Transitive Group Actions: "Where There's a Will, There's a Way!"

Groups are to mathematics what verbs are to language. In other words, they do things! A group isn't merely a collection of objects sitting stagnantly in the pages of your textbook. It encodes an action. (Sometimes a very literal, physical action! See the stop sign here?) In fact, one of the main uses of a group $G$ is that it can act* on a set $X$ by shuffling its elements around. This allows us to obtain information about that set. We say the action is transitive if one of the following equivalent definitions holds:
• (Definition #1) There is a single orbit, i.e. $\text{orb}(x)=\{g\cdot x:g\in G\}=X$ for every $x\in X$.
• (Definition #2) For any $y_1,y_2\in X$, there is a $g\in G$ such that $g\cdot y_1=y_2$.
Now perhaps you're thinking, "That's nice and all, but what does it mean?" Simply put, if $G$ acts transitively on $X$ then $G$ shuffles the elements of $X$ among themselves. That is, given any element $x$ in $X$, you have the ability to obtain any other element $y$. Think of it this way: Suppose $X$ is a finite set and its elements are points. Let's say you're standing on a point $x$, but the grass is greener on the other side at $y$ (sigh...), and so you wish to move there. But is such a move possible?

Yes! The transitivity of $G$'s action ensures us that there is a $g\in G$ - a bridge, if you like - that will kindly escort $x$ (and you!) over to $y$. So the phrase, "a group action is transitive" is just the mathematician's way of saying

"Where there's a will, there's a way!"

In the next section, we'll visually explore the definition of transitivity in hopes of shedding a little more light on its meaning.

The Set-Up

Let's begin by establishing some visual notation. We'll continue to work with a finite** set $X$ and represent its elements by dots. Again let $G$ be a group that acts on our set $X$. For any $x,y\in X$, let's draw an arrow pointing from $x$ to $y$ if there is a $g\in G$ so that $g(x)=y$. (Figure (a)) Notice the notational change! I'm replacing the usual group action dot $"g\cdot x"$ with parentheses $"g(x)"$ which I think is more suggestive: $g$ moves $x$ to $y$. (In this way, $g$ behaves almost like a function $g:x\mapsto g(x)=y$.) Now since every element in a group has an inverse, we also have the ability to go backwards from $y$ to $x$ via $g^{-1}$. We'll denote this by an arrow pointing from $y$ to $x$ (Figure (b)). Since we have the freedom to go either way, we could simply connect $x$ and $y$ by a straight line (Figure (c)), but we'll save this notation for later.

The Details

Assume $G$ acts transitively on $X$ so that there is a single orbit. This is definition #1 in the introduction. Let's show that this implies definition #2 and vice versa. Since there is a single orbit, there exists an $x\in X$ such that $X=\{g(x):g\in G\}$. In other words, for each $y\in X$ there is a $g_y\in G$ such that $g_y(x)=y$. (The subscript on $g_y$ reminds us that a different element $y'$ may require a different "bridge" $g_{y'}$). For a set with four elements, the statement "there is a single orbit" is illustrated by the diagram on the right.
Now let $y_1$ and $y_2$ be any two elements in $X$. We wish to produce a $g\in G$ such that $g(y_1)=y_2$. But how? Let's make an observation. By assumption there exist $g_1, g_2\in G$ such that $g_1(x)=y_1$ and $g_2(x)=y$. This gives us one path between $x$ and $y_1$ and another path between $x$ and $y_2$. So to get from $y_1$ to $y_2$, we simply follow the route illustrated on the left:
1. Start at $y_1$.
2. Walk down to $x$ (via $g_1^{-1}$).
3. And walk across to $y_2$ (via $g_2$).
"But," you exclaim, "that was too much work! We could've walked directly from $y_1$ to $y_2$!" And you are exactly right! And the bridge which takes us there is the desired element $g$. That is, let $g=g_2g_1^{-1}$. Then $g$ takes us from $y_1$ to $y_2$, i.e. $g(y_1)=y_2$, since $$g(y_1)=g_2(g_1^{-1}(y_1))=g_2(x)=y_2.$$ This proves that definition #1 implies definition #2. I claim the other direction (defintion #2 implies definition #1) is easy! So, like all lovable math texts, I'll leave it as an exercise.***

Notice $y_1$ and $y_2$ were arbitrary. This means that we can use the same reasoning as above to move from $y_1$ to $y_3$ or from $y_2$ to $y_3$ as indicated in Figures (d) and (e) below.

