# Normal core of closed subgroup is closed

## Statement

Suppose $G$ is a topological group and $H$ is a closed subgroup of $G$. Then, the normal core of $H$ in $G$ (i.e., the largest normal subgroup of $G$ contained in $H$, also defined as the intersection of all the conjugate subgroups of $H$) is also a closed subgroup of $G$.

## Proof

Given: A topological group $G$, a closed subgroup $H$ of $G$.

To prove: The normal core of $H$ in $G$ is also a closed subgroup of $G$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 For any $g \in G$, the conjugation map $x \mapsto gxg^{-1}$ is a self-homeomorphism of $G$. Definition of topological group $G$ is a topological group PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
2 For any $g \in G$, the conjugate subgroup $gHg^{-1}$ is closed in $G$ self-homeomorphisms send closed subsets to closed subsets $H$ is closed in $G$ Step (1) Step-fact-given combination direct
3 The normal core of $H$ in $G$, which is the intersection $\bigcap_{g \in G} gHg^{-1}$, is closed in $G$. arbitrary intersection of closed subsets is closed Step (2) Step-fact combination direct.