Integral homology of a finite cyclic group

Put $R=\mathbb{Z}{C}_m$ and $N=1+g+\cdots +g^{m-1}$. Use one copy of R in each nonnegative degree, augmentation $ϵ(\sum a_i{g}^i)=\sum a_i$, differential multiplication by $g-1$ in odd degrees and by N in positive even degrees. Since $(g-1)N=g^m-1=0$ and $ϵ(g-1)=0$, this is an augmented chain complex.

For $a={\sum }_{i=0}^{m-1}{a}_i{g}^i$, the coefficient of $g^i$ in $(g-1)a$ is $a_{i-1}-a_i$ (indices modulo m). Thus its kernel consists exactly of constant coefficient vectors, namely $\mathbb{Z}N$. Also $Na=({\sum }_i{a}_i)N$, so the image of multiplication by N is $\mathbb{Z}N$ and its kernel is $\ker ϵ$. If $\sum a_i=0$, then $a={\sum }_{i=1}^{m-1}{a}_i(g^i-1)=(g-1){\sum }_{i=1}^{m-1}{a}_i(1+\cdots +g^{i-1})$. Hence $\ker ϵ=(g-1)R$, proving exactness in every degree. This also covers m=1: the two maps are zero and identity and the displayed sums are empty.

Each term is free, so step 2.1 makes this complex $P\to \mathbb{Z}$ a projective resolution of the trivial left $R$-module. Let $P^{\mathrm{s}\mathrm{u}\mathrm{p}}\to \mathbb{Z}$ be the supplied resolution used in [F2]. By [F3], $P$ and $P^{\mathrm{s}\mathrm{u}\mathrm{p}}$ are chain-homotopy equivalent over $\mathbb{Z}$. The additive functor $\mathbb{Z}{\otimes }_R-$ carries the comparison maps and their homotopies to comparison maps and homotopies, so $H_{\ast }(\mathbb{Z}{\otimes }_{R}P)\cong H_{\ast }(\mathbb{Z}{\otimes }_R{P}^{\mathrm{s}\mathrm{u}\mathrm{p}})=H_{\ast }(C_m;\mathbb{Z})$. Tensoring $P$ makes multiplication by $g-1$ zero and multiplication by N multiplication by m. Thus degree zero is $\mathbb{Z}$, every odd degree has kernel $\mathbb{Z}$ modulo $m\mathbb{Z}$, and every positive even degree has kernel of $m:\mathbb{Z}\to \mathbb{Z}$, which is zero.

Multiplication by m on $\mathbb{Z}\mathrm{/}m\mathbb{Z}$ is zero, and on the zero groups it is zero. This explicitly verifies the conclusion of F1. Degree zero is excluded: $m\cdot 1\mathrm{\ne }0$ in $\mathbb{Z}$. For m=1 all positive groups vanish.