Let $f: G \to G'$ be a homomorphism of group schemes. Then we get a natural action of $G$ on $G'$, given on points by $(g,g') \mapsto f(g) \cdot g'$. Then it is claimed that this action is free if and only if $Ker(f)$ is trivial. I have problem seeing why this is true.
Recall that given an action $\rho: G \times G' \to G'$, we define the graph morphism $$\psi = (\rho, pr_2): G\times G' \to G' \times G'$$ and $\rho$ is said to be free if $\psi$ is a monomorphism of schemes.
It's clear that on points, the map $(g, g') \to (f(g)g', g')$ is injective if and only if $ker(f)$ is trivial. I don't know what to do from here though.