Exercise 4.5.10 (Sylow $2$-subgroup of $\mathrm{SL}_2(\mathbb{F}_3)$)

Question

Prove that the subgroup of $\mathrm{SL}_2(\mathbb{F}_3)$ generated by $\begin{pmatrix} 0 & -1 \ 1 & 0\end{pmatrix}$ and $\begin{pmatrix} 1 & 1 \ 1 & -1\end{pmatrix}$ is the unique Sylow $2$-subgroup of $\mathrm{SL}_2(\mathbb{F}_3)$ (see Exercise 10, Section 2.4).

Richard Ganaye · Accepted Answer

\begin{proof} 
By Exercise 2.4.9,
$$\mathrm{SL}_2(\mathbb{F}_3) = \langle A,B angle,$$
where
$$A = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix},\qquad B = \begin{pmatrix} 1 & 0 \ 1 & 1 \end{pmatrix}.$$
Consider the subgroup of $\mathrm{SL}_2(\mathbb{F}_3)$ defined by
$$H = \left \langle I, J ight angle,$$
where
$$I = \begin{pmatrix} 0 & -1 \ 1 & 0\end{pmatrix}, \qquad J = \begin{pmatrix} 1 & 1 \ 1 & -1\end{pmatrix}.$$
In Exercise 2.4.10, we have proved that
$$H \simeq Q_8,$$
so $H$ is a Sylow $2$-subgroup of $\mathrm{SL}_2(\mathbb{F}_3)$, where $|\mathrm{SL}_2(\mathbb{F}_3)| = 24 = 2^3 \cdot 3$.

Moreover (see Exercise 2.4.10), if $K = IJ = \begin{pmatrix} -1 & 1 \ 1 & 1\end{pmatrix}$, then 
\begin{align*}
H &= \{I_2, I , J , K ,-I_2, -I,-J,-K\} \
\end{align*}
We check that $H \unlhd \mathrm{SL}_2(\mathbb{F}_3)$:
\begin{align*}
A I A^{-1} & = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 \ 1 & 0\end{pmatrix} \begin{pmatrix} 1 & -1 \ 0& 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 1 & -1\end{pmatrix}= J \in H,\
B I B^{-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}= -K \in H,\
A J A^{-1} &=  \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \ 1 & -1\end{pmatrix} \begin{pmatrix} 1 & -1 \ 0& 1 \end{pmatrix} = \begin{pmatrix} -1 & 1 \ 1 & 1\end{pmatrix}= K \in H,\
B J B^{-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} 0 & 1 \- 1 & 0 \end{pmatrix}= -I \in H,\
\end{align*}
Since $\mathrm{SL}_2(\mathbb{F}_3) = \langle A,B angle$ and $H = \left \langle I, J ight angle$, this shows that $H \unlhd \mathrm{SL}_2(\mathbb{F}_3) $. Therefore $H$ is the unique Sylow $2$-subgroup of $\mathrm{SL}_2(\mathbb{F}_3) $.

\end{proof}

Exercise 4.5.10 (Sylow $2$-subgroup of $\mathrm{SL}_2(\mathbb{F}_3)$)

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Exercise 4.5.10 (Sylow $2$-subgroup of $\mathrm{SL}_2(\mathbb{F}_3)$)

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