You can take the profinite completion $\widehat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$ of $\mathbb{Z}$, then open subgroups of $G( \widehat{\mathbb{Z}})$ correspond to congruence subgroups in $G(\mathbb{Z})$.

The identification is easy:

$$ \Gamma \; congruence \leftrightarrow K \; open$$

$\rightarrow$: Assume you have a congruence subgroup $\Gamma \subset \Gamma(N)$, then we can consider $$ \Gamma / \Gamma(N) \subset SL_2(\mathbb{Z} / N) = \bigoplus_{p^k || N} SL_2(\mathbb{Z} / p^k ).$$ Define $K = K(\Gamma)$ as the pullback of $\Gamma / \Gamma(N)$ along the surjection $$p: \bigoplus_{p} SL_2(\mathbb{Z}_p) \rightarrow \bigoplus_{p^k || N} SL_2(\mathbb{Z} / p^k )$$

$\leftarrow$: Pick $\Gamma = K \cap SL_2(\mathbb{Q})$, where $SL_2( \mathbb{Q})$ is diagonal subgroup $\prod\limits_p SL_2(\mathbb{Q}_{p})$.

You can take the profinite completion $\widehat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$ of $\mathbb{Z}$, then open subgroups of $G( \widehat{\mathbb{Z}})$ correspond to congruence subgroups in $G(\mathbb{Z})$.

The identification is easy:

$$ \Gamma \; congruence \leftrightarrow K \; open$$

"$\rightarrow$": Assume you have a congruence subgroup $\Gamma \subset \Gamma(N)$, then we can consider $$ \Gamma / \Gamma(N) \subset SL_2(\mathbb{Z} / N) = \prod\limits_{p^k || N} SL_2(\mathbb{Z} / p^k ).$$ Define $K = K(\Gamma)$ as the pullback of $\Gamma / \Gamma(N)$ along the surjection $$p: \prod\limits_{p} SL_2(\mathbb{Z}_p) \rightarrow \prod\limits_{p^k || N} SL_2(\mathbb{Z} / p^k )$$

"$\leftarrow$": Pick $\Gamma = K \cap SL_2(\mathbb{Q})$, where $SL_2( \mathbb{Q})$ is diagonal subgroup $\prod\limits_p SL_2(\mathbb{Q}_{p})$.

You can take the profinite completion $\widehat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$ of $\mathbb{Z}$, then open subgroups of $G( \widehat{\mathbb{Z}})$ correspond to congruence subgroups in $G(\mathbb{Z})$.

This is completely standard, when you want to translate classical automorphic stuff to adelic automorphic stuff, and you are right that this makes everything computational more pleasant.

You can take the profinite completion $\widehat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$ of $\mathbb{Z}$, then open subgroups of $G( \widehat{\mathbb{Z}})$ correspond to congruence subgroups in $G(\mathbb{Z})$.

The identification is easy:

$$ \Gamma \; congruence \leftrightarrow K \; open$$

$\rightarrow$: Assume you have a congruence subgroup $\Gamma \subset \Gamma(N)$, then we can consider $$ \Gamma / \Gamma(N) \subset SL_2(\mathbb{Z} / N) = \bigoplus_{p^k || N} SL_2(\mathbb{Z} / p^k ).$$ Define $K = K(\Gamma)$ as the pullback of $\Gamma / \Gamma(N)$ along the surjection $$p: \bigoplus_{p} SL_2(\mathbb{Z}_p) \rightarrow \bigoplus_{p^k || N} SL_2(\mathbb{Z} / p^k )$$

$\leftarrow$: Pick $\Gamma = K \cap SL_2(\mathbb{Q})$, where $SL_2( \mathbb{Q})$ is diagonal subgroup $\prod\limits_p SL_2(\mathbb{Q}_{p})$.