I am just trying to learn what transitive group action is. It seems I understand the definition, but I still have an issue.
We read on Wikipedia that (emphasis added): "The group action is transitive if and only if it has only one orbit, i.e. if there exists x in X with G.x = X. This is the case if and only if G.x = X for all x in X".
Direct implication of another definition I have (from the source I learn from) is that group action is transitive if for every x in X : G.x = X. According to Wikipedia (and later on in my source), it seems to be enough only for one x to have G.x = X to deduce transitivity.
In short, I don't see how G.x = X for one x in X => G.x = X for every x in X.
P.S: I have yet to learn what orbit is. So please avoid using orbits to explain the issue.
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$\begingroup$Suppose that $\exists x\in X$ $\forall y \in X \exists g\in G \ : g(x)=y $.
Now let $y_1$, $y_2 \in X$. Choose $g_1$, $g_2$ such that $g_1(x)=y_1$ and $g_2(x)=y_2$ which you can do by original assumption. Set $g=g_2 g_1 ^ {-1}$. Clearly $g(y_1)=y_2$.
Therefore the two definitions of transitivity are in fact equivalent, even though one of them seems weaker at first.
The way to think about transitivity is that by acting with a group, you can get anywhere in your set from any starting point. Orbits of the group are precisely sets of points that you can reach with your group action from some starting point. Hence transitivity means that there is only one orbit. For example if you consider group of rotations with respect to fixed point acting on Euclidean space, orbits are just spheres of constant radius with respect to that point. Group action restricted to these spheres is transitive.
$\endgroup$ $\begingroup$Suppose $Gx=X$ for some $x\in X$.
Let $y\in X$. Then there exists $g\in G$ such that $y=gx$. Now let $z\in X$ be any element.
Then there exists $h\in G$ such that $z=hx=hg^{-1}gx=hg^{-1}y\in Gy$. Thus $X=Gy$.
$\endgroup$ $\begingroup$Intuitively, the implication you're asking about is true because groups have inverses.
By definition, we have $$Gx=\{gx:g\in G\}$$ Assume that $Gx_0=X$ for some element $x_0\in X$. Now choose any particular $x\in X$. By assumption, there exists an $h\in G$ such that $x=hx_0$. Thus $$\begin{align*} Gx&=\{gx:g\in G\}\\ &=\{ghx_0:g\in G\} \end{align*}$$ But as $g$ ranges over all the elements of $G$, so does $gh$; that is, for every element $k\in G$, there exists some $g\in G$ such that $k=gh$, namely $g=kh^{-1}$. Therefore $$\begin{align*} Gx&=\{gx:g\in G\}\\ &=\{ghx_0:g\in G\}\\ &=\{kx_0:k\in G\}\\ &=Gx_0\\ &=X \end{align*}$$
$\endgroup$ $\begingroup$In fact the two statements $\exists x\in x:X=G.x$ and $\forall x\in x:X=G.x$ are not equivalent. However the difficulty is not where you think it lies; the former does imply the latter. The former statement says there is some point of $X$ from where you can reach any point of $X$ by acting with a suitable element of $G$, while the latter says you can get from any point of$~X$ to any other point of $X$ by acting with a suitable element of $G$.
Given the way groups and actions are defined, you can reverse any step made (by acting with the inverse element), and combine two successive steps into one (by acting with the product of the group elements); therefore the former implies the latter by using the given point from where all of $X$ is reachable as a hub: you can get from anywhere to the hub and then from the hub to anywhere.
However the other implication is not valid, because the later statement does not imply that there is any point of $X$ to begin with. Indeed the latter statement is vacuously true when $X$ is empty (since there is no way to choose two points of $X$), but the former statement (like any statement that starts with "there is some point of $X$") is false when $X$ is empty.
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