Proving an Intersection of Subgroups is Normal

Question

Let $N$ be a subgroup of the group $G$.

Show that $$X =\bigcap_{g \ \in \ G} \ g^{-1}Ng $$ is normal in $G$.

It is easy to show $X$ is a subgroup in itself, by just showing $g^{-1}Ng $ is a subgroup for arbitrary $g$ then the intersection of subgroups is a subgroup in itself.

But I'm struggling to show that it is normal, I have tried showing that $$g^{-1}xg \in X \quad \forall \ g \in G, \ x \in X $$

and also directly from the definition $$Xg=gX \quad \forall \ g \in G$$ Yet have had no luck. It woud be greatly appreciated if someone could give me a helping hand to see where I'm going wrong.

Answer 1

Fix $x\in X$ and $g\in G$. We'll show that $g^{-1}xg\in X$. Since $x\in X$, it follows that $x\in h^{-1}Nh$ for all $h\in G$. Now, fix a particular $t\in G$. Our goal is to show that $g^{-1}xg\in t^{-1}Nt$. (This will prove that $g^{-1}xg\in X$, since $t$ is arbitrary.) Note that $tg^{-1}\in G$. Hence, by assumption, $x\in (tg^{-1})^{-1}N(tg^{-1})=gt^{-1}Ntg^{-1}$. Thus, $g^{-1}xg\in t^{-1}Nt$, as desired.

Answer 2

Taking your former approach: for any $x \in X$ and any $g\in G$, we need to show that $g^{-1}xg$ is in $h^{-1}Nh$ for all $h \in H$. But, in particular, $x \in gh^{-1}Nhg^{-1}$ (since $hg^{-1}\in G$, and $(hg^{-1})^{-1} = gh^{-1}$), so $g^{-1}xg\in g^{-1}gh^{-1}Nhg^{-1}g = h^{-1}Nh$, hence the result.