Are double cosets of cyclic subgroups separable in a special linear group? Let $A,B \in \mathrm{SL}_3(\mathbb{Z})$. Set
$$S = \langle A \rangle \cdot \langle B \rangle = \{A^mB^n : m,n \in \mathbb{Z}\}.$$

Is $S$ closed in the profinite topology on
  $\mathrm{SL}_3(\mathbb{Z})$ ?

Equivalently (using the congruence subgroup property), I am asking whether for every $C \in \mathrm{SL}_3(\mathbb{Z})$ for which $$C \equiv A^{m_k}B^{n_k} \pmod k$$ holds for any $k$, we necessarily have $C = A^mB^n$ for some $m,n \in \mathbb{Z}$.
 A: Although this is an older question, I think that the answer may still be of interest.
I realised that a positive answer follows from the work of Grunewald and Segal  [Grunewald, Fritz; Segal, Daniel, Conjugacy in polycyclic groups, Commun. Algebra 6, 775-798 (1978).] In fact, the following is true:
If H and K are virtually polycyclic subgroups of $SL_n(\mathbb{Z})$ then the double coset $HK$ is closed in the congruence topology.
The proof is essentially contained in Exercise 13 on p. 63 of the book of Dan Segal Polycyclic groups, Cambridge Tracts in Mathematics, 82. Cambridge University Press, Cambridge, 1983.
The idea is as follows: consider $SL_n(\mathbb{Z})$ as a subset of the $n \times n$ integer matrices $M_n(\mathbb{Z})$. Here $M_n(\mathbb{Z})$ is a group under matrix addition, so it is isomorphic to $\mathbb{Z}^{n^2}$. A key observation is that the profinite topology on $M_n(\mathbb{Z})$ induces precisely the congruence topology on its subset $SL_n(\mathbb{Z})$ (see Exercise 12 on p. 62 in Segal's book).
Now, consider the action of the direct product $H \times K$ on $M_n(\mathbb{Z})$ defined as follows: $$(h,k) \circ m=h m k^{-1}, \text{ for all } (h,k) \in H \times K,~m \in M_n(\mathbb{Z}).$$ The resulting semidirect product $G=M_n(\mathbb{Z}) \rtimes (H \times K)$ is a virtually polycyclic group, so it is conjugacy separable by a theorem of Remeslennikov/Formanek. Note that the conjugacy class of the identity matrix $i \in M_n(\mathbb{Z})$ in $G$ is precisely the subset $HK \subseteq SL_n(\mathbb{Z}) \subseteq M_n(\mathbb{Z})$. It follows that $HK$ is closed in the profinite topology on $M_n(\mathbb{Z})$, hence it is also closed in the congruence topology on $SL_n(\mathbb{Z})$.
Finally , let me note that this does not extend to triple cosets: indeed, in
[Lennox, John C.; Wilson, John S., On products of subgroups in polycyclic groups, Arch. Math. 33, 305-309 (1980). ] the authors construct an example of 3 cyclic subgroups of $SL_3(\mathbb{Z})$ whose product is not profinitely closed.
