Group theory - proof check about index and quotient group I'm studying Cayley's Theorem on the Humpreys "A Course in Group Theory" and i did not understand a passage in a preposition. (pag 86 Corollary 9.23).
It claims:
"Let $H \leq G$ with finite index $n$. Then there exists a normal subgroup $N$ of $G$ contained in $H$ with $n$ dividing $|G:N|$ and $|G:N|$ dividing $n!$.
the proof is pretty clear:
Apply prep. 9.22 
(it allows you to create, through permutations, a normal subgroup of $G$ using a subgroup $H$, and the normal subgroup is the kernel of a particular homomorphism called $\vartheta$. The kernel is $\bigcap_{x \in G} xHx^{-1}$) 
and let $N$ be the kernel of $\vartheta$. Then $N \unlhd G$ and $N \subseteq H$ so that $H/N$ is a subgroup of $G/N$ of index $n$ by the Theorem of Correspondence.
the proof continues, but my doubt are here
1) $N \subseteq H$ is due to the definition of $N$ ? if $N := \bigcap_{x \in G} xHx^{-1}$ I choose $x= 1_G$ and this leads to $N \subseteq H$? It is correct?
2) " $H/N$ is a subgroup of $G/N$ of index $n$ by the Theorem of Correspondence." I didn't know that the Theorem said something about index too, in fact in the statement or the demonstration of the theorem i didn't find passages that justify this assertion. so I tried to find a demonstration on my own.
I want to demonstrate that "If $|G:H|=n$ and $N \unlhd G$ and $N \subseteq H$ then $|G/N : H/N|=n$"
I think it is an obvious consequence of the Second Isomorphism Theorem but i want to follow a different road.
Let $gN H/N$ be a coset of $H/N$ in $G/N$, so by definition ($*$ is the operation of the group $G$) :
$gN H/N = \lbrace gN* hN : h \in H \rbrace$ $\in gH$ because $NhN=Nh \subseteq H$
$gH = \lbrace g * h : h \in H \rbrace = \lbrace g * (h*n^{-1}*n, n \in N \rbrace \in g H/N$ because $h*n^{-1} \in H$.
So I show a correspondence between cosets and therefore the index must be the same. It is correct?
 A: The "complete" Correspondence Theorem is imo both one of the most astonishing, beautiful theorems in the whole of mathematics (and in group theory, in particular), and also one of the most overseen/forgotten/unknown/"what?" ones. I think we could put it as follows:
Theorem: Let $\,G\, $ be a group, $\,N\lhd G\,\;$ .  Then, there exists a $\,1-1\,$ correspondence between the subgroups of the quotient $\,G/N\,$ and the subgroups of $\,G\,$ that contain $\,N\,$ given by
$$\bar H\le G/N\mapsto H:=\{\;x\in G\;;\;xN\in \bar H\;\}\;,\;\;\text{and its inverse correspondence}$$
$$N\le H\le G\mapsto \bar H:=\{\;xN\in G/N\;;\;x\in H\;\}\;.\;\;\text{ and s.t.: }$$
$$\begin{align*}(1)&\bar H\lhd G/N\iff H\lhd G\\
(2)&[G/N:\bar H]=[G:H]\end{align*}$$
Thus, we can write $\,\bar H=H/N:=\{\;hN\;;\;h\in H\;\}$
In the above theorem we could talk of a general group epimorphism $\,\phi : G\to K\,\;,\;\;N:=\ker\phi$ ,
and then subgroups of $\,K\,$ are in $\,1-1\,$ correspondence with subgroups of $\,G\,$ that contain $\,N\,$  and this correspondence keeps normality and index.
The above addresses, I think, your doubts about (2) in your question.
BTW, the demonstration of the Correspondence Theorem is painfully simple...
