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?
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1 Answer
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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.)
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2$\begingroup$ One has to be careful about the "torus" quotient. More specifically, this argument shows that one gets at least ${\rm{SL}}_2(\mathbf{Z}_{\ell})$, so the answer is actually the image of ${\rm{GL}}_2(\mathbf{Z})$ in the abelian ${\rm{GL}}_2(\mathbf{Z}_{\ell})/{\rm{SL}}_2(\mathbf{Z}_{\ell}) = \mathbf{Z}_{\ell}^{\times}$ (via det), and this image is $\mathbf{Z}^{\times}$. So the normal closure is generated by ${\rm{SL}}_2(\mathbf{Z}_{\ell})$ and diag($u, 1$) for $u \in \mathbf{Z}^{\times}$, with quotient $\mathbf{Z}_{\ell}^{\times}/\mathbf{Z}^{\times}$. $\endgroup$– grghxyCommented Jul 22, 2015 at 16:34
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$\begingroup$ Also, a softer way to handle the squares (which works with more general integer rings in place of $\mathbf{Z}$) is to note that since the subset of squares in $\mathbf{Z}_{\ell}$ is open, the $\mathbf{Z}$-submodule they generate inside $\mathbf{Z}_{\ell}$ is open and hence closed, so it is a $\mathbf{Z}_{\ell}$-submodule. But it contains 1, so that $\mathbf{Z}$-submodule coincides with $\mathbf{Z}_{\ell}$. $\endgroup$– grghxyCommented Jul 22, 2015 at 16:39
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$\begingroup$ Since ${\rm{diag}}(t,1/t)=u^+(t)u^{-}(-1/t)u^+(t-1)u^{-}(1)u^+(-1)$ for units $t$ and standard parameterizations $u^{\pm}$ of standard root groups of ${\rm{SL}}_2$, for any ring $R$ and simply connected Chevalley $R$-group $G$ the subgroup of $G(R)$ generated by $U_a(R)$'s for roots $a$ contains $\Omega(R)$ for an open cell $\Omega$. If $R$ is local then $\ker(G(R)\rightarrow G(k))\subset \Omega(R)$ (as $\Omega$ is an open subscheme of $G$), and translates of $\Omega$ by Weyl-representatives cover $G$ (enough on geometric fibers!), so $U_a(R)$'s generate $G(R)$ (no Euclidean stuff). $\endgroup$– grghxyCommented Jul 22, 2015 at 16:54
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$\begingroup$ The Euclidean argument works; yes I'm aware there are plenty of variants and generalizations, and since the question was about $\mathrm{SL}_2$ I chose the most elementary argument I had in mind. $\endgroup$– YCorCommented Jul 22, 2015 at 17:09
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$\begingroup$ (But it's true that both $R$ local or $R$ Euclidean imply $\mathrm{SL}_2(R)$ generated by elementary matrices, and the argument when $R$ is local is the simpler one.) $\endgroup$– YCorCommented Jul 22, 2015 at 18:47