The intersection of all conjugates of a subgroup is of index at most $n!$ 
Show that if $H$ is a subgroup of a group $G$ of index $n$, then the normal subgroup
$$
N:=\bigcap_{g \in G} g H g^{-1} \subseteq G
$$
has index $\leq n!$.

This is an exercise of my algebra course.
I know that the conjugacy class of $H$ has $[G:N_G(H)]\leq n$ elements and $[G:H_1\cap H_2]\leq [G:H_1][G:H_2]$, which is not enough. I have no idea how to proceed. Any hint is welcomed.
 A: In fact $N$ is known as the core (or normal core) of $H$. Your claim can be proven by constructing an appropriate isomorphism, and appealing to the first isomorphism theorem, as follows:
Note that $G$ acts on elements of $G/H$ by left multiplication; in other words, for each $g\in G$, we have the bijection $\varphi_g:G/H\rightarrow G/H$ defined by $\varphi_g(xH):=(gx)H$. We therefore have the group ismomorphism $\Phi:G\rightarrow S_{G/H}$ defined by $\Phi(g):=\varphi_g$ (where $S_{G/H}$ is the group of permutations on $G/H$). We have the following chain of equivalences
\begin{equation}
\begin{split}
g\in\ker(\Phi)&\iff \varphi_g=\text{id}\\
&\iff gxH=xH\text{ for all }x\in G\\
&\iff x^{-1}gxH=H\text{ for all }x\in G\\
&\iff x^{-1}gx\in H\text{ for all }x\in G\\
&\iff g\in xHx^{-1}\text{ for all }x\in G\\
&\iff g\in N\\
\end{split}
\end{equation}
and thus $N=\ker(\Phi)$. By the first isomorphism theorem, we have that
$$\text{Im}(\Phi)\cong G/\ker(\Phi)=G/N$$
and thus
$$[G:N]=|\text{Im}(\Phi)|\leq |S_{G/H}|=[G:H]!$$
as desired.
A: This is the normal core. It's the largest normal subgroup contained in $H$.
It's easy to show that it's the kernel of the homomorphism induced by the action of $G$ by left multiplication on the cosets $G/H$.
Thus it's the kernel of a homomorphism into $S_n$, and the result follows.
