# Smooth action of a lie group on a manifold?

Let $G$ be a Lie group and $M$ be a $C^\infty$ manifold. My textbook defines a differentiable action of $G$ on $M$ as a map $$G\times M\longrightarrow M, (g, p)\longmapsto g\cdot p,$$ such that:

(i) For all $g\in G$ the map $$L_g:M\longrightarrow M, p\longmapsto g\cdot p,$$ is a diffeomorphism.

(ii) $g\cdot (h\cdot p)=(gh)\cdot p$ for all $g, h\in G$ and $p\in M$.

I have a few questions as to this definition. I was expecting the condition $$e\cdot p=p,\ \forall p\in M,$$ where $e$ is the identity of $G$.

Indeed, if $L_g$ is only $C^\infty$ in $(i)$ and the above condition holds then $L_g$ ia a diffeomorphism.

However, if $L_g$ is a diffeomorphism as $(i)$ then I can't conclude $e$ acts trivially on $M$, can I?

Furtheremore, the smoothness of $(g, p)\longmapsto g\cdot p$ isn't enough to ensure the smoothness of $L_g$?

The condition $ep = p$ really should be there, but it is implied by the given conditions. The reason is that $e^2 = e$, so (ii) implies that $L_e^2 = L_e$, but the only idempotent invertible map is the identity. This also implies that $L_{g^{-1}} = L_g^{-1}$, which also really should be there. (If you had this and $ep = p$ you would already know that each $L_g$ is a diffeomorphism.)
If the multiplication map $G \times M \to M$ is smooth, then its evaluation $L_g : M \to M$ at a fixed $g \in G$ is smooth. There's no difficulty here.