Group Actions

Definition. Let G be a group, and let X be a set. A left action of G on X is a function $\theta: G \times X \rightarrow
   X$ satisfying:

(a) $\theta(g_1, \theta(g_2, x)) =
   \theta(g_1g_2, x)$ for all $g_1, g_2 \in G$ and $x \in X$ .

(b) $\theta(1, x) = x$ for all $x \in X$ .

It's customary to write "$g
   \cdot x$ " or "$gx$ " for $\theta(g,x)$ ; the equations above then become

$$g_1(g_2x) = (g_1g_2)x \quad\hbox{and}\quad 1\cdot x = x.$$

If G acts on X, X is a G-set.

Lemma. If X is a G-set, there is a homomorphism $\phi: G \rightarrow
   S_X$ , where $S_X$ is the group of permutations of X. Conversely, any homomorphism $\phi: G \rightarrow
   S_X$ gives rise to a G-set structure on X.

Proof. Let X be a G-set. Define $\phi: G \rightarrow S_X$ by

$$\phi(g)(x) = gx.$$

Since $\phi(g^{-1})\phi(g)(x) =
   x$ and $\phi(g)\phi(g^{-1})(x) = x$ , $\phi(g)$ and $\phi(g^{-1})$ are inverses. Therefore, $\phi(g)$ is a bijection, so $\phi(g) \in S_X$ .

To see that $\phi$ is a homomorphism, note that

$$\phi(g_1)\phi(g_2)(x) = g_1(g_2x) = (g_1g_2)x = \phi(g_1g_2)(x).$$

Conversely, suppose $\phi: G
   \rightarrow S_X$ is a homomorphism. Define an action of G on X by

$$g\cdot x = \phi(g)(x).$$

Then

$$\phi(g_1g_2)(x) = \phi(g_1)\phi(g_2)(x) \quad\hbox{so}\quad (g_1g_2)x = g_1(g_2x).$$

This verifies property~1.

Since $\phi$ is a homomorphism, $\phi(1) = \id$ . Hence, $1\cdot x = \phi(1)(x) = \id(x) =
   x$ . This verifies property~2. Hence, X is a G-set.


Example. Any group G acts on any set X by defining $g\cdot x = x$ for all $g \in G$ and all $x \in X$ .


Example. Let $\integer$ act on $\real$ by translation. That is, if $n \in \integer$ and $x \in
   \real$ , define $n\cdot x = x + n$ . Now

$$n_1\cdot (n_2\cdot x) = n_1\cdot (x + n_2) = x + n_2 + n_1 = (n_1 + n_2)\cdot x$$

$$0\cdot x = x + 0 = x$$

Therefore, this is an action of $\integer$ on $\real$ .


Example. Let

$$G = \left\{\left[\matrix{1 & 0 \cr 0 & 1 \cr}\right], \left[\matrix{1 & 0 \cr 0 & -1 \cr}\right], \left[\matrix{-1 & 0 \cr 0 & 1 \cr}\right], \left[\matrix{-1 & 0 \cr 0 & -1 \cr}\right]\right\}.$$

Make G into a group via matrix multiplication. G acts on $\real^2$ by multiplication: e.g.

$$\left[\matrix{1 & 0 \cr 0 & -1 \cr}\right] \left[\matrix{x \cr y \cr }\right] = \left[\matrix{x \cr -y \cr }\right].$$

Notice that

$$\left[\matrix{1 & 0 \cr 0 & 1 \cr}\right] \quad\hbox{fixes everything},$$

$$\left[\matrix{1 & 0 \cr 0 & -1 \cr}\right] \quad\hbox{fixes the x-axis},$$

$$\left[\matrix{-1 & 0 \cr 0 & 1 \cr}\right] \quad\hbox{fixes the y-axis},$$

$$\left[\matrix{-1 & 0 \cr 0 & -1 \cr}\right] \quad\hbox{fixes the origin}.\quad\halmos$$


Example. $S_3$ acts on $\{1, 2, 3\}$ . And in general, $S_n$ acts on $\{1, 2,\ldots, n\}$ .


Definition. Let X be a G-set, and let $x \in X$ . The orbit of x under the action is

$$Gx = \{gx \mid g \in G\}.$$

The set of orbits of X under G is denoted $X/G$ .

If $X = Gx$ for some $x \in
   X$ , the action is transitive. (In this case, X is said to be a transitive G-set.)

