# Topological generators for the Sylow pro-$p$ subgroup of $\mathrm{SL}_2(\mathbf{Z}_p)$

$$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$$Let $$G_2(\mathbf{Z}_p):=\begin{pmatrix} 1+p\mathbf{Z}_{p} & \mathbf{Z}_{p}\\ p\mathbf{Z}_{p} & 1+p \mathbf{Z}_{p} \end{pmatrix}$$. Then it is a Sylow pro-$$p$$ subgroup of $$\GL_2(\mathbf{Z}_p$$) and $$S_2(\mathbf{Z}_p):=G_2(\mathbf{Z}_p) \cap \SL_2(\mathbf{Z}_p)$$ is a Sylow pro-$$p$$ subgroup of $$\SL_2(\mathbf{Z}_p$$). It's well-known that $$S_2(\mathbf{Z}_p)$$ is a $$2$$-generator pro-$$p$$ group. For exmaple, the set $$\{\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} ,\begin{pmatrix} 1& 0 \\ p& 1 \end{pmatrix} \}$$ is a topological generating set. Moreover, in Prop. 3.1.1 of The image of Galois representations attached to elliptic curves with an isogeny it's proved that if $$A,B\in S_2(\mathbf{Z}_p)$$ such that the image of $$A$$ in $$\GL_2(\mathbf{F}_p)$$ is non-trivial and the image of $$B$$ in $$\GL_2(\mathbf{F}_p)$$ is trivial and the image of $$B$$ in $$\GL_2(\mathbf{Z}/p^2\mathbf{Z})$$ is not upper triangular, then the set $$\{A,B\}$$ topologically generates $$S_2(\mathbf{Z}_p)$$.

Question: Can $$S_2(\mathbf{Z}_p)$$ be generated by two matrices $$A,B$$ such that the images of $$A,B$$ in $$\GL_2(\mathbf{F}_p)$$ are both non-trivial?

As pointed out in the answer, the answer to question is No in a trivial way. What about the case assuming further that the eigenvalues of $$A,B$$ are all $$1$$?

If $$x$$ and $$y$$ generate a group then so do $$x$$ and $$xy$$, so take $$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$ and $$B = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 0 \\ p & 1 \end{pmatrix}$$.