Page 110 - Ellingham, Mark, Mariusz Meszka, Primož Moravec, Enes Pasalic, 2014. 2014 PhD Summer School in Discrete Mathematics. Koper: University of Primorska Press. Famnit Lectures, 3.

3.4 Some extension theory

Does every factor set induce an extension? Let A be a G -module and φ : G × G → A a
factor set. Let E (φ) be (as a set) G × A, with the operation

(x , a )(y ,b ) = (x y , a y + b + (x , y )φ).

E (φ) becomes a group with identity element (1, −(1, 1)φ) and inversion rule (x , a )−1 =
(x −1, −a x −1 −(1, 1)φ −(x , x −1)φ). Define µ: A → E (φ) by the rule a µ = (1, a −(1, 1)φ), and
ε: E (φ) → G by the rule (x , a )ε = x . Then we have

A / µ / E (φ) ε / / G .

Proposition 3.4.7 Let A be a G -module and φ : G ×G → A a factor set. Then the extension
A / µ / E (φ) ε / / G

induces the given G -module structure. There exists a transversal τ: G → E (φ) such that φ
is the factor set for this extension with respect to τ.

PROOF. Let g ∈ G , a ∈ A. Note that (g , 0)ε = g . By definition, the G -module structure
induced by the extension is given by (a ◦ g )µ = (g , 0)−1a µ(g , 0) = (1, a g − (1, 1)φ) = (a g )µ,
which gives the first part. For the second part, define τ: G → E (φ) by g τ = (g , 0). This is
a transversal function and x τy τ = (x y )τ((x , y )φ)µ.
By looking at factor sets, how can we determine which extensions are equivalent? Let A
be a fixed G -module and let

A / µi / Ei εi / / G , i = 1, 2

be two extensions realizing this module structure. Choose transversal functions τi and
let φi be the resulting factor sets.

First suppose these extensions are equivalent:

A / µ1 / E1 ε1 / /G

1 θ1

/ µ2 / ε2 / /G
A E
2

Then τ2 = τ1θ is a transversal for the second extension. Applying θ to

x τ1 y τ1 = (x y )τ1 ((x , y )φ1)µ1 ,

we get x τ2 y τ2 = (x y )τ2 ((x , y )φ1)µ2 , hence τ2 determines the factor set φ1 for the second
extension. As the factor sets of τ2 and τ2 belong to the same coset of B 2(G , A), we get

φ1 + B 2(G , A) = φ2 + B 2(G , A).