A closed subgroup $G$ of $\operatorname{GL}_2 \mathbb{Z}_\ell$ which surjects onto $\operatorname{GL}_2 \mathbb{F}_\ell$

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?

• 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 $\mathrm{GL}_2(\mathbf{F}_\ell)$.
– YCor
Commented Sep 8, 2021 at 14:42
• @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?
– zom
Commented Sep 8, 2021 at 16:16
• 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).
– YCor
Commented Sep 8, 2021 at 16:47
• In your modified question, what is $K$, and what is the relationship of $G_K$ to $E$? Commented Sep 8, 2021 at 16:51
• (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... Commented Sep 9, 2021 at 12:41

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.
• 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.