Elements of order 2 in the special linear group I am trying to show that the unique Sylow $2$-subgroup of the special linear group $SL(2,\mathbb{F_{3}})$ is isomorphic to the quaternion group $Q_{8}$.
Call the unique Sylow $2$-subgroup $P$, and note $|P|=8$, so $P$ is isomorphic to one of $C_{8},\ C_{2} \times C_{2}\times C_{2},\ C_4 \times C_{2}$ or $Q_{8}$.
My plan of attack is to show that $SL(2,\mathbb{F_{3}})$ has only one element of order $2$, in order to eliminate the possibility of being isomorphic to one of $C_{8},\ C_{2} \times C_{2}\times C_{2},\ C_4 \times C_{2}$, as these each have more than one element of order 2. But I don't know how to show there is only one such in $SL(2,\mathbb{F_{3}})$.
Any help would be appreciated!
 A: Suppose
$$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\;,\;\;ad-bc=1\;\;(*)\;\;,\;\;A^2=I\implies$$
$$I=A^2=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix}\implies$$
i) If $\;a=-d\;$ , then $\;a^2=d^2\;$ , so $\;a^2+bc=1\;(**)\;$ . If $\;a=0\;$ , then $\;bc=1\;$ but also $\;bc=-1\;$ , impossible . So $\;a\neq 0\;$ , but again
$$1+bc\stackrel{(*)}=-a^2\stackrel{(**)}=-1+bc\implies\;-1=1\;$$  and this is impossible , so
ii) It must be $\;b=0\;$ or $\;c=0\;$, and then
$$ad\stackrel{(*)}=1\;,\;\;a^2\stackrel{(**)}=1\implies a=d\implies\;\text{also}\;\;c=0$$
and we get two options
two options:
$$A=\begin{pmatrix}1&0\\0&1\end{pmatrix}\;,\;\;\;A=\begin{pmatrix}\!\!-1&0\\0&\!\!-1\end{pmatrix}$$
so indeed only one unique element of order two.
A: I don't know if this will help, but given an element of order 2, we can find other elements of order 2 by finding conjugates of that element. In $SL_2(\mathbb {F_3})$ consider $2I$ where $I$ denotes the Identity matrix. Note that this commutes with everything, and therefore is part of the centre of the group. Any conjugates will lead right back to itself. Further, since all Sylow 2-subgroups will be related by conjugation, this is an element in all the 2-Sylow subgroups. 
A: Suppose $A$ is an element of the group such that $A^{2} = I$. Suppose that $A \neq \pm I$. Then the minimal polynomial of $A$ is not $x-1$ nor $x+1$ and so it must be $(x-1)(x+1)$ by Cayley-Hamilton Theorem and by dimension consideration this is also the characteristic polynomial and so the eigenvalues of $A$ are $1, -1$, but then $\det A = -1 \cdot 1 = -1$, which means that $A$ is not in the group. Hence there is exactly only one element of order 2, namely $A = -I$.
