----- > [!proposition] Proposition. ([[connected component of the identity is a subgroup of a topological group]]) > Let $G$ be a [[topological group]]. The [[connected component]] of the identity $1 \in G$ is a [[subgroup]] of $G$. > [!proof]- Proof. ([[connected component of the identity is a subgroup of a topological group]]) > Denote the [[connected component]] by $C$. Since [[open sets translate to open sets in topological groups|multiplication and inversion induce homeomorphisms in topological groups]], we know that once $g \in G$ is fixed, the set $gC$ must be a [[connected component]] of $G$ by [[homeomorphisms preserves connected components]]. In fact, $gC=C$. For if not, $gC$ would be disjoint from $C$ (each is a distinct [[connected component]] of $G$). But $g \in C$, contradiction. This implies $C$ is stable under multiplication: that $\{ gc : c\in C \}=\{ c' : c' \in C \}$ implies $gc=c'$ for come $c' \in C$. > For inverses, we proceed analogously. Since $1 \in C$ and $1 \in C^{-1}$, it can't be the case that the inversion map (a [[homeomorphism]] that preserves connected components) takes $C$ anywhere other than itself. Therefore, $C=C^{-1}$. So every element in $C$ has an inverse in $C$. ^e9c16a ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```