Theorem - Let $G$ be a topological group and $e$ its identity element. The connected component of $e$ is a closed normal subgroup of $G$ .

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Proof: Let $F$ be the connected component of $e$ . All components of a topological space are closed, so $F$ is closed.

Let $a \in F$ . Since the multiplication and inversion functions in $G$ are continuous, the set $a F^{-1}$ is also connected, and since $e \in aF^{-1}$ we must have $aF^{-1} \subseteq F$ . Hence, for every $b \in F$ we have $ab^{-1} \in F$ , i.e. $F$ is a subgroup of $G$ .

If $g$ is an arbitrary element of $G$ , then $g^{-1}Fg$ is a connected subset containing $e$ . Hence $g^{-1}Fg \subset F$ for every $g \in G$ , i.e. $F$ is a normal subgroup. $\square$