On 2-cocycles defined on direct product Edited after answer: Let $G$ be a group acts trivially on an abelian group $A$. Let
$\varepsilon $ be a normalized 2-cocycle in $ Z^{2}(G,A)$. Assume
that $G=G_{1} \times G_{2}$ be the direct product of $G_{1}$ and
$G_{2}$. 


*

*Is it true that $\varepsilon $ can be expressed as
$$\varepsilon((h,g),(h',g'))=\varepsilon_{1}(h,h')\times
\varepsilon_{2}(g,g')\times \alpha(g,h')$$ such that $\varepsilon_{1}=res_{G_{1}\times
G_{1}}(\varepsilon) \in Z^{2}(G_{1},A)$, $\varepsilon_{2}=res_{G_{2}\times
G_{2}}(\varepsilon) \in Z^{2}(G_{2},A)$ and $\alpha\in
Hom(G_{2}\times G_{1},A)$.

*If the answer is No, what is the possible expressions of $\varepsilon$ depending on $\varepsilon_{1}$ and $\varepsilon_{2}$. 


The reader can see (Tahara, Proposition 1)
Any help would be appreciated so much. Thank you all.
 A: This is not even true up to coboundaries (that is, it's not even true that a $2$-cocycle is cohomologous to a cocycle that can be split up like this). Take, for example, $G_1 = G_2 = A = C_p$ to be the cyclic group of order $p$. There is a nontrivial central extension of $C_p \times C_p$ by $C_p$ is given by the Heisenberg group of $3 \times 3$ matrices
$$H(\mathbb{F}_p) = \left\lbrace \left[ \begin{array}{ccc} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array} \right] : a, b, c \in \mathbb{F}_p \right\rbrace$$
over $\mathbb{F}_p$, and the restriction of this extension to either of the $C_p$ factors is trivial. 
More generally, by the cohomological Kunneth theorem we have
$$H^2(G_1 \times G_2, F) \cong \bigoplus_{i+j=2} H^i(G_1, F) \otimes H^j(G_2, F)$$
over a field $F$ (e.g. a finite field $\mathbb{F}_p$); this sum has two terms $H^0(G_1, F) \otimes H^2(G_2, F) \cong H^2(G_2, F)$ and $H^2(G_1, F) \otimes H^0(G_1, F) \cong H^2(G_1, F)$ corresponding to the $H^2$ of the factors, but it also has an additional term $H^1(G_1, F) \otimes H^1(G_2, F)$ coming from the external cup product which does not, and in the above example that's where new nontrivial central extensions come from, which don't arise from a central extension of the factors. 
A: I seriously doubt it. A group extension corresponding to $\varepsilon_1\times \varepsilon_2$ would be of the form $E_1\times E_2$, where $E_i$ is the group extension corresponding to $\varepsilon_i$, if I'm not mistaken.
Now, let $(\mathbb{Z}/2\mathbb{Z})^2$ act on $\mathbb{Z}/2\mathbb{Z}$ trivially. The quaternion group $Q_8$ is a central extension of $(\mathbb{Z}/2\mathbb{Z})^2$ by $\mathbb{Z}/2\mathbb{Z}$, which is not isomorphic to a direct product (easy exercise, since a non trivial factor should have order $2$ or $4$). 
So, the answer is NO.
A: Let $P(G_{1},G_{2},A)$ denotes the abelian group of pairings $G_{1}
\times G_{2}\rightarrow A$ (maps that are homomorphisms in both
variables). Then, by Theorem 2.3 (See
Kar),
we have  $$H^2(G_{1} \times G_{2}, A) \cong H^2(G_{1}, A) \times
H^2(G_{2}, A) \times P(G_{1},G_{2},A)$$  Hence,
$$\varepsilon((h,g),(h',g'))=\varepsilon_{1}(h,h')\times
\varepsilon_{2}(g,g')\times
\alpha_{\varepsilon}(h,g')$$ such that $\alpha_{\varepsilon}$ is
given by
$$\alpha_{\varepsilon}(h,g')=\varepsilon(h,g')\varepsilon(g',h)^{-1} $$
