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{\large
{\bf SUBGROUPS AND MODULAR INVARIANTS OF \\ } \vspace{7pt}
{\bf 2-DIMENSIONAL SPECIAL LINEAR GROUPS \\ } \vspace{20pt}
BY \\ \vspace{20pt}
MENG-GEN TSAI \\ \vspace{40pt}
SUPERVISED BY \\ \vspace{20pt}
MING-CHANG KANG \\ \vspace{200pt}
GRADUATE INSTITUTE OF MATHEMATICS \\ \vspace{7pt}
NATIONAL TAIWAN UNIVERSITY \\ \vspace{7pt}
TAIPEI, TAIWAN, REPUBLIC OF CHINA \\ \vspace{40pt}
JUNE, 2006
}
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{\bf SUBGROUPS AND MODULAR INVARIANTS OF \\ } \vspace{3pt}
{\bf 2-DIMENSIONAL SPECIAL LINEAR GROUPS \\ } \vspace{30pt}
{\bf Abstract \\ }
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In this paper, we find out all finite subgroups of $SL(2,F)$ where
$F$ is an algebraically closed field of characteristic $p$.
Besides, we compute the invariants of finite subgroups of $SL(2,F)$ where
$p$ divides the order of the groups.

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\begin{center}
{\bf Acknowledgements \\ }
\end{center}

First of all, I want to thank my advisor for his guiding,
teaching in these years and advice on this thesis.

Finally, I thank my family for their and support in my life.

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\noindent{{\bf \LARGE{Contents} }} \\

\noindent{{\bf 1 \hspace{3pt} Introduction \hfill 1 \\ }}
\noindent{{\bf 2 \hspace{3pt} Definitions and Terminology \hfill 1 \\ }}
\noindent{{\bf 3 \hspace{3pt} Finite Subgroups of $SL(2,F)$ \hfill 2 \\ }}
\noindent{{\bf 4 \hspace{3pt} Modular Invariants \hfill 11 \\ }}
\noindent{{\bf References \hfill 16 \\ }}

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\section{Introduction}
\hspace{13pt} Let $F$ be an algebraically closed field of characteristic $p$.
$p$ is zero or a prime number.
Consider the two-dimensional special linear groups $SL(2,F)$,
and we will determine all finite subgroups of $SL(2,F)$.
Although this problem was solved by Dickson in his book \cite{Di},
we compute the set of generators for each subgroups.
It will be used later.

Let $G$ be a finite subgroup of $SL(2,F)$.
Since all possible finite subgroups of $SL(2,F)$ are determined completely,
we can compute the invariants of $G$.
If $p \nmid |G|$, then the invariants of $G$ are
determined completely by Springer in his book \cite{Sp}.
If $p \mid |G|$, it is hard to get the invariants of $G$.
We need to consider five cases.
The invariants of $SL(2,\Bbb{F}_q)$ where $q$ is a power of $p$ were
computed by Dickson, and Wilkerson \cite{Wi} gave a new proof for this result.
We use Wilkerson's methed \cite{Wi} to determine the invariants of other four cases.

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\section{Definitions and Terminology}
\hspace{13pt} Let $F$ be an algebraically closed field of characteristic $p$.
Let $V$ be the two-dimensional vector space over $F$,
and consider a fixed basis of $V$.  Let $L = SL(V)$.
Let $\zeta_n$ be a primitive $n$-th root of unity in $F$.
\vspace{10pt}

\begin{definition}
Let $d_\omega = \left(\begin{array}{cc} \omega & 0 \\ 0 & \omega^{-1} \end{array} \right)$
where $\omega \in F^\times$,
$t_\lambda = \left(\begin{array}{cc} 1 & \lambda \\ 0 & 1 \end{array} \right)$
where $\lambda \in F$, and
$w = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right)$.
Let $D = \{ d_\omega \mid \omega \in F^\times \}$,
$T = \{ t_\lambda \mid \lambda \in F \}$, $H = DT$, and
$Z = Z(L)$ be the center of $L$.
\end{definition}
\vspace{10pt}

\begin{definition}
If $G$ is a finite subgroup of $SL(2,F)$, then
we define the invariants of $G$ to be
\[
    F[x,y]^G = \{ a \in F[x,y] \mid g \cdot a = a \mbox{ for all } g \in G \}
\]
where
\[
    g \cdot a(x,y)
    = \left(\begin{array}{cc} \alpha & \beta \\ \gamma & \delta \end{array} \right) \cdot a(x,y)
    = a(\alpha x + \beta y, \gamma x + \delta y).
\]
\end{definition}

\section{Finite Subgroups of $SL(2,F)$}
The aim of this section is finding all the finite subgroups of two-dimensional
special linear groups over algebraically closed fields.
\vspace{10pt}

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\noindent {\bf Theorem 3.1.}
{\it Let $V$ be the two-dimensional vector space over an algebraically
closed field $F$ of characteristic $p$ ($p \geq 0$).  Let $L = SL(V)$.
Any finite subgroup $G$ of $L$ is isomorphic to one of the groups
in the following list.}

\noindent Case I: {\it $p \nmid |G|$}

(1) A cyclic group $\langle d_{\zeta_n} \rangle$.

(2) $G = \langle d_{\zeta_{2n}}, w \rangle
\cong \langle x,y \mid x^n = y^2, y^{-1}xy = x^{-1} \rangle$.

(3) $\bigg\langle d_{\zeta_4}, w,
\left(\begin{array}{cc} \frac{-1+\sqrt{-1}}{2} & \frac{-1+\sqrt{-1}}{2} \\
\frac{1+\sqrt{-1}}{2} & \frac{-1-\sqrt{-1}}{2} \end{array} \right) \bigg\rangle
\cong SL(2,3)$.

(4) $\bigg\langle d_{\zeta_8}, w,
\left(\begin{array}{cc} \frac{-1+\sqrt{-1}}{2} & \frac{-1+\sqrt{-1}}{2} \\
\frac{1+\sqrt{-1}}{2} & \frac{-1-\sqrt{-1}}{2} \end{array} \right) \bigg\rangle$.

(5) $\bigg\langle
\left(\begin{array}{cc} \zeta^3 & 0 \\ 0 & \zeta^2 \end{array} \right), w,
\left(\begin{array}{cc}
\frac{\zeta^4-\zeta}{\sqrt{5}} & \frac{\zeta^2-\zeta^3}{\sqrt{5}} \\
\frac{\zeta^2-\zeta^3}{\sqrt{5}} & \frac{\zeta^-\zeta^4}{\sqrt{5}} \end{array} \right)
\bigg\rangle \cong SL(2,5)$ where $\zeta = \zeta_5$.

\noindent Case II: {\it $p \mid |G|$.  Let $Q$ be an $p$-Sylow subgroup of $G$,
and let $|Q| = q$.}

(6) {\it $G = N_G(Q) = \{ g \in G \mid g^{-1}Qg = Q \} =
\langle t_\lambda, d_{\zeta_n} \rangle$ where $n \mid q-1$.}

(7) {\it $p = 2$ and $G = \langle d_{\zeta_n}, w \rangle$
is a dihedral group of order $2n$ with odd $n$.}

(8) {\it $p = 3$ and $G = \bigg\langle
t_1, \left(\begin{array}{cc} -\sqrt{-1} & 0 \\ 1 & \sqrt{-1} \end{array} \right),
d_{\zeta_4} \bigg\rangle \cong SL(2,5)$.}

(9) $SL(2,\Bbb{F}_q)$.

