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! 



$G$ cocycle split and trivialized to a coboundary in $J$, given a group homomorphism $J \overset{r}{\rightarrow} G$
Here is an answer for question 2 on which homomorphisms $J\xrightarrow{r} \mathbb{Z}\_2$ will trivialize $\omega\_3^G$. 




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Your cocycle takes values in the two-element subgroup $\{1,-1\}\subset U(1)$ (aka $\{0,\frac{1}{2}\}\subset \mathbb{R}/\mathbb{Z}$). Since all two-element groups are isomorphic, you could of course identify this subgroup with $\mathbb{F}\_2=\{0,1\}$ under addition. But why would you help? Well, $\mathbb{F}\_2=\{0,1\}$ also has a *multiplication* (that's why I'm using a different letter for it), and so you can ask whether your cocycle factors as a **cup product** of two smaller cocycles. In this case, we'll see that it does! Let's see what happens in this case (this will also illustrate what a cup product is, if you're not familiar with this).


To start, we need to rewrite your formula
$$\omega\_3^{G}(g\_1,g\_2,g\_3) = 
\exp \left( \frac{2 \pi i}{2^{2}} \;
g\_1(g\_2 +g\_3 -[(g\_2 +g\_3)\mod 2]) \right)\in U(1)$$
to take values in $\mathbb{F}\_2=\{0,1\}$. Note that the inputs $g\_i$ in your formula live in $\mathbb{Z}\_2=\{0,1\}$. What we get (for now) is
$$\omega\_3(g\_1,g\_2,g\_3)=g\_1\cdot \frac{g\_2 +g\_3 -[(g\_2 +g\_3)\mod 2]}{2}\in \{0,1\}=\mathbb{F}\_2.$$


The property we want to highlight is that this is a **product** of two simpler functions:


$$\omega\_3(g\_1,g\_2,g\_3)=\alpha(g\_1)\cdot \beta(g\_2,g\_3)$$ where $$\alpha(g\_1)=g\_1\quad\text{ and }\quad\beta(g\_2,g\_3)=\frac{g\_2 +g\_3 -[(g\_2 +g\_3)\mod 2]}{2}.$$




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What does this tell us? Well mathematically, we would write $r^\*(\omega\_3^G)\in H^2(J)$ for what you are calling $\omega\_3^J$. And if $w\_3\in H^3(\mathbb{Z}\_2;\mathbb{F}\_2)$ is the cohomology class represented by $\omega\_3^G$, then $r^\*(w\_3)\in H^3(J;\mathbb{F}\_2)$ is represented by $\omega\_3^J$: so your question about trivializing the cocycle is whether $r^\*(w\_3)\overset{?}{=}0\in H^3(J;\mathbb{F}\_2)$. 


Note that $\alpha$ and $\beta$ are both cocycles, so they represent cohomology classes $a\in H^1(\mathbb{Z}\_2;\mathbb{F}\_2)$ and $b\in H^2(\mathbb{Z}\_2;\mathbb{F}\_2)$. The factorization $$\omega\_3(g\_1,g\_2,g\_3)=\alpha(g\_1)\cdot \beta(g\_2,g\_3)$$ then tells us that $w\_3$ is the **cup product** of the cohomology classes $a$ and $b$: we have $w\_3=a\cdot b$ (see my [answer here](https://mathoverflow.net/a/688/250 "explict formula for cup product in group cohomology") for details).


Cup product is an operation on cohomology that takes $x\in H^i$ and $y\in H^j$ and combines them to give $x\cdot y\in H^{i+j}$. It satisfies $y\cdot x = \pm x\cdot y$ where $\pm=(-1)^{ij}$.




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The reason this is useful is that cup products are always preserved by $r^\*$: there is a universal formula
$$r^\*(x\cdot y)=r^\*(x)\cdot r^\*(y)\quad\text{in cohomology}.$$
This means that ONE way you could hope to show that $r^\*(w\_3)=0$ would be to show that either $r^\*(a)=0$ or $r^\*(b)=0$. Actually in 1st cohomology $r^\*(a)$ will never be 0 unless the map $J\xrightarrow{r}\mathbb{Z}\_2$ is trivial (an uninteresting case, as you discuss in comments), but you could hope that $r^\*(b)=0$. [Side note: in your particular example in Q1, I checked and $r^\*(b)\neq 0$.]




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But actually there is a much better way to look at this. It comes from the fact that you wrote your cocycle in a way that hides its symmetry. I claim that if you work out the values of your cocycle $$\omega\_3(g\_1,g\_2,g\_3)=g\_1\cdot \frac{g\_2 +g\_3 -[(g\_2 +g\_3)\mod 2]}{2}\in \{0,1\}=\mathbb{F}\_2$$ on the 8 different possible inputs, you'll see that it can instead be written as 
$$\omega\_3(g\_1,g\_2,g\_3)=\begin{cases}1&\text{ if $g\_1=1$, $g\_2=1$, and $g\_3=1$}\\0&\text{else}\end{cases}\in \{0,1\}=\mathbb{F}\_2.$$
In other words, $$\omega\_3(g\_1,g\_2,g\_3)=g\_1\cdot g\_2\cdot g\_3=\alpha(g\_1)\cdot \alpha(g\_2)\cdot \alpha(g\_3).$$ This means that $w\_3=a\cdot a \cdot a=a^3$. So not only can your cocycle be written in a simpler form, it actually is the cup product of a *single* class in 1st cohomology with itself three times.