Lemma. X is a transitive G-set if and only if for all $x, y \in X$ , there exists $g \in G$ such that $gx =
   y$ .

Proof. ($\Rightarrow$ ) Since X is transitive, $X = Gz$ for some $z \in X$ . Let $x = g_1z$ and let $y =
   g_2z$ , where $g_1, g_2 \in G$ . Then $g_1^{-1}x
   = z$ , so $y = g_2g_1^{-1}x$ .

($\Leftarrow$ ) Suppose that for all $x, y \in X$ , there exists $g \in G$ such that $gx =
   y$ . Take any $x \in X$ . If $y \in X$ , then $gx = y$ for some $g \in
   G$ , so $y \in Gx$ . Thus, $X
   \subset Gx$ , so $X = Gx$ , and X is a transitive G-set.

Definition. Let X be a G-set, and let $x \in X$ . The isotropy group of x is

$$G_x = \{ g \in G \mid gx = x\}.$$

If $G_x = \{1\}$ for all $x \in X$ , the action is free. (In this case, X is called a free G-set.)


Example. Let $D_3$ , the dihedral group of order 6, act on the vertices $\{1,2,3\}$ of an equilateral triangle. If $m_1$ denotes reflection through the line through vertex 1, then the isotropy group of vertex 1 is $\{\id, m_1\}$ (and similarly for the other two vertices).


Lemma. Let X be a G-set, and let $x \in X$ . Then $G_x < G$ .

Proof. First, $1\cdot x = x$ , so $1 \in G_x$ . Suppose $g \in
   G_x$ . Then $gx = x$ , so $x =
   g^{-1}x$ . Therefore, $g^{-1} \in G_x$ .

Finally, suppose $g, h \in G_x$ . Then

$$(gh)x = g(hx) = gx = x.$$

Hence, $gh \in G_x$ . Therefore, $G_x < G$ .


Example. Consider the translation action of $\integer$ on $\real$ defined above. An orbit has the form

$$x + \integer = \{x + n \mid n \in \integer\}, \quad\hbox{where}\quad 0 \le x < 1.$$

Thus, the space of orbits $\real/\integer$ is equivalent to $S^1$ , the circle. (Of course, the two are actually isomorphic as groups!)

The action is free, since the only $n \in \integer$ which fixes $x \in \real$ is 0. On the other hand, the action is not transitive: There is no integer translation carrying 0 to $\dfrac{1}{2}$ , for instance.


Example. Consider the action of the group

$$G = \left\{\left[\matrix{1 & 0 \cr 0 & 1 \cr}\right], \left[\matrix{1 & 0 \cr 0 & -1 \cr}\right], \left[\matrix{-1 & 0 \cr 0 & 1 \cr}\right], \left[\matrix{-1 & 0 \cr 0 & -1 \cr}\right]\right\}$$

on the plane $\real^2$ .

The isotropy groups of points under this action are:

$$G_{(0,0)} = G,$$

$$G_{(x,0)} = \left\{\left[\matrix{1 & 0 \cr 0 & 1 \cr}\right], \left[\matrix{1 & 0 \cr 0 & -1 \cr}\right]\right\} \quad\hbox{for}\quad x \ne 0,$$

$$G_{(0,y)} = \left\{\left[\matrix{1 & 0 \cr 0 & 1 \cr}\right], \left[\matrix{-1 & 0 \cr 0 & 1 \cr}\right],\right\} \quad\hbox{for}\quad y \ne 0,$$

$$G_{(x,y)} = \left\{\left[\matrix{1 & 0 \cr 0 & 1 \cr}\right]\right\} \quad\hbox{for}\quad x, y \ne 0.$$

Obviously, the action is not free.

The orbits of points under the action are:

$$\{(0,0)\},$$

$$\{(x,0), (-x,0)\} \quad\hbox{for}\quad x \ne 0,$$

$$\{(0,y), (0,-y)\} \quad\hbox{for}\quad y \ne 0,$$

$$\{(x,y), (-x,y), (x,-y), (-x,-y)\} \quad\hbox{for}\quad x, y \ne 0.$$

$$\hbox{\epsfysize=1.75in \epsffile{group-actions1.eps}}$$

Since there is more than one orbit, the action is not transitive.

The set of orbits $\real^2/G$ looks like the first quadrant $\{(x,y) \mid x, y \ge 0\}$ .


