§4 Group Actions

§4.1 Motivation for Group Actions

As one of the greatest mathematicians once said, let mathematics speak for itself—you should not feel that everything needs to be motivated. If something is beautiful, it motivates itself. That being said, I do not really want to motivate group actions for you, but their history is actually quite interesting, so it is worth discussing.

Suppose you were a French or German mathematician in the middle of the nineteenth century. Your definition of a group would not have been the one we gave above. In fact, a group was essentially understood simply as a matrix group $GL_n(\mathbb{R})$, or perhaps over $\mathbb{C}$ if you wanted to work in a world where everything was especially beautiful. Naturally, we have a bijection

where $V$ is a real vector space of dimension $n$.

Thus, rather than blindly performing row reduction, it is natural to study how elements of $GL_n(\mathbb{R})$ behave through the bijection above. The "group action" appearing in the automorphism description (4.1) actually gives rise to the more abstract theory of linear algebra that you have already studied! This is what is called a group representation: whenever you have a group homomorphism from $G$ to the group of linear automorphisms of some vector space, you have a group representation, which we will study shortly.

The important point is that group actions arise naturally from studying the automorphisms in (4.1), so what we are studying is not an entirely artificial construction. To understand how we move from studying linear automorphisms of vector spaces to studying automorphisms of sets, note that the notion of a group action occurs universally outside linear algebra as well. So why not "generalize" the theory of group actions from vector spaces to sets?

This type of process turns out to be extremely important in algebra. You look at what structure you have, try to remove whatever structure is not actually necessary for studying the "abstract theory", and then see whether an interesting theory remains. It was essentially this process that led Emmy Nöther, in the early twentieth century, toward the modern abstract notion of a group.

§4.2 Definition of a Group Action

Definition 4.1 Let $X$ be a set and $G$ a group. A group action of $G$ on $X$ is a homomorphism

Definition 4.2 A left group action of $G$ on $X$ is a map

such that

• $1_Gx=x$.

• For $g,h\in G$,

§4.3 Examples of Group Actions

Example 4.1 Let

Define

where

(rotation by $z$).

§4.4 Propositions about Group Actions

Proposition 4.1 A group action determines a map of sets

where we denote the value at $(g,x)$ by $gx$.

This map satisfies

(a)

(b)

Conversely, any map

satisfying (a) and (b) determines a group action.

Proof.

Given

$$\phi:G\to \mathrm{Aut}_{\mathrm{set}}(X),$$

write

$$\phi(g)=\phi_g.$$

Define

$$G\times X\to X$$

by

$$(g,x)\mapsto\phi_g(x).$$

(a) Since $\phi_{1_G}=\mathrm{id}_X$ (because $\phi$ is a homomorphism),

$$\begin{aligned} (1,x)\mapsto\phi_1(x) &=\mathrm{id}_X(x)\\ &=x. \end{aligned}$$

(b) Since $\phi$ is a group homomorphism,

$$\phi(g_1g_2)=\phi(g_1)\phi(g_2).$$

Hence

$$\phi_{g_1g_2}(x) = \phi_{g_1}\circ\phi_{g_2}(x) \qquad \forall x.$$

Using our notation,

$$\begin{aligned} (g_1g_2)(x) &=\phi_{g_1}(g_2x)\\ &=g_1(g_2x). \end{aligned}$$

Conversely, suppose we are given a map

$$G\times X\to X$$

satisfying (a) and (b). Restrict the map to the subset ${g}\times X\subset G\times X$.

The map ${g}\times X\to X$ may be identified with

$$\begin{aligned} \psi_g:X\cong\{g\}\times X&\to X\\ x\mapsto(g,x)&\mapsto gx. \end{aligned}$$

We claim that $\psi_g$ is a bijection because of (a) and (b).