(10) {\it $\langle SL(2,\Bbb{F}_q), d_\sigma \rangle$ where $\sigma \in
\Bbb{F}_{q^2} \setminus \Bbb{F}_q$ and $\langle \sigma^2 \rangle = \Bbb{F}_q^\times$.}
\vspace{10pt}

To begin with, we state some lemmas which will be used later.
The proof of all of the following lemmas can be found in Suzuki
\cite[Chapter 3, \textsection 6]{Su}.
\vspace{10pt}

\noindent {\bf Lemma 3.2.} (1) {\it The center $Z = \{ d_1 \}$ if $p = 2$. Otherwise,
$Z = \{ d_1, d_{-1} \}$. The group $L/Z$ is simple.}

(2) {\it For any $\omega, \omega' \in F^\times$ and $\lambda, \mu \in F$,
we have $d_\omega d_{\omega'} = d_{\omega \omega'}$,
$t_\lambda t_\mu = t_{\lambda+\mu}$, $d_\omega^{-1} t_\mu d_\omega = t_{\omega^2 \mu}$,
and $w^{-1} d_\omega w = d_\omega^{-1}$.}

(3) {\it The sets $D$, $T$, and $H = DT$ are subgroups of $L$,
and we have $D \cong F^\times$, $T \cong F$, and $T$ is normal in $H$.}
\vspace{10pt}

\noindent {\bf Lemma 3.3.} (1) {\it Any element of $L$ is conjugate to either
$d_\omega$ for some $\omega \in F^\times$ or to $\pm t_\lambda$ for some $\lambda \in F$.}

(2) {\it If $p \neq 2$, then $L$ contains a unique element of order $2$.}

(3) {\it Let $x \in L$ of finite order $n$.  Then, $x$ is conjugate to
$\pm t_\lambda \in L$ ($\lambda \neq 0$) if and only if $p > 0$ and if $p \mid n$.
In this case, $n = p$ or $2p$.}
\vspace{10pt}

\noindent {\bf Lemma 3.4.} (1) {\it If $x^{-1} t_\lambda x = t_\mu$ for some
$x \in L$ and $\mu \neq 0$, then $x \in H$.  If $\lambda = \mu$ in addition,
then $x \in T \times Z$. Thus, if $\mu \neq 0$, then $C_L(t_\mu) = T \times Z$.}

(2) {\it If $y^{-1} d_\omega y = d_\rho$ for some $y \in L$ and
$\omega \neq \pm 1$, then $\rho = \omega$ or $\omega^{-1}$ and $y \in \langle D,w \rangle$.
If $\omega \neq \pm 1$, then $C_L(d_\omega) = D$.  If $D_1$ is a subgroup of $D$ such
that $|D_1| \geq 3$, then $N_L(D_1) = \langle D,w \rangle$.}
\vspace{10pt}

\noindent {\bf Lemma 3.5.}
\noindent {\it The centralizer of an element $x$ in $L$, $C_L(x)$,
is abelian unless $x \in Z$.}
\vspace{10pt}

\noindent {\bf Lemma 3.6.}
\noindent {\it Let $G$ be a finite subgroup of $L = SL(V)$, and let $\frak{M}$
be the set of all maximal abelian subgroups of $G$.  Assume that $G \supset Z$.}

(1) {\it If $x \in G \setminus Z$, then $C_G(x) \in \frak{M}$.}

(2) {\it For any distinct subgroups $A$ and $B$ of $\frak{M}$, $A \cap B = Z$.}

(3) {\it An element $A$ of $\frak{M}$ is either a cyclic group with
$p \nmid |A|$, or of the form $Q \times Z$ where $Q$ is an $p$-Sylow subgroup of $G$.}

(4) {\it If $A \in \frak{M}$ and $p \nmid |A|$, then $|N_G(A):A| \leq 2$.
If $|N_G(A):A| = 2$, then there is $y \in N_G(A)$ such that
$y^{-1}xy = x^{-1}$ for any element $x \in A$.}

(5) {\it Let $Q$ be an $p$-Sylow subgroup of $G$.  If $Q \neq \{1\}$,
then there is a cyclic subgroup $K$ of $G$ such that $N_G(Q) = QK$.  If
$|K| > |Z|$, then we have $K \in \frak{M}$.}
\vspace{10pt}

The proof of Theorem 3.1 is based on Suzuki \cite[Chapter 3, \textsection 6]{Su}.
To begin with, we write down some basic facts.

Let $G$ be a finite subgroup of $L$ containing $Z$.
Let $|Z| = e$ and $|G| = ge$.  Let $Q$ be an $p$-Sylow subgroup of $G$,
and let $|Q| = q$ where $q$ is a power of $p$, and $|N_G(Q):Q| = ek$.
Let $\frak{M}$ be the set of all maximal abelian subgroups of $G$.
Then, $\frak{M}$ contains all the conjugate subgroups of $Q \times Z$.
The rest subgroups of $\frak{M}$ are cyclic groups
whose orders are prime to $p$, and we partition them into several conjugate
classes $\mathcal{C}_1, \mathcal{C}_2, \cdots \mathcal{C}_{s+t}$ such that
$N_G(G_i) = G_i \mbox{ for } i \leq s$, and
$|N_G(G_j):G_j| = 2  \mbox{ for } s < j \leq s+t$
where $G_i$ is a representation of $\mathcal{C}_i$ for all $1 \leq i \leq s+t$.
Moreover, let $|G_i| = eg_i$ for all $1 \leq i \leq s+t$.  Every element of
$G$ is in some element of $\frak{M}$, and any two distinct elements of $\frak{M}$
have only the element(s) of $Z$ in common.

Fixed $i$, the number of noncentral elements containing in some conjugate subgroup
of $G_i$ is $\frac{e(g_i-1)eg}{eg_i\epsilon} = \frac{e(g_i-1)}{g_i\epsilon}$
where $\epsilon = 1$ if $i \leq s$ and $\epsilon = 2$ if $s < i \leq s+t$.
Similarly, the number of noncentral elements containing in some conjugate subgroup
of $Q \times Z$ is $\frac{e(q-1)eg}{eqk} = \frac{e(q-1)g}{qk}$.
Thus, we get
$eg = e + \frac{e(q-1)g}{qk} + \sum_{1 \leq i \leq s+t} \frac{e(g_i-1)}{g_i\epsilon}$,
or
\begin{align}
    1 = \frac{1}{g} + \frac{q-1}{qk} + \sum_{1 \leq i \leq s}\frac{g_i-1}{g_i}
    + \sum_{s+1 \leq i \leq s+t}\frac{g_i-1}{2g_i}. \tag{ 1 }
\end{align}
Note that $q = 1$ or a power of $p$, and every $g_i$ is an integer $\geq 2$.
Hence, $1 > \frac{s}{2} + \frac{t}{4}$.
Thus, we have one of the following 6 cases: (I) $(s,t) = (1,0)$; (II) $(s,t) = (1,1)$;
(III) $(s,t) = (0,0)$; (IV) $(s,t) = (0,1)$; (V) $(s,t) = (0,2)$; (VI) $(s,t) = (0,3)$.
We discuss each case separately.
\vspace{10pt}

%%%%  I  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%  I  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Theorem 3.7.}
{\it For (I), $Q \subsetneq G$ is an elementary abelian normal subgroup of $G$.
The factor group $G/Q$ is a cyclic group with $|G/Q| \mid q-1$.}
\vspace{10pt}

\noindent {\it Proof.} By Equation $(1)$,
\begin{align}
    \frac{1}{qk} + \frac{1}{g_1} = \frac{1}{g} + \frac{1}{k}. \tag{ 2 }
\end{align}
If $q = 1$, then $g = g_i$, or $G = G_1$.  Hence, $G$ is cyclic, i.e.,
$G$ is generated by $d_{\zeta_n}$ (up to isomorphism).

