# 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 $G$.

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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 $e\in a{F}^{-1}$ 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. $\mathrm{\square}$

Title | component of identity^{} of a topological group is a closed normal subgroup |
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Canonical name | ComponentOfIdentityOfATopologicalGroupIsAClosedNormalSubgroup |

Date of creation | 2013-03-22 18:01:42 |

Last modified on | 2013-03-22 18:01:42 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 6 |

Author | asteroid (17536) |

Entry type | Theorem |

Classification | msc 22A05 |