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 $m,n \in \mathbb{Z}$.



Are double cosets of cyclic subgroups separable in a special linear group?
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**](http://dx.doi.org/10.1080/00927877808822268), 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**](https://doi.org/10.1017/CBO9780511565953), 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**](http://dx.doi.org/10.1007/BF01222760), 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.


