Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology
It's called the congruence topology, and is (obviously) always at least as coarse as the profinite topology. If your group has the Congruence Subgroup Property (the modular group doesn't, but $SL_n(\mathbb{Z})$ does for $n>2$) then it's the same as the profinite topology.
A google search found, for instance, Section 7.3 of Algebraic theory of the Bianchi groups by Benjamin Fine.
To expand Henry Wilton's concise answer, the Congruence Subgroup Problem has a distinguished history including important work by Serre and a number of others (exploiting effectively the congruence topology). See for example: MR0272790 (42 #7671) 14.50, Serre, Jean-Pierre, Le problème des groupes de congruence pour SL2. (French) Ann. of Math. (2) 92 1970 489–527.
This sort of topology on a group originates earlier, but the application here is highly original.
ADDED: Like many other journal articles, the one mentioned here by Serre is available in PDF format but only through JSTOR (or other library resource). There is a lot of literature, including my 1980 Springer Lecture Notes 789 Arithmetic Groups which cover some of the background as well as an expository account of Matsumoto's thesis.
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 $\prod\limits_p SL_2(\mathbb{Q}_{p})$.