Let us write $R\in H^1(J;\mathbb{F}\_2)$ for $R=r^\*(a)$. Then we have $r^\*(w\_3)=r^\*(a)^3=R^3$, so the question of whether $r^\*(w3)=0$ is simply the question of whether $R\in H^1(J;\mathbb{F}\_2)$ satisfies
$$R^3=0\in H^3(J;\mathbb{F}\_2).$$


Moreover, an element $R\in H^1(J;\mathbb{F}\_2)$ is *equivalent* to a homomorphism $J\xrightarrow{r}\mathbb{Z}\_2$ (that's why I used the same letter). So for a given $J$, your question is equivalent to finding all $R\in H^1(J;\mathbb{F}\_2)$ satisfying $R^3=0$. Now, that's not going to necessarily be easy! But it is more likely to be something you can look up somehow. (And in some cases, you may be able to show that $R^2=0$, which would be enough.)




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One last comment: you might be concerned that I was working with $\mathbb{F}\_2$ coefficients here, whereas you had general $U(1)$ coefficients. Here is an argument why this is not so bad. Say we're given some class $z\in H^1$. If we take $x=z$ and $y=z$ in the formula $y\cdot x = \pm x\cdot y$ from above, it tells us $z\cdot z = - z\cdot z$. In other words, we always have $z^2+z^2=0\in H^2$. This means that if you have coefficients in $\mathbb{Q}$ or $\mathbb{Z}\_3\subset U(1)$ or anywhere you can divide by 2, then $z^2$ would ALWAYS be 0 (and thus $z^3$ would be as well). This isn't quite a proof that you can always stick to $\mathbb{F}\_2$ coefficients (for example, you might need to worry about $\mathbb{Z}/4\mathbb{Z}\subset U(1)$ as well) but as a general rule it argues that you can probably ignore factors other than $2$. 




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EDIT: Here is a concrete example of how to apply this to $Q\_8$. According to (eq uation 2.4.2) of [this survey by Benson](http://library.msri.org/books/Book51/files/01benson.pdf), the cohomology ring of $Q\_8$ is
$$H^\*(Q\_8;\mathbb{F}\_2)\cong \mathbb{F}\_2[x,y,z]/(\,x^2+xy+y^2,\,x^2y+xy^2\,)$$
where $x\in H^1$, $y\in H^1$, and $z\in H^4$.


Therefore $H^3(Q\_8;\mathbb{F}\_2)$ is spanned by the four elements $x^3$, $x^2y$, $xy^2$, and $y^3$ subject to the three relations $x^3+x^2y+xy^2=0$, $x^2y+xy^2+y^3=0$, and $x^2y+xy^2=0$. (The first two are $x(x^2+xy+y^2)$ and $y(x^2+xy+y^2)$.) Subtracting the last relation from the first two gives a basis of relations as $x^3=0$, $y^3=0$, and $x^2y+xy^2=0$.


You can now read off easily exactly which $R\in H^1(Q\_8;\mathbb{F}\_2)$ have $R^3=0$. Namely, if $R=x$ we have $R^3 = x^3 = 0$; if $R=y$ we have $R^3 = y^3 = 0$; and if $R=x+y$ we have $R^3 = (x+y)^3 = x^3 +3x^2y+3xy^2+y^3 = x^3 + (x^2y+xy^2) + y^3 = 0$. Therefore we conclude that *every* homomorphism $Q\_8\to \mathbb{F}\_2$ splits your cocycle!




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You can do this for lots of other groups, if you can find where someone has already computed the $\mathbb{F}\_2$-cohomology ring. For example, (eq. 2.8.1) of that same survey says that the "semidihedral group $SD\_{2^n}$ of order $2^n$", whatever that is, has (for $n\geq 4$) the cohomology ring
$$H^\*(SD\_{2^n};\mathbb{F}\_2)\cong \mathbb{F}\_2[x,y,z,w]/(\,xy,\,y^3,\,yz,\,z^2+x^2w\,)$$
where $x\in H^1$, $y\in H^1$, $z\in H^3$, and $w\in H^4$. We find that $H^3(SD\_{2^n};\mathbb{F}\_2)$ is spanned by $x^3,x^2y,xy^2,y^3,z$ but with relations $x^2y=0$, $xy^2=0$, and $y^3=0$. Therefore we conclude that $R=x$ has $R^3=x^3\neq 0$ and $R=x+y$ has $R^3 = (x+y)^3 \equiv x^3 \neq 0$ but $R=y$ has $R^3=y^3=0$. So only one of the three nontrivial homomorphisms $SD\_{2^n}\xrightarrow{r}\mathbb{Z}\_2$ will split your cocycle.




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Finally, you **only have to look at groups** $J$ **whose order is a power of 2**. This follows from a theorem of Cartan-Eilenberg, although it's a bit tricky to explain how (see Theorem 4.1 of [these lectures by Adem](http://homepages.math.uic.edu/~bshipley/ConMcohomology1.pdf)). In any case, here is a precise statement you can use:


**Theorem**: Fix a homomorphism $J\xrightarrow{r}\mathbb{Z}\_2$. For any subgroup $H<J$ whose order is a power of 2, you can consider the restriction $H\xrightarrow{r|\_H}\mathbb{Z}\_2$. Then the homomorphism $J\xrightarrow{r}\mathbb{Z}\_2$ splits your cocycle $\iff$ for every subgroup $H<J$ whose order is a power of 2, the homomorphism $H\xrightarrow{r|\_H}\mathbb{Z}\_2$ splits your cocycle.


