Group homomorphism, the uniqueness of k for g' = gk Group homomorphism is  $ \Phi: G \rightarrow H $
Show, that for all $ h \in H $ and all $ g,g' \in \Phi^{-1}(\{h\}) $ there exists a unique $k \in \ker(\Phi) $, so that $g'=gk$.
$$
\forall h \in H, \forall g,g' \in \Phi^{-1}(\{h\})\exists! k \in \ker(\Phi): g'=gk
$$
I do not want a solution for this problem, just maybe a hint or a first step into the right direction, because I really don't know where to start.

Notice that $k=g^{-1}g'$ is in $\ker \Phi$ and $g'=gk$ and by the unicity of its form $k$ is unique
