Let $G$ be a locally compact group, $\Gamma$ a (uniform) lattice in $G$. Let $H \leq G$ be a closed subgroup of $G$. Is $H \cap \Gamma$ a (uniform) lattice in $H$?
In the discrete case, this is easy to prove by passing from $\Gamma$, which is a finite-index subgroup, to a finite index normal subgroup, and using the isomorphism theorems, but this reduction cannot be done in the continuous case.
What if we assume $G$ and $H$ to be compactly generated? I figure this may help at least in the question about uniform lattices.
Best Answer
For the sake of removing this from the "unanswered" list: the easiest counterexample is probably in $G=\mathbb{Z}$, $H=3\mathbb{Z}$, $\Gamma=2\mathbb{Z}$. As YCor mentioned, you can also have an example where $G$ and $H$ are connected, say if $G=\Bbb{R}^2$, $H$ is a line with irrational slope, and $\Gamma=\mathbb{Z}^2$. Observe that in both examples $\Gamma$ is a uniform lattice, but the intersection is not a lattice at all.