$\indent$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.