# What does the $p$-adic closure of an arithmetic lattice look like?

Let $\Gamma$ be an arithmetic lattice in a linear algebraic $\mathbb{Q}$-group $\mathbf{G}$, that is, $\Gamma$ is a subgroup of $\mathbf{G}(\mathbb{Q})$ that is commensurable with $\mathbf{G}(\mathbb{Z})$.

For a prime $p$, we can consider $\Gamma$ as a subspace of $\mathbf{G}(\mathbb{Q}_p)$. My question is:

What does the closure of $\Gamma$ in $\mathbf{G}(\mathbb{Q}_p)$ with respect to the $p$-adic topology look like?

The closure of $\mathbb{Z}$ in $\mathbb{Q}_p$ is the ring of $p$-adic integers $\mathbb{Z}_p$. So it seems plausible to me that, for example, the closure of the lattice $\Gamma = SL(n,\mathbb{Z})$ in the $p$-adic topology of $SL(n,\mathbb{Q}_p)$ would be $SL(n,\mathbb{Z}_p)$. Is this correct?

Also, what about other lattice, for example, what is the closure of a congruence subgroup $$\Gamma(c) := \{ g \in SL(n,\mathbb{Z}) : g - I_n \equiv 0 \;\text{ mod } c\, \} \subset SL(n,\mathbb{Z})$$ in the $p$-adic topology of $SL(n,\mathbb{Q}_p)$?

Suppose $G$ is $\mathbb Q$ simple (i.e. has no connected normal algebraic subgroups which are defined over $\mathbb Q$) and is simply connected (i.e. $G(\mathbb C)$ is simply connected). Assume also that $G(\mathbb R)$ is not compact. With these assumptions, the closure of an arithmetic lattice in $G({\mathbb Z}_p)$ is an open subgroup. This statement is known as strong approximation.
More generally, if $G(\mathbb Z)$ is Zariski dense in $G$, and $G(\mathbb C)$ is connected and simply connected, then the closure of a finite index subgroup of $G(\mathbb Z)$ is open in $G(\mathbb Z _p)$.
Examples are $G=SL_n$ and $Sp_{2n}$. But not $PGL_n$ (this is not simply connected).
In your example, the closure of $SL(n,\mathbb Z)$ is indeed $SL(n,\mathbb Z _p)$; this can be proved by using the fact that $SL(n,\mathbb Z), SL(n,\mathbb Z _p)$ are generated by unipotent elements.