This is part $4$ of a $4$ part question involving various properties of the group action. The link to part $1$ and hence the nexus to all other parts is linked below.
Part $1$ and links to other parts
The body of the problem asks me to show:
Let $G \times X \to X$ be an action. Prove that the action of $G$ on $X$ is transitive if and only if $G \cdot x = X, \text{for some}\ x \in X$
For this problem I have an outline of a proof for the '$\Leftarrow$' part of the proof and would like some feedback on its validity or suggestions on improvement. For the '$\Rightarrow$' portion I have some rough idea of where I want to go/what I want to use for this part but it's quite vague and I'm having trouble really seeing the proof.
Obviously as an 'if an only if' statement we need to prove the conclusion from the predicate ($\Rightarrow$) and the predicate from the conclusion ($\Leftarrow$).
$\Leftarrow$: For this part I would like to show equality by two-way inclusion, i.e., that $X \subset G \cdot x$ and $G \cdot x \subset X$. To that end, assume that the action of $G$ on $X$ is transitive. That means that $$\forall x,y \in X,\ \exists g \in G \mid g \cdot x = y$$ Take $y \in X$ by transitivity we have $g \cdot x = y$ for some $g \in G$ and $x \in X$ hence $y \in G \cdot x$ and since $y$ was an arbitrary element we can conclude that $X \subset G \cdot x$. Now, let $y' \in G \cdot x$ this means there exists $g' \in G$ and $x \in X$ such that $g' \cdot x = y'$ but by the definition of transitivity above we know that $x, y' \in X$ since $y'$ was an arbitrary element of $g \cdot x$ we can conclude that $G \cdot x \subset X$. Hence $G \cdot x = X$.
$\Rightarrow$: Assume $G \cdot x = X,\ \text{for some}\ x \in X$. Here I was inclined to appeal to Part $3$ and use the fact that the set of orbits of $X$ form a partition of $X$ and use the fact that, by definition of a partition, the union of all equivalence classes (cells of the partition, orbits here) is the whole set. However, that's referring to the set of equivalence classes where the orbit of $x$, $G \cdot x$, is just one equivalence class. So here this is saying for some element of the set, $x$, the equivalence class of that element is the whole set.
I was hoping to show that this would imply that since $X$ is a partition then every element of $X$ is sent to a distinct equivalence class, where the equivalence class of an element $x$ would be $\{ y \in X \mid x \sim y \} = \{ y \in X \mid g \cdot x = y,\ \text{for some}\ g \in G \}$. However, again, the assumption of $G \cdot x = X$ for some $x \in X$ is implying that every element of the set would be sent to the same equivalence class of some element $x \in X$. Some help here would be greatly appreciated!
Your labels are backwards as I write this. Also, as noted, there is an unspoken assumption that $X$ is nonempty.
In the $\Rightarrow$ direction, we want to prove that if the action is transitive, then there exists $x\in X$ such that $G\cdot x= X$. Your argument is correct, though note that for any action we necessarily have $G\cdot x\subseteq X$ for all $x\in X$, just by definition of what an action is, so you really only need to establish $X\subseteq G\cdot x$ to prove the equality.
In the $\Leftarrow$ direction, assuming that there exists an $x\in X$ such that $G\cdot x = X$, you want to prove transitivity. To that end, let $x',y'\in X$ and you want to show that there exists $g\in G$ such that $g\cdot x'=y'$.
Well, we know that $x',y'\in X = G\cdot x$, so there exist $g_1,g_2\in G$ such that $g_1\cdot x = x'$ and $g_2\cdot x = y'$. Can you come up with an element of $G$ that will map $x'$ to $y'$, perhaps by going from $x'$ to $x$ to $y'$?