First,

$$\begin{aligned} \psi_{1_G}:X\cong\{1_G\}\times X&\to X\\ x\mapsto(1_G,x)&\mapsto1_G\cdot x. \end{aligned}$$

By (a),

$$1_G\cdot x=x$$

and hence

$$\psi_{1_G}(x)=x.$$

Therefore,

$$\psi_{1_G}=\mathrm{id}_X.$$

Next, note that $\psi_g$ is a bijection for every $g$.

Injectivity:

$$\begin{align*} &&\psi_g(x)&=\psi_g(y) &&\text{notation}\\ &\Rightarrow&\qquad gx&=gy &&\text{by the given map }G\times X\to X\\ &\Rightarrow&\qquad g^{-1}(gx)&=g^{-1}(gy) &&\text{by (b)}\\ &\Rightarrow&\qquad1_Gx&=1_Gy\\ &\Rightarrow&\qquad x&=y &&\text{by (a)}. \end{align*}$$

Surjectivity: If $x\in X$, let

$$y=g^{-1}x.$$

Then

$$\begin{aligned} \psi_g(y) &=g(g^{-1}x)\\ &=(gg^{-1})x\\ &=x. \end{aligned}$$

Thus

$$\psi_g\in\mathrm{Aut}_{\mathrm{set}}(X)$$

for every $g\in G$.

Finally,

$$g\mapsto\psi_g$$

is a homomorphism, because

$$g_1g_2\mapsto\psi_{g_1g_2},$$

and

$$\begin{aligned} \psi_{g_1g_2}(x) &=(g_1g_2)x\\ &=g_1(g_2x)\\ &=\psi_{g_1}(\psi_{g_2}(x)) \qquad\forall x. \end{aligned}$$

Therefore,

$$\psi_{g_1g_2} = \psi_{g_1}\circ\psi_{g_2}.$$

$$~\tag*{$\square$}$$

Question: How do we prove that two functions $f,g:A\to B$ are equal?

We show that

for every $a\in A$. This is the definition of $f=g$.

Thus you may find it more difficult to think of a group action as a map

satisfying (a) and (b), rather than as a homomorphism

§4.5 Orbits

Philosophy: Given a group action

we can decompose $X$ into orbits.

Definition 4.3 Let $G$ act on a set $X$. For every $x\in X$, the orbit of $x$ is the set

Example 4.2 Let

be given by

(the trivial action).

Then

Example 4.3 Let $G=S^1$, $X=\mathbb{C}$, and

Then

which is the circle of radius $|z|$. Thus

Also note that

Remark. Let $\mathcal{P}(X)$ denote the power set of $X$, the set of all subsets of $X$. A group action determines a map

Of course, this map does not hit every element of $\mathcal{P}(X)$—for example, it never hits the empty set. But it does hit certain subsets.

Definition 4.6 The orbit set, or orbit space, of a group action is the image of the map

We denote it by

as though we were dividing $X$ by $G$.

If $y=gx$, then $\mathcal{O}_x=\mathcal{O}_y$. Thus $x$ and $y$ have the same image in $\mathcal{P}(X)$, i.e. they correspond to the same element of $X/G$.

Proposition 4.2

(1) For every $x\in X$,

(2)

for some $g\in G$.

Proof.

(1)

$$x=1_Gx,$$

so

$$x\in\mathcal{O}_x.$$

(2)

$$\mathcal{O}_x=\mathcal{O}_y \Longleftrightarrow y\in\mathcal{O}_x \Longleftrightarrow y=gx$$

for some $g\in G$.

$$~\tag*{$\square$}$$

If $X$ is finite, we can count its elements orbit by orbit.

Example 4.4 Let $G$ be a group. For every $g\in G$, we have a bijection

This is a bijection because

• if $y\in G$, then

• if

then

Moreover,

Hence the map

is a homomorphism.

In other words, every group acts on itself.

If

is a subgroup, then

defines a group action. More explicitly, for every $h\in H$,

Proposition 4.3 Let $H$ be a subgroup of $G$. We have just seen that $H$ acts on $G$. Moreover, for every $x,y\in G$,

Proof.

