##### Department of Mathematics,

University of California San Diego

****************************

### Algebra Seminar

## David Ben-Ezra

#### UCSD

## The Congruence Subgroup Problem for ${\rm Aut}(F_2)$

##### Abstract:

$\indent$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.

Alireza Salehi Golsefidy

### October 23, 2017

### 3:00 PM

### AP&M 7321

****************************