Let $p$ be a prime number, and let $F$ be a finite extension of $\mathbf Q_p$ with integer ring $\mathcal O_F$, maximal ideal $\mathfrak p$, and residue field $k$. I want to prove that
The kernel of $\operatorname{GL}_2(\mathcal O_F)\to\operatorname{PGL}_2(k)$ is solvable.
The kernel is $$K=\left\{ \begin{pmatrix}a&b\\c&d\end{pmatrix}\in \operatorname{GL}_2(\mathcal O_F):a-d\equiv b\equiv c\equiv 0\pmod{\mathfrak p} \right\}.$$
My plan was to directly construct a chain $K\supset K_1\supset K_2\dots\supset K_n=\{1\}$ such that $K_{i+1}$ is a normal subgroup of $K_i$ and $K_i/K_{i+1}$ is abelian.
Let $K_1$ be the kernel of $\det$, then
$$K_1=K\cap\operatorname{SL}_2(\mathcal O_F)$$ and $K/K_1$ is a subgroup of $\mathcal O_F^\times$.
Let $K_2=\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}\in K_1:a=1\pmod{\mathfrak p} \right\}$ then $K_2$ is the kernel of $K_1\to\operatorname{SL}_2(k)$ and $K_1/K_2$ is a subgroup of $k^\times$.
So now I need to show that $$K_2=\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\operatorname{SL}_2(\mathcal O_F): a\equiv d\equiv 1,\,c\equiv b\equiv 0\pmod{\mathfrak p} \right\}$$ is solvable.
The kernel is not solvable. By intersecting $K$ with $\operatorname{SL}_2(\mathbf Z)$ we get a group of finite index in $\operatorname{SL}_2(\mathbf Z)$, which is shown here to not be solvable.