I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.(By the way, Kazuo Matsuzaka was one of the most famous elementary mathematics writers in Japan.)
There is the following problem in this book with many hints.
With the hints, I can solve this problem, but I want to know a typical solution for this problem.
p.78 Ex.9
$G$ is a group.
$H$ is a subgroup of $G$.
$[G:H]$ is finite.Show that there exists a normal subgroup $N$ of $G$ such that $N \subset H$ and $[G:N]$ is finite.
I will write the hints and my answer:
p.65 Ex.12
If $S \subset G$, then $N(S) = \{g \in G | g S = S g\}$ is a subgroup of $G$.
p.65 Ex.13
If $H$ is a subgroup of $G$, then $H$ is a (normal) subgroup of $N(H) = \{g \in G | g H = H g\}$.
p.60 Ex.4(Hint: use Ex.2 and Ex.3 on p.60)
If $H_1, \cdots, H_n$ are subgroups of a group $G$ and
$[G:H_i] < \infty$ for all $i \in \{1, \cdots, n\}$,
then, $[G:H_1 \cap \cdots \cap H_n] < \infty$ and $[G:H_1 \cap \cdots \cap H_n] \leq [G:H_1] \cdots [G:H_n]$.
p.77 Ex.6
Let $G$ be a group.
Let $S$ be a subset of $G$.
Let $[G:N(S)] < \infty$.Then, $\#\{a S a^{-1} | a \in G\} = [G:N(S)]$.
p.77 Ex.8
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $N = \bigcap_{a \in G} a H a^{-1}$.(1)
Show that $N$ is a normal subgroup of $G$.(2)
Show that if $M$ is a normal subgroup of $G$ and $M \subset H$, then, $M \subset N$.
And my solution using these hints is here:
$[G:H] < \infty$ by assumption and $H$ is a subgroup of $N(H)$ by p.65 Ex.13.
Obviously $[G:N(H)] < \infty$ because $N(H)$ is larger than $H$.
By p.77 Ex.6, $\#\{a H a^{-1} | a \in G\} = [G:N(H)] =: n<\infty$.
Obviously $[G:H] = [G:a H a^{-1}]$ for all $a \in G$ because $H$ and $a H a^{-1}$ have the same algebraic structure in $G$.
Let $\{a H a^{-1} | a \in G\} = \{a_1 H a_1^{-1}, \cdots, a_n H a_n^{-1}\}$.
$[G:a_1 H a_1^{-1} \cap \cdots \cap a_n H a_n^{-1}] \leq [G:a_1 H a_1^{-1}] \cdots [G:a_n H a_n^{-1}] < \infty$ by p.60 Ex.4.
$N := a_1 H a_1^{-1} \cap \cdots \cap a_n H a_n^{-1}$ is a normal subgroup of $G$ by p.77 Ex.8. and $N \subset H$ and $[G:N] < \infty$.
Best Answer
$G$ acts on $G/H$, this defines a morphism $G\rightarrow Sym(G/H)$ consider the kernel$N$ of this morphism.