Prove that $N$ is normal 
Let $H$ be a subgroup of $G$. Consider the set $N=\cap_{x\in G}xHx^{-1}$. Prove that $N$ is normal subgroup of $G$.

Using the fact that any (finite or infinite) intersection of subgroups is a subgroup I am able to prove $N$ is a subgroup of $G$ and even $H$.
 A: You can solve the problem by defining the following action as well. So let define an action of group $G$ on the set of $X$ of left cosets of $H$ by $$g\cdot(xH)=(gx)H.$$ Now try to show that this is a group action inducing a homomorphism $$\phi:G\to S_X.$$ What is its kernel? Indeed, it is:
$$\ker\phi=\bigcap_{x\in G}x^{-1}Hx. $$
A: Hints for the proof:
1) Establish that each conjugate $xHx^{-1}$ is a subgroup of $G$.
2) Establish that the intersection of any collection of subgroups is a subgroup. 
3) Deduce that $N$ is a subgroup. 
4) To prove normality, use the criterion that $N$ is normal if, and only if, $yNy^{-1}= N$ for all $y\in G$. 
5) Compute the conjugate $yNy^{-1}$. First show that it too can be represented as an intersection, namely of $yxNx^{-1}y^{-1}$. Now argue that each such conjugate is precisely one of the original conjugate, and conclude the two intersections agree. That shows that any conjugate of $N$ is $N$. QED.
A: For each given $n\in N$, $\forall x\in G$ there exists an $h_x\in H$ s.t. $xh_xx^{-1}=n$. Now for each $g\in G$, we have $h_{g^{-1}x}\in H$ s.t. $n=g^-1xh_{g^{-1}x}x^{-1}g$, then$$gng^{-1}=g(g^-1xh_{g^{-1}x}x^{-1}g)g^{-1}=xh_{g^{-1}x}x^{-1}\in xHx^{-1},$$Thus $gng^{-1}\in N$.Hence, $N$ is a normal subgroup of $G$.
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 18 on p.48 in Herstein's book.
I solved this problem as follows:

Let $y\in N$.
Let $a\in G$.
Since $y\in N\subset a^{-1}H(a^{-1})^{-1}$, we can write $y=a^{-1}h(a^{-1})^{-1}$ for some $h\in H$.
Then, $aya^{-1}=aa^{-1}h(a^{-1})^{-1}a^{-1}=h\in H$.
Let $b\in G$.
Let $x$ be an arbitrary element of $G$.
$byb^{-1}=x(x^{-1}b)y(x^{-1}b)^{-1}x^{-1}\in xHx^{-1}$ since $(x^{-1}b)y(x^{-1}b)^{-1}\in H$.
So, $byb^{-1}\in N$.
So, $N$ is a normal subgroup of $G$.

