Page 108 - Ellingham, Mark, Mariusz Meszka, Primož Moravec, Enes Pasalic, 2014. 2014 PhD Summer School in Discrete Mathematics. Koper: University of Primorska Press. Famnit Lectures, 3.
P. 108
3.4 Some extension theory
Suppose that we have fixed τ. Then the elements {g τ : g ∈ G } act on M by conjuga-
tion. Since µ: A → M is an isomorphism, we can define g χ ∈ Aut(A) by the rule
(a g χ )µ = (g τ)−1a µ(g τ)
for a ∈ A and g ∈ G . We obtain a function χ : G → Aut(A). We prove that χ does not
depend on the choice of τ. Here we will use the fact that A is abelian. Suppose that τ
is another transversal function. Then (g τ(g τ )−1)ε = g τε(g τ ε)−1 = 1, hence g τ = g τm g
for some m g ∈ M . If τ induces χ : G → Aut(A) as above, then
(a g χ )µ = (g τ )−1a µ(g τ ) = m g−1((g τ)−1a µ(g τ))m g ,
hence g χ = g χ . Thus χ is uniquely defined. We claim that χ is a homomorphism. Let
g 1, g 2 ∈ G . Then (g 1g 2)τ ≡ g τ g τ mod M . Thus (g 1 g 2)χ = g χ g χ , hence χ is a homomor-
1 2 1 2
phism. We have proved:
Proposition 3.4.3 Each extension A / µ / E ε / / G , where A is abelian, determines a
unique homomorphism χ : G → Aut(A) which arises by conjugation in im µ by elements
of E .
Let χ : G → Aut(A) be a homomorphism. Then χ induces a G -action A given by a · g =
a g χ . We say that A is a G -module. More precisely, let g ∈ G and x ∈ E such that x ε = g .
Then
(a g )µ = x −1a µx
for a ∈ A (well defined, since A is abelian). Note that this action is trivial precisely when
im µ is central in E , i.e., when the corresponding extension is a central extension.
Theorem 3.4.4 Equivalent extensions of A by G , where A is abelian, induce the same G -
module structure on A.
PROOF. Suppose we have equivalent extensions
A / µ /E ε / /G
1 β1
/ µ ε / /G
A /E
Let χ and χ be the respective homomorphisms G → Aut(A). Choose a transversal func-
tion τ: G → E . Let τ = τβ . Then τε = τβ ε = τε = 1G , hence τ is a transversal function
for the second extension. Then (a g χ )µ = (g τ)−1a µ(g τ) and (a g χ )µ = (g τ)−1a µ(g τ) for
a ∈ A and g ∈ G . Applying β to the first equation and using the fact that µβ = µ, we get
(a g χ )µ = (g τβ )−1a µβ (g τβ ) = (a g χ )µ and thus g χ = g χ .
Suppose that we have fixed τ. Then the elements {g τ : g ∈ G } act on M by conjuga-
tion. Since µ: A → M is an isomorphism, we can define g χ ∈ Aut(A) by the rule
(a g χ )µ = (g τ)−1a µ(g τ)
for a ∈ A and g ∈ G . We obtain a function χ : G → Aut(A). We prove that χ does not
depend on the choice of τ. Here we will use the fact that A is abelian. Suppose that τ
is another transversal function. Then (g τ(g τ )−1)ε = g τε(g τ ε)−1 = 1, hence g τ = g τm g
for some m g ∈ M . If τ induces χ : G → Aut(A) as above, then
(a g χ )µ = (g τ )−1a µ(g τ ) = m g−1((g τ)−1a µ(g τ))m g ,
hence g χ = g χ . Thus χ is uniquely defined. We claim that χ is a homomorphism. Let
g 1, g 2 ∈ G . Then (g 1g 2)τ ≡ g τ g τ mod M . Thus (g 1 g 2)χ = g χ g χ , hence χ is a homomor-
1 2 1 2
phism. We have proved:
Proposition 3.4.3 Each extension A / µ / E ε / / G , where A is abelian, determines a
unique homomorphism χ : G → Aut(A) which arises by conjugation in im µ by elements
of E .
Let χ : G → Aut(A) be a homomorphism. Then χ induces a G -action A given by a · g =
a g χ . We say that A is a G -module. More precisely, let g ∈ G and x ∈ E such that x ε = g .
Then
(a g )µ = x −1a µx
for a ∈ A (well defined, since A is abelian). Note that this action is trivial precisely when
im µ is central in E , i.e., when the corresponding extension is a central extension.
Theorem 3.4.4 Equivalent extensions of A by G , where A is abelian, induce the same G -
module structure on A.
PROOF. Suppose we have equivalent extensions
A / µ /E ε / /G
1 β1
/ µ ε / /G
A /E
Let χ and χ be the respective homomorphisms G → Aut(A). Choose a transversal func-
tion τ: G → E . Let τ = τβ . Then τε = τβ ε = τε = 1G , hence τ is a transversal function
for the second extension. Then (a g χ )µ = (g τ)−1a µ(g τ) and (a g χ )µ = (g τ)−1a µ(g τ) for
a ∈ A and g ∈ G . Applying β to the first equation and using the fact that µβ = µ, we get
(a g χ )µ = (g τβ )−1a µβ (g τβ ) = (a g χ )µ and thus g χ = g χ .
