Bourbaki

Example 4

Tag 0DYK, permanent. Cite it as [Bourbaki, Tag 0DYK].

Algebra, chapter X, § 6. Utilisation de résolutions non canoniques

Let MM be a Z\mathbf{Z}-module, considered as a right Z(G)\mathbf{Z}^{(G)}-module by making GG operate trivially. The group H1(G,M)H_1(G,M) is isomorphic to the quotient of M⊗ZZ(G)M\otimes_{\mathbf{Z}}\mathbf{Z}^{(G)} by the sub-Z\mathbf{Z}-module generated by the elements m⊗(eg1g2−eg1−eg2)m\otimes(e_{g_1g_2}-e_{g_1}-e_{g_2}) for mm in MM, g1,g2g_1,g_2 in GG; it follows easily that H1(G,M)H_1(G,M) is isomorphic to M⊗Z(G/(G,G))M\otimes_{\mathbf{Z}}(G/(G,G)).

Let us denote by σ\sigma the anti-automorphism of Z(G)\mathbf{Z}^{(G)} defined by σ(eg)=eg−1\sigma(e_g) = e_{g^{-1}} for g∈Gg \in G. Every left Z(G)\mathbf{Z}^{(G)}-module may be considered as a right Z(G)\mathbf{Z}^{(G)}-module by means of σ\sigma, and conversely. This makes it possible, for example, to define the groups Hq(G,M)H_q(G, M) for a left Z(G)\mathbf{Z}^{(G)}-module MM, by setting Hq(G,M)=Hq(G,σ∗(M))=TorqZ(G)(Z,M)H_q(G, M) = H_q(G, \sigma_*(M)) = \mathrm{Tor}_q^{\mathbf{Z}^{(G)}}(\mathbf{Z}, M).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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