It is well known (see e.g. K. Brown, "Cohomology of groups") that a degree-3 cohomology class of a group G with coefficients in a module A can be thought of as an equivalence class of crossed modules, in the sense that every crossed module gives a double extension of G by A. Now, it is also well known that degree-3 cohomology of Z/n with coefficients in R/Z (with a trivial action of Z/n) is isomorphic to Z/n. My question is, how does one actually construct a crossed module corresponding to an element of Z/n?



Crossed modules and degree-3 group cohomology
Here is a general procedure to construct a crossed module $\mathcal{G} = (G\_1 \stackrel{d}{\to} G\_0)$ from a normalized 3-cocycle $\alpha \in Z^3(G; \mathbb{R}/\mathbb{Z})$ which is additive in the third variable.


First, we get a cocycle $\omega \in Z^2(G; G^{\ast})$, where $G^{\ast}$ is the Pontryagin dual of $G$ equipped with the trivial $G$-action, by $\omega(a,b)(c) = \alpha(a,b,c)$. We use this cocycle to get an extension
$$0 \to G^{\ast} \to G\_0 \to G \to 0.$$


We will construct the $G\_0$-module $G\_1$ as an extension of $G\_0$-modules
$$0 \to \mathbb{R}/\mathbb{Z} \to G\_1 \to G^{\ast} \to 0.$$
Since $\mathbb{R} / \mathbb{Z}$ is divisible, this sequence splits as an exact sequence of abelian groups. Thus, $G\_1 \cong \mathbb{R} / \mathbb{Z} \times G^{\ast}$ as abelian groups. The action of $G\_0$ factors through an action of $G$:
$$a: G \times (\mathbb{R} / \mathbb{Z} \times G^{\ast}) \to \mathbb{R} / \mathbb{Z} \times G^{\ast}$$
$$a(g, (\theta, \gamma)) = (\theta + \gamma(g), \gamma)$$


The map $G\_1 \stackrel{d}{\to} G\_0$ is just the composite $G\_1 \to G^{\ast} \to G\_0$.


OK, let's specialize now to the case $G = \mathbb{Z} / n\mathbb{Z}$. We can realize the class in $H^3(\mathbb{Z} / n\mathbb{Z}; \mathbb{R}/\mathbb{Z})$ corresponding to $m \in \mathbb{Z} / n\mathbb{Z}$ by the normalized cocycle 
$$\alpha(a, b, c) = \begin{cases}
0 & \text{if $a + b < n$,}\\
cm & \text{otherwise.}
\end{cases}$$ (Here we are taking the image of the cocycle as living in $\mathbb{Z} / n\mathbb{Z}$ as a subgroup of $\mathbb{R}/\mathbb{Z}$, and we will likewise identify $\mathbb{Z} / n\mathbb{Z}$ with its Pontryagin dual by sending $a \in \mathbb{Z} / n\mathbb{Z}$ to $\frac{a \cdot -}{n} \in (\mathbb{Z} / n\mathbb{Z})^\ast$.) So explicitly, the group $G\_0$ is the set $\mathbb{Z} / n\mathbb{Z} \times \mathbb{Z} / n\mathbb{Z}$ equipped with the operation 
$$(a, b) \oplus (c, d) = \begin{cases}
(a + c, b + d) & \text{if $b + d < n$,}\\
(a + c + m, b + d) & \text{otherwise.}
\end{cases}$$
The group $G\_1$ is just $\mathbb{R} / \mathbb{Z} \times \mathbb{Z} / n\mathbb{Z}$. The action of $G\_0$ on $G\_1$ is given by
$$(a, b) \rhd (\theta, c) = (\theta + \frac{bc}{n}, c).$$
The map $G\_1 \stackrel{d}{\to} G\_0$ is the composite of projection onto the second factor followed by inclusion into the first.


