Let $(G,*)$ a group and $H \leqslant G$ and $K= \bigcap_{g \in G} gHg^{-1}$.Prove that $K$ is a proper normal subgroup of $G$.
I'm not very familiar with group theory and i began studying some basic consepts of normal subgroups to use them in a graduate course of Galois theory.
This is my solution which is not complete:
It is very easy to see that $K$ is a subset of $G$
We'll prove that $K$ is a subgroup of $G$.Let $a,b \in K$.
Then from the definition of $K$,
$$a=gh_1g^{-1}$$
$$b=gh_2g^{-1}$$
$\forall g \in G$ nad for some $ h_1,h_2 \in H$
Then $ab^{-1}=gh_1h_2^{-1}g^{-1}$ $\forall g \in G$ and $h_1h_2^{-1} \in H $ form hypothesis, thus $ab^{-1} \in K$
Therefore $K \leqslant G$
Suppose that $K=G=\bigcap_{g \in G} gHg^{-1}$
Then $gHg^{-1}=G$ $\forall g \in G$
$g^{-1}gHg^{-1}g=g^{-1}Gg=G$ $\forall g \in G$,hence $G=H$
Now to prove that $K$ is a normal subgroup we need to show that $sKs^{-1}=K$ $\forall s \in G$
Let $s \in G$ and $x \in sKs^{-1}$,then $x=sghg^{-1}s^{-1}=(sg)h(sg)^{-1}$
$\forall g \in G$ and some $h \in H$
Can someone help me to proceed and finish the proof ?
Thank you in advance!
Notice that $$x^{-1}Kx=x^{-1}(\bigcap_{g\in G}g^{-1}Hg)x$$ $$=\bigcap_{g\in G}x^{-1}g^{-1}Hgx$$ $$=\bigcap_{g\in G}(gx)^{-1}H(gx)$$
For a fixed $x$, if $g$ changes over $G$, then $gx$ changes over $G$.
$$=\bigcap_{t\in G}(t)^{-1}H(t)=K$$
Note: $K\leq e^{-1}He=H$. Hence, If $H$ is proper in $G$, so is $K$.