Let $M$ be a manifold, $\Phi_t, t\in \mathbb R$ a one parameter group of diffeomorphisms and $X$ a vector field on $M$ definied by $$X_z:=\frac{d}{dt}_{|t=0} \Phi_t(z).$$ Show that $\Phi_t$ is the flow of $X$.
own Ideas:
Okay, since $\Phi_t$ is a one parameter group of diffeomorphisms we have
(1): $\Phi_a(\Phi_b(z))=\Phi_{a+b}(z)$ for $a,b \in \mathbb R$
(2): $\Phi_0(z)=$id$(z)$
We have to show that $$\frac{d}{dt}\Phi_t(z)=X_{\Phi_t(z)}.$$
So since I can not insert $\Phi_t(z)$ into $X$ I think we have to look at $X_{\Phi_c(z)}$ with $c\in \mathbb R$. Then we get: $$X_{\Phi_c(z)}=\frac{d}{dt}_{|t=0} \Phi_t(\Phi_c(z)) \overset{(1)}{=}\frac{d}{dt}_{|t=0} \Phi_{t+c}(z)$$
I do not know how to continue nor if this is the right way. Pls help.
Write instead $$X_{\Phi_t(z)}=\frac{d}{ds}_{|s=0} \Phi_s(\Phi_t(z))=\frac{d}{ds}_{|s=0} \Phi_{s+t}(z)=\frac{d}{dt} \Phi_t(z).$$