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 $aF^{-1}$ 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. $\\square$

单词	ComponentOfIdentityOfATopologicalGroupIsAClosedNormalSubgroup
释义	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 $aF^{-1}$ 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. $\\square$
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