If $G$ is a finite group and $A$ is a $\Bbb ZG$-module such that the underlying abelian group is finitely generated, then I want to show that $H^n(G,A)$ is also finitely generated.

All $\Bbb ZG$-modules in the standard resolution $$\cdots P_2\to P_1\to P_0 \to \Bbb Z\to 0$$ have finitely generated underlying abelian group since $G$ is finite. Why are all the groups occuring in $$\cdots\operatorname{Hom}_{\Bbb ZG}(P_2,A)\leftarrow \operatorname{Hom}_{\Bbb ZG}(P_1,A)\leftarrow \operatorname{Hom}_{\Bbb ZG}(P_0,A)\leftarrow 0$$

finitely generated?

Does it follow from the fact that $\operatorname{Hom}_{R}(A,B)$ is a finitely generated $R$-module whenever $A,B$ are finitely generated $R$-modules? ($R$ Noetherian)


The standard resolution of $\mathbb{Z}$ by $\mathbb{Z}G$-modules that you mention is the bar-resolution that has $\mathbb{Z}[G^{n+1}]$ in degree $n$. One sees that these are free $\mathbb{Z}G$-modules generated by symbols $[g_1\mid g_2 \mid \cdots \mid g_n]$, where $$[g_1 \mid g_2 \mid \cdots \mid g_n] = (1, g_1, g_1 g_2, \ldots, g_1 g_2 \cdots g_n) \in \mathbb{Z}[G^{n+1}].$$ So thanks to this we have $$\operatorname{Hom}_{\mathbb{Z}G} (\mathbb{Z}[G^{n+1}], A) \cong \operatorname{Map} (G^n, A).$$ The underlying abelian group is the one induced by the addition in $A$. Now, as $G^n$ is a finite set and $A$ is a finitely generated abelian group, $\operatorname{Map} (G^n, A)$ is a finitely generated abelian group. So the bar-resolution leads to a complex of finitely generated abelian groups.

(Bonus: using, for instance, the bar-resolution, one can see that the order of $G$ annihilates $H^n (G,A)$ for $n > 0$, so in our case these are finitely generated torsion groups, hence finite.)


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