Let $G$ be a countably infinite group.
Is it true that $H^2(G, \mathbb{C}^G)=0$ ?
Here, $\mathbb{C}^G$ denotes all functions from $G$ to $\mathbb{C}$ and is treated as a $G$-module, i.e. $(gf)(g'):=f(g^{-1}g')$ for all $g, g'\in G$ and $f\in \mathbb{C}^G$.
The second cohomology group is defined as the quotient of the set of all 2-cocycles by the subset of all 2-coboundaries.
More precisely, I am asking the following.
If $c: G\times G\to \mathbb{C}^G$ is any map satisfying $c(g_1, g_2)+c(g_1g_2, g_3)=g_1c(g_2, g_3)+c(g_1, g_2g_3)$ for all $g_i\in G$, can be write $c$ as $c(g_1, g_2)=b(g_1)-b(g_1g_2)+g_1b(g_2)$ for some function $b: G\to \mathbb{C}^G$ and all $g\in G$?