The connected component of the identity is a normal subgroup

Let $𝐺$ be a Topological group and $[𝑒] \sim \subseteq 𝐺$ be the (path) connected component containing the identity $𝑒$ , where $\sim$ is the (path) connectedness relation. Then $𝐺 𝑒 = [𝑒] \sim ⊴ 𝐺$ is a normal subgroup of $𝐺$ , called the (path) connected subgroup of $𝐺$ #m/thm/group

Proof

$𝑒 \in 𝐺 0$ by construction. Let $𝑎, 𝑏 \in [𝑒] \sim$ , so $𝑎 \sim 𝑏 \sim 𝑒$ . We will use The continuous image of a connected space is connected. $(-) - 1$ so continuous $𝑏 - 1 \sim 𝑒 - 1 \sim 𝑒 \sim 𝑏$ . Right-multiplication by $𝑏 - 1$ is continuous so $𝑎 𝑏 - 1 \sim 𝑏 𝑏 - 1 = 𝑒$ . Thus $𝑎 𝑏 - 1 \in [𝑒] \sim$ , and $[𝑒] \sim$ is a subgroup by One step subgroup test. For any $𝑥 \in 𝐺$ , conjugation by $𝑥$ is continuous. Hence $𝑥 𝑦 𝑥 - 1 \sim 𝑥 𝑒 𝑥 - 1 = 𝑒$ for any $𝑦 \in [𝑒] \sim$ . Therefore $[𝑒] \sim ⊴ 𝐺$ is a normal subgroup.

Properties

The cosets of $𝐺 𝑒$ are the connected components of $𝐺$ , i.e. $ℎ [𝑔] \sim = [ℎ 𝑔] \sim$

Proof of 1

Since multiplication is continuous and hence preserves connected components

ℎ \sim 𝑔 ⟺ 𝑔 - 1 ℎ \sim 𝑔 - 1 𝑔 = 𝑒 ⟺ 𝑔 - 1 ℎ \in 𝐺 𝑒 ⟺ ℎ \in 𝑔 𝐺 𝑒

proving ^P1.

#state/tidy | #lang/en | #SemBr