I like to ask a simple question: **how to trivialize a cup-product 2-cocycle of $G$ into a 2-coboundary of $J$ in a larger group $J$.**


Let us take a nontrivial 2-cocycle $\omega\_3^G(g\_a, g\_b) \in H^2(G,\mathbb{R}/\mathbb{Z})$ in the cohomology group of $G$ with $U(1)=\mathbb{R}/\mathbb{Z}$ coefficient. In otherwords, here the 2-cocycle $\omega\_2^G$ is a complex $U(1)=\mathbb{R}/\mathbb{Z}$ function with the norm $|\omega\_2^G|=1$ but with a $U(1)$ complex phase satisfying the cocycle condition. Here $g\_a, g\_b \in G$.


How can we trivialize the 2-cocycle $\omega\_2(g\_a, g\_b)$ of $G$ into 2-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  

$$\omega\_2^J(j\_a, j\_b)=\omega\_2^G(r(j\_a),r(j\_b))=\omega\_2^G(g\_a, g\_b)=\delta v\_1(j\_a, j\_b)=\frac{v\_1(j\_a) v\_1(j\_b)}{v\_1(j\_a j\_b)} \text{ is trivial in } H^3(J,\mathbb{R}/\mathbb{Z}).$$


Namely, $\omega\_2^J(j\_a, j\_b) =\delta v\_1(j\_a, j\_b)$ is a 2-coboundary in $J$.
I like to take $G=\mathbb{Z}\_2 \oplus \mathbb{Z}\_2$, and consider the group element $g\_a=(g\_{a1}, g\_{a2}) \in (\mathbb{Z}\_2, \mathbb{Z}\_2)=\mathbb{Z}\_2 \oplus \mathbb{Z}\_2 =G$, similarly, $g\_b=(g\_{b1},g\_{b2}) \in G$.


Let me focus on the 2-cocycle



> 
> $$
> \omega\_2^G(g\_a, g\_b)=\exp[i \pi \cdot g\_{a1} \cdot g\_{b2} ]=(-1)^{g\_{a1} \cdot g\_{b2} } \in H^2(G,\mathbb{R}/\mathbb{Z}),
> $$
>  is in a cup product form of $g\_a, g\_b$.
> 
> 
> 


**Question: What are examples and restrictions of $J$? Whether dihedral group $D\_8$ and quaternion $Q\_8$ work for $J$? What is the explicit form of 1-cochain $v\_1(j)$?**



Trivialize a cup-product 2-cocycle of $G$ in a larger group $J$
The given particular 2-cocycle $\omega\_2 \in H^2(G,\mathbb{R}/\mathbb{Z})$ of $G=(\mathbb{Z}\_2)^2$ is
$$
\omega\_2(g\_a , g\_b)=(-1)^{[g\_{a\_1}]\_2 [g\_{b\_2}]\_2 }, \;\;\;\;
$$
where $g\_a=(g\_{a\_1},g\_{a\_2}) \in G=(\mathbb{Z}\_2)^2$, and similarly for $g\_b$.
Below I aim to show that both $Q\_8 \to (\mathbb{Z}\_2)^2$ and $D\_8 \to (\mathbb{Z}\_2)^2$ provide a 1-cochain trivilize the $\omega\_2$ into 2-coboundary in $J$:



> 
> Ans 1: True for $D\_8 \to (\mathbb{Z}\_2)^2$
> 
> 
> 


Write the dihedral group
$$D\_8=\langle x, R |x^2=R^4=1, x Rx=R^{-1}\rangle$$  

We can choose that the $\{1,x,R,xR\}$ in $D\_8$ maps to the $(\mathbb{Z}\_2)^2$. The $\{R^2 ,xR^2, R^3,x R^3\}$ in $Q\_8$ also maps to the $(\mathbb{Z}\_2)^2$. This gives us the following group homomorphism:
$$D\_8 \to (\mathbb{Z}\_2)^2.$$


The 1-cochain for $j \in J= D\_8$ 
$$
\beta\_1(j) =(-1)^{f(j)}
$$
satisfies the desired 2-cocycle splitting property. Here
we define the function $f$:
$$
f(1)=f(x)=f(R) =f(xR)=0,
$$
$$ 
f(R^2)=f(x \cdot R^2)=f(R \cdot R^2) =f(xR \cdot R^2)=1. 
$$
I find that $\omega\_2= \delta(\beta\_1)$.



> 
> Ans 2: True for $Q\_8 \to (\mathbb{Z}\_2)^2$
> 
> 
> 


Write the quaternion 
$$Q\_8=\langle x,y|x^2=y^2, xyx^{-1}=y^{-1},x^4=y^4=1\rangle$$ 
so that each element in the group we can write uniquely as $x^p y^{q}$ with $p\in\{0,1\}$ and $q \in\{0,1,2,3\}$.


We can choose that the $\{1,x,y,xy\}$ in $Q\_8$ maps to the $(\mathbb{Z}\_2)^2$. The $\{y^2 ,x\cdot y^2,y\cdot y^2,xy \cdot y^2\}$ in $Q\_8$ also maps to the $(\mathbb{Z}\_2)^2$. This gives us the following group homomorphism:
$$
Q\_8 \to (\mathbb{Z}\_2)^2.
$$
The 1-cochain for $j \in J= Q\_8$  

$$
\beta\_1(j)=\beta\_1(x^{p} y^{q}) = i^{(p+q)}.
$$
satisfies the desired 2-cocycle splitting property. I find that $\omega\_2= \delta(\beta\_1)$.


So to summarize, I give an answer through examples of $J$. I show that dihedral group and quaternion trivialize $G$, with the explicit form of 1-cochain given.


