Free action of Lie group on a manifold

Question

Let $G$ be a compact Lie group acting on a manifold $M$. So for each vector $X \in TeG$ we have $X^{\#}$ the vector field on $M$ defined at each point $p \in M$ by the curve $exp(tX) \cdot p$. If the action is free we have that $X^{\#}_p=0 \iff X=0$, I get the instinctive idea but I can not proof it rigorously.

Answer 1

Define $\phi_t(x)=exp(tx).x$, $\phi_t$ is the flow of $X^{\#}$, we deduce that $X^{\#}_P=0$ implies that $\phi_t(x)=exp(tX).x=x$ for every $t$. This contradicts the fact that the action is free.

N.B. The flow $\psi_t$ of $Y$ verifies the differential equation

(1) ${d\over {dt}}\psi_t=X(\psi_t)$,

so if $Y(x)=0$ the constant solution $\psi_t(x)$ satisfies $(1)$, so $\psi_t(x)=x$.