Show that $N$ is a normal subgroup in $G$ when $N$ is the intersection of normal subgroups in $G$ QUESTION : Let $G$ be a group, let $X$ be a set, and let $H$ be a subgroup of $G$. Let $$N = \bigcap_{g\in G} gHg^{-1}$$ Show that $N$ is a normal subgroup of $G$ cointained in $H$.
MY ATTEMPT: I began by asking myself precisely what $\bigcap_{g\in G} gHg^{-1}$ means. I concluded that it must mean that if $g_1, g_2, g_3, ... , g_n \in G$ then $N$ might just be $$ g_1H{g_1}^{-1} \cap g_2H{g_2}^{-1} \cap g_3H{g_3}^{-1} \cap ... g_nH{g_n}^{-1}$$ which I figured is actually just $H$ because that's the only element that all those elements have in common.
Therefore I figured that $N=H$.
Now my problem comes in showing that $H$ is a normal subgroup of $G$. I've never been good at that.
 A: Hint: $H\leq G$ and let $\Omega$ be the set of all $Ha$ where $a\in G$. Define an action like: $$(Ha)^x=Hax,~~ Ha,Hax\in\Omega;~~x\in G$$ By this action we see that the stabilizer of $Ha$ for example is $a^{-1}Ha$. Now try to show that the map $x\mapsto\bar{x},~~\bar{x}(Ha)=Hax$ is a homomorphism with the kernel $N$.
A: You already have several good answers, and have probably completed the question yourself, but I'd like to provide my point of view.
Begin with two trivial observations:


*

*Since one of the components of the intersection is $H$ conjugated by the identity, $N\subseteq H$.

*Any intersection of subgroups is a subgroup.


Together these facts give you that $N$ is a subgroup of $H$.  Therefore, the main part of the proof is normality.  Now, ask yourself the following question:

What are you gonna conjugate $N$ by so that the result isn't in $N$?

In particular, my claim is that when we conjugate $N$ by some $x\in G$, we are simply inducing a permutation of the components of the intersection, which of course does not change the resulting content.  Can you see why this is true?

 $\displaystyle N^x=\left(\bigcap_{g\in G}H^g\right)^x=\bigcap_{g\in G}(H^g)^x=\bigcap_{g\in G}H^{gx}=\bigcap_{k\in Gx}H^{k}=\bigcap_{y\in G}H^y=N$

It's easy to visualize: like a pinwheel, permuting the leaves does not change the bulb.
Now let's look at this from a different angle.
We know that we can't form a quotient group from $H$ unless $H$ is normal.  But let's try anyway and see what we get.

Let $G$ act on the right coset space $G\backslash H$ by right multiplication.  What is the kernel of this action?

Explicitly, this means "given the homomorphism $\theta:G\rightarrow \operatorname{Sym}(G\backslash H)$ by $\theta(g)=\theta_g$ where $\theta_g(Ha)=H(ag)$, what is $\operatorname{ker}(\theta)$?"  or, more simply, "for which $k\in G$ does $Hak=Ha$ hold for all $a\in G$?

 $Hak=Ha$ if and only if $Haka^{-1}=H$ if and only if $aka^{-1}\in H$.

If this is true for all $a\in G$, then in particular it is true for $a=\operatorname{id}_G$, so $k\in H$.  Furthermore, the set of these $k$ must be normal in $G$ (as expected, since kernels are always normal).  Since all such $k$ are contained in $\operatorname{ker}(\theta)$, we must then have that $N=\operatorname{ker}(\theta)$ is precisely the largest normal subgroup of $G$ contained in $H$.
In this way, we see that $G/N$ is the best we can do when trying to make a quotient group from a not-necessarily-normal subgroup $H$.
Hopefully this provides some motivation and/or intuition towards this problem and why it matters.
A: Your $N$ is normally denoted as $core_G(H)$, the largest normal subgroup of $G$ contained in $H$.
