# Computation of $H^n(G,A)$for a finite cyclic group $G$ in Rotman's 'An Introduction to Homological Algebra'

I am reading Rotman's 'An Introduction to Homological Algebra, Second Edition'.

On page 522 he computes $H^n(G,A)$ for a finite cyclic group $G=\langle x \rangle$ of order $k$.

For that purpose, he uses the $G$-free resolution $\rightarrow \mathbb{Z}G\overset{D}{\longrightarrow}\mathbb{Z}G\overset{N}{\longrightarrow} \mathbb{Z}G\overset{D}{\longrightarrow}\mathbb{Z}G\overset{\epsilon}{\longrightarrow} \mathbb{Z}\longrightarrow 0$. In this resolution, $\epsilon$ is the augmentation map and $D=x-1$ and $N=\sum_{i=0}^{k-1} x^i$ (those elements, when treated as maps, mean 'multiplication by').

He then says that he applies $\text{Hom}_G(\mathbb{Z},-)$ to this resolution and takes homology (the proof is not very detailed). From my understanding of how group cohomology is computed using a projective resolution, I would expect him to apply $\text{Hom}_G(-,A)$ instead of $\text{Hom}_G(\mathbb{Z},-)$ to the resolution before taking homology.

I tried it, and I got the result he claims to get. Initially I was sure it's a mistake, but I couldn't find it in the book's online errata. Is this a mistake in the book?

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Yes, that's a mistake. One should apply $\mathrm{Hom}_{\mathbf{Z}[G]}(-,A)$ to a $\mathbf{Z}[G]$-projective resolution of the trivial module $\mathbf{Z}$ to get $H^i(G,A)$. If one started with an injective resolution of $A$, then one would apply $\mathrm{Hom}_{\mathbf{Z}[G]}(\mathbf{Z},-)$ to get $H^i(G,A)$.