Prove that if a subgroup $H$ of $G$ contains all of its conjugates, it is a normal subgroup.

Question

It is known the conjugate closure of a subgroup $H$, $H^G$, contains $H$. But the conjugate closure can be interpreted as the union of all the conjugates of $H$. The question is

If it is given for all $x \in G$, $xHx^{-1} \subset H$, then is it true that $xHx^{-1} = H$ for all $x \in G$?

I am not sure this is the case.

Answer 1

This is true. Take any $h \in H$. Which element of $H$ gets mapped to $h$ via conjugation? It is the element $y = x^{-1}hx$.

Now, how do we know that $y$ is actually in $H$? Because

$$h^{-1} \in H$$

$$\implies x h^{-1} x^{-1} \in H$$

$$\implies (x h^{-1} x^{-1})^{-1} \in H$$

$$\implies x^{-1}hx \in H$$

Answer 2

If $h \in H$, then for all $g \in G$ we are given that $g^{-1}hg \in H$. So consider $$h=ehe=gg^{-1}(h)gg^{-1}=g(g^{-1}hg)g^{-1} \in gHg^{-1}.$$ Thus $H \subset gHg^{-1}$