# Decomposition of group-valued “skew-symmetric” map.

Let $X$ be a set, and $G$ a group.

Consider a map $g:X\times X \to G$ such that for any $x,y\in X$: $$g(x,y) = g(y,x)^{-1}.$$

Is it always true that there is a map $h:X\to G$ such that for any $x,y\in X$, $$g(x,y) = h(x)h(y)^{-1}\quad ?$$

If yes/no, how do you prove it? There is probably something trivial that I'm missing anyway... Thanks in advance.

The answer is no. Let $X=\{1,2,3\}$, $G=(\mathbb{Z},+)$ and let $g(x,y)=\begin{cases} 0, \textrm{ if } x=y \\ 1 , \textrm{ if } x<y \\ -1, \textrm{ if } x>y \end{cases}$. Clearly, $g$ satisfies your assumptions. If there was such an $h$, we would have:
$h(1)-h(2)=1$,
$h(1)-h(3)=1$,
$h(2)-h(3)=1$,