Index of the projection of a subgroup on a quotient by finite normal subgroup.

Question

Given a finitely generated group $G$ and a finite normal subgroup $N \leq G$. I am trying to compare finite index subgroups in $G$ and $G/N$. I know that $H$ is a subgroup of $G$ iff $H/N$ is a subgroup of $G/N$, but does it hold that both have same index, i.e. $[G:H] = [G/N: H/N]$? I tried to construct a bijection from cosets of $H/N$ to cosets of $H$ that maps $aH/N$ to $aH$. This map is surjective and injective since if we take $aH/N$ in the kernel of this map we have that $$aH = e_HH \Rightarrow a \in H \Rightarrow aH/N = e_HH/N$$

Answer 1

First of all, the right statement is that each subgroup of $G/N$ can be written as $H/N$, for a unique $H$ such that $N \le H \le G$.

And then, let $T$ be a complete set of representatives for the cosets $a H$. So each element $g \in G$ can be written uniquely as $a h$, for $a \in T$ and $h \in H$. Then each element $g N \in G/N$ can be written as $aN \cdot h N$. Are the $a N$ unique here? Well, if $$ a N \cdot h N = b N \cdot k N, $$ for $a, b \in T$ and $h, k \in H$, then $$ a h N = b k N, $$ and thus there are $n, m \in N$ such that $$ a (h n) = b (k m), $$ with $h n, k m \in H$, hence $a = b$.

So $T' = \{ aN : a \in T \}$ is a complete set of representatives for the cosets of $H/N$ in $G/N$, and $$\begin{align}T &\to T'\\a &\mapsto a N\end{align}$$ is a bijection.