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Let $\ell \ge 5$ be a prime and $G$ be a closed subgroup of $\operatorname{GL}_2 \mathbb{Z}_\ell$ whose image in $\operatorname{GL}_2 \mathbb{F}_\ell$ is $\operatorname{GL}_2 \mathbb{F}_\ell$. Then $G = \operatorname{GL}_2 \mathbb{Z}_\ell$?

In Serre's "Abelian $l$-adic representations and elliptic curves", the author shows its $\operatorname{SL}_2 \mathbb{Z}_\ell$-version. How can I show the highlighted statement from it?

It suffices to show that $H := G \cap \operatorname{SL}_2 \mathbb{Z}_\ell$ surjects onto $\operatorname{SL}_2 \mathbb{F}_\ell$.

EDIT:

In comments YCor pointed that this is false. I ask this question for the following:

Let $E$ be an elliptic curve over a number field $K$, $\ell \ge 5$ a prime. Assume that the mod $\ell$ representation $G_K \to \operatorname{GL}_2 \mathbb{Z}/\ell$ is surjective. Then is the $\ell$-adic representation $G_K \to \operatorname{GL}_2 \mathbb{Z}_\ell$ surjective?

In the proof of lemma 2.4 of González-Jiménez and Najman - An algorithm for determining torsion growth of elliptic curves, the authors say that this is true by the lemma of Serre that I mentioned above.

How can I show this without first highlighted statement?

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    $\begingroup$ It fails in $\mathrm{GL}_1$, so the answer is expected to be no. And indeed, consider a subgroup $H$ of $\mathbf{Z}_\ell^*$ of cardinal $\ell-1$ (this exists). Consider the subgroup of $\mathrm{GL}_2(\mathbf{Z}_\ell)$ generated by $\mathrm{SL}_2(\mathbf{Z}_\ell)$ and the diagonal matrices $(t,1)$ with $t\in H$. That is the same as the set of $2\times 2$ matrices with entries in $\mathbf{Z}_\ell$ with determinant in $H$; it surjects onto $\mathrm{GL}_2(\mathbf{F}_\ell)$. $\endgroup$
    – YCor
    Commented Sep 8, 2021 at 14:42
  • $\begingroup$ @YCor Thanks for your comment. I want to show that for an elliptic curve $E$ over a number field, if the mod $\ell$ representation $G_K \to \operatorname{Aut}E[\ell]$ is surjective, then so are $G_K \to \operatorname{Aut}E[\ell^n]$ for all $n$. In a paper arxiv.org/abs/1904.07071, the authors say this true. (lemma 2.4) (And also they use the highlighted statement in the proof of lemma 2.3.) Is this also false? $\endgroup$
    – zom
    Commented Sep 8, 2021 at 16:16
  • $\begingroup$ Possibly (I haven't checked), the quoted lemma allows to show you that the image contains $\mathrm{SL}_2(\mathbf{Z}_\ell)$. If so, you are reduced to proving that composing with the determinant map $G_K\to\mathrm{GL}_2(\mathbf{Z}_\ell)\to\mathbf{Z}_\ell^*$ is surjective, which sounds easier to check (or contradict). $\endgroup$
    – YCor
    Commented Sep 8, 2021 at 16:47
  • $\begingroup$ In your modified question, what is $K$, and what is the relationship of $G_K$ to $E$? $\endgroup$
    – LSpice
    Commented Sep 8, 2021 at 16:51
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    $\begingroup$ (1) Lemma 2.3 states for an arbitrary number field but weirdly uses in the statement the action of the Galois group of $\mathbb Q$, which doesn't exist in general. I think the proof of Lemma 2.3 is fine if you just replace $\mathbb Q$ everywhere with $K$. (2) Yeah, I think the statement in your question is true assuming also surjectivity of the determinant. Let me think of a good proof... $\endgroup$
    – Will Sawin
    Commented Sep 9, 2021 at 12:41

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The statement you are trying to prove is false.

In particular, in the comments YCor gave an example of an open proper subgroup $H$ of ${\rm GL}_{2}(\mathbb{Z}_{\ell})$ that surjects onto ${\rm GL}_{2}(\mathbb{F}_{\ell})$. If you take an elliptic curve $E/\mathbb{Q}$ whose $\ell$-adic image is ${\rm GL}_{2}(\mathbb{Z}_{\ell})$ (like $y^{2} + y = x^{3} - x$), let $\rho_{E,\ell^{\infty}}$ be the $\ell$-adic Galois representation attached to $E$, and let $K$ be the fixed field of $\rho_{E,\ell^{\infty}}^{-1}(H)$. Then the map $G_{K} \to {\rm Aut}~ E[\ell]$ is surjective, but $G_{K} \to {\rm Aut}~E[\ell^{2}]$ is not.

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  • $\begingroup$ The example I provided is not open since it's the kernel of a homomorphism from $\mathrm{GL}_2(\mathbf{Z}_p)\to\mathbf{Z}_p$. However composing with the canonical surjection $\mathbf{Z}_p\to\mathbf{Z}/p\mathbf{Z}$ we get an open one. $\endgroup$
    – YCor
    Commented Sep 9, 2021 at 7:40

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