If $(G : H) = r$ and there exists $K \triangleleft G$ contained in $H$ such that $(G : K) = r!$ [duplicate]

Question

Possible Duplicate:
How to prove that if $G$ is a group with a subgroup $H$ of index $n$, then $G$ has a normal subgroup $K\subset H$ whose index in $G$ divides $n!$

I'm trying to prove the following: "If $G$ is an infinite group and $H < G$ satisfies $(G:H) = r$, then there exists a normal subgroup $K$ of $G$ contained in $H$ such that $(G:K) \leq r!$". The equality is trivial if we if we find $K \triangleleft G$ such that there exists an injective action of $K$ on the cosets of $H$. However, I can't think of any such action, considering $K < H$.

Answer 1

I think you're very near the following argument. The whole group $G$ acts by left translation on the set of cosets $G/H$, and we can view this [see Theorem 1.6 of Keith Conrad's handout] action as a homomorphism $G \to \operatorname{Perm}(G/H) \approx S_r$. Let $K$ be the kernel of this homomorphism. Elements of $K$ must in particular fix the coset $H$, so what does this tell you about $K$? What do you know about the group $G/K$?

[Note that we don't need $G$ to be infinite, although the corollary, "Every finite index subgroup contains a subgroup which is normal and of finite index in the whole group," is less surprising in the finite case.]