Lemma 1

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

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

Let $M$ be a $\mathbf{Z}$ -module; let $M^G$ denote the group $\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}^{(G)}, M)$ endowed with its natural structure of left $\mathbf{Z}^{(G)}$ -module. Then:

H^i(G, M^G) = 0 \quad \text{for} \quad i \geqslant 1 .

It follows indeed from Prop. 8, b) (X, p. 109), applied with $A = N = \mathbf{Z}^{(G)}$ and $B = Q = \mathbf{Z}$ , that one has a canonical isomorphism:

\mathrm{Ext}_{\mathbf{Z}^{(G)}}(\mathbf{Z}, M^G) \to \mathrm{Ext}_{\mathbf{Z}}(\mathbf{Z}, M)

whence the lemma.

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

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