Topological group

The connected component of the identity is a normal subgroup

Let 𝐺 be a Topological group and [𝑒]∼ ⊆𝐺 be the (path) connected component containing the identity 𝑒, where ∼ is the (path) connectedness relation. Then 𝐺𝑒 =[𝑒]∼ ⊴𝐺 is a normal subgroup of 𝐺, called the (path) connected subgroup of 𝐺 group

Proof

𝑒 ∈𝐺0 by construction. Let 𝑎,𝑏 ∈[𝑒]∼, so 𝑎 ∼𝑏 ∼𝑒. We will use The continuous image of a connected space is connected. ( −)−1 so continuous 𝑏−1 ∼𝑒−1 ∼𝑒 ∼𝑏. Right-multiplication by 𝑏−1 is continuous so 𝑎𝑏−1 ∼𝑏𝑏−1 =𝑒. Thus 𝑎𝑏−1 ∈[𝑒]∼, and [𝑒]∼ is a subgroup by One step subgroup test. For any 𝑥 ∈𝐺, conjugation by 𝑥 is continuous. Hence 𝑥𝑦𝑥−1 ∼𝑥𝑒𝑥−1 =𝑒 for any 𝑦 ∈[𝑒]∼. Therefore [𝑒]∼ ⊴𝐺 is a normal subgroup.

Properties

1. The cosets of 𝐺𝑒 are the connected components of 𝐺, i.e. ℎ[𝑔]∼ =[ℎ𝑔]∼

Proof of 1

Since multiplication is continuous and hence preserves connected components

ℎ∼𝑔⟺𝑔−1ℎ∼𝑔−1𝑔=𝑒⟺𝑔−1ℎ∈𝐺𝑒⟺ℎ∈𝑔𝐺𝑒

proving ^P1.

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