Index of a Subgroup with normal subgroup

Question

I have read in a book, about the index that $|G:H|=|G/N:H/N|$, with $G$ some group, $H\subset G$ a subgroup and $N$ is an normal subgroup of $G$ and $H$ contains $N$. So why this equation holds, if $G$ ist finite this is easy to proof by Lagrange but is there an argument for infinit groups? Straight without proof directly $|G:H|\leq |G/N:H/N|$ and $|G:H|\geq |G/N:H/N|$.

Answer 1

The index $|G : H|$ is the number of cosets of $H$ (assume this is finite, even if $G$ and $H$ are not).

$\overline{H} = H/N$ is a subgroup of $\overline{G} = G/N$, and $|\overline{G} : \overline{H}|$ is the number of cosets.

You want to show a bijection between the two sets of cosets. For a coset $gH$ (of $H$ in $G$), the natural obvious choice is to associate it with the coset $\bar{g} \overline{H}$ (of $\overline{H}$ in $\overline{G}$, where $\bar{g}$ is the image of $g$ in $G/N$). Again, you want to show this is a bijection (without appealing to the finiteness of $H$ or $G$).

It is obvious that this map is surjective, since the map $G \to G/N$ is surjective.

To show it's injective, suppose that $\bar{x} \overline{H} = \bar{y} \overline{H}$ (and we want to conclude that $xH = yH$). The fact that $\bar{x}$ and $\bar{y}$ represent the same coset means that $\bar{x} \bar{y}^{-1} \in \overline{H}$. In other words, if $\pi \colon G \to G/N$ is the quotient map, then $\pi(xy^{-1}) \in \overline{H}$. If $\pi(xy^{-1}) = \bar{h} \in \overline{H}$, then $xy^{-1} \in hN$ (where $h \in H$). Since $N \subset H$, this shows that $xy^{-1} \in H$, and so $xH = yH$.