When does an affine subset of an orbit have affine preimage under the orbit map? I have an algebraic group $G$ acting on an affine variety $X$, the orbit $O(m)$ of an element $m \in X$, and an affine curve $C$ contained in the Zariski closure $\overline{O(m)}$ of $O(m)$, such that $m \in C$. 
If we define $C^\prime = C \cap O(m) $ then it's not hard to see that $C^\prime$ is open in $C$. And therefore $C^\prime$ is again affine (follows for example from an exercise in Hartshorne). 
Let $p: G \to O(m)$ defined by $p(g) = g \cdot m$ be the orbit map. I want to prove that there is an affine subset $G^\prime$ of $G$ such that $p(G^\prime)$ is dense in $C^\prime$. 
Is the preimage of $X^\prime$ affine? If so why? If not, how do I construct such a $G^\prime$? 
For reference: The statement that such an $G^\prime$ exists can be found in the article 
A characterization of orbit closure and applications (MR0944153) on the first page of the proof of Theorem 1.2. Clearly he has more assumptions on $X$ and $G$, but I do not see anything else that could be useful here. 
 A: I was able to figure out an elementary answer with some help: 
It seems like a good hint is to look at abstract varieties or even easier use that every  quasi-affine variety is an abstract variety. 
The preimage is presumably not affine, but it is a locally closed subset of $G$ since $C$ is locally closed. Therefore $p^{-1}(C)$ is a quasi-affine variety and as such it has an open cover $U_i$ where each $U_i$ has the structure of an affine varieties such that for each intersection $U_i \cap U_j$ the induced variety structures are isomorphic (see for example Hartshorne - Algebraic Geometry Section I.8 page 58 where this is stated without proof). So let 
$p^{-1}(C) = U_1 \cup \dots \cup U_\alpha$.
Since $p$ is surjective we have 
$C = p(U_1) \cup \dots \cup p(U_\alpha)$.
Since $C$ is irreducible it can't be the union of two proper closed subsets and therefore some $p(U_i)$ has to be dense which finishes the proof. 
A: I do not know if the pre-image of $ C^{'} $ (I assume that you mean $ C^{'} $ not $ X^{'} $) is affine.  However, here is an idea for how to construct such a $ G^{'} $.  For notational purposes I will relabel $ m $ as $ x $ so that people don't confuse $ G_{m} $ with $ \mathbb{G}_{m} $.  When you said that $ x $ was an element of $ X $, I assumed that you meant that it is a closed point of $ X $.
If $ G_{x} $ is the stabilizer of the point $ x \in X $, then $ G_{x} $ acts on $ G $.  Let us assume that $ G $ is quasi-projective.  If we denote the maximal ideal of a closed point $ y $ by $ \mathfrak{m}_{y} $, then let $ \mathcal{L} $ be an invertible sheaf on $ O(x) $ such that $ H^{0}(O(x), \mathcal{L}) $ contains a non-constant global section $ s_{1} $ such that $ s_{1} \notin \mathfrak{m}_{x} \mathcal{L}_{x} $.  Since $ O(x) $ is a sub-variety of an affine variety such an $ \mathcal{L} $ must exist.
The pull-back $ p^{\ast}(\mathcal{L}) $ is a $ G_{x} $ linearized invertible sheaf $ \mathcal{M} $ with a non-constant global section $ s $ such that $ s \notin \mathfrak{m}_{e_{G}} \mathcal{M} $.  Let $ U $ be the points $ g \in G $ such that $ s_{g} \notin \mathfrak{m}_{g} \mathcal{M}_{g} $.  The neighborhood $ U $ is an open, affine $ G_{x} $-stable sub-scheme of $ G $.  Assume that $ U \cong \operatorname{Spec}(A) $.
Since $ U $ is $ G_{x} $ stable, there is a map $ p_{1}: U \to U//G_{x} $ (here $ U//G_{x} $ denotes $ \operatorname{Spec}(A^{G_{x}}) $).  By the universal property of categorical quotients there is a map $ f: U//G_{x} \to p(U) $ such that $ f \circ p_{1} = p $.  Since quotients are unique, $ U//G_{x} \cong p(U) $.  The map $ p_{1} $ is affine, so since $ p(U) \cap C^{'} $ is non-empty as $ x \in p(U) \cap C^{'} $, it is an open sub-scheme of $ C^{'} $.  Therefore, $ U $ should be the affine set you seek.
