The index of a subgroup of finite index under a homomorphsim

Question

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.

Answer 1

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 $[G:H]=m$.

Answer 2

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 $ker(\phi) \subseteq H$.