Imagine we have a group $G$ acting properly and freely (as a group action $\Phi: G \times M \rightarrow M$) on a manifold $M$, then $M/G$ is a manifold and there is a smooth submersion $\pi: M \rightarrow M/G.$
I now want to show that we can actually identify $\operatorname{ker}(D\pi(p)) = T_p(\Phi(G,p)) $ for every $p$ in $M$.
Could anybody show me how to do this?
I assume $G$ is a Lie group and it's a smooth action, otherwise the question doesn't make sense.
Pick a curve $\gamma: [0,1] \to G$, with $\gamma(0) = e$. Then we get a new curve $\gamma_p: [0,1] \to M$, given by $\gamma_p(t) = \gamma(t) \cdot p$. This obviously lies in the orbit of $p$ under $G$, and hence dies in the quotient. So $\gamma_p'(0) \in \text{ker}(D\pi(p))$. So $T_p(\Phi(G,p)) \subset \text{ker}(D\pi(p))$. By dimensional considerations the two subspaces are equal.