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I'm trying to understand the proof of Lemma 1 of Swinnerton-Dyer, On $\ell$-adic representations and congruences for coefficients of modular forms.

The aim is to show that if $G$ is a closed subgroup of $\mathrm{GL}_2(\mathbb Z_\ell)$ ($\ell>3$) whose image under reduction mod $\ell$ contains $\mathrm{SL}_2(\mathbb F_\ell)$, then $G$ contains $\mathrm{SL}_2(\mathbb Z_\ell)$.

Since $G$ is closed, it is enough to show for each $n>0$ that $G_n\supset\mathrm{SL}_2(\mathbb Z/\ell^n\mathbb Z)$, where $G_n$ is the image of $G$ in $\mathrm{GL}_2(\mathbb Z/\ell^n\mathbb Z)$. This holds by hypothesis for $n=1$.

Let $H_n$ be the kernel of the map $$\mathrm{SL}_2(\mathbb Z/\ell^n\mathbb Z)\to\mathrm{SL}_2(\mathbb Z/\ell^{n-1}\mathbb Z).$$

The proof claims that the result will follow by induction if we can prove that for each $n>1$, $H_n\subset G_n.$ I am struggling to understand why this is the case.

We have the following diagram, where the rows are exact.

enter image description here

The right downard arrow is an isomorphism by induction. The left downward arrow is an isomorphism if we assume that $H_n\subset G_n$. If the top-right map were surjective, then the result would follow by the five-lemma.

But a priori, I don't see why this should be the case, since the image of the intersection of two groups is not necessarily the intersection of the images.

Is there something obvious that I'm missing?

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  • $\begingroup$ Consider the group $G'=G\cap \text{SL}_2(\mathbb{Z}_\ell)$ and try to prove that it is dense in $\text{SL}_2(\mathbb{Z}_\ell)$ by the same method. $\endgroup$
    – Uri Bader
    Nov 29, 2016 at 13:46
  • $\begingroup$ @UriBader There still seems to be a jump required there. A priori, the image of $G'$ in $\mathrm{SL}_2(\mathbb F_\ell)$ is a possibly proper subgroup of $G_1\cap \mathrm{SL}_2(\mathbb F_\ell)$. So even if you can prove the result with $\mathrm{SL}_2$ more is still needed to prove the result for $\mathrm{GL}_2$. That seems to be Serre's approach (page iv-23 of Abelian l-adic representations and elliptic curves - Lemmas 2,3,5), but I was hoping there was a way of making sense of the above statement. $\endgroup$ Nov 29, 2016 at 14:09
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    $\begingroup$ Yes, you will have to prove that $G'\to \text{SL}_2(\mathbb{F}_\ell)$ is onto. This follows for example by the fact the latter is perfect (for $\ell\neq 2$) and by the fact that the commutator group of $G$ is contained in $G'$. $\endgroup$
    – Uri Bader
    Nov 29, 2016 at 16:23
  • $\begingroup$ @UriBader Ah that makes sense and greatly simplifies the proof I have of this lemma. But that argument still wouldn't show that the top-right map in the diagram is surjective, since $\mathrm{SL}_2(\mathbb Z/\ell^{n-1}\mathbb Z)$ is (I think!) not perfect. Is there a way of making sense of Swinnerton-Dyer's argument? $\endgroup$ Nov 29, 2016 at 17:22
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    $\begingroup$ But ${\rm SL}_{2}(\mathbb{Z}/\ell^{n-1} \mathbb{Z})$ is perfect (at least if $\ell \geq 5$), because it is generated by elements $x = S$, $y = ST$ where $x$ has order $4$, $y$ has order $6$ and $xy = -T$ has order $\ell^{n-1}$. $\endgroup$ Nov 29, 2016 at 18:21

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