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.

  • $\begingroup$ An orbit is just a subset of $X$ that equals $Gx$ for some $x\in X$. $\endgroup$ – Zev Chonoles Sep 6 '15 at 8:23
  • $\begingroup$ thanks for the tip, but it still does not answer my question. "I don't see how G.x = X for one x in X => G.x = X for every x in X." $\endgroup$ – d_e Sep 6 '15 at 8:26
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    $\begingroup$ I would recommend looking at orbits to understand the issue. While it may be possible to answer your question without using orbits, it really misses the whole point. $\endgroup$ – Morgan Rodgers Sep 6 '15 at 8:47
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    $\begingroup$ I second the comment from @MorganRodgers. It makes no sense to try to understand a statement about orbits without knowing what an orbit is. $\endgroup$ – Andreas Blass Sep 6 '15 at 8:54
  • $\begingroup$ I will look into orbits probably today - I already got some spoilers in comments and answers. my source gave the definition without mention of orbits. So it was only after I read in Wiki , that I understood the issue is extremely related to orbits. $\endgroup$ – d_e Sep 6 '15 at 9:13

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.


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 $X=Gy$.


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 $$\begin{align*} Gx&=\{gx:g\in G\}\\ &=\{ghx_0:g\in G\}\\ &=\{kx_0:k\in G\}\\ &=Gx_0\\ &=X \end{align*}$$


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