Before stating my question, let me recall the Congruence Subgroup Property/Problem: Given simply connected absolutely and almost simple algebraic group $G$ with fixed realization as a matrix group one has the following short exact sequence $$1\rightarrow C \rightarrow \widehat{G(\mathbb Z)} \rightarrow \overline{G(\mathbb Z)}\rightarrow 1$$ where $\widehat{G(\mathbb Z)}$ is the profinite completion of $G(\mathbb Z)$ and $\overline{G(\mathbb Z)}$ is the congruence completion of $G(\mathbb Z)$ (i.e. the completion with respect to the finite-index subgroup $Ker (G(\mathbb Z)\rightarrow G(\mathbb Z/n\mathbb Z))$). The group $C$ is called the congruence kernel. One says that $G$ has the Congruence subgroup Property if $C$ is a finite group. (for the interested reader- I think this is a great subject- I recommend a book by B.Sury about it)

For my question. Assume $G$ has the congruence subgroup property. As $C$ is discrete in $\widehat{G(\mathbb Z)}$ there exist a finite-index subgroup of $\widehat{G(\mathbb Z)}$ that intersects $C$ trivially. Equivalently, there exist a finite-index subgroup $\Gamma$ of $G(\mathbb Z)$ such that $\widehat \Gamma \cong \overline{\Gamma}$.

My question is:

Does there exist a uniform bound $M$ such that given a subgroup $\Lambda$ which is commensurable to $G(\mathbb Z)$, there exist a subgroup $\Lambda'<\Lambda$ of index at most $M$, such that $\widehat {\Lambda'} =\overline{\Lambda }$?

(commensurable = the intersection has finite index in both groups)

I think I have a complicated argument for $SL_n$ using the so-called Kubota Homomorphism as introduced in the article by Bass, milnor and Serre, but I don't see how do it for all other groups, and I have a feeling that there should be a simpler argument to show that such a bound exist.

I'd be grateful for any thought you have! Menny