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Let $G=\bigg\{\begin{pmatrix}1&b\\0&1 \end{pmatrix}\colon b\in\mathbb{Z}\bigg\}$ and $H\leq G$ a subgroup with finite index. Define by $\pi:\text{SL}_2(\mathbb{Z})\to\text{SL}_2(\mathbb{Z}/N\mathbb{Z})$ the reduction homomorphism mapping $\gamma\to\gamma\mod(N)$. I am wondering how I can relate the index $(\pi(H):\pi(G))$ to $(H:G)$. Probably, they won't be equal as $\pi$ has non-trivial kernel, but I couldn't find the transformation property, also because $G$ and $H$ are not finite, but their images under $\pi$ are.

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    $\begingroup$ Maybe. Now compare and contrast with what happens when you replace $Nk$ with, say $(N+1)k$. (You might as well consider the case where $N$ is an odd prime first, as it'll give you the most insight, I expect.) $\endgroup$ – John Hughes Mar 16 '20 at 19:44
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    $\begingroup$ Probably I'm wrong, but I don't understand why we consider $G\subset SL_2(\mathbb{Z})$; it seems that we can consider only the group $\mathbb{Z}\cong G$ and the projection map $\pi:\mathbb{Z} \rightarrow \mathbb{Z} / N\mathbb{Z}$. Now we are asking if $\pi$ conserve the index of a generic subgroup $m\mathbb{Z} \subset \mathbb{Z}$ and it is in general false: take $m$ and $N$ coprime and you obtain thtat $[G:H]=m$ and $[\pi(G):\pi(H)]=1$ $\endgroup$ – Menezio Mar 16 '20 at 19:48
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    $\begingroup$ I am not asking whether $\pi$ conserves indices, I understand it does not as your counter example perfectly shows. No I am wondering if there is more to be said about $(G:H)$ and $(\pi(G):\pi(H))$, is there a function relating them? $\endgroup$ – user680806 Mar 16 '20 at 19:52
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    $\begingroup$ Well if you agree with the semiplificaiton of the problem, I think we can conclude in this case: taking a generic subgroup $m\mathbb{Z}$, thanks to the Bezout's Lemma you can write $g.c.d.(N,m) = aN+bm$ for some $a,b\in \mathbb{Z}$ then $[\pi(G):\pi(H)] = g.c.d.(m,N)$ and $[G:H]=m$ $\endgroup$ – Menezio Mar 16 '20 at 20:07
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    $\begingroup$ I agree on your simplification, and the fact that $(G:H)=m$. But how do we use Bézout's Lemma to conclude that $(\pi(G):\pi(H))=\gcd(m,N)$? $\endgroup$ – user680806 Mar 16 '20 at 20:14
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According to the comment above, we can consider only the group $\mathbb{Z}\cong G$ and the projection map $\pi:\mathbb{Z} \rightarrow \mathbb{Z} / N\mathbb{Z}$. Now we are asking how $\pi$ modify the index of a generic subgroup $m\mathbb{Z} \subset \mathbb{Z}$. Thanks to the Bezout's Lemma you can write $\gcd(N,m) = aN+bm$ for some $a,b\in \mathbb{Z}$, then: $$\pi(H) = (m\mathbb{Z} + N\mathbb{Z}) / N\mathbb{Z} = \gcd(m,N) \mathbb{Z} / N\mathbb{Z}$$ Then you have $$\pi(G) / \pi(H) = (\mathbb{Z} / N\mathbb{Z}) / (\gcd(m,N) \mathbb{Z} / N\mathbb{Z}) \cong \mathbb{Z} / \gcd(m,N) \mathbb{Z} $$ Hence $[\pi(G):\pi(H)] = \gcd(m,N)$ and $[G:H]=m$.

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If $\phi$ is a homomorphism of $G$, then $G/ker(\phi) \cong \phi(G)$. So, if $H \leq G$ with $|G:H|$ being finite, also $|\phi(G):\phi(H)|$ is finite, since $Hker(\phi)/ker(\phi) \cong H/(H \cap ker(\phi)) \cong \phi(H)$. Hence $|\phi(G):\phi(H)|=|G:Hker(\phi)|$, which divides $|G:H|$. And we see that there is equality if and only if $ker(\phi) \subseteq H$.

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