component of identity of a topological group is a closed normal subgroup

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

$$

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 $e\in a{F}^{-1}$ we must have $a{F}^{-1}\subseteq F$. Hence, for every $b\in F$ we have $a{b}^{-1}\in F$, 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. $\mathrm{\square }$