This question is similar to the question "How to find a p-Sylow subgroup of $GL_2(F_p)$?", which is relatively easy. Since if they share the same prime p, then we can quickly conclude the order of any p-Sylow subgroup is p. So this allows us simply to choose some element with order p like $\begin{array}{ccc}1&1\\0&1\\\end{array}$ to generate a group with order p. However, in the current case, G has 48 elements so the 2-Sylow will have order 16. Which seems had quite different approaches.
And I know that $SL_2(F_3)$ has order 24 and can use exclusion (based on 3-Sylow's number) to get the result. However, I think it doesn't work here. So quite confused which method is a suitable one to solve this question.
Best Answer
$|GL(2,q)|=(q^2-1)q(q-1)$.
Let $q$ be odd prime power, and let $2^{m+1}$ be the highest power of $2$ dividing $q^2-1$.
So if $q-1$ is not divisible by $4$ (but it is divisible by $2$), then $2^{m+2}$ is the order of Sylow-$2$; this is found as follows:
Since $|\mathbb{F}_{q^2}^*|=q^2-1$, so there is element $\zeta$ of order $2^{m+1}$ in $\mathbb{F}_{q^2}^*$; note that this can not lie in $\mathbb{F}_q^*$ ($4\nmid (q-1)$). Consider the matrix $$A= \begin{bmatrix} 0 & 1\\1 & \zeta+\zeta^q\end{bmatrix} $$ and although $\zeta\notin \mathbb{F}_q$, we have $\zeta+\zeta^q\in\mathbb{F}_q$. It is easy to see (linear algebra) that this matrix has order $2^{m+1}$ (equal to order of $\zeta$). Let $B$ be the matrix $$B=\begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}.$$ Then check that $BAB^{-1}=A^{2^m-1}$; this gives us that $\langle A,B\rangle$ is subgroup of order $2^{m+2}$, Sylow-$2$ of $G$ (when $q\not\equiv 1\pmod{4}$).