If $q > 1$, then $k > 1$ by Equation (2).  $k = g_1$ by Lemma 3.6(5).
Hence $n = qk$ and $G = N_G(Q) = \{ g \in G \mid g^{-1}Qg = Q \}$.
The centralizer of any element $x \in Q \setminus \{1\}$
is $Q \times Z$ by Lemma 3.6(3).
Thus, each element of $Q \setminus \{1\}$ has exactly $k$ conjugate elements in
$N_G(Q) = G = QG_1$.  Thus, $|G/Q| \mid q-1$.

Next we suppose $q > 1$, and we will determine the generators of $G$.
Let $|Q| = q = p^s$ and $Q = \langle t_{\lambda_1}, \cdots, t_{\lambda_s} \rangle$.
We may assume $\lambda_1 = 1$ (Consider $d_\omega^{-1} Q d_\omega$ instead of $Q$
where $\omega = \sqrt{1/\lambda_1}$).
Let $W$ be a finite dimensional vector space over
$\Bbb{F}_p$ with basis $\langle \lambda_1, \cdots, \lambda_s \rangle$.
Thus, $W \subset F$ and $Q \cong W$.  Let
$x = \left(\begin{array}{cc} \alpha & \gamma \\ \beta & \delta \end{array} \right) \notin Z$
be a generator for $G_1$.  Consider $x^{-1} t_1 x \in Q$,
$x^{-1} t_1 x =$
$\left(\begin{array}{cc} \delta & -\gamma \\ -\beta & \alpha \end{array} \right)$
$\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$
$\left(\begin{array}{cc} \alpha & \gamma \\ \beta & \delta \end{array} \right)$
$= \left(\begin{array}{cc} 1 + \beta\delta & \delta^2 \\ -\beta^2 & 1 - \beta\delta \end{array} \right)$.
Thus, $\beta = 0$ and
$x = \left(\begin{array}{cc} \alpha & \gamma \\ 0 & \alpha^{-1} \end{array} \right)$
where $\alpha^k = 1$ and $k \mid q-1$. Note that $t_{\mu}^{-1} x t_{\mu} = d_\alpha$
and $t_{\mu}^{-1} t_{\lambda_i} t_{\mu} = t_{\lambda_i}$
where $\mu = \frac{\gamma}{\alpha^{-1}-\alpha}$.  Thus, we may choose $x = d_{\zeta_k}$.
Thus, $x^{-1} t_1 x = t_{\zeta_k^{-2}}$ and $\zeta_k^{-2} \in W$.
$G = \langle d_{\zeta_k}, t_{\lambda_1}, \cdots, t_{\lambda_s} \rangle$ where $k \mid q-1$.
$\Box$
\vspace{10pt}

%%%%  II %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%  II %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Theorem 3.8.}
{\it For (II), $p \nmid |G|$, and $G$ is (isomorphic to) either the group
of order $4n$ defined by $\langle x,y \mid x^n = y^2, y^{-1}xy = x^{-1} \rangle$
where $n$ is odd, or $\langle d_{\zeta_4}, w, t \rangle \cong SL(2,3)$.}
\vspace{10pt}

\noindent {\it Proof.}
By Equation $(1)$,
$\frac{1}{g_1} + \frac{1}{2g_2} = \frac{1}{2} + \frac{1}{g} + \frac{q-1}{qk}$.
If $q > 1$, then the last term $\frac{q-1}{qk} \geq \frac{1}{2k}$.  Thus,
$\frac{1}{2g_2} - \frac{1}{2k} > \frac{1}{2} - \frac{1}{g_1} \geq 0$.
Thus $k > g_2$, and $k = g_1$ by Lemma 3.6(5).  Thus,
$\frac{1}{2g_1} + \frac{1}{2g_2} > \frac{1}{2}$, and it is absurd.
Thus, $q = 1$, and
$\frac{1}{g_1} + \frac{1}{2g_2} = \frac{1}{2} + \frac{1}{g} > \frac{1}{2}$.
Thus, we have one of the following two cases:
(i) $g_1 = 2, g_2 = \frac{1}{2}g$; (ii) $g_1 = 3, g_2 = 2, g = 12$.

For (i), $G = N_G(G_2)$.  Thus, $G = \langle d_{\zeta_{2n}}, w \rangle
\cong \langle x,y \mid x^n = y^2, y^{-1}xy = x^{-1} \rangle$.  Moreover,
$n$ is odd since each element of order 4 is conjugate to $w$.

For (ii), $p \neq 2$ and $e = 2$ since $2 \mid g$.  Thus, $|G| = 24$.
Let $x = d_{\zeta_4}$ be a generator for $G_2$.
There is an element $y = w$ of $G' = N_G(G_2)$ such that $y^{-1} x y = x^{-1}$.
We will show that $G'$ is normal in $G$.
$\langle y \rangle$ is conjugate to $G_2$ since the order of $y$ is 4.
Similarly, all three cyclic subgroups of order 4 of $G'$ are conjugate.
Since $|G:G'| = 3$ and no other conjugate subgroups of $G_2$ exist,
we get $G'$ is normal in $G$.  Let $t$ be a generator for $G_1$.
Since $t$ is conjugate to $d_{\zeta_3}$ and the trace of $d_{\zeta_3}$ is
$\zeta_3 + \zeta_3^{-1} = -1$, we may assume
$t = \left(\begin{array}{cc} \alpha & \gamma \\ \beta & -1-\alpha \end{array} \right)$.
Consider the element $t^{-1}xt \in G'$. $t^{-1}xt$
$= \left(\begin{array}{cc} (-2\alpha^2-2\alpha-1)\sqrt{-1} & -2(\alpha+1)\gamma\sqrt{-1} \\
-2\alpha\beta\sqrt{-1} & (2\alpha^2+2\alpha+1)\sqrt{-1} \end{array} \right)$.
If $t^{-1}xt$ is of the form $\left(\begin{array}{cc} * & 0 \\ 0 & * \end{array} \right)$, then $-2\alpha\beta\sqrt{-1} =
-2(\alpha+1)\gamma\sqrt{-1} = 0$, and it is absurd.  Thus, $t^{-1}xt$ is of the form $\left(\begin{array}{cc} 0 &
* \\ * & 0 \end{array} \right)$.  Thus, $t^{-1}xt = \pm \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right), \pm
\left(\begin{array}{cc} 0 & \sqrt{-1} \\ \sqrt{-1} & 0 \end{array} \right)$. Thus, $(-2\alpha^2-2\alpha-1)\sqrt{-1} = 0$, and
$\alpha = \frac{-1 \pm \sqrt{-1}}{2}$. Hence, we have
$t = \left(\begin{array}{cc} \frac{-1+\sqrt{-1}}{2} & \frac{-1+\sqrt{-1}}{2} \\
\frac{1+\sqrt{-1}}{2} & \frac{-1-\sqrt{-1}}{2} \end{array}\right)$ and $t^{-1}xt = xy$.
In fact, there are all eight possible choices of $t$.  If $t'$ is any other choice,
then $t' \in \langle x, y, t \rangle$.  Hence, $G = \langle d_{\zeta_4}, w, t \rangle$.
Define a group homomorphism $\phi$ from $\langle d_{\zeta_4}, w, t \rangle$ to $SL(2,3)$
by
$\phi(x) = \left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right)$,
$\phi(y) = \left(\begin{array}{cc} 1 & -1 \\ -1 & -1 \end{array} \right)$,
and $\phi(t) = \left(\begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array} \right)$.
Hence, $\phi$ is an isomorphism, i.e., $G = \langle d_{\zeta_4}, w, t \rangle
\cong SL(2,3)$. $\Box$
\vspace{10pt}

