Bourbaki

Exercise 3, § 7

Algebra, X, p. 198

Let GG be a group, AA the algebra Z(G)\mathbf{Z}^{(G)}, j:G→Aj:G\to A the canonical injection. We endow AA with the augmentation π:A→Z\pi:A\to\mathbf{Z} defined by π(eg)=1\pi(e_g)=1 for all g∈Gg\in G. Let MM be a left AA-module.

a) Let 0→M(π)→B→pA→00\to M_{(\pi)}\to B\xrightarrow{p}A\to0 be an extension of AA by the (A,A)(A,A)-bimodule M(π)M_{(\pi)} (X, p. 196, exercise 27). We denote by Γ\Gamma the subset of BB formed by the elements bb of BB such that p(b)∈j(G)p(b)\in j(G). Show that Γ\Gamma, endowed with the multiplication induced by that of BB, is a group, extension of GG by the GG-module MM.

b) The preceding construction defines a mapping from ExZ(A,M(π))Ex_{\mathbf Z}(A,M_{(\pi)}) into Ex(G,M)Ex(G,M) (VIII, § 13). Show that the diagram

ExZ(A,M(π))⟶Ex(G,M)↓c↓H2(A,M(π))→φH2(G,M) \begin{array}{ccc} Ex_{\mathbf Z}(A,M_{(\pi)})&\longrightarrow&Ex(G,M)\\ \downarrow c&&\downarrow\\ H^2(A,M_{(\pi)})&\xrightarrow{\varphi}&H^2(G,M) \end{array}

is commutative.

Solution

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Machine translation of the English, by gpt-5-6. Not checked by a person.