A subgroup problem If $H$ is of finite index in $G$ prove that there is a subgroup $N$ of $G$, contained in $H$, and of finite index in $G$ such that $aNa^{-1}=N$ for all $a \in G$.
This problem is from Herstein's introductory exercise to subgroups. There seems to be a solution involving quotienting (pg.10, Q.20), but Herstein has yet not introduced quotienting in the text. Is there an alternate solution?

One approach might be to let $G$ act on $G/H$ by left-multiplication. We can think of the action as a group homomorphism from $G$ to the symmetric group on the finite set $G/H$. Let $N$ be the kernel of this homomorphism. It is a normal subgroup of $G$ since it is a kernel. Also, since every element of $N$ fixes the identity coset, we see that $N$ is a subgroup of $H$. Also, the group $G/N$ is isomorphic to a subgroup of the (finite) symmetric group of $G/H$ by the First Isomorphism Theorem, so $G/N$ is finite. In other words, the index of $N$ is finite. 