Example. Consider the action of $S_3$ on $\{1, 2,
   3\}$ . The action is transitive: If $i, j \in \{1, 2, 3\}$ , $i \ne j$ , the transposition $(i\ j)$ carries i to j. (If $i = j$ , there's nothing to do.)

The isotropy groups are

$$G_{\{1\}} = \{\id, (2\ 3)\}$$

$$G_{\{2\}} = \{\id, (1\ 3)\}$$

$$G_{\{3\}} = \{\id, (1\ 2)\}\quad\halmos$$


Proposition. Let X be a G-set. Then X is partitioned by the orbits.

Proof. If $x \in X$ , then $x \in Gx$ . Hence, X is certainly the union of the orbits. I need to show that two orbits are either disjoint or identical.

Suppose that $Gx \cap Gy \ne
   \emptyset$ . Find $z \in Gx \cap Gy$ . Then $z = g_1x$ and $z = g_2y$ for some $g_1,
   g_2 \in G$ . Hence, $g_1x = g_2y$ , so $x =
   g_1^{-1}g_2y$ and $y = g_2^{-1}g_1x$ .

Now let $gx \in Gx$ . Then $gx = gg_1^{-1}g_2y \in Gy$ . This proves $Gx \subset Gy$ . Likewise, if $gy \in Gy$ , then $gy =
   gg_2^{-1}g_1x \in Gx$ , so $Gy \subset Gx$ . Therefore, $Gx
   = Gy$ . Hence, two orbits which intersect nontrivially must in fact coincide. This proves that the orbits partition X.

There is a group-theoretic relationship between isotropy groups of points in the same orbit.

Proposition. Let X be a G-set, let $x \in X$ , and let $g \in
   G$ . Then

$$G_{gx} = gG_xg^{-1}.$$

Proof. Suppose $h \in G_{gx}$ , so $hgx = gx$ . Then $g^{-1}hgx = x$ , so $g^{-1}hg =
   g_0 \in G_x$ . Hence, $h = gg_0g^{-1} \in gG_xg^{-1}$ . This shows that $G_{gx} \subset gG_xg^{-1}$ .

Conversely, suppose $gg_0g^{-1}
   \in gG_xg^{-1}$ , where $g_0 \in G_x$ . Since $g_0x =
   x$ ,

$$gg_0g^{-1}(gx) = gg_0x = gx.$$

This shows that $gg_0g^{-1} \in
   G_{gx}$ , so $gG_xg^{-1} \subset G_{gx}$ . Therefore, $G_{gx} = gG_xg^{-1}$ .

Corollary. Let X be a G-set. If x and y are in the same G-orbit, then $G_x$ and $G_y$ are conjugate subgroups. In particular, $G_x \approx G_y$ .

The following result not only serves as a "counting theorem" in the finite case, but also says that up to equivalence, every G-set "looks like" a union of sets of cosets of G.

Theorem. Let X be a G-set, and let $x \in X$ .

(a) There is a bijection $Gx
   \leftrightarrow G/G_x$ .

(b) If G is finite, $|Gx| =
   |G/G_x| = (G:G_x)$ : the order of the orbit equals the index of the isotropy group.

Proof. Define $\phi: Gx \rightarrow G/G_x$ by

$$\phi(gx) = gG_x.$$

I must check that $\phi$ is well-defined.

Suppose then that $g_1x = g_2x$ . Then $g_2^{-1}g_1x = x$ , so $g_2^{-1}g_1 \in G_x$ . Hence, $g_1G_x = g_2G_x$ . Now

$$\phi(g_1x) = g_1G_x = g_2G_x = \phi(g_2x).$$

Therefore, $\phi$ is indeed well-defined.

Define $\psi: G/G_x \rightarrow
   Gx$ by

$$\psi(gG_x) = gx.$$

I must check that $\psi$ is well-defined.

Suppose that $g_1G_x = g_2G_x$ . Then $g_2^{-1}g_1G_x = G_x$ , so $g_2^{-1}g_1 \in G_x$ . This means that $g_2^{-1}g_1x = x$ , or $g_1x = g_2x$ . It follows that

$$\psi(g_1G_x) = g_1x = g_2x = \psi(g_2G_x).$$

Hence, $\psi$ is well-defined.

