Exercise 2.4.9 (Generators of $\mathrm{SL}_2(\mathbb{F}_3)$)

Question

Prove that $\mathrm{SL}_2(\mathbb{F}_3)$ is the subgroup of  $\mathrm{GL}_2(\mathbb{F}_3)$ generated by $\begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \ 1 & 1 \end{pmatrix}$. [Recall from Exercise 9 of Section 1 that $\mathrm{SL}_2(\mathbb{F}_3)$ is the subgroup of matrices of determinant $1$. You may assume this subgroup has order $24$ -- this will be an exercise in Section 3.2.]

Richard Ganaye · Accepted Answer

ewcommand{\F}{\mathbb{F}}

I have not found such an exercise in Section 3.2, so I prove it here.
\begin{proof} 
We show first that $|\mathrm{GL}_2(\mathbb{F}_3)| = 48$. To obtain a matrix $\begin{pmatrix} a & b \ c & d \end{pmatrix}$ in  $\mathrm{GL}_2(\mathbb{F}_3)$, we choose first the first row $(a,b)$ among the nonzero vectors of $\F_3^2$. There are $3^2 - 1 = 8$ possibilities. Then we choose $(c,d)$ such that $(c, d)$ is not collinear to $(a,b)$. There are exactly $3$ vectors $\lambda (a,b)$ with $\lambda \in \F_3$, so it remains $9 - 3 = 6$ possibilities for $(c,d)$. This gives $8 \cdot 6 = 48$ matrices in $\mathrm{GL}_2(\mathbb{F}_3)$, so
$$|\mathrm{GL}_2(\mathbb{F}_3)| = 48.$$
Consider now the map
$$
\varphi \ 
\left\{
\begin{array}{ccl}
\mathrm{GL}_2(\mathbb{F}_3) &	o& \{-1,1\}\
A & \mapsto & \det(A)
\end{array}
ight.
$$
where $\{-1,1\}\subset \F_3^*$ is a subgroup of $ \F_3^*$
Then
\begin{itemize}
\item $\varphi$ is a homomorphism, because $\varphi(AB) = \det(AB) = \det(A) \det(B) = \varphi(A) \varphi(B)$.

\item $\varphi$ is surjective, since $\det(I_2) = 1$, where $I_2 \in   \mathrm{GL}_2(\mathbb{F}_3)$, and $\det (M)= -1$, where $ M = \begin{pmatrix} -1 & 0 \ 0 & 1 \end{pmatrix}  \in \mathrm{GL}_2(\mathbb{F}_3)$.
\item Let $A \in \mathrm{GL}_2(\mathbb{F}_3)$. Then $A \in \ker(\varphi)$ if and only if $\det(A) = 1$, so
$$\ker(\varphi) = \mathrm{SL}_2(\mathbb{F}_3).$$
\end{itemize}
The first isomorphism Theorem gives $\mathrm{GL}_2(\mathbb{F}_3)/ \mathrm{SL}_2(\mathbb{F}_3) \simeq \{-1,1\}$, thus 
$$|\mathrm{SL}_2(\mathbb{F}_3) | = |\mathrm{GL}_2(\mathbb{F}_3) |/ 2 = 24.$$
\end{proof}