%%%% III %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%% III %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Theorem 3.9.}
{\it For (III), $G = Q \times Z$.}
\vspace{10pt}

\noindent {\it Proof.} Trivial. $\Box$
\vspace{10pt}

%%%%  IV %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%  IV %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Theorem 3.10.}
{\it For (IV), either $p = 2$ and $G$ is a dihedral group of order $2n$ with odd $n$,
or $p = 3$ and $G \cong \langle d_{\zeta_4}, w, t \rangle \cong SL(2,3)$.}
\vspace{10pt}

\noindent {\it Proof.}
By Equation $(1)$, $\frac{1}{2} + \frac{1}{2g_1} = \frac{1}{g} + \frac{q-1}{qk}$.
$\frac{q-1}{qk} \geq \frac{1}{2}$ since $g \geq 2g_1$.  Thus, $q > 1$ and $k = 1$.
Thus, $\frac{1}{q} + \frac{1}{2g_1} = \frac{1}{2} + \frac{1}{g}$.
The proof is the same as the proof of case (ii) of Theorem 3.8.
Moreover, if $p = 3$ then we can put $G = SL(2,3)$ into $F$.
In this situation, $G$ is generated by
$\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right)$,
$\left(\begin{array}{cc} 1 & -1 \\ -1 & -1 \end{array} \right)$, and
$\left(\begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array} \right)$.
$\Box$
\vspace{10pt}

%%%%  VI %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%  VI %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Theorem 3.11.}
{\it For (VI), $G$ is (isomorphic to) one of the following three groups:
$\langle x,y \mid x^n = y^2, y^{-1}xy = x^{-1} \rangle$,
$\langle d_{\zeta_8}, w, t \rangle$, or $SL(2,5)$.}
\vspace{10pt}

\noindent {\it Proof.}
By Equation $(1)$,
\begin{align}
    \frac{1}{2g_1} + \frac{1}{2g_2} + \frac{1}{2g_3} =
    \frac{1}{2} + \frac{1}{g} + \frac{q-1}{qk}. \tag{ 3 }
\end{align}
If $q > 1$, then $k = g_i$ for some $i$ by Lemma 3.6 (5).
Thus, the last term of Equation (3) $\frac{q-1}{qk} \geq \frac{1}{2g_i}$,
and it is absurd.
Thus, $q = 1$, and $\frac{1}{2g_1} + \frac{1}{2g_2} + \frac{1}{2g_3} =
\frac{1}{2} + \frac{1}{n} > \frac{1}{2}.$

We may assume that $1 < g_1 \leq g_2 \leq g_3$ without loss of generality.
Thus, we have one of the following four cases:

(i) $g_1 = 2, g_2 = 2, g_3 = \frac{1}{2}g$.

(ii) $g_1 = 2, g_2 = 3, g_3 = 3, g = 12$.

(iii) $g_1 = 2, g_2 = 3, g_3 = 4, g = 24$.

(iv) $g_1 = 2, g_2 = 3, g_3 = 5, g = 60$.

For (i), $G = N_G(G_3)$.
Thus, $G = \langle d_{\zeta_{2n}}, w \rangle
\cong \langle x,y \mid x^n = y^2, y^{-1}xy = x^{-1} \rangle$.
Moreover, $n$ is even since $G_1$ is not conjugate to $G_2$.

For (ii), $G_2$ is conjugate to $G_3$ by Sylow's theorem, and it is absurd.

For (iii), $p \neq 2$ and $e = 2$ since $2 \mid g$.  Thus, $|G| = 48$.
Let $x = d_{\zeta_8}$ be a generator for $G_3$.
Similar to Proof of case (ii) of Theorem 3.8,
we have $G = \langle d_{\zeta_8}, w, t \rangle$ where
$t = \left(\begin{array}{cc} \frac{-1+\sqrt{-1}}{2} & \frac{-1+\sqrt{-1}}{2} \\
\frac{1+\sqrt{-1}}{2} & \frac{-1-\sqrt{-1}}{2} \end{array} \right)$.

For (iv), $p \neq 2$ and $e = 2$ since $2 \mid g$.  Thus, $|G| = 120$.
Now we determine the structure of $G$.
A 2-Sylow subgroup of $G$ is the quaternion group, i.e.,
$G \cong \langle x, y \mid x^2 = y^2, y^{-1}xy = x^{-1} \rangle$.
Each 2-Sylow subgroup $G'$ contains exactly 3 conjugate subgroups of $G_1$, and
$G'$ is the normalizer of each of these 3 conjugate subgroups.
Since $G$ has exactly 15 conjugate subgroups of $G_1$, there are 5 2-Sylow subgroups.
By Sylow's theorem, there is a homomorphism from $G$ into $S_5$,
the symmetric group of degree 5.
But any element of $G \setminus Z$ moves some 2-Sylow subgroup.
Therefore, $G/Z$ is isomorphic to a subgroup of $S_5$.
Since $|G/Z| = 60$ and the image of $G/Z$ is a normal subgroup of $S_5$,
$G/Z \cong A_5$, the alternating group of degree 5.
Thus, we have $G \cong SL(2,5)$.

In this case, we want to determine the generators of $G$.  Note that
$G/Z \cong A_5 =$ $\langle a, b, c \mid a^5 = b^2 = c^2 = 1, bcb^{-1} = c \rangle$.
So we can choose
$a = \left(\begin{array}{cc} \zeta^3 & 0 \\ 0 & \zeta^2 \end{array} \right)$,
$b = w$, and
$c = \left(\begin{array}{cc} \frac{\zeta^4-\zeta}{\sqrt{5}} & \frac{\zeta^2-\zeta^3}{\sqrt{5}} \\
\frac{\zeta^2-\zeta^3}{\sqrt{5}} & \frac{\zeta^-\zeta^4}{\sqrt{5}} \end{array} \right)$
where $\zeta = \zeta_5$.
$\Box$
\vspace{10pt}

%%%%  V  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%  V  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Theorem 3.12.}
{\it For (V), we have one of the following three cases:}

(i) {\it $G \cong SL(2,\Bbb{F}_q)$.}

(ii) {\it $G \cong \langle SL(2,\Bbb{F}_q), d_\pi \rangle$ where
$\pi \in \Bbb{F}_{q^2}$ and $\langle \pi^2 \rangle = \Bbb{F}_q^\times$.}

(iii) {\it $p = q = 3$ and $G \cong SL(2,5)$.}
\vspace{10pt}

By Equation $(1)$,
$\frac{1}{2g_1} + \frac{1}{2g_2} = \frac{1}{g} + \frac{q-1}{qk} > \frac{q-1}{qk}$.
$q > 1$ trivially.
Since $\frac{q-1}{qk} \geq \frac{1}{2}$ and
$\frac{1}{2g_1} + \frac{1}{2g_2} \leq \frac{1}{2}$, we have $k > 1$.
Thus, $k = g_1$ without loss of generality.
Rewrite the above equation, and we have
\begin{align}
    \frac{1}{2g_2} = \frac{1}{g} + \frac{1}{2g_1} - \frac{1}{qg_1}. \tag{ 4 }
\end{align}
Thus, $g_1 < g_2$.
Also, $k \mid q-1$ by the proof of Theorem 3.7.
Let $ag = 2g_1g_2q$, and multiply $ag$ on both sides of Equation (4).
$g_2 q + g_1 q = a + (q-1)g_2$, $g_1 q = a + (q-2) g_2$.
Thus, $a$ is a positive integer, $g_2 \equiv a (\mbox{mod } g_1)$, and
$g_1 > \frac{(q-2)g_2}{q}$.
$a \mid g_1$ since $\frac{g}{2g_2} = |G:N_G(G_2)|$.

