Intuitive meaning of transitive group action.

Question

My idea (not read) is that a transitive group action enforces surjective mapping.

The definition of transitive action is that there exists an $x \in X$ such that for any $y \in X$, there exists a $g \in G$ with $g \star x = y.$

$\exists x \in X,\,\forall y\in X, \, \exists g \in G,\, \mid\, g\star x = y$

I want to ask for an example that shows why more than one orbit makes it necessary that transitive group action property is not followed.

Edit : In view of comments by @Qiaochu Yuan, and other want to add:

It's just not the mapping $G \times X \to X$ that transitivity is making surjective, that's always surjective. But transitivity says there exists an $x$ such that the mapping $G \to X$ given by $g \mapsto g \star x$ is surjective. Equivalently, the action has one orbit.

Answer 1

First, notice that orbits partition the given set into disjoint subsets. That is, any two orbits are either same or disjoint, and the union of all the orbits is the whole space.

Now, suppose $O_1$ and $O_2$ are two distinct (and hence, disjoint) orbits for the action. That would mean that if we start with any two elements $x \in O_1$ and $y \in O_2$, we cannot find any $g \in G$ such that $g * x = y$. To write in quantifiers, we have

$\forall x \in X, \exists y \in X \text{ such that } \forall g \in G, g * x \neq y$.

This is the definition of an action not being transitive. Also, for every $x \in X$, the element must be in some orbit, and by assumption, there are at least two distinct orbits so that the existence of such a $y$ is also guaranteed.

I hope this answers your question.

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Edit (Example):

To look at an example illustrating that existence of two more more orbits forces the action not to be transtive, consider the usual action of $SO(2)$ on $\mathbb{R}^2$. From basic linear algebra, we know that this action is precisely rotation of $\mathbb{R}^2$. The orbits in this case are concentric circles. Now, if we take two points on distinct circles, they cannot be brought to each other using rotations, since rotations do not change the length of vectors (and points being on distinct concentric circles require their lengths to be different). I hope you can write the details mathematically for this intuition.

Answer 2

Let us consider a group action of $G$ on a set $X$ defined by $(g, x) =g\star x$

Let $x\in X$ , then orbit of $x$ is defined as $\mathcal{O}_x=\{g\star x:g\in G\}$

Claim:

1. The relation $\sim$ on $X$ defined by $x\sim y\iff \exists g\in G$ such that $y=g\star x$ is an equivalence relation.

2. $[x]=\mathcal{O}_x$

3. $x\sim y\iff y\in \mathcal{O}_x$

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Claim 2: The following two statements are equivalent :

1. $\exists x \in X, \forall y\in X, \exists g \in G \, \mid \, g\star x = y$

2. $\forall y\in X \implies y\in \mathcal{O}_x$

A group action is called transitive if there is a single orbit i.e $\exists x\in X$ such that $\mathcal{O}_x=X$

So from this equivalent (standard) definition it's clear that more than one orbit means non transitive action.

Examples:

Transitive action: Consider the group action of $G=\{r_{120},r_{240},r_{360}=e\}$ (the group of rotational symmetries of an equilateral triangle ABC) on the set $X=\{A,B,C\}$ (all vertices of ABC) is a transitive action. (action is the rotation of vertices. For an example $r_{120} \star A=B$, $r_{240}\star B=A$).

Nontransitive action: Let $G=S_3$ and consider the action of $G$ on itself by conjugation i.e $(\pi, \sigma) =\pi \sigma \pi^{-1}$

Then $\mathcal{O}_{(1) }=\{(1)\}$ $\mathcal{O}_{(12) }=\{(12),(13),(23)\}$

$\mathcal{O}_{(123) }=\{(123),(132)\}$

There is more than one orbit!

Now verify nontransitivity by $1$.