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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 thing that conjugation is a homemorphism, and therefore maps components to components. since $geg^{-1} = e$ it must map the component containing the identity into itself. – mike May 2 '12 at 17:47
@DylanMoreland You're right. I was on crack. =P – Jonathan Gleason May 2 '12 at 23:10
Indeed, conjugation is always an automorphism of a group (topological or not). – Jonathan Gleason May 2 '12 at 23:10
We're all on crack at times. – Dylan Moreland May 2 '12 at 23:11
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). – Jonathan Gleason May 2 '12 at 23:12
up vote 4 down vote accepted

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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+1 To the point. As usual. – t.b. May 2 '12 at 22:25

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.

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