%%%%  g \geq 4  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

If $q \geq 4$, then $2g_1 > g_2 > g_1$ and $g_2 = a + lg_1$ for some integer $l$.
Thus, $l = 1$ and $g_2 = g_1 + a$. $g_1 q = a + (q-2)g_2 = a + (q-2)(g_1 + a)$.
Thus, $2g_1 = a(q-1)$, $2g_2 = a(q+1)$, and $2g = aq(q^2-1)$.
Note that $\frac{2}{a} = \frac{q-1}{g_1}$ is a positive integer, $a = 1, 2$.
Let $d = \frac{2}{a}$.
Thus, we have $g_1 = \frac{q-1}{d}$, $g_2 = \frac{q+1}{d}$,
and $g = \frac{q(q^2-1)}{d}$ ($d = 1$ or $2$).
Next we need to determined the structure of $G$.
\vspace{10pt}

\noindent{\bf{Lemma 3.13.}}
{\it If $d = 2$, then $G \cong SL(2,\Bbb{F}_q)$.}
\vspace{10pt}

\noindent {\it Proof.}
Let $Q$ be a $p$-Sylow subgroup of $G$ and $N_G(Q) = QG_1$ where $|G_1| = q-1$.
Thus, $Q \subset T$, $G_1 \subset D$, and $t_1 \in Q$ by the proof of Theorem 3.7.
Let $d_{\zeta_{q-1}} = d_\sigma$ be a generator for $G_1$.
Thus, $\sigma^{q-1} = 1$.  $\sigma \in \Bbb{F}_q^\times$.
Thus, $d_\sigma \in SL(2,\Bbb{F}_q)$.
Since $|N_G(G_1):G_1| = 2$, there is $y = d_\beta w \in G$.
Now we let $W = \{ \lambda \mid t_\lambda \in Q \}$.
We will show that $W = \Bbb{F}_q$.
Observe that $G = N_G(Q) \cup N_G(Q) y Q$.
In fact, $|N_G(Q) y Q| = |N_G(Q)||Q:Q \cap y^{-1} N_G(Q) y|$ and
$Q \cap y^{-1} N_G(Q) y = \{1\}$.  Thus, $|N_G(Q) \cup N_G(Q) y Q| = |G|$.
In particular, for any $t_\lambda \in Q \setminus \{1\}$,
there exist $d_\omega t_\mu \in N_G(Q)$ and $t_\nu \in Q$ such that
$y^{-1} t_\lambda y = d_\omega t_\mu y t_\nu$.
$\left(\begin{array}{cc} 1 & 0 \\ \beta^{-2}\lambda & 1 \end{array} \right)$
$= \left(\begin{array}{cc} -\beta^{-1}\omega\mu & \beta\omega - \beta^{-1}\omega\mu\nu \\
-\beta^{-1}\omega^{-1} & -\beta^{-1}\omega^{-1}\nu \end{array} \right)$.
Thus, $\omega\lambda = \beta$.
We may put $\lambda = 1$ since $t_1 \in Q$, and
we get $\beta = \omega \in \Bbb{F}_q^\times$.
Thus, $w = d_\beta^{-1}y \in N_G(G_1)$ and we may re-choose $y = w$.
Thus by the above equation, $\lambda = -\omega^{-1}$ for all $\lambda \in W$.
Hence $W \subset \Bbb{F}_q$, and $W = \Bbb{F}_q$ by observing $|W| = q = |\Bbb{F}_q|$.
Thus, $G = \langle t_\lambda, d_\pi, w \mid \lambda \in \Bbb{F}_q,
\sigma \in \Bbb{F}_q^\times \rangle = SL(2,\Bbb{F}_q)$.
$\Box$
\vspace{10pt}

\noindent{\bf{Lemma 3.14.}}
{\it If $d = 1$, then $G \cong \langle SL(2,\Bbb{F}_q), d_\pi \rangle$ where $\pi \in
\Bbb{F}_{q^2}$ and $\langle \pi^2 \rangle = \Bbb{F}_q^\times$.}
\vspace{10pt}

\noindent {\it Proof.}
It is similar to the proof of Lemma 3.13.
With the same notation, $|G_1| = 2(q-1)$.
Let $d_{\zeta_{2(q-1)}} = d_\sigma$ be a generator for $G_1$.
Thus, $\sigma^{2(q-1)} = 1$.
$\sigma^2 \in \Bbb{F}_q^\times$ and $\sigma \in \Bbb{F}_{q^2}^\times$.
Thus, $d_\sigma^2 \in SL(2,\Bbb{F}_q)$ and $d_\sigma \in SL(2,\Bbb{F}_{q^2})$.
Similarly, we have $G = \langle t_\lambda, d_\sigma,
w \mid \lambda \in \Bbb{F}_q,
\sigma \in \Bbb{F}_{q^2}^\times, \sigma^2 \in \Bbb{F}_q^\times \rangle
= \langle SL(2,\Bbb{F}_q), d_\sigma \rangle$.
$\Box$
\vspace{10pt}

If $q \leq 3$, then $q = 3$ and $g_1 = 2$.
Thus, $2 < g_2 < 6$.
$g_2 = 4$ or $g_2 = 5$ since $p \nmid g_2$.
\vspace{10pt}

\noindent{\bf{Lemma 3.15.}}
{\it If $q = 3$, $g_1 = 2$, $g_2 = 4$ and $g = 24$,
then $G \cong \langle SL(2,\Bbb{F}_3), d_\sigma \rangle$ where
$\sigma \in \Bbb{F}_9$ and $\langle \sigma^2 \rangle = \Bbb{F}_3^\times$.}
\vspace{10pt}

\noindent {\it Proof.}
The proof is the same as the proof of Lemma 3.13. $\Box$
\vspace{10pt}

\noindent{\bf{Lemma 3.16.}}
{\it If $q = 3$, $g_1 = 2$, $g_2 = 5$ and $g = 60$, then $G \cong SL(2,5)$.}
\vspace{10pt}

\noindent {\it Proof.}
The proof is the same as the proof of case (iv) of Theorem 3.11.
Note that $A_5 = \langle a, b, c \mid
a^3 = b^2 = c^2 = (ab)^3 = (bc)^3 = (ac)^2 = 1 \rangle$,
we can choose $a = t_1$,
$b = \left(\begin{array}{cc} -\sqrt{-1} & 0 \\ 1 & \sqrt{-1} \end{array} \right)$,
and $c = d_{\zeta_4}$ as a set of generators of $G$.
$\Box$
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\it Proof of Theorem 3.1.}

By Theorem 3.7 to 3.12,  it remains to prove this theorem when $G \nsupseteq Z$.
In this situation we have $p \neq 2$ and $|G|$ is odd.
Thus, $GZ = G \times Z$.  Running through the cases of (I) to (VI),
we see that $G$ is cyclic or $Q$ is normal in $G$.
Therefore, we have (I) or (VI).
$\Box$
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Modular Invariants}
\hspace{13pt} In this section, we will determine the invariants of finite subgroups
of $SL(2,F)$ where $p$ divides the order of the groups.
We may say that these invariants are called modular invariants.
To begin with, recall Definition 2.1,
and let $F$ be an algebraically closed field of characteristic $p > 0$.
Let $G$ be a finite group of $SL(2,F)$ such that $p \mid |G|$.
\vspace{10pt}

\noindent {\bf Lemma 4.1.}
{\it If $G$ is a finite group of $SL(2,F)$ and $H$ is a normal subgroup of $G$,
then $F[x,y]^G = (F[x,y]^H)^{G/H}$.}
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Lemma 4.2.}
{\it Let $G = \langle d_{\zeta} \rangle$ be a cyclic subgroup of $SL(2,F)$ where
$\zeta = \zeta_n$ and $n \nmid p$.  Then $F[x,y]^G = F[xy, x^n, y^n]$
with the relation $x^n y^n - (xy)^n = 0$.}
\vspace{10pt}

\noindent {\it Proof.}
$d_{\zeta}: x \mapsto \zeta x$, and $d_{\zeta}: y \mapsto \zeta^{-1} y$.
Since $\zeta^n = 1 \in F$, $F[x,y]^G = F[xy, x^n, y^n]$.
The relation is $x^n y^n - (xy)^n = 0$.
$\Box$
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Lemma 4.3.}
{\it Let $G = \langle d_{\lambda} \mid \lambda \in \Bbb{F}_q \rangle$
be a subgroup of $SL(2,F)$ of order $q$ where $q$ is a power of $p$.
Then $F[x,y]^G = F[y, x(x^{q-1} - y^{q-1})]$.}
\vspace{10pt}

\noindent {\it Proof.}
Let $f = x(x^{q-1} - y^{q-1})$.
For lucidity, we consider the following diagram and
divide the proof into four steps.

