2-Sylow subgroup of $\operatorname{GL}(2,3)$ Consider the group $\operatorname{GL}(2,3)$, the group of invertible $2 × 2$ matrices over the field of 3 elements. It is of order $48 = 2^4 \cdot 3$. A 3-Sylow subgroup is easily seen to be the Heisenberg group (unitary upper triangular matrices).
What about the 2-Sylow subgroups? Is there a nice way to identify one of them?
 A: The Sylow $2$-subgroups of the finite classical groups were described by Carter and Fong (J. Algebra, 1964).
Your example is a special case of $G = \operatorname{GL}_2(\mathbb{F}_q)$ with $q \equiv 3 \mod{4}$. Let $2^s$ be the largest power of $2$ dividing $q+1$, so a Sylow $2$-subgroup of $G$ has order $2^{s+2}$.
The generators Carter and Fong give for a $2$-Sylow subgroup $P < G$ are
$x = \begin{pmatrix} 0 & 1 \\ 1 & \varepsilon + \varepsilon^q \end{pmatrix}$ and $y = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, where $\varepsilon$ is a primitive $2^{s+1}$th root of unity in $\mathbb{F}_{q^2}$. For $a = x$ and $b = yx$ we have the relations $a^{2^{s+1}} = b^2 = 1$ and $bab^{-1} = a^{2^s - 1}$. Hence it turns out that $P$ is a semidihedral group: $$P \cong \langle a, b | a^{2^{s+1}} = b^2 = 1, bab^{-1} = a^{2^s - 1} \rangle.$$
So that is fine, but you might wonder how one could come up with these matrices.
For this, we can consider $V = \mathbb{F}_{q^2}$ as a $2$-dimensional vector space over $\mathbb{F}_q$, so then we can identify $G$ with $\operatorname{GL}(V)$. We first want to find an element of order $2^{s+1}$. For this, take $f \in G$ to be the map defined by $v \mapsto \varepsilon v$, where $\varepsilon$ is a primitive $2^{s+1}$th root of unity. With respect to the basis $\{1, \varepsilon \}$ of $V$, you can verify that the matrix of $f$ becomes the matrix $x$ defined above. We also have the Frobenius map $g \in G$ defined by $v \mapsto v^q$. You can check that the matrix of $g$ with respect to the basis $\{1, \varepsilon \}$ is the matrix $yx$.
