A pro-$p$ group $G$ is called uniform if it is finitely generated, torsion-free with $G/\overline{G^p}$ abelian.
Let $\Gamma$ be a $p$-adic Lie group. Then it can be shown that $\Gamma$ has a uniform open subgroup $H$. Now let $H_1$, $H_2$ be two uniform open subgroups of $\Gamma$, can we find another uniform open subgroup $H_3$ containing both $H_1$ and $H_2$?
A first thought is that we can take $H_3$ to be the subgroup of $\Gamma$ generated by the union of the (topological) generators of $H_1$ and $H_2$. However, I don't know how to prove it is uniform.
No: there is no guarantee that the subgroup generated by $H_1$ and $H_2$ is still torsion-free. For a simple example, let $\Gamma=\mathbb{Z}_p\times\mathbb{Z}/(p)$, let $H_1$ be the closed subgroup generated by $(1,0)$ and let $H_2$ be the closed subgroup generated by $(1,1)$ (explicitly, $H_1=\mathbb{Z}_p\times \{0\}$ and $H_2=\bigcup_{n=0}^{p-1}(n+p\mathbb{Z}_p)\times\{n\}$). Then $H_1$ and $H_2$ are both uniform open subgroups but together they generate all of $\Gamma$ which has torsion.