\Diagram
\qquad\qquad\qquad\qquad & & & F[x,y] & \rLine & \rLine & F(x,y) & & \\
\qquad\qquad\qquad\qquad & & & \uInto &        &        & \uInto & & \\
\qquad\qquad\qquad\qquad & & & F[x,y]^G & \rLine & \rLine & F(x,y)^G & & \\
\qquad\qquad\qquad\qquad & & & \uInto &        &        & \uBar & & \\
\qquad\qquad\qquad\qquad & & & F[y,f] & \rLine & \rLine & F(y,f) & & \\
\endDiagram

\noindent \underline{Step 1.} $F[x,y]^G$ is integral over $F[y,f]$.

It suffices to show that $F[x,y]$ is integral over $F[y,f]$.
$x \in F[x,y]$ satisfies the equation
$h(T) = \prod_{\lambda \in \Bbb{F}_q}(T - (x+\lambda y)) = 0$.
Direct computing shows that
$\prod_{\lambda \in \Bbb{F}_q}(T - (x+\lambda y)) = T^q - y^{q-1}T - f
\in F[y,f][T]$.  Hence $x$ is integral over $F[y,f]$.
Also, $y$ and $F$ are integtal over $F[y,f]$ trivially.
Hence $F[x,y]$ is integral over $F[y,f]$ by
Atiyah-Macdonald \cite[Corollary 5.3]{AM}.

\noindent \underline{Step 2.} $F(x,y)^G = F(y,f)$.

\noindent \underline{Step 3.}

\noindent \underline{Step 4.}

$\Box$
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Theorem 4.4.}
{\it Let $F$ be an algebraically closed field with characteristic $p > 0$.
Let $G = N_G(Q) = \{ g \in G \mid g^{-1}Qg = Q \} =
\langle t_\lambda, d_{\zeta_n} \rangle$ be a finite group of $SL(2,F)$
where $Q$ is a $p$-Sylow subgroup of $G$ of order $q = p^s$ for some
positive integer $s$ and $n \mid q-1$, then
\[ F[x,y]^G = F[xy(x^{q-1}-y^{q-1}), x^n(x^{q-1}-y^{q-1})^n, y^n] \]
with the relation $(xy(x^{q-1}-y^{q-1}))^n - (x^n(x^{q-1}-y^{q-1})^n)(y^n) = 0$.}
\vspace{10pt}

\noindent {\it Proof.}
Let $Q = \langle t_\lambda \mid \lambda \in \Bbb{F}_q \rangle$ be a normal
subgroup of $G$ by definition.
By Lemma 4.3, $F[x,y]^Q = F[y,f]$ where $f = x(x^{q-1} - y^{q-1})$.
By Lemma 4.1, $F[x,y]^G = (F[x,y]^Q)^{G/Q} = F[y,f]^{\langle d_\zeta \rangle}$.
Direct computing shows that
$d_\zeta \cdot y = \zeta^{-1}y$,
$d_\zeta \cdot f = \zeta f$. (Note that $n \mid q-1$, and $\zeta^{q-1} = 1$.)
Hence
$F[y,f]^{\langle d_\zeta \rangle} = F[yf, y^n, f^n]$ with the relation
$(y^n)(f^n) - (yf)^n = 0$.
$\Box$
\vspace{10pt}

\noindent {\bf Theorem 4.5.}
{\it Let $F$ be an algebraically closed field with characteristic $p = 2$.
Let $G = \langle d_{\zeta_n}, w \rangle \subset SL(2,F)$ be a dihedral group
of order $2n$ with odd $n$, then \[ F[x,y]^G = F[xy, x^n+y^n] \].}
\vspace{10pt}

\noindent {\it Proof.}
Let $H = \langle d_{\zeta_n}, w \rangle$ be a normal subgroup of $G$.
By Lemma 4.1, $F[x,y]^H = F[xy,x^n,y^n] = F[f_1,f_2,f_3]$ where
$f_1 = xy$, $f_2 = x^n+y^n$ and $f_3 = y^n$.
By Lemma 4.2, $F[x,y]^G = (F[x,y]^H)^{G/H} = F[f_1,f_2,f_3]^{\langle w \rangle}$.
Direct computing shows that
$w \cdot f_1 = f_1$,
$w \cdot f_2 = f_2$,
$w \cdot f_3 = x^n = f_2 + f_3$.
Hence $F[x,y]^G = F[f_1,f_2,f_3(f_2+f_3)] = F[xy,x^n+y^n,x^n y^n] = F[xy,x^n+y^n]$.
$\Box$
\vspace{10pt}

\noindent {\bf Theorem 4.6.}
{\it Let $F$ be an algebraically closed field with characteristic $p > 0$.
Let $G = SL(2,\Bbb{F}_q)$ be a subgroup of $SL(2,F)$ where $q$ is a power of $p$,
then \[ F[x,y]^G = F[x^qy - xy^q, \frac{x^{q^2}y-xy^{q^2}}{x^qy-xy^q}] \].}
\vspace{10pt}

\noindent {\it Proof.}
See \cite{Wi}.  $\Box$
\vspace{10pt}

\noindent {\bf Theorem 4.7.}
{\it Let $F$ be an algebraically closed field with characteristic $p > 0$.
Let $G = \langle SL(2,\Bbb{F}_q), d_\pi \rangle$  be a subgroup of $SL(2,F)$
where $\pi \in \Bbb{F}_{q^2} \setminus \Bbb{F}_q$ and
$\langle \pi^2 \rangle = \Bbb{F}_q^\times$, then
\[ F[x,y]^G = F[(x^qy - xy^q)^2, \bigg(\frac{x^{q^2}y-xy^{q^2}}{x^qy-xy^q}\bigg)^2,
x^{q^2}y-xy^{q^2}] \] with the relation
$(x^qy - xy^q)^2 \big(\frac{x^{q^2}y-xy^{q^2}}{x^qy-xy^q}\big)^2-
(x^{q^2}y-xy^{q^2})^2 = 0$.}
\vspace{10pt}

\noindent {\it Proof.}
Let $H = SL(2,\Bbb{F}_q)$ be a normal subgroup of $G$..
By Theorem 4.6, $F[x,y]^H = F[f_1,f_2]$ where $f_1 = x^qy - xy^q$,
$f_2 = \frac{x^{q^2}y-xy^{q^2}}{x^qy-xy^q}$.
By Lemma 4.1, $F[x,y]^G = (F[x,y]^H)^{G/H} = F[f_1,f_2]^{\langle d_\pi \rangle}$.
By $\pi^{q-1} = -1$, $d_\pi \cdot f_1 = -f_1$, $d_\pi \cdot f_2 = -f_2$.
Hence, $F[f_1,f_2] = F[f_1^2, f_2^2, f_1f_2]$.
The relation is $(f_1^2)(f_2^2) - (f_1f_2)^2 = 0$. $\Box$
\vspace{10pt}


