336: Closure of Normal Subgroup of Topological Group Is Normal Subgroup

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A description/proof of that closure of normal subgroup of topological group is normal subgroup

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Topics

About: topological group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of normal subgroup.
• The reader admits the proposition that the closure of any subgroup of any topological group is a subgroup.
• The reader admits the proposition that the conjugation map of any topological group is a homeomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that the closure of any normal subgroup of any topological group is a normal subgroup.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any topological group, $T_1$, and any normal subgroup, $T_2\subseteq T_1$, the closure, $\overline{T_2}$, of $T_2$ is a normal subgroup.

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2: Proof

By the proposition that the closure of any subgroup of any topological group is a subgroup, $\overline{T_2}$ is a subgroup.

It is about $p_0\overline{T_2}{p_0}^{-1}=\overline{T_2}$ for any $p_0\in T_1$.

Let us prove that $p_0\overline{T_2}{p_0}^{-1}\subseteq \overline{T_2}$. By the proposition that the conjugation map of any topological group is a homeomorphism, the conjugation map, $f:T_1\to T_1$, $p\mapsto p_{0}p{p_0}^{-1}$, is continuous. For any point, $p\in \overline{T_2}$, and any neighborhood, $N_{f(p)}\subseteq T_1$, of $f(p)$, there is a neighborhood, $N_p\subseteq T_1$, of $p$ such that $f(N_p)\subseteq N_{f(p)}$. Whether $p$ is a point on $T_2$ or an accumulation point of $T_2$, there is a point, $p^{\prime }\in N_p\cap T_2$. $f(p^{\prime })\in N_{f(p)}$. $f(p^{\prime })=p_0{p}^{\prime }{p_0}^{-1}\in T_2$ as $T_2$ is a normal subgroup. So, $N_{f(p)}\cap T_2\ne \mathrm{\varnothing }$. So, $f(p)$ is a point on $T_2$ or an accumulation point of $T_2$, so, $f(p)=p_{0}p{p_0}^{-1}\in \overline{T_2}$.

Let us prove that $\overline{T_2}\subseteq p_0\overline{T_2}{p_0}^{-1}$. By the previous paragraph, ${p_0}^{-1}\overline{T_2}{p}_0\subseteq \overline{T_2}$. So, $\overline{T_2}\subseteq p_0\overline{T_2}{p_0}^{-1}$.

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References

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