Of course, each of these red paths have inverses, so for example if we want to go from $y_2$ to $y_1$, our bridge is simply the inverse of $g_2g_1^{-1}$, namely $g_1g_2^{-1}$. We can summarize this observation as follows: if we represent the elements of $X$ as vertices of a polygon, then any two vertices can be connected by a line if $G$ acts transitively on $X$! In other words, we have what the graph theorists like to call a complete graph. In the case of a set with 4 elements, this looks like Figure (f) above where the loops indicate the identity element $e\in G.$ (Recall $e(x)=x$ for all $x\in X$ and so $x$ doesn't move anywhere!) And if our set $X$ has 5, 6, 7, or 8 elements, we get the figures below!

Cool huh? This is really just neat rather than practical, but the visual notation we established above does come in handy when proving things about group actions. Take, for instance, the little fact about group actions that you see on the right. (For an element $a$ in a set $X$, we denote the stabilizer of $a$ by $G_a$.) The proof is pretty easy when you use dots and arrows! (Click the image to go to that post!)

An Example and a Non-Example

We'll close by looking at a quick example and non-example. Let $G=\mathbb{R}\smallsetminus\{0\}$ and observe that $G$ forms a group under multiplication. Let's take our set $X$ to also be $\mathbb{R}\smallsetminus\{0\}$, and let $G$ act on $X$ by multiplication: $$\text{for g\in G and x\in X, define g\cdot x=gx.}$$ (Yes, I've reverted back to the usual "group action dot" notation.) The action of $G$ on $X$ is transitive since $\text{orb}(x)=\{gx:g\in G\}=X=\mathbb{R}\smallsetminus\{0\}$ for every $x\in X$. Indeed, for any nonzero real numbers $x$ and $y$, there is another nonzero real number $g$ such that $y=gx$, namely $g=y/x$! (And this $g$ is in $G=\mathbb{R}\smallsetminus\{0\}$ since $x\neq 0$.)

On the other hand, if we let $Q$ be the group $\mathbb{Q}\smallsetminus\{0\}$ and let $Q$ act on $X$ by multiplication, then the action is not transitive. That is, there exists an $x\in X$ such that $\text{orb}(x)\subsetneq X$ -- any irrational number will do! Take $x=\sqrt{2}$ for instance. Then $\text{orb}(\sqrt{2})=\{q\sqrt{2}:q\in Q\}$ is a proper subset of $\mathbb{R}\smallsetminus\{0\}$ since $\text{orb}(\sqrt{2})$ doesn't contain any integers!

For more examples (and a wealth of facts about transitive group actions) check out Keith Conrad's blurb.

Footnotes

* Here's the formal definition: a group action of a group $G$ on a set $X$ is a map from $G\times X$ to $X$, denoted by $g\cdot x$ for all $g\in G$ and $x\in X$, such that

• $g_1\cdot(g_2\cdot x)=(g_1g_2)\cdot x$ for all $g_1,g_2\in G$ and $x\in X$, and
• $e\cdot x=x$ for all $x\in X$ (where $e\in G$ is the identity).

**Only for the sake of illustration - I don't have enough ink to draw infinitely many dots!

*** Okay okay, here's the proof: Suppose for all $y_1,y_2\in X$ there is a $g\in G$ such that $g(y_1)=y_2$. We want to show there is a single orbit, i.e. there exists an $x\in X$ such that $\text{orb}(x)=\{g(x):g\in G\}=X$. Choose $x=y_1$! Then $\text{orb}(y_1)=X$ since for any $y\in X$ there is, by assumption, a $g\in G$ such that $g(y_1)=y$ which means $y\in \text{orb}(y_1).$ QED.

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