}(g)w_{1}^{\prime}+\delta_{13}(g)v_{0}).\\\
g(z)=\beta_{1}z+\beta_{2}(w_{1}^{\prime})^{p}+\beta_{3}v_{0}^{p}+$mixed terms
in
$\\{w_{1}^{\prime},v_{0}\\}.\indent\beta_{1},\beta_{2},\beta_{3}\in\mathbb{F}_{p}.\\\
g(w_{1}^{\prime})=\lambda_{1}(g)w_{1}^{\prime}+\delta_{23}(g)v_{0}.\\\
\end{array}\right.$

Hence by comparing coefficient of $w_{2}^{p}$ in $g(z)$ , we get
$\beta_{1}=\lambda_{2}(g)^{p}.$ Let $m=(z,w_{1}^{\prime},v_{0})$ be the graded
maximal homogenous ideal of $S(V)^{T(G)}.$ So $g$ acts linearly on $m/m^{2}$
and is represented by the matrix:

$\left(\begin{array}[]{ccc}\lambda_{2}(g)^{p}&0&0\\\
0&\lambda_{1}(g)&\delta_{23}(g)\\\
0&0&\lambda(g)\end{array}\right)=\left(\begin{array}[]{ccc}\lambda_{2}(g)&0&0\\\
0&\lambda_{1}(g)&\delta_{23}(g)\\\ 0&0&\lambda(g)\end{array}\right).$

Hence by [Theorem 1.2] $S(V)^{G}$ is Gorenstein.

∎

######  Corollary 7.2.

If case (ii) holds.Then $S(V)^{G}$ is Gorenstein.

###### Proof.

Since $|T(G)|=p^{2},$ it is elementary abelian,hence

$T(G)=\left(\begin{array}[]{ccc}1&\mathbb{F}_{p}&\mathbb{F}_{p}\\\ 0&1&0\\\
0&0&1\end{array}\right)\indent or\indent
T(G)=\left(\begin{array}[]{ccc}1&0&\mathbb{F}_{p}\\\ 0&1&\mathbb{F}_{p}\\\
0&0&1\end{array}\right).$

Assume firstly that,

$T(G)=\left(\begin{array}[]{ccc}1&\mathbb{F}_{p}&\mathbb{F}_{p}\\\ 0&1&0\\\
0&0&1\end{array}\right).$

Hence by Proposition 4.3, $S(V)^{T(G)}=S(V)^{Fix_{T(G)}(W)}=S(W)[t]$ is a
polynomial ring where $t=w_{2}^{p^{2}}+q_{1}w_{2}^{p}+q_{2}w_{2},$ and
$q_{1},q_{2}\in\mathbb{F}_{p}[w_{1},v_{0}]$ is homogenous of degrees
$p^{2}-p,p^{2}-1$ respectively.
Let $m=(t,w_{1},v_{0})$ be the graded maximal homogenous ideal of
$S(V)^{T(G)}.$By a similar discussion to the one in (7.d) we get that $g$ acts
linearly on $m/m^{2}$ and is represented by the matrix:

$\left(\begin{array}[]{ccc}\lambda_{2}(g)^{p^{2}}&0&0\\\
0&\lambda_{1}(g)&\delta_{23}(g)\\\
0&0&\lambda(g)\end{array}\right)=\left(\begin{array}[]{ccc}\lambda_{2}(g)&0&0\\\
0&\lambda_{1}(g)&\delta_{23}(g)\\\ 0&0&\lambda(g)\end{array}\right).$

Consequently,by [Theorem 1.2] $S(V)^{G}$ is Gorenstein.

Assume now that

$T(G)=\left(\begin{array}[]{ccc}1&0&\mathbb{F}_{p}\\\ 0&1&\mathbb{F}_{p}\\\
0&0&1\end{array}\right).$

Hence by the Corollary 6.5, $S(V)^{T(G)}=\mathbb{F}_{p}[y,x,v_{0}]$ is a
polynomial ring, where $y=w_{2}^{p}-v_{0}^{p-1}w_{2}$ and
$x=w_{1}^{p}-v_{0}^{p-1}w_{1}.$ We have,using $F=\mathbb{F}_{p}$ (so
$\lambda(g)^{p-1}=1$):

