Is a Lie group invarint set differentiable? I'm a physics student and my question may be silly in a math society. I apologize for that.
I want to find the most general set, say $X_G$, that is invariant under action of a Lie group $G$ (of dimension $N$). I choose a base point, say $x_0$, and write any other element of $X_G$ as 
$X_G \ni x(\theta)=\exp[\theta^i Y_i] \circ x_0 \quad ,   \qquad (1)$
where $\theta^i $s are (real) continuous group parameters, $Y_i \in g$, the Lie algebra of $G$, and $\circ$ is group operation on the set. The group action has two properties:
$\exists e\in G ,\quad e\circ x = x  \quad,  \quad \forall x\in X_G \quad ,   \qquad (2) \\
(G_1 * G_2 )\circ x = G_1 \circ (G_2 \circ x)   \quad , \forall G_1, G_2 \in G \quad \forall x \in X_G \quad .   \qquad (3)$ 
To be practical, I want firstly to find a set $X_g$ that is invariant under the action of Lie algebra $g$, and secondly construct an element of $X_G$ by the formula (1). 
The compatibility conditions for relations (2) and (3) at the identity element of group imply that
$e \circ x_0 = x_0  \quad ,  \qquad (4)   $
$[ \frac{\partial G_1}{\partial \theta^i} * G_2 +  G_1 * \frac{\partial G_2}{\partial \theta^i} ] \circ x_0 = \frac{\partial G_1}{\partial \theta^i} \circ (G_2 \circ x_0) + G_1 \circ ( \frac{\partial G_2}{\partial \theta^i} \circ x_0) \qquad (5.a)\\
\to (Y_1 + Y_2)\circ x_0 = Y_1 \circ x_0 + Y_2 \circ x_0 \quad . \qquad (5.b)$
My problem is that I didn't prove the following steps.
1) The elements of $X_G$ can be differentiated (Eq. (5.a)).
2) If so, is the differentiation above correct.
3)  The elements of $X_g$  can be added to each other (right hand-side of Eq. (5.b)).
I think that the three statements above need the set $X_G$ to be partially ordered and possesses other properties that I'm not aware of. 
Please tell me what I should prove for the statements above to be valid. 
Any comment or suggesting reference is appreciated. 
 A: Here is a sketch, I do not have much time for the details:
Consider a smooth Lie group action $G\times M\to M$ of a Lie group $G$ on a smooth manifold $M$. Pick a point $p\in M$ and consider its $G$-orbit $S=Gp$ (which you denote $X_G$), i.e. the set of images of $p$ under the action of elements  of $G$. Let $H< G$ denote the stabilizer of $p$ in $G$, i.e. the subgroup consisting of $h\in G$ such that $h(p)=p$. We have a natural bijective continuous map $f: G/H\to S$, where $G/H$ is a topological space equipped with the quotient topology. Then: 
Theorem. 1. $H$ is a Lie subgroup of $G$. 2. The space $G/H$ has the natural structure of a smooth manifold such that the projection map $G\to G/H$ is smooth. 3. The map $f$ is an immersion. 
In other words, $S$ is an "injectively immersed manifold" in $M$. The key to the proof is the "Constant Rank Theorem" which implies that locally the map  $f$ has the form of a linear projection from $R^k$ to $R^n$ (where $k$ is the dimension of $G$ and $k-n$ is the dimension of $H$) followed by a linear embedding $R^n\to R^m$, where $m=dim(M)$. The fact that $df$ has constant rank (independent of a point where we compute the derivative) is quite easy to check, it follows from the Chain Rule.      
