Group actions

A group action is a representation of the elements of a group as symmetries of a set. Many groups have a natural group action coming from their construction; e.g. the dihedral group $D_4$ acts on the vertices of a square because the group is given as a set of symmetries of the square. A group action of a group on a set is an abstract generalization of this idea, which can be used to derive useful facts about both the group and the set it acts on.

Formally, a group action of a group $G$ on a set $X$ is a function $f \colon G \times X \to X$ satisfying the following properties:

(1) $f(e_G,x) = x$ for all $x\in X$

(2) $f(gh,x) = f\big(g,f(h,x)\big)$ for all $g,h \in G$ and $x\in X.$

When the action is clear, the function $f(g,x)$ is often written as $g \cdot x.$ With this notation, the axioms become

(1) $e_G \cdot x = x$

(2) $g \cdot (h \cdot x) = (gh) \cdot x$.

The standard example of a group action is when $G$ equals the symmetric group $S_n$ $($or a subgroup of $S_n)$ and $X = \{1,2,\ldots,n\}$. Then $G$ acts on $X$ by the formula $g \cdot x = g(x).$ The properties are clear: $e \cdot x = e(x) = x$ when $e$ is the identity of $S_n,$ and $g \cdot (h \cdot x) = g \cdot h(x) = g\big(h(x)\big) = (g \circ h)(x).$

Here are several basic concepts related to group actions.

Let $G$ be a group acting on a set $X.$

A fixed point of an element $g \in G$ is an element $x \in X$ such that $g \cdot x = x.$
The stabilizer $G_x$ of a point $x \in X$ is the set of elements $g \in G$ such that $x$ is a fixed point of $g.$
The orbit of an element $x \in X$ is the set of elements $y \in X$ such that $g \cdot x = y$ for some $g \in G.$

Let $G = {\mathbb Z}_2 = \{e,g\}$ and $X = {\mathbb Z}.$
Let $G$ act on $X$ by the formulas $e\cdot x = x$ and $g\cdot x = -x.$

Find the fixed points of every element of $G.$
Find the stabilizer of every element of $X.$
Find the orbits of the elements of $X.$

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Every element of $X$ is a fixed point of $e,$ but $g$ has exactly one fixed point, namely $0.$

The stabilizer of $0$ is all of $G,$ but the stabilizer of any other element of $X$ is the trivial subgroup $\{e\}$ of $G.$

The orbit of $0$ is $\{0\},$ and the orbit of any other element $x$ is a two-element set $\{-x,x\}.$ $_\square$

Note that $X$ is partitioned as a union of its distinct and disjoint orbits. This is true in general; it follows from the fact that the relation defined by $x\sim y$ if $y$ is in the orbit of $x$ is an equivalence relation.

Certain common properties related to these definitions arise in many group actions.

Let $G$ be a group acting on a set $X.$

The action is transitive if there is only one orbit: for any $x,y \in X,$ there is an element $g \in G$ such that $g\cdot x = y.$
The action is faithful if the intersection of the stabilizers $G_x$ for $x \in X$ consists only of the trivial element $e_G.$

The axioms of a group action give a group homomorphism $G \to \text{Sym}(X),$ where $\text{Sym}(X)$ is the group of permutations of the elements of $X.$ A faithful action, then, is one for which this homomorphism is injective, as its kernel is trivial.

There are many examples where a group $G$ acts on objects related to $G.$ Here are some examples:

(1) Every group acts on itself by left multiplication: $G$ acts on $G$ via the formula $g \cdot x = gx.$ This is a transitive and faithful action; there is one orbit, and in fact the stabilizer of any element $x$ is trivial: $gx=x$ if and only if $g$ is the identity.

(2) Every group acts on itself by conjugation: $G$ acts on $G$ via the formula $g \cdot x = gxg^{-1}.$ The orbits of this action are called conjugacy classes, and the stabilizer of an element $x$ is called the centralizer $C_G(x).$

(3) If $H$ is a subgroup of $G,$ then $G$ acts on the set of cosets $G/H$ by left multiplication. The action is transitive, since $\big(kg^{-1}\big)(gH) = kH$ for any $g,k \in G.$ This induces a map $G \to \text{Sym}(G/H)$ whose kernel is equal to the intersection of all the conjugates of $H,$ which is the largest normal subgroup contained in $H.$ So if $H$ does not contain any nontrivial normal subgroups, the action is faithful. This construction is quite useful in the analysis of simple groups.

