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I am interested in some interesting examples of $\mathbb{Z} / m $ - modules, as I am reading about the group cohomology of cyclic groups in Weibel's Introduction to homological algebra right now for a seminar talk. Do you have some interesting examples for $\mathbb{Z} / m $- modules to compute $H^* ( \mathbb{Z} , A)$?

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Assuming you mean a representation of the cyclic group $C_m$ of order $m$ (your use of the term "$\mathbb{Z}/m$-module is confusing because it's ambiguous whether $\mathbb{Z}/m$ is being regarded as a group or a ring), these correspond to modules over the group algebra

$$\mathbb{Z}[C_m] \cong \mathbb{Z}[x]/(x^m - 1)$$

where $x$ is a generator of $C_m$. The easiest examples of such modules are ones which are also vector spaces over some field $k$, which if you prefer can be $\mathbb{F}_p$ or $\mathbb{Q}$ if you want examples which are technically still just abelian groups of a particularly simple form. Then we can instead work over the group algebra $k[x]/(x^m - 1)$, which is easier to understand. By the Chinese remainder theorem, it splits up as a product

$$\prod k[x]/p_i(x)^{m_i}$$

where $x^m - 1 = \prod p_i(x)^{m_i}$ is the irreducible factorization of $x^m - 1$ over $k$. For example, if $k = \mathbb{Q}$, this is the factorization $x^m - 1 = \prod_{d | m} \Phi_d(x)$ into cyclotomic polynomials, and in general these factors will further factor. As an extreme example, if $m = p$ is prime and $k = \mathbb{F}_p$, then $x^m - 1 = (x - 1)^p$.

By the structure theorem for finitely generated modules over a PID (in this case $k[x]$), finitely generated modules over the ring above are finite direct sums of modules of the form $k[x]/p_i(x)^m$ where $m \le m_i$. They can be written down explicitly as vector spaces of the appropriate dimension where $x$ acts by the companion matrix of the polynomial $p_i(x)^m$.

Another case to try to understand is actions of $C_m$ on finitely generated free abelian groups $\mathbb{Z}^n$. Here we can act by the companion matrix of any cyclotomic polynomial, since these have integer coefficients, and take finite direct sums of these, but because the Chinese remainder theorem is not available here (because we can't divide by some integers we need to divide by; there is already a problem when $m = 2$) I don't know if every representation has this form, although I would guess not.

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