Intuitive meaning of transitive group action.

abstract-algebragroup-actionsgroup-theorysoft-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.

Best Answer

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

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 $1$.

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