I'm trying to wrap my head around group cosmology, and I wanted to inquire what would be the most straight forward way to compute $H^{2}(\mathbb{Z}_{n},\mathbb{Z}_{k})$? Would it be to use the fact that, $$H^{2}(\mathbb{Z}_{n},\mathbb{Z}_{k}) \simeq \textrm{Ext}_{\mathbb{Z}\mathbb{Z}_{n} }(\mathbb{Z},\mathbb{Z}_{k})?$$ And then compute the ext, or is there a more straight forward way?
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1For group cohomology of cyclic groups, one way that usually works pretty easily is to just take a projective resolution to compute the ext. – Asvin Apr 02 '22 at 22:23
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If the action $\theta\colon \Bbb Z_n\rightarrow {\rm Aut}(\Bbb Z_k)$ is injective then $H^s(\Bbb Z_n,\Bbb Z_k)=0$ for all $s\ge 1$. – Dietrich Burde Apr 03 '22 at 10:50
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For a straightforward way see also this duplicate. – Dietrich Burde Apr 03 '22 at 11:08