Intersection of conjugate subgroups is normal Is there a better (more direct or intuitive) proof for this proposition than I have come up with below? I am not sure whether it could be simiplified:
Let $G$ be a group with $H \leq G$. Then $K = \bigcap_{g \in G} gHg^{-1}$ is normal in $G$.
Let $a \in K$. Then $a \in gHg^{-1}$ for all $g \in G$. Therefore for all $g_1,g \in G$, $g_1ag_1^{-1} \in g_1gHg^{-1}g_1^{-1} = (g_1g)H(g_1g)^{-1}$ and so $g_1ag_1^{-1} \in K$ since $g_1g \in G$. Then $K$ is normal in $G$.

For every $g\in G$, we have
$$
gKg^{-1}=g\left(\bigcap_{g'\in G} g'Hg'^{-1}\right)g^{-1}=\bigcap_{g'\in G} gg'Hg'^{-1}g^{-1}=\bigcap_{g'\in G} gg'H(gg')^{-1}
$$
$$
=\bigcap_{g'\in G}g'Hg'^{-1}=K.
$$
Note that the less trivial step is the second one. It is due to the fact that $x\longmapsto gxg^{-1}$ is injective for $\supseteq$. The inclusion $\subseteq$ is straightforward.