Let

$$h=x^{-1}y\in G.$$

Then define

$$\begin{align*} \mathcal{O}_x&\to\mathcal{O}_y\\ gx&\mapsto gxh &&\text{which lies in }\mathcal{O}_y \text{ because }gxh=gx(x^{-1}y)=gy,\\ \mathcal{O}_y&\to\mathcal{O}_x\\ gy&\mapsto gyh^{-1}. \end{align*}$$

These maps are inverses of each other because

$$\begin{aligned} gx&\mapsto gxh\mapsto gxhh^{-1}=gx,\\ gy&\mapsto gyh^{-1}\mapsto gyh^{-1}h=gy. \end{aligned}$$

$$~\tag*{$\square$}$$

§4.6 Lagrange's Theorem

Proposition 4.4

Moreover,

Proof.

(1) Take any element $x\in X$.

By the definition of an orbit, $x$ belongs to the orbit $\mathcal{O}_x$ containing it.

Therefore,

$$x\in \bigcup_{\mathcal{O}\in X/G}\mathcal{O}.$$

Since this is true for every $x\in X$, we have

$$X= \bigcup_{\mathcal{O}\in X/G}\mathcal{O}.$$

(2) For any map of sets

$$f:A\to B,$$

we know

$$A=\bigcup_{b\in B}f^{-1}(b)$$

and

$$f^{-1}(b)\cap f^{-1}(b^{\prime})=\varnothing$$

whenever $b\neq b^{\prime}$.

Here,

$$\mathcal{O}=f^{-1}(\mathcal{O}),$$

so

$$\mathcal{O}\neq\mathcal{O}^{\prime} \quad\Rightarrow\quad \mathcal{O}\cap\mathcal{O}^{\prime}=\varnothing.$$

$$~\tag*{$\square$}$$

Corollary 4.5

Corollary 4.6

Proof.

Corollary 4.7

for any $x\in X$.

Proposition 4.8

Proof.

$$\begin{aligned} \mathcal{O}_{\mathrm{id}_G} &= \{y\mid y=h\cdot\mathrm{id}_G, ~\text{for some}~h\in H\}\\ &= \{y\mid y=h, ~\text{for some}~h\in H\}\\ &=H. \end{aligned}$$

$$~\tag*{$\square$}$$

We have shown that if $G$ is finite, then

for every $x\in G$.

Therefore,

and

This means that $|H|$ divides $|G|$.

This is Lagrange's theorem:

Theorem 4.9 (Lagrange's Theorem) Let $G$ be finite. Then $|H|$ divides $|G|$.

§4.7 Cosets and Normal Subgroups

Definition 4.7 Let $H\subset G$ be a subgroup. We define

to be the right coset of $H$ determined by $g$.

We have

for some $h\in H$.

Definition 4.8 We define

to be the left coset of $H$ determined by $g$.

We will also use $G/H$ to denote this set, and whenever possible we will try not to discuss right cosets further.

Example 4.6 Let

Let

Then

whereas

Thus, in general,

Definition 4.9 A subgroup $H\subset G$ is called normal if for every $g\in G$,

The left-hand side is also written

Thus $H$ is normal if and only if

We write

Proposition 4.10

(1)

if and only if there exists $h\in H$ such that

The left coset $gH$ is an orbit for a right group action

Thus our definition of $G/H$ is the same—the same elements and the same construction.

(2)

for every $g\in G$ if and only if

Proof.