\begin{proof}(of Exercise 2.4.9.)
Put 
$$A = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix},\qquad B = \begin{pmatrix} 1 & 0 \ 1 & 1 \end{pmatrix}.$$
Then $A, B \in \mathrm{SL}_2(\mathbb{F}_3)$ have order $3$, and $C = AB=\begin{pmatrix} -1 & 1 \ 1 & 1 \end{pmatrix}$ has order $4$.
The list of matrices $A^i B^j C^k\ (0\leq i \leq 3, 0 \leq j \leq 3, 0 \leq k \leq 4)$ has $36$ (not distinct) items. After removing duplicates, we obtain $24$ distinct matrices in $\mathrm{SL}_2(\mathbb{F}_3)$, given by
\begin{align*}
&\left(\begin{array}{rr}
0 & 1 \
-1 & 0
\end{array}ight), \left(\begin{array}{rr}
0 & 1 \
-1 & 1
\end{array}ight), \left(\begin{array}{rr}
0 & 1 \
-1 & -1
\end{array}ight), \left(\begin{array}{rr}
0 & -1 \
1 & 0
\end{array}ight), \left(\begin{array}{rr}
0 & -1 \
1 & 1
\end{array}ight),
 \left(\begin{array}{rr}
0 & -1 \
1 & -1
\end{array}ight),\
&\left(\begin{array}{rr}
1 & 0 \
0 & 1
\end{array}ight),
\left(\begin{array}{rr}
1 & 0 \
1 & 1
\end{array}ight), \left(\begin{array}{rr}
1 & 0 \
-1 & 1
\end{array}ight), \left(\begin{array}{rr}
1 & 1 \
0 & 1
\end{array}ight), \left(\begin{array}{rr}
1 & 1 \
1 & -1
\end{array}ight), \left(\begin{array}{rr}
1 & 1 \
-1 & 0
\end{array}ight),\
& \left(\begin{array}{rr}
1 & -1 \
0 & 1
\end{array}ight), \left(\begin{array}{rr}
1 & -1 \
1 & 0
\end{array}ight), \left(\begin{array}{rr}
1 & -1 \
-1 & -1
\end{array}ight), \left(\begin{array}{rr}
-1 & 0 \
0 & -1
\end{array}ight), \left(\begin{array}{rr}
-1 & 0 \
1 & -1
\end{array}ight), \left(\begin{array}{rr}
-1 & 0 \
-1 & -1
\end{array}ight),\
& \left(\begin{array}{rr}
-1 & 1 \
0 & -1
\end{array}ight), \left(\begin{array}{rr}
-1 & 1 \
1 & 1
\end{array}ight), \left(\begin{array}{rr}
-1 & 1 \
-1 & 0
\end{array}ight), \left(\begin{array}{rr}
-1 & -1 \
0 & -1
\end{array}ight), \left(\begin{array}{rr}
-1 & -1 \
1 & 0
\end{array}ight), \left(\begin{array}{rr}
-1 & -1 \
-1 & 1
\end{array}ight).
\end{align*}

This shows that $$\mathrm{SL}_2(\mathbb{F}_3) = \langle A, B angle.$$
\end{proof}

With Sagemath
\begin{verbatim}
sage: F = GF(3)
sage: gens = [matrix(F, 2, [1,1, 0,1]), matrix(F, 2, [1,0,1,1])]
sage: G = MatrixGroup(gens)
sage: A, B = G(gens[0]), G(gens[1])
sage: C = A * B
sage: l = [A^i*B^j*C^k for i in range(3) for j in range(3) for k in range(4)]
sage: import numpy as np
....: unique_array = np.unique(l)
....: unique_list = unique_array.tolist()
....: unique_list
....:

[
[0 1]  [0 1]  [0 1]  [0 2]  [0 2]  [0 2]  [1 0]  [1 0]  [1 0]  [1 1]
[2 0], [2 1], [2 2], [1 0], [1 1], [1 2], [0 1], [1 1], [2 1], [0 1],

[1 1]  [1 1]  [1 2]  [1 2]  [1 2]  [2 0]  [2 0]  [2 0]  [2 1]  [2 1]
[1 2], [2 0], [0 1], [1 0], [2 2], [0 2], [1 2], [2 2], [0 2], [1 1],

[2 1]  [2 2]  [2 2]  [2 2]
[2 0], [0 2], [1 0], [2 1]
]
\end{verbatim}
Of course, we may obtain directly the list of elements of $G$ by
\begin{verbatim}
G.list()
\end{verbatim}
to obtain the same result (curiously in exactly the same order!).

Exercise 2.4.9 (Generators of $\mathrm{SL}_2(\mathbb{F}_3)$)

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Exercise 2.4.9 (Generators of $\mathrm{SL}_2(\mathbb{F}_3)$)

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