The congruence subgroup problem has a positive answer for $SL_n(\mathbb Z)$ for $n\geq 3$ - every finite index subgroup contains a subgroup of the form $\{A\mid A\equiv I_n\pmod N\}$ for some $N$.

For $n=2$, this is no longer the case - $SL_2(\mathbb Z)$ has many non-congruence finite index subgroups, in fact most (in asymptotic sense) of them are.