$\indent$the classical congruence subgroup problem asks whether every finite quotient of $g={\rm gl}_{n}(\mathbb{z})$ comes from a finite quotient of $\mathbb{z}$. i.e. whether every finite index subgroup of $g$ contains a principal congruence subgroup of the form $g(m)= ker(g\to{\rm gl}_{n}(\mathbb{z}/m\mathbb{z})$ for some $m\in\mathbb{n}$? if the answer is affirmative we say that $g$ has the congruence subgroup property (csp). it was already known in the $19^{\underline{th}}$ century that ${\rm gl}_{2} (\mathbb{z})$ has many finite quotients which do not come from congruence considerations. quite surprising, it was proved in the sixties that for $n\geq3, {\rm gl}_{n} (\mathbb{z})$ does have the csp. observing that ${\rm gl}_{n} (\mathbb{z}) \cong {\rm aut} (\mathbb{z}^{n})$, one can generalize the congruence subgroup problem as follows: let $\gamma$ be a group. does every finite index subgroup of $g = {\rm aut}(\gamma)$ contain a principal congruence subgroup of the form $g(m) = \ker(g\to{\rm aut}(\gamma/m)$ for some finite index characteristic subgroup $m\leq\gamma$? very few results are known when $\gamma$ is not abelian. for example, we do not know if ${\rm aut} (f_{n})$ for $n\geq3$ has the csp. but, in 2001 asada proved, using tools from algebraic geometry, that ${\rm aut} (f_{2})$ does have the csp, and later, bux-ershov-rapinchuk gave a group theoretic version of asada's proof (2011). on the talk, we will give an elegant proof to the above theorem, using basic methods of profinite groups and free groups.