Central extension of an abelian group

Let $(𝐴, +)$ be an abelian group and consider a central extension

1 \to 𝐶 e x p ↪ ˆ 𝐴 𝜋 ↠ 𝐴 \to 1

Then $ˆ 𝐴$ is nilpotent with its commutator subgroup central #m/thm/group since

[ˆ 𝐴, ˆ 𝐴] ⊴ e x p (𝐶) ⊴ 𝑍 (ˆ 𝐴)

and given a $𝖲 𝖾 𝗍$ -section $𝑠 (-) : 𝐴 ↪ ˆ 𝐴$ of $𝜋$ we have the associated commutator map

𝑐 0 : 𝐴 \times 𝐴 \to 𝐶 (𝑎, 𝑏) \mapsto l n [𝑠 𝑎, 𝑠 𝑏]

which is alternating $ℤ$ -bilinear and independent of $𝑠 (-)$ .¹

Properties

In what follows, if $𝐵 \leq 𝐴$ is a subgroup let $ˆ 𝐵 = 𝜋 - 1 𝐵$ .

𝑅 = {𝑎 \in 𝐴 : 𝑐 0 (𝑎, 𝐴) = 0}

Then $ˆ 𝑅 = 𝑍 (ˆ 𝐴)$ . 3. The associated 2-cycle $𝜀 0 : 𝐴 \times 𝐴 \to 𝐶$ and associated commutator $𝑐 0 : 𝐴 \times 𝐴 \to 𝐶$ are related by

𝑐 0 (𝑎, 𝑏) = 𝜀 0 (𝑎, 𝑏) - 𝜀 0 (𝑏, 𝑎)

that is, $𝑐 0$ is the antisymmetrization of $𝜀 0$ .

Proof

Note $ˆ 𝐵 = 𝑠 𝐵 e 𝐶$ , and $[𝑠 𝐵 e 𝐶, s 𝐵 e 𝐶] = [𝑠 𝐵, 𝑠 𝐵]$ , from which one easily verifies ^P1.

Assume $𝑠 𝑎 e 𝑝 \in ˆ 𝑅$ , so $𝑎 \in 𝑅$ . Then $𝑐 0 (𝑎, 𝐴) = l n [𝑠 𝑎, 𝑠 𝐴] = 0$ . Given any $𝑠 𝑏 e 𝑞 \in ˆ 𝐴$ ,

𝑠 𝑎 e 𝑝 𝑠 𝑏 e 𝑞 = e 𝑞 𝑠 𝑎 𝑠 𝑏 e 𝑝 = e 𝑞 𝑠 𝑏 𝑠 𝑎 e 𝑝 = 𝑠 𝑏 e 𝑞 𝑠 𝑎 e 𝑝

so $𝑠 𝑎 e 𝑝 \in 𝑍 (ˆ 𝐴)$ . Similarly, if $𝑠 𝑎 \in 𝑍 (ˆ 𝐴)$ then $𝑐 0 (𝑎, 𝐴) = l n [𝑠 𝑎, 𝐴] = 0$ . Therefore $ˆ 𝑅 = 𝑍 (ˆ 𝐴)$ , proving ^P2.

Finally, noting that $e x p$ is a group monomorphism,

e 𝑐 0 (𝑎, 𝑏) = [𝑠 𝑎, 𝑠 𝑏] = 𝑠 𝑎 𝑠 𝑏 𝑠 - 1 𝑎 𝑠 - 1 𝑏 = 𝑠 𝑎 + 𝑏 e 𝜀 0 (𝑎, 𝑏) 𝑠 - 1 𝑏 + 𝑎 e - 𝜀 0 (𝑏, 𝑎) = e 𝜀 0 (𝑎, 𝑏) - 𝜀 0 (𝑎, 𝑏)

proving ^P3.

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