About topological group and a connected component

Question

If $G$ is a toplogical group and $H$ is a connected component of $G$ containing the identity $e$ of $G$, then how can we show that $H$ is a normal subgroup of $G$?

I get that $xHx^{-1}$ is a connected component of $G$. Now I need to know how to show that $H \subseteq xHx^{-1}$. Because then $H= xHx^{-1}$ due to the fact that $H$ is maximal connected. So $H$ is normal... So how can I show that $H \subseteq xHx^{-1}$?

Answer 1

Sketch of Proof: A subgroup $H$ is normal if $gHg^{-1} \subset H$ for all $g \in G$. Note, however, that the map $x \mapsto gxg^{-1}$ is continuous. So, $gHg^{-1}$ must be a connected component of $G$ that contains $e$.

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Actually, we never showed that $H$ is necessarily a subgroup of $G$. To that end: consider any $g\in H$.

Claim: for any $g \in H$, $gH = H$. Similarly, $Hg = H$.

Proof: Note that the map $$ x \mapsto gx $$ is a homeomorphism, so $gH$ is a connected component. Moreover, $gH$ contains $ge = g$. So, $gH$ is a connected set with $gH \cap H \neq \emptyset$. It follows that $gH \subset H$, so that $gH = H$. $\square$

Claim: The map $\phi(x) = x^{-1}$ satisfies $\phi(H) = H$.

Proof: same argument as above, noting that $\phi(H)$ is a connected component containing $e$.