## Exhibit all of the Sylow 3-subgroups of a special linear group

Exhibit all Sylow 3-subgroups of $SL_2(\mathbb{F}_3)$.

By this previous exercise, $[GL_2(\mathbb{F}_3) : SL_2(\mathbb{F}_3)] = 2$. We also know that $|GL_2(\mathbb{F}_3)| = (3^2-1)(3^2-3) = 2^4 \cdot 3$. Thus $|SL_2(\mathbb{F}_3)| = 2^3 \cdot 3$. By Sylow’s Theorem, the number $n_3$ of Sylow 3-subgroups is congruent to 1 mod 3 and divides 8. Thus $n_3 \in \{1,4\}$.

Let $A \in SL_2(\mathbb{F}_3)$, with $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.

Note that $A^2 = \begin{bmatrix} a^2+bc & ab+bd \\ ac+dc & bc+d^2 \end{bmatrix}$ and $A^3 = \begin{bmatrix} a^3+2abc+bcd & a^2b+b^2c+d^2b+abd \\ a^2c+c^2b+d^2c+adc & d^3+2bcd+abc \end{bmatrix}$. If $A^3 = I_3$, then the entries of $A$ must satisfy the following four equations. Note that $\alpha^3 = \alpha$ for all $\alpha \in \mathbb{F}_3$.

1. $a+2abc+bcd = 1$
2. $a^2b+b^2c+d^2b+abd = 0$
3. $a^2c+c^2b+d^2c+adc = 0$
4. $d+2bcd+abc = 1$

Suppose $c = 0$. Then equations (1) and (4) give $a = d = 1$. Now $b \in \mathbb{F}_3$, but $b \neq 0$ since otherwise $A = I_3$. Thus $b \in \{1,2\}$. In either case, we generate the Sylow 3-subgroup $H_1 = \left\langle \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \right\rangle = \left\{ I, \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \right\}.$

Now suppose $b = 0$. Then equations (1) and (4) give $a = d = 1$. Now $c \in \mathbb{F}_3$, but $c \neq 0$ since otherwise $A = I_3$. Thus $c \in \{1,2\}$. In either case, we generate the Sylow 3-subgroup $H_2 = \left\langle \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \right\rangle = \left\{ I_3, \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \right\}.$

Henceforth we may assume that $c$ and $b$ are not zero.

Suppose $c = 1$.

• If $a = 0$, then equation (1) gives us $bd = 1$. Clearly then $b = d$, as otherwise $bd = 2$. Moreover, equation (2) gives $b^2(b+1) = 0$. Since $b \neq 0$, we have $b = 2 = d$. Thus $A$ generates the Sylow 3-subgroup $H_3 = \left\langle \begin{bmatrix} 0 & 2 \\ 1 & 2 \end{bmatrix} \right\rangle = \left\{ I_3, \begin{bmatrix} 0 & 2 \\ 1 & 2 \end{bmatrix}, \begin{bmatrix} 2 & 1 \\ 2 & 0 \end{bmatrix} \right\}.$
• If $a = 1$, then equation (1) gives us $b(2+d) = 0$. Since $b \neq 0$, we have $d = 1$. Then equation (3) gives $b = 0$, a contradiction.
• If $a = 2$, then equation (1) gives us $b(1+d) = 2$. Then $b \neq 1+d$, as otherwise $b(1+d) = 1$. If $b = 1$, then $1+d=2$, so that $d = 1$. Then equation (2) gives us $2 = 0$, a contradiction. Thus $b = 2$, so that $1+d = 1$ and we have $d = 0$. Thus $A$ generates the Sylow 3-subgroup $H_4 = \left\langle \begin{bmatrix} 2 & 2 \\ 1 & 0 \end{bmatrix} \right\rangle = \left\{ I_3, \begin{bmatrix} 2 & 2 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 2 & 2 \end{bmatrix} \right\}.$

Suppose $c = 2$.

• If $a = 0$, then equation (1) gives us $2bd = 1$, so that $bd = 2$. If $b = 2$, then $d = 1$ and equation (2) gives us $1 = 0$, a contradiction. Thus $b = 1$ and $d = 2$; in this case $A$ generates $H_4$.
• If $a = 1$, then equation (1) gives us $b(1+d) = 0$. Since $b \neq 0$, we have $1+d = 0$, so that $d = 2$. But then equation (1) gives $b = 0$, a contradiction.
• If $a = 2$, then equation (1) gives us $b(1+d) = 1$. Then $b = 1+d$, as otherwise $b(1+d) = 2$. If $b = 2$, then $1+d = 2$, so that $d = 1$. Then equation (3) gives $1 = 0$, a contradiction. Thus $b = 1$, so that $d = 0$, and $A$ generates $H_3$.

Thus $n_3 = 4$, and we have exhausted the Sylow 3-subgroups of $SL_2(\mathbb{F}_3)$.