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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!

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    $\begingroup$ So you have to include that $H \neq G$ since otherwise your statement can't be true (indeed, you used this). However as far as I see you are actually done. You have proven that $x \in sg H (sg)^{-1}$ for all $g$ (and this particular $s$). But since the map $g \to sg$ is bijective you can say that $x \in t H t^{-1}$ for all $t$, i.e. $x \in K$. Hence you showed that $s K s^{-1} \subseteq K$ and so you are done (it is a general fact that this already implies that $s K s^{-1} = K$). $\endgroup$ – M.U. Nov 16 '16 at 19:12
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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 $K$.

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