Let $N = \left(\begin{matrix} 1 & * \\ 0 & 1 \end{matrix}\right)$ be the upper triangular unipotent subgroup in ${\rm GL}_2(\mathbf{Q}_p)$ and $K_n = 1 + p^n M_2(\mathbf{Z}_p)$ the usual compact opens in ${\rm GL}_2(\mathbf{Q}_p)$.
I am trying to show that the subgroup generated by $N$ and $K_r$ would contain ${\rm SL}_2(\mathbf{Q}_p)$ for every $r \gg 0$. This would prove the claim in the title.
I believe that, I could make this work if I showed that the group $\langle N, K_r \rangle$ contains $K_{r-1} \cap N^t$ where $N^t$ is the lower triangular unipotent subgroup. I can't quite think of the right matrices from $N$ and $K_r$ to multiply to get the power of $p$ in the lower right hand corner to decrease...
Any thoughts would be appreciated.
Let $a,x \in \mathbf{Q}_p$ with $x \neq 0$. A short computation shows $$\left(\begin{matrix} 1 & -1/x \\ 0 &1 \end{matrix} \right) \left(\begin{matrix} 1 & 0 \\ x &1 \end{matrix} \right) \left(\begin{matrix} 1 & -1/x \\ 0 &1 \end{matrix} \right)=\left(\begin{matrix} 0 & -1/x \\ x & 0 \end{matrix} \right)$$ and $$\left(\begin{matrix} 0 & -1/x \\ x &0 \end{matrix} \right) \left(\begin{matrix} 1 & a \\ 0 &1 \end{matrix} \right) \left(\begin{matrix} 0 & 1/x \\ -x & 0 \end{matrix} \right)=\left(\begin{matrix} 1 & 0 \\ -a x^2 &1 \end{matrix} \right).$$ Taking $x=p^n$ and letting $a$ range over $\mathbf{Q}_p$ shows that the subgroup generated by $N$ and $K_n$ contains both $N$ and the opposite unipotent subgroup, which together generate $\mathrm{SL}_2(\mathbf{Q}_p)$.