Action of a Lie group on a smooth manifold

Suppose a Lie group acts smoothly (but not necessarily properly or freely) on a smooth manifold $$M$$. Show that each orbit is an immersed submanifold of $$M$$; which is embedded if the action is proper.

This is problem 15 of chapter 21 of Lee's smooth manifolds book I was searching to solve the above problem and I'm now a little confused writing an easy argument for this. I found the following argument but this is not easy enough for me to understand:

1. For each $$p\in M$$, the isotropy group $$G_p = \{g\in G: g\centerdot p=p\}$$ is a closed subgroup of $$G$$.
2. The quotient space $$G/G_p$$ has a unique smooth manifold structure, and $$G$$ acts transitively on this smooth manifold by left multiplication.
3. The orbit map $$\mathscr O_p\colon G\to M$$ given by $$\mathscr O_p(g) = g\centerdot p$$ descends smoothly to the quotient $$G/G_p$$ to give a smooth map $$F_p\colon G/G_p\to M$$ that is a bijection from $$G/G_p$$ onto the orbit $$G\centerdot p$$.
4. The map $$F_p$$ is equivariant with respect to the left actions of $$G$$ on $$G/G_p$$ and $$M$$, and therefore it has constant rank.
5. Because it is a constant-rank injection, $$F_p$$ is a smooth immersion, and therefore its image $$G\centerdot p$$ is a smoothly immersed submanifold of $$M$$.

Could anyone help me prove this?

So I am going to do this step by step. So I belive it's clear that $$G_p$$ is a closed subgroup of $$G$$, and we can consider the proper free action $$G\times G_p\rightarrow G_p$$ by left translation , and since $$G_p$$ is a closed subgroup it also has a manifold structure and we can consider the space $$G/G_p$$. Now we get a bijection from $$G/G_p$$ to the orbit $$g.p$$ of $$p$$ since we are doing the quotient by the points that would fix the orbit. Then you have that $$F_p$$ is an equivariant map because $$g.F_p(hG_p)=g(hp)=(gh)p=(gh)G_p p=F_p(ghG_p)$$. Using the fact that the action of $$G$$ in $$G_p$$ is transitive and the equivariant rank theorem you have that $$F_p$$ must have constant rank and then using the fact that locally for suitable coordinates $$F_p$$ looks like $$(x_1,...,x_r)\rightarrow (x_1,...,x_k,0,...,0)$$ where $$k$$ is the rank of $$F_p$$, but since $$F_p$$ is injective this forces $$k=r$$ and so you get that it's an immersion, and so since the image of $$F_p$$ is the orbit of $$p$$ and that $$F$$ is an injective immersion you get the desired result.
• If the action is proper will you have that it's an embedded submanifold , since we can consider again the injective immersion from $G/G_p\rightarrow M$, and if we prove that this is a proper map we have that the orbit is actually embedded. But I belive this just comes from using the commutative diagram and the fact that $\phi$ is proper we have that $\hat \phi^{-1}_p(K)=\pi\circ (\phi^{-1}_p)(K)$ @Reza. Jan 18 '21 at 8:27