Bourbaki

Exercise 8, § 6

Algebra, I, 149

Bourbaki marks this one as harder than the rest.

With the notation of the above exercise, C(G,A)C(G, A) is used to denote the group of mappings of GG into AA. GG operates on C(G,A)C(G, A) by the formula

(gϕ)(g′)=g(ϕ(g−1g′)). (g \phi)(g') = g(\phi(g^{-1}g')) .
If σ\sigma denotes the corresponding homomorphism of GG into Aut(C(G,A))\mathrm{Aut}(C(G, A)), the semi-direct product C(G,A)×σGC(G, A) \times_\sigma G is denoted by E0E_0.

(a) Let ε:A→C(G,A)\varepsilon : A \to C(G, A) be the mapping which associates with each a∈Aa \in A the constant mapping equal to aa. Show that ε\varepsilon is an injective homomorphism, compatible with the action of GG.

(b) Let A→iE→pGA \xrightarrow{i} E \xrightarrow{p} G be an extension of GG by AA such that i(ga)=xi(a)x−1i(ga) = xi(a)x^{-1} if a∈A,x∈Ea \in A, x \in E and g=p(x)g = p(x). Let ρ\rho be a mapping G→EG \to E such that p∘ρ=IdGp \circ \rho = \mathrm{Id}_G. For all x∈Ex \in E, let ϕx\phi_x be the mapping of GG into AA such that

xρ(p(x−1)g)=i(ϕx(g))ρ(g)for all g∈G. x \rho(p(x^{-1})g) = i(\phi_x(g)) \rho(g) \quad \text{for all } g \in G .
Show that
ϕxy=ϕx+p(x)ϕyif x,y∈E. \phi_{xy} = \phi_x + {}^{p(x)}\phi_y \quad \text{if } x, y \in E .
Deduce that the mapping Φ:E→E0\Phi : E \to E_0 defined by
Φ(x)=(ϕx,p(x)) \Phi(x) = (\phi_x, p(x))
is a homomorphism which makes the following diagram commutative:

A→iE→pG∣∣∣εΦIdGC(G,A)⟶E0⟶G. \begin{array}{ccccc} A & \xrightarrow{i} & E & \xrightarrow{p} & G \\ | & & | & & | \\ \varepsilon & & \Phi & & \mathrm{Id}_G \\ C(G, A) & \longrightarrow & E_0 & \longrightarrow & G . \end{array}

(c) A GG-mean on AA is a homomorphism

m:C(G,A)→A m : C(G, A) \to A
satisfying the two following conditions:
(c1)m∘ε=IdA(c2)m(gϕ)=gm(ϕ)if g∈G,ϕ∈C(G,A). \begin{align*} (c_1) & \quad m \circ \varepsilon = \mathrm{Id}_A \\ (c_2) & \quad m(g \phi) = {}^g m(\phi) \quad \text{if } g \in G, \phi \in C(G, A) . \end{align*}

Let mm be a G-mean on A. Show that, if ϕ\phi is a crossed homomorphism and a=m(ϕ)a = m(\phi), then ϕ=θ−a\phi = \theta_{-a} (cf. Exercise 7). In particular,

Z(G,A)=B(G,A). Z(G, A) = B(G, A).

Show that the mapping (ϕ,g)↦(m(ϕ),g)(\phi, g) \mapsto (m(\phi), g) is a homomorphism of E0E_0 into A×τGA \times_{\tau} G. Deduce that every extension E of G by A satisfying the condition in (b) is isomorphic to A×τGA \times_{\tau} G (use the composition E→ΦE0→A×τGE \xrightarrow{\Phi} E_0 \to A \times_{\tau} G) and hence admits a section. Show that two such sections are transformed one into the other by an inner automorphism of E defined by an element of A (use Exercise 7 and the fact that Z(G,A)=B(G,A)Z(G, A) = B(G, A)).

(d) Suppose that G is finite of order n and that the mapping α↦nα\alpha \mapsto n\alpha is an automorphism of A. For all ϕ∈C(G,A)\phi \in C(G, A), let m(ϕ)m(\phi) denote the unique element of A such that

n⋅m(ϕ)=∑g∈Gϕ(g). n \cdot m(\phi) = \sum_{g \in G} \phi(g).

Show that mm is a G-mean on A.

This applies in particular when A is finite of order prime to n.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.