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?
 A: 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. 
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is Problem 20 on p.48 in Herstein's book.
I solved this problem as follows:

I referred to "An Introduction to Algebraic Systems" (in Japanese) by Kazuo Matsuzaka.
The following problem is in this Matsuzaka's book:
Problem 3 (on p.60)
Suppose $K,L$ be subgroups of $G$ and $i_G(L)<+\infty$.
Then, $i_K(K\cap L)<+\infty$ and $i_K(K\cap L)\leq i_G(L)$ holds.
My solution to Problem 3 in Matsuzaka's book:
$K/K\cap L\ni k(K\cap L)\mapsto kL\in G/L$ is well-defined since $k^{-1}k^{'}\in K\cap L$ implies $k^{-1}k^{'}\in L$.
Suppose that $kL=k^{'}L$.
Then $k^{-1}k^{'}\in L$.
Since $k,k^{'}\in K$, $k^{-1}k^{'}\in K\cap L$.
So, $k(K\cap L)=k^{'}(K\cap L)$.
So, this well-defined mapping is injective.
So, $i_K(K\cap L)<+\infty$ and $i_K(K\cap L)\leq i_G(L)$ must hold.



My solution to Problem 20 in Herstein's book:
We use the result of Problem 18 on p.48 and Problem 19 on p.48 in Herstein's book.
By Problem 19 on p.48, there are only a finite number of distinct subgroups in $G$ of the form $aHa^{-1}$.
Let $\{H,a_1Ha_{1}^{-1},\dots,a_{n-1}Ha_{n-1}^{-1}\}$ be the set of all distinct subgroups in $G$ of the form $aHa^{-1}$.
$n=\#G/N(H)$.
Let $H_0:=H$ and $H_i:=a_iHa_{i}^{-1}$ for $i\in\{1,\dots,n-1\}$.
Obviously, $i_G(H_i)=i_G(H)$ for all $i\in\{0,\dots,n-1\}$.
Then, $i_G(\cap_{i=0}^k H_i) < +\infty$ holds for $k\in\{0,1,\dots,n-1\}$.
We prove this.
$i_G(\cap_{i=0}^0 H_i) < +\infty$ holds by assumption of Problem 20.
Assume that $i_G(\cap_{i=0}^k H_i) < +\infty$ holds.
Then, by Problem 3 in Matsuzaka's book, $i_{\cap_{i=0}^k H_i}(\cap_{i=0}^{k+1} H_i)<+\infty$ holds since $i_G(H_{k+1})<+\infty$.
(Let $K:=\cap_{i=0}^{k} H_i$ and $L:=H_{k+1}$ in Problem 3.)
$i_G(\cap_{i=0}^{k} H_i)<+\infty$ holds by assumption.
So, $i_G(\cap_{i=0}^{k+1} H_i)=i_G(\cap_{i=0}^{k} H_i)i_{\cap_{i=0}^k H_i}(\cap_{i=0}^{k+1} H_i)<+\infty$ holds.
$N:=\cap_{i=0}^{n-1} H_i$ is a normal subgroup of $G$ by Problem 18 on p.48 in Herstein's book.
$N\subset H$ is obvious.
So, $N$ is a normal subgroup of $G$ and $i_G(N)<+\infty$ and $N\subset H$.
And $i_G(H)^{i_G(N(H))}$ is an upper bound for $i_G(N)$.