Let ${\mathbb P}^1_{\mathbb C}$ be the projective complex line: this can be thought of as the set of complex numbers plus the point $\infty.$ Then the group $GL_2({\mathbb C})$ of invertible $2\times 2$ matrices with complex entries acts on ${\mathbb P}^1$ via the formula

$$\begin{pmatrix} a&b \\ c&d \end{pmatrix} \cdot z = \frac{az+b}{cz+d},$$

with the understanding that $-d/c$ maps to $\infty$ and $\infty$ maps to $a/c.$ $($If $c=0,$ then $\infty$ maps to $\infty.)$

Checking that this is a group action boils down to the following computation for the second axiom:

$$\begin{aligned} \begin{pmatrix} a&b\\c&d \end{pmatrix} \cdot \left( \begin{pmatrix} e&f\\g&h \end{pmatrix} \cdot z \right)&= \begin{pmatrix} a&b\\c&d \end{pmatrix} \cdot \frac{ez+f}{gz+h} \\ &= \frac{a\left(\frac{ez+f}{gz+h}\right) +b}{c\left(\frac{ez+f}{gz+h}\right)+d} \\ &= \frac{a(ez+f)+b(gz+h)}{c(ez+f)+d(gz+h)} \\ &= \frac{(ae+bg)z+(af+bh)}{(ce+dg)z+(cf+dh)} \\ &= \begin{pmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{pmatrix} \cdot z \\ \end{aligned}$$

and the matrix $\begin{pmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{pmatrix}$ is the product $\begin{pmatrix} a&b\\c&d \end{pmatrix} \begin{pmatrix} e&f\\g&h \end{pmatrix},$ which shows that this gives a group action.

Note that the matrix $\begin{pmatrix} a&0 \\ 0&a \end{pmatrix}$ acts trivially on the projective line, since $\frac{az}{a} = z.$ So the action of $GL_2({\mathbb C})$ turns into an action of $GL_2({\mathbb C})/N,$ where $N$ is the subgroup of matrices which are constant multiples of the identity. This group is called the projective general linear group $PGL_2({\mathbb C});$ the action of $PGL_2({\mathbb C})$ is faithful and also transitive. (In fact, it is 3-transitive: any list of three distinct points can be sent to any other list of three distinct points.)

There is a natural relationship between orbits and stabilizers of a group action.

Let $G$ be a group acting on a set $X.$ Fix a point $x\in X$ and consider the function $f_x \colon G \to X$ given by $g \mapsto g \cdot x.$ In fact, this gives a function $h_x \colon G/G_x \to X$ given by the same formula, where $G_x$ is the stabilizer of $x.$ Here $G/G_x$ is the set of left cosets of $G_x.$ Note that $G_x$ is not necessarily a normal subgroup. So $G/G_x$ is just a set of cosets, not a quotient group.

The function $h_x$ is injective, and its image is precisely the orbit of $x.$ The conclusion is as follows:

There is a bijection between the set $G/G_x$ and the orbit of $x.$

When the orbit of $x$ is finite, this gives the following formula:

Let $G$ be a group acting on a set $X.$ Let $G_x$ be the stabilizer of an element $x \in X.$ Suppose that the orbit $O_x$ of $x$ is finite. Then the index $[G:G_x]$ is finite and equal to $|O_x|.$ If $G$ is finite, then

$$|G_x| \cdot |O_x| = |G|.$$

This theorem is used to prove many useful facts in group theory, including Burnside's lemma and the class formula. Here are two explicit examples of the theorem in action.

Let $G = S_n$ and let $X = \{ 1,\ldots,n\}.$ Let $x = n.$ The orbit of $x$ is all of $X$ (the action is transitive), so the theorem says that the index $[G:G_x]$ is equal to $n.$ Indeed, $G_x$ is isomorphic to $S_{n-1},$ the group of permutations of $\{ 1, \ldots, n-1 \}.$ So $|G|=n!,$ $|G_x| = (n-1)!,$ and $[G:G_x] = |G|/|G_x| = n!/(n-1)! = n.$

How many rotational symmetries does a cube have? (Extra credit: what is the group of symmetries of the cube?)

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The question is asking for the number of ways to permute the vertices (or faces, or edges) of a cube by twisting and turning (but not reflecting) the cube in 3D-space. The orbit-stabilizer theorem can be used to solve this problem in three different ways.

Let $G$ be the group of symmetries of the cube. Then $G$ acts on the set $F$ of faces of the cube. There are six faces, and the action is transitive, so the size of an orbit $O_f$ of a given face is $6.$ The order of the stabilizer $G_f$ of a face is 4, because there are four rotations of a cube that fix a face (the rotations around the axis perpendicular to the face). So by the orbit-stabilizer theorem, $|G| = 4 \cdot 6 = 24.$

But $G$ also acts on the vertices of a cube. There are eight vertices, and again the action is transitive. What is the order of the stabilizer $G_x$ of a vertex? The only rotations of a cube that leave one vertex fixed are the three rotations of $120,240,$ and $360$ degrees around the body diagonal of the cube. So $|G_x|=3$ and $|O_x|=8,$ so by the orbit-stabilizer theorem $|G| = 3\cdot 8 = 24.$

And $G$ also acts on the edges of a cube. There are twelve edges, and the action is transitive. The stabilizer $G_e$ of an edge has order two; there is the identity of $G,$ and the unique element of $G$ which switches the vertices of the edge. So again $|G| = 2 \cdot 12 = 24.$ $_\square$

In fact, $G$ is isomorphic to the symmetry group $S_4,$ and the isomorphism uses yet another action of $G,$ namely the action of $G$ on the body diagonals, of which there are four. This action gives a homomorphism $G \to S_4,$ and it is straightforward to show that this homomorphism is injective (the action is faithful), and the groups have the same size, so it must be an isomorphism.

• Burnside's Lemma

• Conjugacy Classes