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:

*

*For each $p\in M$, the isotropy group $G_p = \{g\in G: g\centerdot p=p\}$ is a closed subgroup of $G$.

*The quotient space $G/G_p$ has a unique smooth manifold structure, and $G$ acts transitively on this smooth manifold by left multiplication.

*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 $G\centerdot p$.

*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.

*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?
 A: 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.
I don't know what much could be added , is there anything that needs clarification ?
