topological closure of normal subgroup

Question

Let G be any topological group. Let N be the normal subgroup of G. Is it true that closure of N normal? I know the definition of topological group and have done for subgroup but i dont have idea where i should start for normal case.

Answer 1

The mapping $\phi_y:G\to G, x\mapsto yxy^{-1}$ is continuous. It follows that the set $\displaystyle\bigcap_{y\in G}\{x\in G: yxy^{-1}\in \overline N\} = \bigcap_{y\in G}{\phi_y}^{-1}(\overline N)$ is closed and contains $N$, hence it contains $\overline N$. This yields $yxy^{-1}\in \overline N$ for all $y\in G$ and $x\in \overline N$.

Answer 2

Yes, it is true.

Let $x\in G$ and let $a\in\overline{N}$. Let us show that $xax^{-1}\in\overline{N}$, that is that any neighborhood of $xax^{-1}$ has non-trivial intersection with $N$. For that purpose, let $W$ be a neighborhood of $xax^{-1}$ in $G$. Since $g\mapsto xgx^{-1}$ is continuous, $x^{-1}Wx$ is a neighborhood of $a$. Hence, there exists $U$ an open set in $G$ containing $a$ such that $xUx^{-1}\subset W$. Moreover, since $a\in\overline{N}$, there exists $n\in N\cap U$. Therefore, using the fact that $N$ is normal, one has $xnx^{-1}\in N\cap (xUx^{-1})\subseteq N\cap W$. Finally, $xax^{-1}\in\overline{N}$.