$\langle A,B \rangle$ is the unique Sylow $2$-subgoup of $SL_2(F_3)$ Prove that the subgroup of $SL_2(F_3)$ generated by $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and $\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ is the unique Sylow $2$-subgoup of $SL_2(F_3)$.
Proof: Let $G=SL_2(F_3)$. $|G|=24=2^3\cdot 3$. Let
$A =\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
and
$B =\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$
By the Sylow Theorems, let $n_2$ be the number of Sylow $2$-subgroup. $n_2\equiv 1 \mod 2$ and $n_2 | 3 \Rightarrow n_2=1,3$, and Sylow $2$-subgroup(s) have order $8$. We need to check $|\langle A,B \rangle | = 8$. Note $A^4=B^4=I$. We find
$$
\langle A,B \rangle 
= \left\{ \pm I, \pm A, \pm B, \pm \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}\right\}
$$
What is left to show is $\langle A,B\rangle$ is the only one. We can do that by show $\langle A,B \rangle$ is normal.
i.e. $gng^{-1} \in \langle A,B \rangle$ $\forall g=\begin{pmatrix}
a & b \\ c & d \end{pmatrix}$ and $n\in N$.
So this is where I am stuck. I know I can do the computation for each element of $\langle A,B \rangle$ but that would suck. Is there a way to avoid that or not?
 A: Let $P = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$.  Then $P^3 = I$ and $P^{-1} = P^2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$.  Since $| \langle P \rangle | = 3$ and $| \langle A, B \rangle | = 8$ we have $\langle P \rangle \cdot \langle A, B \rangle = \langle P, A, B \rangle = SL_2(\mathbb{F}_3)$.
Calculating $PAP^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = -B \in \langle A, B \rangle$,
and
$PBP^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}= AB \in \langle A, B \rangle$.
Of course $P$ commutes with $\pm I$ and since these elements generate $\langle A, B \rangle$ we have that conjugation by $P$ fixes $\langle A, B \rangle$.  Since $\langle A, B \rangle$ is closed under conjugation by every element of itself, and by $P$, and since these elements generate $SL_2(\mathbb{F}_3)$, we have that $\langle A,B \rangle$ is closed under conjugation by any element, and thus is normal.  Since all Sylow 2-subgroups are conjugate, there is only one, namely $\langle A, B \rangle$.
A: Here is another way to see that the $2$-Sylow subgroup of $SL_2({\bf F}_3)$ is normal.
The group $GL_2({\bf F}_3)$ acts on the points of the projective line  via fractional linear transformations. Up to conjugacy there are three points with field of definition equal to ${\bf F}_9$. They are $\pm i$, $1\pm i$ and $-1\pm i$. Since $GL_2({\bf F}_3)$ preserves them, we get a  homomorphism $f:GL_2({\bf F}_3)\rightarrow S_3$.
It is easy to see that the matrix $\pmatrix{1&1\cr0&1\cr}$ is mapped to a $3$-cycle, while  $\pmatrix{0&1\cr1&0\cr}$ is mapped to a $2$-cycle.  Therefore $f$ is surjective.  It also shows that $f$ induces an isomorphism $GL_2({\bf F}_3)/SL_2({\bf F}_3)\rightarrow S_3/A_3$. Therefore  the subgroup $SL_2({\bf F}_3)$ is mapped onto the alternating group $A_3$. So the kernel of $f$ is a $2$-Sylow subgroup of $SL_2({\bf F}_3)$.
