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Could you please help me to solve this one:

The connected component of the identity of a topological group is a normal subgroup? I also need a hint to show path-connected components are normal subgroups. I am not familiar with deep properties of topological groups other than the definition.

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    I was thinking of the fact that the map that sends $g$ to the map that is conjugation by $g$ is not injective (for example, in an abelian group).2012-05-02

2 Answers 2

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Let $H$ be this connected component. You first need to show that this is even a subgroup. You have a continuous map $H \times H \to G$ given by $(x, y) \mapsto xy^{-1}$, and you want to show that the image is contained in $H$. What are all of the definitions involved?

  • A product of two connected spaces is connected.
  • The image of a connected space under a continuous map is connected.
  • A point is in the connected component of $e$ if and only if there exists a connected subset of $G$ containing that point and $e$.

I think your idea for proving that $H$ is normal is a fine one. Proving all of this for the path component of $e$ can be very similar.

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For proving that $H$ is normal, let's observe that $gHg^{-1}$ is a connected subset containing $e$. Since $H$ is the connected component of $e$, then $gHg^{-1}\subseteq H$. The proof for the path component of $e$ is similar.