It's well known that $SL\_2(\widehat{\mathbb{Z}})$ contains $SL\_2(\mathbb{Z})$ as a dense and finitely generated subgroup. However, $GL\_2(\mathbb{Z})$ is not dense in $GL\_2(\widehat{\mathbb{Z}})$, since $GL\_2(\mathbb{Z})$ is contained in the closed subgroup of matrices with determinant $\pm 1$, which is very far from the entirety of $GL\_2(\widehat{\mathbb{Z}})$ (whose determinant map surjects onto $\widehat{\mathbb{Z}}^\times$)


Is there a finitely generated dense subgroup of $GL\_2(\widehat{\mathbb{Z}})$?



Does $GL_2(\widehat{\mathbb{Z}})$ contain a dense finitely generated subgroup?
The answer is no. If $GL\_2(\widehat{\mathbb Z})$ is topologically finitely generated, then so is the quotient $\widehat {\mathbb Z}^\*$ (quotient via the determinant map). The latter has quotient $\widehat {\mathbb Z}^\*/ (\widehat{\mathbb Z}^\*)^2=\prod \_p {\mathbb Z}\_p^\*/({\mathbb Z}\_p^\*)^2$, where the product is over all primes. This group has as quotient 
$\prod \_p {\mathbb Z}/2{\mathbb Z}$. The latter is an infinite dimensional vector space over the finite field ${\mathbb F}\_2$ of two elements and is hence not topologically finitely generated. 


