# The index of a subgroup of finite index under a homomorphsim

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.

• 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.) – John Hughes Mar 16 '20 at 19:44
• 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 $[\pi(G):\pi(H)]=1$ – Menezio Mar 16 '20 at 19:48
• 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? – user680806 Mar 16 '20 at 19:52
• 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 $[G:H]=m$ – Menezio Mar 16 '20 at 20:07
• I agree on your simplification, and the fact that $(G:H)=m$. But how do we use Bézout's Lemma to conclude that $(\pi(G):\pi(H))=\gcd(m,N)$? – user680806 Mar 16 '20 at 20:14

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 $$[G:H]=m$$.
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 $$ker(\phi) \subseteq H$$.