(1) If

$$gH=g^{\prime}H,$$

then there exist $h_1,h_2\in H$ such that

$$gh_1=g^{\prime}h_2.$$

Hence

$$gh_1h_2^{-1}=g^{\prime}.$$

Let

$$h=h_1h_2^{-1}.$$

Then

$$gh=g^{\prime}.$$

(2)

$$gH=Hg$$

implies that for every $h\in H$ there exists $h^{\prime}\in H$ such that

$$gh=h^{\prime}g.$$

Therefore,

$$ghg^{-1}=h^{\prime},$$

so

$$ghg^{-1}\in H$$

for every $h\in H$. Hence

$$gHg^{-1}\subset H$$

for every $g\in G$, and therefore

$$H\triangleleft G.$$

Conversely, if

$$H\triangleleft G,$$

then

$$ghg^{-1}\in H$$

for every $g\in G$ and $h\in H$. Thus for every $g\in G$ and $h\in H$, there exists $h^{\prime}\in H$ such that

$$ghg^{-1}=h^{\prime}.$$

Therefore,

$$gh=h^{\prime}g.$$

$$~\tag*{$\square$}$$

§4.8 Index

Definition 4.8 Let $H\subset G$ be a subgroup. The index of $H$ in $G$ is the number of elements of $G/H$. It is denoted by

Proposition 4.11 Suppose

are subgroups. ($H$ is a subgroup of $G$, and $K$ is a subgroup of $H$. Note that this also implies that $K$ is a subgroup of $G$.) Then

Proof.

In the proof of Lagrange's theorem, we saw that

$$G = \bigsqcup_{\mathcal{O}_g\in G/H} \mathcal{O}_g.$$

That is, $G$ is the disjoint union of the orbits of the action of $H$ on $G$. Moreover, all the orbits have the same size:

$$|\mathcal{O}_g|=|H|$$

for every $g\in G$.

Therefore,

$$|G|=n\cdot|H|,$$

where $n$ is the number of distinct orbits, namely

$$n=[G:H].$$

Thus

$$[G:K] = |G|/|K| = (|G|/|H|)\cdot(|H|/|K|) = [G:H][H:K].$$

$$~\tag*{$\square$}$$

This proposition connects many of the concepts we have introduced.

Proposition 4.12 In any group, every subgroup of index $2$ is normal.

Proof.

Since the index of $H$ in $G$ is $2$, there are exactly two distinct left cosets of $H$ in $G$:

$$G=H\cup gH,$$

where

$$g\in G\setminus H.$$

Similarly, there are exactly two right cosets:

$$G=H\cup Hg$$

for the same $g$.

We consider two cases depending on whether $g\in H$ or $g\in G\setminus H$.

• If $g\in H$, then

$$gH=H=Hg,$$

since $H$ is a subgroup and is closed under the group operation. Thus the left and right cosets are equal.

• If $g\in G\setminus H$, then $gH$ and $Hg$ are both the unique coset other than $H$. Since there are only two cosets,

$$gH=G\setminus H=Hg.$$

Thus the left and right cosets are again equal.

Hence in both cases,

$$gH=Hg$$

for every $g\in G$. Equality of left and right cosets implies that $H$ is a normal subgroup of $G$.

$$~\tag*{$\square$}$$

§4.9 The Orbit-Stabilizer Theorem

Definition 4.9 Let

be a group action. Given $x\in X$, the stabilizer of $x$ is the subgroup

Proposition 4.13 $G_x$ is a subgroup of $G$.

Proof.

Suppose

$$\phi_g(x)=x$$

and

$$\phi_{g^{\prime}}(x)=x.$$

Then

$$x = \phi_g(x) = \phi_g(\phi_{g^{\prime}}(x)) = \phi_{gg^{\prime}}(x),$$

and

$$\phi_{1_G}(x)=\mathrm{id}_Xx=x.$$

$$~\tag*{$\square$}$$

Thus, given an action of $G$ on $X$, we can associate two objects to each $x\in X$:

Proposition 4.14 The function

is a bijection.

Proof.

