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Let H be the closure of $\{e\}$, where $e$ is the identity element of a topological group $G$. We know that $H$ is a subgroup of $G$. We want to prove that $H$ is normal subgroup of $G$. In Valenza's "Fourier Analysis on Number Fields" the proof is as follows:

"Since $\{e\}$ is a subgroup of $G$ so is its closure $H$. Moreover, it is the smallest closed subgroup of $G$ containing $e$ and therefore normal, since each conjugate of $H$ is likewise a closed subgroup containing $e$."

The bold argument above is the one that I don't understand. I can see that for every $g\in G$, $gHg^{-1}$ is a closed subgroup (because traslations are homeomorphisms and $H$ is closed) and that implies that $H\leq gHg^{-1}$, but I don't have any idea about how to get $gHg^{-1}\leq H$.

Any help would be appreciated.

HeMan
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1 Answers1

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The conjugation $x\mapsto g^{-1}xg$ is a group automorphism and also a homeomorphism. Thus if $H$ is a closed subgroup, also $g^{-1}Hg$ is a closed subgroup.

Now if $H$ is the closure of $\{e\}$, it is contained in every closed subgroup of $G$. Now $g^{-1}Hg$ is a closed subgroup, so $H\subseteq g^{-1}Hg$, for every $g$. Using $g^{-1}$ we conclude that $H\subseteq gHg^{-1}$, which implies $g^{-1}Hg\subseteq H$.

The argument above just uses the fact that each conjugate of $H$ is a closed subgroup, which is the statement you're worrying about.

egreg
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