\noindent {\bf Theorem 4.8.}
{\it Let $F$ be an algebraically closed field with characteristic $p = 3$.
Let $G = \bigg\langle
t_1, \left(\begin{array}{cc} -\sqrt{-1} & 0 \\ 1 & \sqrt{-1} \end{array} \right),
d_{\zeta_4} \bigg\rangle$ be a subgroup of $SL(2,F)$, then
\[ F[x,y]^G = F[x^9y - xy^9, x^{12}-x^{10}y^2 + x^6y^6 - x^2y^{10} - y^{12}]. \]}
\vspace{10pt}

The proof is similar to one of Lemma 4.1.
For simplicity, let
$\alpha = t_1$,
$\beta = \left(\begin{array}{cc} -\sqrt{-1} & 0 \\ 1 & \sqrt{-1} \end{array} \right)$,
and $\gamma = d_{\zeta_4}$.
Let $H = \langle \alpha, \beta \rangle$.  Then
$H = \left(\begin{array}{cc} \sqrt{-1} & 1 \\ 0 & \sqrt{-1} \end{array} \right)
SL(2,\Bbb{F}_3)
\left(\begin{array}{cc} -\sqrt{-1} & 1 \\ 0 & -\sqrt{-1} \end{array} \right)$.
($\Bbb{F}_3 = \{ 0, 1, -1 \}$).
Changing of coordinate and Theorem 4.6 imply that $F[x,y]^H = F[f_1,f_2]$ where
\begin{align*}
    f_1
    &= \left(\begin{array}{cc} -\sqrt{-1} & 1 \\ 0 & -\sqrt{-1} \end{array} \right)
    \cdot (xy^3 - x^3y) \\
    &= xy^3 - x^3y - \sqrt{-1}y^4, \\
    f_2
    &= \left(\begin{array}{cc} -\sqrt{-1} & 1 \\ 0 & -\sqrt{-1} \end{array} \right)
    \cdot (x^6 + x^4y^2 + x^2y^4 + y^6) \;\;\;\;\;\; \\
    &= x^6 + x^4y^2 - \sqrt{-1}x^3y^3 + x^2y^4 + \sqrt{-1}xy^5.
\end{align*}
Now we let
\begin{align*}
    \varphi_1 &= f_1 f_2 = x^9y - xy^9, \\
    \varphi_2 &= -\sqrt{-1}f_1^3+f_2^2 = x^{12}-x^{10}y^2 + x^6y^6 - x^2y^{10} - y^{12}.
\end{align*}
Surely, $\varphi_1$ and $\varphi_2$ are two invariants of $G$, and we will
show that $F[x,y]^G = F[\varphi_1,\varphi_2]$ exactly.

To begin with, let
$\Pi =
\left(\begin{array}{cc} \sqrt{-1} & 1 \\ 0 & \sqrt{-1} \end{array} \right)
SL(2,\Bbb{F}_9)
\left(\begin{array}{cc} -\sqrt{-1} & 1 \\ 0 & -\sqrt{-1} \end{array} \right)$.
($\Bbb{F}_9 = \{ 0, \pm 1, \pm \sqrt{-1}, 1 \pm \sqrt{-1}, -1 \pm \sqrt{-1}\}$).
Note that $G \subsetneq \Pi$.
Changing of coordinate and Theorem 4.6 imply that
$F[x,y]^\Pi = F[\Psi_1,\Psi_2]$ where
\begin{align*}
    \Psi_1 =& x^9y - xy^9, \\
    \Psi_2 =& x^{72} + x^{64}y^8 + x^{56}y^{16} + x^{48}y^{24} + x^{40}+y^{32}
    + x^{32}y^{40} \;\;\\
    &+ x^{24}y^{48} + x^{16}y^{56} + x^8y^{64} + y^{72}.
\end{align*}

By direct computing, the relation between $\varphi_1$, $\varphi_2$, $\Psi_1$, $\Psi_2$
are $\Psi_1 = \varphi_1$ and $\Psi_2 = \varphi_2^6 - \varphi_1^6 \varphi_2$.
Now we consider the following diagram:

\Diagram
\qquad\qquad\qquad\qquad & & & F[x,y] & \rLine & \rLine & F(x,y) & & \\
\qquad\qquad\qquad\qquad & & & \uInto &        &        & \uInto & & \\
\qquad\qquad\qquad\qquad & & & F[x,y]^G & \rLine & \rLine & F(x,y)^G & & \\
\qquad\qquad\qquad\qquad & & & \uInto &        &        & \uInto & & \\
\qquad\qquad\qquad\qquad & & & F[\varphi_1,\varphi_2] & \rLine & \rLine & F(\varphi_1,\varphi_2) & & \\
\qquad\qquad\qquad\qquad & & & \uInto &        &        & \uInto & & \\
\qquad\qquad\qquad\qquad &F[\Psi_1,\Psi_2] & \rEq & F[x,y]^\Pi & \rLine & \rLine & F(x,y)^\Pi & \rEq & F(\Psi_1,\Psi_2) \\
\endDiagram
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Lemma 4.9.} {\it $F[x,y]$ is integral over $F[\varphi_1,\varphi_2]$.}
\vspace{10pt}

\noindent {\it Proof.}
$F[\varphi_1,\varphi_2] \subset F[x,y]^H = F[f_1,f_2] \subset F[x,y]$.
If $F[x,y]$ is integral over $F[x,y]^H$ and
$F[x,y]^H = F[f_1,f_2]$ is integral over $F[\varphi_1,\varphi_2]$, then
$F[x,y]$ is integral over $F[\varphi_1,\varphi_2]$ by the transitivity of
integral dependence \cite[Corollary 5.4]{AM}.

First, it is well-known that $F[x,y]$ is integral over $F[x,y]^H = F[f_1,f_2]$
\cite[Corollary 4.1.2.(i)]{Sp}.
Next, $\varphi_1 = f_1 f_2$ and $\varphi_2 = -\sqrt{-1}f_1^3+f_2^2$ imply that
$f_1$ satisfies the equation $T^5 - \sqrt{-1}\varphi_2 T^2 + \sqrt{-1}\varphi_1^2 = 0$,
and $f_2$ satisfies the equation $T^5 - \varphi_2T^3 - \sqrt{-1}\varphi_1^3 = 0$.
Also, $F \subset F[\varphi_1,\varphi_2]$ is integral over $F[\varphi_1,\varphi_2]$
trivially.  Hence $F[x,y]^H$ is integral over $F[\varphi_1,\varphi_2]$ by
Atiyah-Macdonald \cite[Corollary 5.3]{AM}. $\Box$
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent{\bf{Corollary 4.10.}} {\it $F[x,y]^G$ is integral over
$F[\varphi_1,\varphi_2]$.}
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Lemma 4.11.} {\it $\varphi_1 \in F(x,y)$ and
$\varphi_2 \in F(x,y)$ are algebraically independent over $F$. }
\vspace{10pt}

\noindent {\it Proof.}
If not, then there exists a non-zero polynomial $f(X,Y) \in F[X,Y] = F[X][Y]$
such that $f(\varphi_1,\varphi_2) = 0$.  Write
$f(X,Y) = \sum_{i=0}^N a_i X^i$ with the minimal positive degree $N$ where
$a_i = a_i(Y) = \sum_{j=0}^{N_i} b_{i,j} Y^j \in F[Y]$ ($b_{i,j} \in F$) for all
$0 \leq i \leq N$.

