$\DeclareMathOperator\SL{SL}$ Let $p>3$ and $G$ be an open subgroup of the special linear group $\SL\_2(\mathbf{Z}\_p)$ over the ring $\mathbf{Z}\_p$ of $p$-adic integers. Suppose that $G$ is topologically generated by a finite set $S=\{g\_1,\cdots,g\_r\}$ such that all $g\_i$ have finite order. Is it true that $G=\SL\_2(\mathbf{Z}\_p)$? If we denote the natural surjective morphism $ \SL\_{2}(\mathbf{Z}\_{p})\to \SL\_{2}(\mathbf{F}\_{p}) $ by $ \eta $, then it is equivalent to ask whether $\eta(S)$ generates $\SL\_2(\mathbf{F}\_p)$.



Topological generators for $\mathrm{SL}_2(\mathbf{Z}_p)$
$\def\ZZ{\mathbb{Z}}\def\SL{\text{SL}}\def\Id{\text{Id}}$This seems false to me.


**Lemma:** $e:=\left[ \begin{smallmatrix} 1&p\\0&1 \\ \end{smallmatrix} \right]$ and $f:=\left[ \begin{smallmatrix} 1&0\\p&1 \\ \end{smallmatrix} \right]$ topologically generate an open subgroup of $\SL\_2(\ZZ\_p)$.


**Proof:** Let $G$ be the group topologically generated by $e$ and $f$. Let
$$\Gamma(p^k) = \{ g \in \SL\_2(\ZZ\_p) : g \equiv \Id\_2 \bmod p^k \}.$$
We will show that $\Gamma(p^2) \subset G$.


To this end, note that $G$ contains $e^{p^{k-1}} = \left[ \begin{smallmatrix} 1&p^k\\0&1 \\ \end{smallmatrix} \right]$, $f^{p^{k-1}} = \left[ \begin{smallmatrix} 1&0\\ p^k&1 \\ \end{smallmatrix} \right]$ and $[e, f^{p^{k-2}}] \equiv \left[ \begin{smallmatrix} 1+p^k&0\\ 0&1-p^k \\ \end{smallmatrix} \right] \bmod p^{k+1}$ for all $k \geq 2$. Thus, $G$ contains generators for $\Gamma(p^k)/\Gamma(p^{k+1})$ for all $k \geq 2$, so any element of $\Gamma(p^k)$ can be written as an infinite convergent product of elements of $G$. $\square$


Now, let $\zeta$ be a $(p-1)$-st root of unity in $\ZZ\_p$ other than $\pm 1$. (Here I use $p>3$.) Let $d = \left[ \begin{smallmatrix} \zeta & 0 \\ 0 & \zeta^{-1} \end{smallmatrix} \right]$.


The matrices $d$, $de$ and $df$ all have order $p-1$. (They are all upper or lower triangular with diagonal entries $\zeta$, $\zeta^{-1}$ and, since $\zeta \neq \zeta^{-1}$, they are diagonalizable.) Clearly, the group topologically generated by $d$, $de$, $df$ contains the group topologically generated by $e$ and $f$, so it is open.


However, reducing $d$, $de$ and $df$ modulo $p$ gives diagonal matrices, so the map onto $\SL\_2(\mathbb{F}\_p)$ isn't surjective.


