Inspired by [this question](https://mathoverflow.net/questions/250287/g-cocycle-split-to-a-coboundary-in-j-via-a-group-extension), let us take a nontrivial 3-cocycle $\omega\_3^G(g\_a, g\_b, g\_c) \in H^3(G,\mathbb{R}/\mathbb{Z})$ in the cohomology group of $G$ with $U(1)=\mathbb{R}/\mathbb{Z}$ coefficient. In otherwords, here the 3-cocycle $\omega\_3^G$ is a complex $U(1)=\mathbb{R}/\mathbb{Z}$ function with the norm $|\omega\_3^G|=1$ but with a $U(1)$ complex phase satisfying the cocycle condition. Here $g\_a, g\_b, g\_c \in G$.


We like to ask how can we trivialize the 3-cocycle $\omega\_3(g\_a, g\_b, g\_c)$ of $G$ into 3-coboundary if we lift $G$ into a larger group $J$, and given that we know the group homomorphism $r$:



> 
> $$J \overset{r}{\rightarrow} G,$$
> 
> 
> 


so that  

$$\omega\_3^J(j\_a, j\_b, j\_c)=\omega\_3^G(r(j\_a),r(j\_b),r(j\_c))=\omega\_3^G(g\_a, g\_b, g\_c) \text{ is trivial in } H^3(J,\mathbb{R}/\mathbb{Z}).$$


I like to take $G=\mathbb{Z}\_2 \oplus \mathbb{Z}\_2 \oplus \mathbb{Z}\_2$, and consider the group element $g\_a=(g\_{a1}, g\_{a2}, g\_{a3}) \in (\mathbb{Z}\_2, \mathbb{Z}\_2, \mathbb{Z}\_2)=\mathbb{Z}\_2 \oplus \mathbb{Z}\_2 \oplus \mathbb{Z}\_2 =G$, similarly, $g\_b=(g\_{b1},g\_{b2},g\_{b3}) \in G$ and $g\_{c}=(g\_{c1},g\_{c2},g\_{c3}) \in G$. Let me focus on the 3-cocycle



> 
> $$
> \omega\_3^G(g\_a, g\_b, g\_c)=\exp[i \pi \cdot g\_{a1} \cdot g\_{b2} \cdot g\_{c3}]=(-1)^{g\_{a1} \cdot g\_{b2} \cdot g\_{c3}} \in H^3(G,\mathbb{R}/\mathbb{Z}),
> $$
>  is in a cup product form of $g\_a, g\_b, g\_c$.
> 
> 
> 


Here $\cdot$ is a usual product of multiplication. And $g\_{a1} \cdot g\_{b2} \cdot g\_{c3} \in \{0,1\}= \mathbb{Z}\_2$. 


We wish to find that $\omega\_3^G(r(j\_a),r(j\_b),r(j\_c))$ becomes a 3-coboundary in $H^3(J,\mathbb{R}/\mathbb{Z})$ for the cohomology group of $J$, but $\omega\_3^G(g\_a,g\_b,g\_c)$ originally was not a 3-coboundary but was a 3-cocycle for the cohomology group of $G$. We hope to explicitly write 
$$
\omega\_3^G(g\_a,g\_b,g\_c)=\omega\_3^G(r(j\_a),r(j\_b),r(j\_c))=
\frac{\beta\_2^J(j\_b,\; j\_c)\beta\_2^J(j\_a,\; j\_b j\_c)}{\beta\_2^J(j\_a j\_b,\; j\_c) \beta\_2^J(j\_a,\; j\_b)}. 
$$
Here $\beta\_2^J(j\_a,\; j\_b)$ is a 2-cochain for $j\_a, j\_b, j\_c \in J$, and that 
$g\_a=r(j\_a)$, $g\_b=r(j\_b)$, $g\_c=r(j\_c) \in G$. 



> 
> **Question:** For example, can we find explicitly what is the minimal $J$, so we can trivialize a cup-product 3-cocycle $(-1)^{g\_{a1} \cdot g\_{b2} \cdot g\_{c3}}$ of $G=\mathbb{Z}\_2 \oplus \mathbb{Z}\_2 \oplus \mathbb{Z}\_2$ in a larger group $J$? Here $G$ is order 8 group. Perhaps $J$ is a group of order 16 at least?
> 
> 
> 



Trivialize a cup-product 3-cocycle of $G$ in a larger group $J$
The group $J$ has to be of order divisible by 16, due to the fact that if the order of the group $J$ is $8n$ where $n$ is an odd number, then the inflation of the cocycle will not be trivial (for example due to the fact that the quotient map $r:J\to G$ will split).


There is no one minimal $J$. The reason for this is the following: You can consider your cocycle in the following way: on $\mathbb{Z}\_2$ take the nontrivial element $c$ in $H^1(\mathbb{Z}\_2,\mathbb{Z}\_2)$, lift to to $G$ three times (each time use a different copy of $\mathbb{Z}\_2$), and denote the results by $c\_1, c\_2$ and $c\_3$. Take the cup product $c\_1c\_2c\_3\in H^3(G,\mathbb{Z}\_2)$ and apply the natural embedding of $\mathbb{Z}\_2$ in $\mathbb{R}/\mathbb{Z}$. In this way you get your cocycle $\omega$. Now, if you consider the cocycle $c\_1c\_2\in H^2(\mathbb{Z}\_2\times \mathbb{Z}\_2,\mathbb{Z}\_2)$ This corresponds to an extension of the form $$1\to \mathbb{Z}\_2\to D\_8\to \mathbb{Z}\_2\times\mathbb{Z}\_2\to 1,$$ and the inflation of $c\_1c\_2$ to $D\_8$ is then trivial. For that reason, the map $D\_8\times \mathbb{Z}\_2\to G$ trivializes the cocycle. Of course that you could have chosen $\mathbb{Z}\_2\times D\_8$, with the cocycle $c\_2c\_3$ instead. This also gives us a concrete way to construct $\beta$: After following the trivialization to $c\_1c\_2$ arising from the above extension and by writing elements in $D\_8$ as triples $(x,y,z)$ with $x,y,z\in\mathbb{Z}\_2$ (so that the multiplication is given by $$(x\_1,y\_1,z\_1)(x\_2,y\_2,z\_2) = (x\_1+x\_2+y\_1z\_2,y\_1+y\_2,z\_1+z\_2),$$
we end up with the formula $$\beta(((x\_1,y\_1,z\_1),w\_1)),((x\_2,y\_2,z\_2),w\_2)) = (-1)^{x\_1w\_2}$$ 


