How to use Sylow theorems to prove that a subgroup of $SL_2(\mathbb{F}_3)$ is normal.

Question

I have $G = SL_2(\mathbb{F}_3)$ and $H = \langle i, j \rangle$ where $$ i = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \quad j = \begin{pmatrix} -1 & 1 \\ 1 & 1\end{pmatrix} $$ (In other words, $H$ is the quaaternion group.)

The one part of the exercise I am working on asks:

Show that $H$ is a subgroup of $G$. Use Sylow theory to show that $H$ is normal.

So far, I have managed to prove that $H$ is indeed a subgroup of $G$, but I'm having a hard time figuring out how to proceed to prove that it is a normal subgroup. From the Sylow theorems, I have so far managed to conclude that $H$ is a Sylow 2-subgroup, and that there must either be 1 or 3 Sylow 2-subgroups of $G$, since $n_p \equiv 1 \mod 2$ and $n_p \bigm| |G| = 24$.

As this is for my homework, I'm not asking for full solutions to the problem, I only need some hints to help me along to the correct answer.

Answer 1

An optional strategy: Let $N_G(H)$ denote the normalizer of $H$. You know already that either $N_G(H)=G$ or $N_GH=H$. Hence, it suffices to find one element $g\in G\setminus H$ which satisfies $gHg^{-1}=H$. Pick some $g\in G\setminus H$. to verify the last equality, it is enough to check that $gig^{-1},gjg^{-1}\in H$.

I guess that one advantage of this strategy is that it is simple, and does not involve too many computations. On the other hand, it will be nice to find a proof based on more general arguments.

Answer 2

What about looking at the Sylow 3 subgroup? $A=\left[\begin{array}{cc}1&1\\0&1\end{array}\right]$ has order 3. And $BAB^{-1}\not\in<A>$ for $B=\left[\begin{array}{cc}0&1\\-1&0\end{array}\right]$. Therefore there are four Sylow 3 subgroups. Since $|G|=24$, that doesn't leave enough room for three Sylow 2 subgroups each of order 8.