On 2-cocycles defined on direct product

Question

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 $G_{2}$.

1. 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 $\alpha\in Hom(G_{2}\times G_{1},A)$.
2. 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.

Answer 1

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 $$\alpha_{\varepsilon}(h,g')=\varepsilon(h,g')\varepsilon(g',h)^{-1} $$

Answer 2

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.

Answer 3

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.