Intersection of conjugate subgroups

I have got the following task: Prove that if $H$ is a subgroup of the group $G$, then $K:=\bigcap_{g \in G} g H g^{-1}$ is a normal subgroup in $G$, it lies inside $H$ and contains each normal subgroup of $G$ with lies in $H$.

For the first task, I have already found an answer here. However, I am still unsure about the two remaining ones.

If I want to prove that $K$ lies in $H$, is it enough to use that since $H$ is a subgroup of $G$, multiplying any element of $H$ with an arbitrary $g \in G$ will always remain in $H$?

For the last one, I should maybe use the fact that the intersection of normal subgroups is also a normal subgroup?

Any help appreciated.

• The first one you're asking about has the easiest proof: $K=\bigcap_{g \in G} g H g^{-1}\subseteq eHe^{-1}=H$. No, it doesn't work to take an arbitrary $g\in G$, but a special one works (any $h\in H$ would work, but if $gHg^{-1}$ remains in $H$ for any $g\in G$, then $K=H$ and $H$ is normal itself). – Arthur Feb 17 '18 at 21:59
• Oh yeah, it is really trivial, thanks for this. :) – Atvin Feb 17 '18 at 22:02
• Note that the solution @Arthur gave also solves the last question. – TPace Feb 17 '18 at 22:03
• Why does it solve the last question? I just don't see it :( – Atvin Feb 17 '18 at 22:15

Let $N\subseteq H$ be a normal subgroup of $G$. We want to prove $N\subseteq K$.
We have for any $g\in G$ that $gNg^{-1}\subseteq gHg^{-1}$. But $gNg^{-1}=N$, since $N$ is normal. Therefore $N\subseteq gHg^{-1}$.
Since $N$ is contained in each of the $gHg^{-1}$, it must be contained in their intersection, which is $K$.