In $SL_2(\mathbb{Z}_p)$, is $\pmatrix{1 & 1 \\ 0 & 1}$ in the closure of the subgroup generated by $\pmatrix{1 & e \\ 0 & 1}$ for $(e, p) = 1$? Let $e$ be an integer coprime to $p$. In the profinite group $SL_2(\mathbb{Z}_p)$, is the matrix $\begin{bmatrix}
1 & 1 \\ 0 & 1
\end{bmatrix}$ contained in the closure of the cyclic subgroup generated by 
$\begin{bmatrix}
1 & e \\ 0 & 1
\end{bmatrix}$ ?
 A: As orangeskid commented, the answer is yes. Though the proof isn't just formal.
Let $T := \begin{bmatrix}1 & 1 \\0 &1\end{bmatrix}$. Then $T$ generates an abstract cyclic group $T^\mathbb{Z}\subset SL_2(\mathbb{Z}_p)$. Concretely, this subgroup is just the group of matrices
$$\begin{bmatrix}1 & n \\0 &1\end{bmatrix}\qquad n\in\mathbb{Z}$$
Note that $SL_2(\mathbb{Z}_p) = \varprojlim_k SL_2(\mathbb{Z}/p^k\mathbb{Z})$. 
By Ribes-Zalesskii's Corollary 1.1.8, the closed subgroup generated by $T^\mathbb{Z}$ is the inverse limit of its images in $SL_2(\mathbb{Z}/p^k\mathbb{Z})$, but these images are just $T^\mathbb{Z}/T^{p^k\mathbb{Z}}\cong\mathbb{Z}/p^k\mathbb{Z}$, and so the closed subgroup of $SL_2(\mathbb{Z}_p)$ generated by $T$ is isomorphic to $\varprojlim_k \mathbb{Z}/p^k\cong\mathbb{Z}_p$.
Thus, the (multiplicative) closed subgroup generated by $T$ is isomorphic to the additive group $\mathbb{Z}_p$. Let $\varphi : \mathbb{Z}_p\rightarrow\overline{T^{\mathbb{Z}}}$ denote this isomorphism, where $\varphi(1) = T$, which restricts to an isomorphism $e\mathbb{Z}_p\cong \overline{(T^e)^{\mathbb{Z}}}\cong \overline{T^{e\mathbb{Z}}}$. Then it makes sense to say:
$$\varphi(e) = \varphi(e\cdot 1) = T^e = \begin{bmatrix}1&e\\0&1\end{bmatrix}$$
$$\varphi(e)^{e^{-1}} = \varphi(e^{-1}\cdot e) = \varphi(1) = T$$
And hence $T$ is in $\overline{T^{e\mathbb{Z}}}$.
