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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 $GL_2(\widehat{\mathbb{Z}})$?

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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.

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    $\begingroup$ What do you mean by $\widehat{\mathbb{Z}}^*/(\widehat{\mathbb{Z}}^*)^*$? $\endgroup$ – Will Chen Aug 7 '17 at 6:34
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    $\begingroup$ The last star should be a $2$. $\endgroup$ – Joël Aug 7 '17 at 7:03

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