This weird problem popped up in my research:


Let $\ell$ be a prime. Is there a description of the smallest normal subgroup of $GL\_2(\mathbb{Z}\_\ell)$ containing $GL\_2(\mathbb{Z})$?


Is there a description of the corresponding quotient group?



What is the normal closure of $GL_2(\mathbb{Z})$ inside $GL_2(\mathbb{Z}_\ell)$?
It's the subgroup of these elements in $\mathrm{GL}\_2(\mathbf{Z}\_\ell)$ with determinant in $\pm 1$. Thus the quotient is naturally isomorphic to the quotient of $\mathbf{Z}\_\ell^\times$ by $\{\pm 1\}$. 


Indeed, denote by $e\_{ij}(x)$ (for $i\neq j$) the matrix with entry $(i,j)$ equal to $x$, diagonal entries equal to 1, and other entries 0, and $d\_i(x)$, for $x$ invertible, the diagonal matrix with $(i,i)$ entry equal to $x$, and other diagonal entries equal to $1$; write $s\_{ij}(x)=d\_i(x)d\_j(x^{-1})$. Then, for $\{i,j\}=\{1,2\}$, and $x\in \mathbf{Z}\_\ell^\times$, we have $s\_{ij}(x)e\_{ij}(1)s\_{ij}(x)^{-1}=e\_{ij}(x^2)$. Since any element in $\mathbf{Z}\_\ell$ is a sum of squares (of 2 squares for $\ell$ odd, 4 squares for $\ell=2$, if I remember correctly), it follows that the normal closure of $\mathrm{SL}\_2(\mathbf{Z})$ contains all matrices $e\_{ij}(y)$ for any $y\in\mathbf{Z}\_\ell$. Since such matrices generate $\mathrm{SL}\_2(\mathbf{Z}\_\ell)$ (because it's a Euclidean ring), we deduce that the normal closure of $\mathrm{SL}\_2(\mathbf{Z})$ is all of $\mathrm{SL}\_2(\mathbf{Z}\_\ell)$. (The same simple argument work in $\mathrm{SL}\_d$ for $d\ge 2$.) Now since $\det(\mathrm{GL}\_2(\mathbf{Z})=\{\pm 1\}$, we get the result.


(Thanks to grghxy for a correction, I was initially essentially thinking of $\mathrm{SL}\_2$ and $\mathrm{GL}\_2$ makes an issue with the determinant.)


