Bourbaki

Exercise 7, § 6

Algebra, I, 148

Let G and A be two groups, A being commutative. Let τ\tau be a homomorphism of G into the automorphism group Aut(A) of A; if g∈G,a∈Ag \in G, a \in A, we write ga=τ(g)(a)^g a = \tau(g)(a). A crossed homomorphism of G into A is any mapping ϕ:G→A\phi : G \to A such that

ϕ(gg′)=ϕ(g)+gϕ(g′), \phi(gg') = \phi(g) + ^g \phi(g'),

the group A being written additively. The crossed homomorphisms of G into A form a group under addition denoted by Z(G,A)Z(G, A).

(a) If a∈Aa \in A, the mapping g↦ga−ag \mapsto ^g a - a is denoted by θa\theta_a. Show that a↦θaa \mapsto \theta_a is a homomorphism θ\theta of A into Z(G,A)Z(G, A). The kernel of θ\theta is the subgroup of A consisting of the elements invariant under G. The image of θ\theta is denoted by B(G,A)B(G, A).

(b) Let X=A×τGX = A \times_{\tau} G be the semi-direct product of G by A. If ϕ\phi is a mapping of G into A, show that g↦(ϕ(g),g)g \mapsto (\phi(g), g) is a section of X if and only if ϕ\phi belongs to Z(G,A)Z(G, A). For the sections corresponding to ϕ1,ϕ2∈Z(G,A)\phi_1, \phi_2 \in Z(G, A) to be conjugate by an element of A, it is necessary and sufficient that ϕ1≡ϕ2mod  B(G,A)\phi_1 \equiv \phi_2 \mod B(G, A).

(c) Let F→iE→pGF \xrightarrow{i} E \xrightarrow{p} G be an extension of GG by a group FF whose centre is equal to AA. Suppose that, if y∈Ey \in E and x=p(y)x = p(y), then

i(xf)=y.i(f).y−1 i(xf) = y . i(f) . y^{-1}
for all f∈Af \in A. If ϕ∈Z(G,A)\phi \in Z(G, A), let uϕu_\phi be the mapping of EE into EE given by the formula uϕ(y)=i(ϕ(p(y))).yu_\phi(y) = i(\phi(p(y))) . y. Show that ϕ↦uϕ\phi \mapsto u_\phi is an isomorphism of Z(G,A)Z(G, A) onto the automorphism group Aut(E)\mathrm{Aut}(E) of the extension EE. This isomorphism maps B(G,A)B(G, A) to the group of automorphisms IntE(x)\mathrm{Int}_E(x), where xx runs through AA.

Solution

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Transcribed from the printed volume. Not translated.