$(7.e)\left\\{\begin{array}[]{ll}g(y)=(\lambda_{2}(g)w_{2}^{p}+\delta_{12}(g)w_{1}^{p}+\delta_{13}(g)v_{0}^{p})-v_{0}^{p-1}(\lambda_{2}(g)w_{2}+\delta_{12}(g)w_{1}+\delta_{13}(g)v_{0})=\lambda_{2}(g)y+\delta_{12}(g)x.\\\
g(x)=(\lambda_{1}(g)w_{1}^{p}+\delta_{23}(g)v_{0}^{p})-v_{0}^{p-1}(\lambda_{1}(g)w_{1}+\delta_{23}(g)v_{0})=\lambda_{1}(g)x.\\\
\end{array}\right.$

Let $m=(y,x,v_{0})$ be the graded maximal homogenous ideal of $S(V)^{T(G)}.$
So $g$ acts linearly on $m$ (resp. $m/m^{2}$) and is represented by the
matrix:

$\left(\begin{array}[]{ccc}\lambda_{2}(g)&\delta_{12}(g)&0\\\
0&\lambda_{1}(g)&0\\\ 0&0&\lambda(g)\end{array}\right).$

Consequently,by [Theorem 1.2] $S(V)^{G}$ is Gorenstein.

∎

######  Corollary 7.3.

If case (iii) holds.Then $S(V)^{G}$ is Gorenstein.

###### Proof.

Since $|T(G)|=p^{3},$

$T(G)=\left(\begin{array}[]{ccc}1&\mathbb{F}_{p}&\mathbb{F}_{p}\\\
0&1&\mathbb{F}_{p}\\\ 0&0&1\end{array}\right).$

Hence $T(G)=U_{3}(\mathbb{F}_{p}),$ the group of all upper triangular matrices
with $1^{\prime}$s along the diagonal and entries from the finite field
$\mathbb{F}_{p}.$ This group was shown to have a polynomial ring of invariants
by Bertin [1].

We shall find explicitly the generators of $S(V)^{T(G)}.$
$T(G)=<\sigma_{1},\sigma_{2},\sigma_{3}>,$ where

$\sigma_{1}=\left(\begin{array}[]{ccc}1&0&1\\\ 0&1&0\\\
0&0&1\end{array}\right),\sigma_{2}=\left(\begin{array}[]{ccc}1&1&0\\\ 0&1&0\\\
0&0&1\end{array}\right),\sigma_{3}=\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&1\\\
0&0&1\end{array}\right).$

Consequently, by [3, Lemma 3.13],
$S(V)^{<\sigma_{1}>}=\mathbb{F}_{p}[a,w_{1},v_{0}],$is a polynomial ring where
$a=w_{2}^{p}-v_{0}^{p-1}w_{2}.$Define $u=a^{p}-b^{p-1}a,$ where
$b:=w_{1}^{p}-v_{0}^{p-1}w_{1}.$ Observe that
$\sigma_{3}(b)=b,\sigma_{3}(a)=a,\sigma_{2}(b)=b,\sigma_{2}(a)=a+b,$hence
$\sigma_{2}(u)=u.$ We have, for $i=1,2,3:$

$\left\\{\begin{array}[]{ll}\sigma_{i}(u)=u.\\\ \sigma_{i}(b)=b.\\\
\sigma_{i}(v_{0})=v_{0}.\\\ \end{array}\right.$

The $T(G)-$invariants $\\{u,b,v_{0}\\}$ are algebraically independent over
$\mathbb{F}_{p}.$Since $deg(u)\cdot deg(b)\cdot deg(v_{0})=p^{2}\cdot p\cdot
1=p^{3}=|T(G)|,$we have by [4, Theorem 3.9.4] that
$S(V)^{T(G)}=\mathbb{F}_{p}[u,b,v_{0}]$ is a polynomial ring.Let
$m=(u,b,v_{0})$ be the graded maximal homogenous ideal of $S(V)^{T(G)}.$
Since $g(b)=\lambda_{1}(g)^{p}w_{1}^{p}+s=\alpha b+\beta v_{0}^{p},$ where
$s\in m^{2},$and $\alpha,\beta\in\mathbb{F}_{p},$ we get that
$\alpha=\lambda_{1}(g)^{p}.$Since $g(u)=\lambda_{2}(g)^{p^{2}}u+r=\gamma
u+\delta b^{p}+\pi v_{0}^{p^{2}}+$ mixed terms in $\\{b,v_{0}\\},$ where $r\in
m^{2},$and $\gamma,\delta,\pi\in\mathbb{F}_{p},$ we get that
$\gamma=\lambda_{2}(g)^{p^{2}}.$Hence,

