Topological generators for $\mathrm{SL}_2(\mathbf{Z}_p)$

Question

$\DeclareMathOperator\SL{SL}$ 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 $\SL_2(\mathbf{F}_p)$.

Answer 1

$\def\ZZ{\mathbb{Z}}\def\SL{\text{SL}}\def\Id{\text{Id}}$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.