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

/*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 $g\cdot y_1=y_2$.

*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!"**

*visually*explore the definition of transitivity in hopes of shedding a little more light on its meaning.

## The Set-Up

*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

*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.

- Start at $y_1$.
- Walk
*down*to $x$ (via $g_1^{-1}$). - And walk across to $y_2$ (via $g_2$).

*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.

*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

*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.