The connected component of the identity is a normal subgroup
Let
Proof
by construction. Let , so . We will use The continuous image of a connected space is connected. so continuous . Right-multiplication by is continuous so . Thus , and is a subgroup by One step subgroup test. For any , conjugation by is continuous. Hence for any . Therefore is a normal subgroup.
Properties
- The cosets of
are the connected components of , i.e.
Proof of 1