It is well defined because

$$\begin{aligned} gG_x=g^{\prime}G_x &\Rightarrow g^{\prime}=gh ~\text{for some}~h\in G_x\\ &\Rightarrow \begin{aligned} g^{\prime}x &=(gh)x\\ &=g(hx)\\ &=gx. \end{aligned} \end{aligned}$$

Injectivity:

$$\begin{aligned} g^{\prime}x=gx &\Rightarrow g^{-1}g^{\prime}x=x\\ &\Rightarrow g^{-1}g^{\prime}\in G_x\\ &\Rightarrow g^{\prime}=gh ~\text{for some}~h\in G_x\\ &\Rightarrow gG_x=g^{\prime}G_x. \end{aligned}$$

Surjectivity:

$$x^{\prime}\in\mathcal{O}_x \Rightarrow x^{\prime}=gx$$

for some $g\in G$.

$$~\tag*{$\square$}$$

Corollary 4.15 (Orbit-Stabilizer Theorem) If $|G_x|$ and $|\mathcal{O}_x|$ are finite, then $|G|$ is finite. Moreover,

Proof.

By the proposition, there is a bijection between the left cosets $G/G_x$ and the orbit $\mathcal{O}_x$. Therefore,

$$|G/G_x|=|\mathcal{O}_x|.$$

Each left coset $gG_x$ contains exactly $|G_x|$ elements because $G_x$ is a subgroup of $G$.

Thus the total number of elements of $G$ is the number of cosets multiplied by the size of each coset:

$$|G| = |G/G_x||G_x| = |\mathcal{O}_x||G_x|.$$

Since both $|\mathcal{O}_x|$ and $|G_x|$ are finite, their product $|G|$ is finite as well.

Hence, if $|G_x|$ and $|\mathcal{O}_x|$ are finite, then $G$ is finite and

$$|G|=|G_x||\mathcal{O}_x|.$$

$$~\tag*{$\square$}$$

This is the orbit-stabilizer theorem, and it is extremely useful.

Example 4.6 Let $P_n\subset\mathbb{R}^2$ be the regular $n$-gon centered at the origin. Let $D_{2n}\subset GL_2(\mathbb{R})$ be the group of linear transformations satisfying

This means both

and

but it does not mean

for every $x\in P_n$.

Thus, $D_{2n}$ is the group of linear symmetries of $P_n$.

Proposition 4.16

Definition 4.10 $D_{2n}$ is called the $n$th dihedral group.

The set $P_n$ itself is not very useful here—it has infinitely many points. However, if $D_{2n}$ acts on $P_n$, then it must permute the vertices

Thus $D_{2n}$ acts on the set

Given a vertex $v_i$, we have

Why? Rotation by $\frac{2\pi}{n}$ is linear and sends $P_n$ to itself, since we chose $P_n$ to be centered at the origin. Therefore rotations by

belong to $D_{2n}$, and rotating $v_i$ by these angles reaches every vertex $v_j$.

Question: What is the stabilizer?

Suppose $g\in D_{2n}$ fixes $v_i$. What can it do to $v_{i-1}$ and $v_{i+1}$?

• If

then two linearly independent vectors, $v_i$ and $v_{i-1}$, are fixed by $g$. Hence

• Otherwise,

Then $g$ must be reflection across the line passing through the origin $\vec{O}$ and $v_i$.

Hence exactly two elements of $D_{2n}$ fix $v_i$. Thus the stabilizer of $v_i$ has order $2$ and is therefore isomorphic to $\mathbb{Z}/2\mathbb{Z}$.

By the orbit-stabilizer theorem,

Example 4.7 (Rotational Symmetry) Let $T$ be a regular tetrahedron centered at the origin. Let

be the group of rotations satisfying

Then $G$ acts on the set of vertices of $T$. The tetrahedron has four vertices:

First calculate the stabilizer of some $v_i$. If $g$ is a rotation fixing $v_i$, then it must rotate the face opposite $v_i$ while fixing the line through $v_i$.

There are three possible planar rotations:

Therefore, the stabilizer of $v_i$ is a group of order $3$, and hence isomorphic to $\mathbb{Z}/3\mathbb{Z}$.

What is the orbit? Every vertex.

Indeed, if you want to find a rotation sending one vertex to another, choose a suitable rotation fixing a third vertex.

By the orbit-stabilizer theorem,

We will later identify exactly what this group is.