Consider $\mu:M \times G \to M$ a smooth right action of a Lie group $G$ on a manifold $M$. For each $\alpha\in\mathfrak{g}$ (the Lie algebra of $G$) we can define a vector field on $M$ :
$\alpha_M(x)=\frac{d}{dt}|_{t=0}x\cdot \exp(t\alpha)$.
I'm trying to show that this defines a Lie algebra homomorphism : $\alpha\in \mathfrak{g}\to \alpha_M\in\mathfrak{X}(M)$.
I think that, with the chain rule, I can write: $\frac{d}{dt}|_{t=0}x\cdot \exp(t\alpha) = \frac{d}{dt}|_{t=0}\mu (x , \exp(t\alpha))=d_{(x,e)}\mu(0,\alpha)$, where $e$ is the identity group, but I'm not sure on how to continue.
Many thanks