Let $G$ be a finitely generated group. $n>0$ is a fixed integer. $\{K_\alpha\}$ is the set of all normal subgroups of $G$ with index $n$, that is $[G:K_\alpha]=n$. Consider $\bigcap_\alpha K_\alpha$. It is a normal group. But why is $[G: \bigcap_\alpha K_\alpha]$ finite?
The integer ring $\mathbb Z$ is an example. Its normal subgroup with index $n$ is $n\mathbb Z$.
Best Answer
Let G be a group generated by m generators. $n>0$ is an integer. Then G has at most $(n!)^m$ subgroups of index n.
Proof : Every subgroup K of index n corresponds to a group homomorphism
$\phi_K : G \longrightarrow S(G/K)\cong S_n$
$g \mapsto (gxK \mapsto xK )$
We always let $K$ be the first element in $\{1,2,...n\}$. If H and K are different subgroups of index n, then choose $g \in K-H$. $\phi_K(g)(K) = K, \phi_H(g)(H) = gH \not= H$. Therefore, $\phi_K \not= \phi_H$. Therefore, $K \mapsto \phi_K$ is bijective.
Such a map is defined on $m$ generators of G. For each generators, there are $n!$ choices of images. There are at most $(n!)^m$ such maps, so are subgroups of index n.