A description/proof of that closure of normal subgroup of topological group is normal subgroup
Topics
About: topological group
The table of contents of this article
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.
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.
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.
Main Body
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.
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, \(f: T_1 \rightarrow T_1\), \(p \mapsto p_0 p {p_0}^{-1}\), 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, \(p' \in N_p \cap T_2\). \(f (p') \in N_{f (p)}\). \(f (p') = p_0 p' {p_0}^{-1} \in T_2\) as \(T_2\) is a normal subgroup. So, \(N_{f (p)} \cap T_2 \neq \emptyset\). 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, \(\overline{T_2} \subseteq p_0 \overline{T_2} {p_0}^{-1}\).
