$G$ cocycle split and trivialized to a coboundary in $J$, given a group homomorphism $J \overset{r}{\rightarrow} G$ Consider a generic nontrivial 3-cocycle $\omega_3^G(g_1,g_2,g_3) \in H^3(G,U(1))$ 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.
We like to ask how can we trivialize (or split) the 3-cocycle $\omega_3(g_1,g_2,g_3)$ 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
$$\omega_3^J(j_1,j_2,j_3)=\omega_3^G(r(j_1),r(j_2),r(j_3))=\omega_3^G(g_1,g_2,g_3) \text{ is trivial in  }  H^3(J,U(1)).$$
Namely $\omega_3^G(r(j_1),r(j_2),r(j_3))$ becomes a 3-coboundary in $H^3(J,U(1))$ for the cohomology group of $J$, but $\omega_3^G(g_1,g_2,g_3)$ originally wass not a 3-coboundary but was a 3-cocycle for the cohomology group of $G$. Namely, we can explicitly write 
$$
\omega_3^G(g_1,g_2,g_3)=\omega_3^G(r(j_1),r(j_2),r(j_3))=
\frac{\beta_2^J(j_2,\; j_3)\beta_2^J(j_1,\; j_2\cdot j_3)}{\beta_2^J(j_1 \cdot  j_2,\; j_3) \beta_2^J(j_1,\; j_2)}. 
$$
Here $\beta_2^J(j_1,\; j_2)$ is a 2-cochain for $j_1, j_2, j_3 \in J$, and that 
$g_1=r(j_1)$, $g_2=r(j_2)$, $g_3=r(j_3) \in G$. 

question 1: For example, can we try explicitly that $G=\mathbb{Z}/2
\mathbb{Z}=\mathbb{Z}_2$ the cyclic group, and $J=Q_8$ the quaternion
  group. There are $\mathbb{Z}_2 \times \mathbb{Z}_2$ types of group
  morphism $r(j)=g$ of $J \overset{r}{\rightarrow} G$ as $Q_8 \overset{r}{\rightarrow} \mathbb{Z}_2$. If we know the 
  explicit form of 3-cocycle of $G$, $\omega_3^{G}(g_1,g_2,g_3)    = 
\exp \left( \frac{2 \pi i}{2^{2}} \;
g_1(g_2 +g_3 -[(g_2 +g_3)\mod 2]) \right)$, how can we determine the analytic explicit form of 2-cochain $\beta_2^J(j_1,\; j_2)$ for the quaternion $J=Q_8$ such that it can trivialize the 3-cocycle of $G$ to 3-coboundary of $J$? 
The key question for this post is asking the demonstration of this specific example: What is the explicit form of $\beta_2^{Q_8}(j_1,\; j_2)$?

-

question 2: Comments about the general procedure to trivialize
  (split) 3-cocycle of $G$ to 3-coboundary of $J$, given that $J
 \overset{r}{\rightarrow} G$. One can test that NOT all the $J$ with group homomorphism $J  \overset{r}{\rightarrow} G$ can trivialize $G$-cocycle. What conditions are necessary to find such $J$? 

Any comments and partial answers are more than welcome, please! 
 A: A small remark first: I believe that the denominator in the exponent should be 2, and not $2^2$. 
In any case, write $Q_8=\langle x,y|x^2=y^2, xyx^{-1}=y^{-1},y^4=1\rangle$ so that each element in the group we can write uniquely as $y^ax^b$ with $a\in\{0,1,2,3\}$ and $b\in\{0,1\}$. 
By manipulating a bit the Lyndon Hochschild Serre spectral sequence, one gets the function $$\beta(y^ax^b,y^cx^d) = i^{bc}.$$ This function satisfies the desired property. 
Let me explain how to receive this $\beta$: Let us write $Q_8=G$, the normal subgroup as $N$ and the quotient group as $Q$. By choosing a set-theoretic lifting $s:Q\to G$, we can write each element of $G$ uniquely as a product $ns(q)$ for some $n\in N$ and $q\in Q$. Let us identify in this way $G$ with $N\times Q$ as a set. The key point here is that in the Lyndon Hochschild Serre spectral sequence you have a homomorphism $d_2:H^1(Q,H^1(N,U(1)))\to H^3(Q,U(1))$. Every element in the image of this $d_2$ will be in the kernel of the inflation map. In our case the group $H^1(Q,H^1(N,U(1)))$ is cyclic of order 2. It is generated by the map $f$ which sends the generator of $Q$ to the generator of the cyclic group of order 4, $H^1(N,U(1))$ (which sends a generator of $N$ to $i$). 
The idea now is that you try to cook up a two cocycle out of this element in the following way: $$\beta((n_1,q_1),(n_2,q_2)) = f(q_1^{-1})(n_2).$$
However, this $\beta$ might not be a two cocycle. The obstruction for this is of course $\partial \beta$. By a direct calculation, this is a three cocycle inflated from the quotient group $Q$, and this is exactly $d_2(f)$. 
