Let $G$ be a group, $H$ a subgroup of $G$, and $N:=\cap_{x\in G} \ \ x^{-1}Hx$. Prove, that $N$ is normal subgroup in $G$.
I did this:
Let $g\in G$.
Whether $g(xhx^{-1})g^{-1}\in N$?
Take $f=gx$.
$g(xhx^{-1})g^{-1}=(gx)h(x^{-1}g^{-1})=(gx)h(xg)^{-1}=fhf^{-1}\in N$, because $f\in G$, when $f\in G$ and $gx\in G$
So, $g(xhx^{-1})g^{-1}\in N$, so $N$ is normal subgroup $G$.
Is this argument is ok?
It may help you to see a more verbose proof.
First, we notice that the elements of $N$ are such that no matter which conjugation we apply to $H$, they are going to be there. So: $$ N=\{n\in H:\forall x\in G\ \exists h\in H\ st\ x^{-1}hx=n\}\subset H $$ So now we are all set to prove the normality of $N$. We are going to pick an arbitrary conjugation of $N$ and see that it coincides with $N$.
That means, let $g\in G$ and $n\in N$. Is it true that $g^{-1}ng\in N$?
By our definition of $N$, $g^{-1}ng$ would have to satisfy that $\forall x\in G\ \exists h\in H\ st\ x^{-1}hx=g^{-1}ng$.
Let $x\in G$. Is it true that $\exists h\in H\ st\ x^{-1}hx=g^{-1}ng$?
We notice here that this last condition is equivalent to $\exists h\in H\ st\ gx^{-1}hxg^{-1}=(gx^{-1})h(gx^{-1})^{-1}=n$, which is automatically fulfilled since $n\in N$.
Thus we conclude that $N$ is normal.