$g|m/m^{2}=\left(\begin{array}[]{ccc}\lambda_{2}(g)^{p^{2}}&0&0\\\
0&\lambda_{1}(g)^{p}&0\\\
0&0&\lambda(g)\end{array}\right)=\left(\begin{array}[]{ccc}\lambda_{2}(g)&0&0\\\
0&\lambda_{1}(g)&0\\\ 0&0&\lambda(g)\end{array}\right).$

Consequently,by [Theorem 1.2] $S(V)^{G}$ is Gorenstein.

∎

##  8 Chapter($G$): $V$ is indecomposable and contains (at least) two
$1-$dimensional $G-$ submodules.

Let $Fw_{1},Fw_{2}$ be the two $1-$dimensional $G-$submodules. Hence
$W=Fw_{1}+Fw_{2}$ is $G-$submodule of $V.$ Let $v\in V\backslash W,$ so
$\\{v,w_{1},w_{2}\\}$ is a basis of $V.$ Each $g\in G$ is represented with
respect to this basis by the matrix :

$g=\left(\begin{array}[]{ccc}\lambda(g)&\delta_{12}(g)&\delta_{13}(g)\\\
0&\lambda_{2}(g)&0\\\ 0&0&\lambda_{1}(g)\end{array}\right).$

In particular each transvection $\sigma\in T(G)$ is represented by the matrix
:

$\sigma=\left(\begin{array}[]{ccc}\lambda(\sigma)&\delta_{12}(\sigma)&\delta_{13}(\sigma)\\\
0&1&0\\\ 0&0&1\end{array}\right).$

Hence $T(G)|W=Id.$ Hence

$T(G)\subseteq\left(\begin{array}[]{ccc}1&F&F\\\ 0&1&0\\\
0&0&1\end{array}\right).$

Therefore by Proposition 4.3 $S(V)^{T(G)}=S(W)[z],$ where
$z=v^{p^{t}}+q_{1}v^{p^{t-1}}+...+q_{t}v,$ an homogenous polynomial, where
$q_{i}\in S(W)^{T(G)|W}=S(W)$ and $|T(G)|=p^{t}.$ Let $m=(z,w_{2},w_{1}).$
Then $m/m^{2}=F\bar{z}+F\bar{w_{2}}+F\bar{w_{1}}.$ Now for $g\in G$ we have
$g(z)=\alpha z+x$ where $x$ is in $m^{2}.$ Therefore
$\bar{g}(\bar{z})=\alpha\bar{z},$ where $\bar{g}=gT(G).$ Also
$g(z)=g(v)^{p^{t}}+g(q_{1})g(v)^{p^{t-1}}+...+g(q_{t})g(v)=\lambda(g)^{p^{t}}v^{p^{t}}+$
mixed terms. Hence $\alpha=\lambda(g)^{p^{t}}$ and therefore $\bar{g}$ is
represented on $F\bar{z}+F\bar{w_{2}}+F\bar{w_{1}}$ by the matrix :

$\left(\begin{array}[]{ccc}\lambda(g)^{p^{t}}&0&0\\\ 0&\lambda_{2}(g)&0\\\
0&0&\lambda_{1}(g)\end{array}\right).$

Therefore by [Theorem 1.2] $S(V)^{G}$ is Gorenstein iff
$\lambda(g)^{p^{t}}\cdot\lambda_{2}(g)\cdot\lambda_{1}(g)=1,$ for each $g\in
G.$

######  Corollary 8.1.

Suppose $V=\mathbb{F}_{p}^{3},$ contains two $1-$ dimensional $G-$submodules
and $G\subseteq SL(3,\mathbb{F}_{p}).$ Then $S(V)^{G}$ is Gorenstein.

###### Proof.

$\lambda(g)\in\mathbb{F}_{p}$ hence
$\lambda(g)^{p^{t}}\cdot\lambda_{2}(g)\cdot\lambda_{1}(g)=\lambda(g)\cdot\lambda_{2}(g)\cdot\lambda_{1}(g)=det(g)=1,$
where the last equaliry holds since $g\in SL(3,\mathbb{F}_{p}).$

∎

##  9 References

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