I would like to know the explicit expression of 2-cocycle from a 2nd cohomology group $H^2[Q_8 \times \mathbb{Z}/2\mathbb{Z}]$, or namely $H^2[Q_8 \times \mathbb{Z}_2]$, here we denote the cyclice group $\mathbb{Z}/2\mathbb{Z}=\mathbb{Z}_2$, and the $Q_8$ is the order 8 quaternion group.


What I had computed (from Kunneth formula, Universal Coefficient Theorem) and what I had known is that:
$$H^2[Q_8 \times \mathbb{Z}_2]=\mathbb{Z}_2 \times \mathbb{Z}_2.$$
So **I would like to know the two generators of $H^2[Q_8 \times \mathbb{Z}_2]=\mathbb{Z}_2 \times \mathbb{Z}_2$ in terms of 2-cocycles.** 
Any Reference and any partial answer is welcome.

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    Additional remarks and guides:

What I also computed and knew are:$H^1[Q_8,U(1)]=\mathbb{Z}_2 \times \mathbb{Z}_2$, $H^2[Q_8,U(1)]=0$, $H^3[Q_8,U(1)]=\mathbb{Z}_8$. I also know that the explicit 3-cocycle $\omega_3((A,a),(B,b),(C,c)) \in H^3[Q_8,U(1)]=\mathbb{Z}_8$ as:
$$\omega_3((A,a),(B,b),(C,c))$$  
$$= \exp\left( \frac{2\pi i p}{8}
                 \{ (-)^{B+C}a \left( (-)^C b + c- [(-)^C b + c +2BC] \right) 
                 - 2 ABC \} \right)$$ 
where $p \in \mathbb{Z}_8$.

Here we denote the elements of $Q_8$ by the 2-tuples
$$
(A,a)  :=  X^A R^a  \qquad \qquad \mbox{with $A \in 0,1$ and               
$a \in -1, 0, 1, 2$.   } \qquad
$$
with $R^{2N}  =  e, \qquad X^2 \; = \; R^N, \qquad  XR \; = \; R^{-1} X.$

So for instance the identity is $e=(0,0)$. 
The multiplication law of $Q_8$ then reads
$$
(A,a) \cdot (B,b) = ([A+B], [(-)^B a+b+NAB]),
$$