Bourbaki

Exercise 14, § 6

Algebra, X, p. 191

Let G be a group, H a subgroup of finite index of G. Put n=(G:H)n = (G : H). a) Let Q, M be two Z(G)\mathbf{Z}^{(G)}-modules. Now prove that there exists a Z\mathbf{Z}-homomorphism

t:HomZ(H)(Q,M)→HomZ(G)(Q,M) t : \mathrm{Hom}_{\mathbf{Z}(H)}(Q, M) \to \mathrm{Hom}_{\mathbf{Z}(G)}(Q, M)
such that, for every system of representatives g1,…,gng_1, \ldots, g_n of the classes of G modulo H,
(tu)(q)=∑i=1ngiu(gi−1q)for q∈Q,u∈HomZ(H)(Q,M). (tu)(q) = \sum_{i=1}^n g_i u(g_i^{-1} q) \quad \text{for } q \in Q, \quad u \in \mathrm{Hom}_{\mathbf{Z}(H)}(Q, M).
If j:HomZ(G)(Q,M)→HomZ(H)(Q,M)j : \mathrm{Hom}_{\mathbf{Z}(G)}(Q, M) \to \mathrm{Hom}_{\mathbf{Z}(H)}(Q, M) is the canonical injection, one has t∘j=n.Idt \circ j = n.\mathrm{Id}. b) Let N be a right Z(G)\mathbf{Z}^{(G)}-module. With the notations of a), now prove that there exists a Z\mathbf{Z}-homomorphism θ:N⊗Z(G)Q→N⊗Z(H)Q\theta : N \otimes_{\mathbf{Z}(G)} Q \to N \otimes_{\mathbf{Z}(H)} Q such that
θ(n⊗q)=∑i=1nngi−1⊗giqfor n∈N,q∈Q. \theta(n \otimes q) = \sum_{i=1}^n n g_i^{-1} \otimes g_i q \quad \text{for } n \in N, \quad q \in Q.
If p:N⊗Z(H)Q→N⊗Z(G)Qp : N \otimes_{\mathbf{Z}(H)} Q \to N \otimes_{\mathbf{Z}(G)} Q is the canonical mapping, one has p∘θ=n.Idp \circ \theta = n.\mathrm{Id}. c) Let P be a projective resolution of the Z(G)\mathbf{Z}^{(G)}-module Z\mathbf{Z}, q an integer. Applying a) and b) with Q=PqQ = P_q, define homomorphisms tq:Hq(H,M)→Hq(G,M)t^q : H^q(H, M) \to H^q(G, M) and θq:Hq(G,N)→Hq(H,N)\theta_q : H_q(G, N) \to H_q(H, N). If ii denotes the canonical injection of H into G, one has tq∘iq=n.Idt^q \circ i^q = n.\mathrm{Id} and iq∘θq=n.Idi_q \circ \theta_q = n.\mathrm{Id}. d) Consider the Z(G)\mathbf{Z}^{(G)}-modules M^=HomZ(H)(Z(G),M)\hat{M} = \mathrm{Hom}_{\mathbf{Z}(H)}(\mathbf{Z}^{(G)}, M) and N~=N⊗Z(H)Z(G)\tilde{N} = N \otimes_{\mathbf{Z}(H)} \mathbf{Z}^{(G)} (exercise 13). Taking Q=Z(G)Q = \mathbf{Z}^{(G)}, one obtains from the homomorphisms t and θ\theta of a) and b) homomorphisms tM:M^→Mt_M : \hat{M} \to M and θN:N→N~\theta_N : N \to \tilde{N}. Now prove that these homomorphisms are Z(G)\mathbf{Z}^{(G)}-linear. Now prove that the composed homomorphism
Hq(H,M)→(iq)−1Hq(G,M^)→Hq(G,tM)Hq(G,M) H^q(H, M) \xrightarrow{(i^q)^{-1}} H^q(G, \hat{M}) \xrightarrow{H^q(G, t_M)} H^q(G, M)
(resp. Hq(G,N)→Hq(G,θN)Hq(G,N~)→(i~q)−1Hq(H,N)H_q(G, N) \xrightarrow{H_q(G, \theta_N)} H_q(G, \tilde{N}) \xrightarrow{(\tilde{i}_q)^{-1}} H_q(H, N)) is equal to tqt^q (resp. θq\theta_q).

e) Let (g1,…,gn)(g_1, \ldots, g_n) be a system of representatives of the classes of G modulo H, and let x∈H0(H,M)x \in H^0(H, M). Now prove that one has t0(x)=∑i=1ngixt^0(x) = \sum_{i=1}^n g_i x. Describe t1t^1. f) If one identifies the groups H1(G,Z)H_1(G, \mathbf{Z}) and H1(H,Z)H_1(H, \mathbf{Z}) with G/(G,G)G/(G, G) and H/(H,H)H/(H, H) respectively, the homomorphism t1t_1 is identified with a homomorphism (transfer\text{transfer}) V:G/(G,G)→H/(H,H)V : G/(G, G) \to H/(H, H). If s~:H/G→G\tilde{s} : H/G \to G is a section of the canonical mapping, now prove that V is obtained by passing to the quotient from the mapping Vs:G→HV_s : G \to H such that

Vs(g)=∏x∈H∖Gs(x) g(s(xg))−1for g∈G. V_s(g) = \prod_{x \in H \setminus G} s(x) \, g(s(xg))^{-1} \quad \text{for } g \in G.

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.