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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.

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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:

$h(1)-h(2)=1$,

$h(1)-h(3)=1$,

$h(2)-h(3)=1$,

which can not hold true simultaneously.

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