$f(\varphi_1,\varphi_2) = 0$ implies $\sum_{i=0}^N a_i(\varphi_2) \varphi_1^i = 0$.
Thus, $-\varphi_1\Big(\sum_{i=1}^N a_i(\varphi_2) \varphi_1^{i-1} \Big) =
a_0(\varphi_2)$ $\in F[x,y]$.  Put $y = 0$, and we get $\varphi_1 = 0$
and $\varphi_2 = x^{12}$.  Thus, $0 = a_0(x^{12}) =
\sum_{j=0}^{N_0} b_{0,j} (x^{12})^j = \sum_{j=0}^{N_0} b_{0,j}x^{12^j}$ $\in F[x]$.
Thus, $b_{0,j} = 0$ for all $0 \leq j \leq N_0$. $a_0 = 0 \in F[Y]$.
Thus, $\varphi_1\Big(\sum_{i=1}^N a_i(\varphi_2) \varphi_1^i \Big) = 0$ $\in F[x,y]$.
$\varphi_1 \neq 0$ implies that $\sum_{i=1}^N a_i(\varphi_2) \varphi_1^{i-1} = 0$.
Let $f'(X,Y) = \sum_{i=0}^{N-1} a_{i+1} X^i \in F[X,Y]$.
$f'(\varphi_1,\varphi_2) = 0$ and it is absurd. $\Box$
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent{\bf{Corollary 4.12.}} {\it $F[\varphi_1,\varphi_2]$ is a
polynomial ring in two variables, $\varphi_1$ and $\varphi_2$, over $F$.}
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Lemma 4.13.} {\it If $\Pi'$ is a subgroup of $\Pi$
containing $G$, then either $\Pi' = \Pi$ or $\Pi' = G$. }
\vspace{10pt}

\noindent {\it Proof.}
Since $|\Pi| = 720$ and $|G| = 120$, $|\Pi'| = 120k$ for $1 \leq k \leq 6$.
Now we can apply Main Theorem to get a contradiction when $1 < k < 6$.
Assume $1 < k < 6$.
If $\Pi'$ appears in case (6) of Main Theorem, then
$|\Pi'| = |\langle t_\lambda, d_{\zeta_n} \rangle| = nq$ where
$q$ is a power of $3$ and $n \mid q-1$.  Direct checking shows that it is absurd.
Trivially, $\Pi'$ cannot appear in case (7) or (8) of Main Theorem.
If $\Pi'$ appears in case (9) of Main Theorem, then
$|\Pi'| = |SL(2,\Bbb{F}_q)| = q(q^2-1)$ where $q$ is a power of $3$.  It is absurd.
Similarly, if $\Pi'$ appears in case (10) of Main Theorem, then
$|\Pi'| = |\langle SL(2,\Bbb{F}_q), d_\pi \rangle| = 2q(q^2-1)$ where
$q$ is a power of $3$. It is absurd again.  Hence $|\Pi'| = 720$ or $120$,
i.e., $\Pi' = \Pi$ or $\Pi' = G$. $\Box$
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Lemma 4.14.} {\it $F(x,y)^G = F(\varphi_1,\varphi_2)$.}
\vspace{10pt}

\noindent {\it Proof.}
$F(x,y)$ is finite Galois extension of $F(x,y)^\Pi$, and its Galois group is $\Pi$
\cite[Chapter VI, \textsection 1, Theorem 1.8]{La}.
Since $F(\varphi_1,\varphi_2)$ is an intermediate field,
$F(x,y)^\Pi \subseteq F(\varphi_1,\varphi_2) \subseteq F(x,y)$,
$F(\varphi_1,\varphi_2) = F(x,y)^{\Pi'}$ for some subgroup $\Pi'$ of $\Pi$
\cite[Chapter VI, \textsection 1, Theorem 1.1]{La}.

$G \subseteq \Pi' \subseteq \Pi$ since
$F(x,y)^G \subseteq F(\varphi_1,\varphi_2) = F(x,y)^{\Pi'}$.
$\Pi' = \Pi$ or $\Pi' = G$ by Lemma 4.13.  If $\Pi' = G$, then we are done.
Now we assume that $\Pi' = \Pi$.  If we can show that
$F[\varphi_1,\varphi_2] = F[\Psi_1,\Psi_2] = F[x,y]^\Pi$, then we get a contradiction.

\Diagram
\qquad\qquad\qquad\qquad & & & F[x,y] & \rLine & \rLine & F(x,y) & & \\
\qquad\qquad\qquad\qquad & & & \uInto &        &        & \uInto & & \\
\qquad\qquad\qquad\qquad & & & F[\varphi_1,\varphi_2] & \rLine & \rLine & F(x,y)^{\Pi'} & \rEq & F(\varphi_1,\varphi_2) & & \\
\qquad\qquad\qquad\qquad & & & \uInto &        &        & \uBar & & \\
\qquad\qquad\qquad\qquad &F[\Psi_1,\Psi_2] & \rEq & F[x,y]^\Pi & \rLine & \rLine & F(x,y)^\Pi & \rEq & F(\Psi_1,\Psi_2) \\
\endDiagram

Now $F[x,y]$ is integral over $F[\Psi_1,\Psi_2]$ \cite[Corollary 4.1.2.(i)]{Sp},
and hence $F[\varphi_1,\varphi_2]$ is integral over $F[\Psi_1,\Psi_2]$.
Also, the fraction field of $F[x,y]^\Pi$ is $F(x,y)^\Pi$, and
$F[x,y]^\Pi$ is integrally closed in $F(x,y)^\Pi$ (See Benson
\cite[Proposition 1.1.1]{Be}).

Given $f \in F[\varphi_1,\varphi_2] \subset F(\varphi_1,\varphi_2) = F(\Psi_1,\Psi_2)$.
$f$ is integral over $F[\Psi_1,\Psi_2]$, and hence $f \in F[\Psi_1,\Psi_2]$.
Thus, $F[\varphi_1,\varphi_2] = F[\Psi_1,\Psi_2]$ and it is absurd.
$\Box$
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\it Proof of Theorem 4.8.}

\Diagram
\qquad\qquad\qquad\qquad & & & F[x,y] & \rLine & \rLine & F(x,y) & & \\
\qquad\qquad\qquad\qquad & & & \uInto &        &        & \uInto & & \\
\qquad\qquad\qquad\qquad & & & F[x,y]^G & \rLine & \rLine & F(x,y)^G & & \\
\qquad\qquad\qquad\qquad & & & \uInto &        &        & \uBar & & \\
\qquad\qquad\qquad\qquad & & & F[\varphi_1,\varphi_2] & \rLine & \rLine & F(\varphi_1,\varphi_2) & & \\
\endDiagram

$F(x,y)^G = F(\varphi_1,\varphi_2)$ by Lemma 4.14.
$F[x,y]^G$ is integral over $F[\varphi_1,\varphi_2]$ by Corollary 4.10.
$F[\varphi_1,\varphi_2]$ is a polynomial ring in two variables,
$\varphi_1$ and $\varphi_2$, over $F$ by Corollary 4.12, and hence
$F[\varphi_1,\varphi_2]$ is integrally closed. (See Lang
\cite[Chapter VII, \textsection 1, Proposition 1.7]{La}).
Similar to the proof of Lemma 4.14, we have
$F[x,y]^G = F[\varphi_1,\varphi_2]$. $\Box$
\vspace{10pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent {\bf Remark 4.15.}

Theorem 4.8 implies that $G$ is generated by pseudo-reflections
(See Benson \cite[Theorem 7.2.1]{Be}).
Precisely, $G = \langle \alpha, \alpha^2\beta, \alpha\gamma\beta\alpha \rangle$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\newpage

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\end{document}