Since $\phi$ and $\psi$ are inverses, they are both bijections, and $Gx$ and $G/G_x$ are in 1-1 correspondence.

The second statement follows immediately from the first.

Definition. Let X be a G-set, and let $S \subset G$ . The fixed point set of X under S is

$$X^S = \{x \in X \mid gx = x \quad\hbox{for all}\quad g \in S\}.$$

If $g \in G$ , then $X^g$ denotes the elements of X fixed by g: i.e. $X^g = \{x \in X \mid gx
   = x\}.$


Example. Consider the action of $S_3$ on the set $\{1,
   2, 3\}$ . Here is a picture of the isotropy groups and the fixed point sets:

$$\matrix{ & \id & (1\ 2) & (1\ 3) & (2\ 3) & (1\ 2\ 3) & (1\ 3\ 2) \cr & & & & & & \cr 1 & \star & \cdot & \cdot & \star & \cdot & \cdot \cr & & & & & & \cr 2 & \star & \cdot & \star & \cdot & \cdot & \cdot \cr & & & & & & \cr 3 & \star & \star & \cdot & \cdot & \cdot & \cdot \cr}$$

A $\star$ in a position means that the element of $\{1, 2, 3\}$ which labels the row is fixed by the element of $S_3$ which labels the column.


Example. Consider the group of symmetries of the hexagon below:

$$\hbox{\epsfysize=1.75in \epsffile{group-actions2.eps}}$$

Since the hexagon is not regular, the symmetry group has only 4 elements: the identity map (id), reflection across the x-axis ($m_x$ ), reflection across the y-axis ($m_y$ ), and rotation by $\pi$ (r). The symmetry group acts on the set of vertices by permutations.

$$\matrix{ & \id & m_x & m_y & r \cr & & & & \cr 1 & \star & \cdot & \star & \cdot \cr & & & & \cr 2 & \star & \cdot & \cdot & \cdot \cr & & & & \cr 3 & \star & \cdot & \cdot & \cdot \cr & & & & \cr 4 & \star & \cdot & \star & \cdot \cr & & & & \cr 5 & \star & \cdot & \cdot & \cdot \cr & & & & \cr 6 & \star & \cdot & \cdot & \cdot \cr}$$

This action is not transitive (unlike the action in the preceding example). The orbits are $\{1,
   4\}$ and $\{ 2, 3, 5, 6\}$ .


The following "counting theorem",has important combinatorial applications; it relates the number of orbits under a group action to the size of the fixed point sets.

Theorem. (Burnside) Let X be a finite G-set, where G is a finite group. Then

$$|X/G| = \dfrac{1}{|G|} \sum_{g\in G} |X^g|.$$

Proof. Observe that $g \in G_x$ if and only if $x \in X^g$ . It follows that

$$\sum_{g \in G} |X^g| = \sum_{x \in X} |G_x|.$$

Now just compute:

$$\dfrac{1}{|G|} \sum_{g \in G} |X^g| = \dfrac{1}{|G|} \sum_{x \in X} |G_x| = \sum_{x \in X} \dfrac{|G_x|}{|G|} = \sum_{x \in X} \dfrac{1}{(G:G_x)} =$$

$$\sum_{x \in X} \dfrac{1}{|Gx|} = \sum_{K \in X/G} \sum_{x \in K} \dfrac{1}{|K|} = \sum_{K \in X/G} 1 = |X/G|.\quad\halmos$$


Example. Each side of a square may be colored red, green, or blue. Colors may be repeated, and the square may be "turned over". How many distinct squares are there?

Note that since the square may be "turned over", the following colored squares are considered the same:

$$\hbox{\epsfysize=1in \epsffile{group-actions3.eps}}$$

I'll temporarily nail the square to the wall so it can't move. There are then $3^4 = 81$ ways to paint the square. Two such paintings are "the same" if one can be carried into the other by a symmetry of the square --- i.e., by an element of $D_4$ .

Thus, X is the set of paintings of the fixed square. $G = D_4$ acts by symmetries. An equivalence class of paintings is simply an orbit under this action, and I want the number of orbits. According to Burnside, I simply count the number of fixed-square paintings (elements of X) fixed by each element of G.

$$\matrix{\hbox{element} & id & r_1 & r_2 & r_3 & m_x & m_y & m_{+} & m_{-} \cr & & & & & & & & \cr \hbox{fixed points} & 81 & 3 & 9 & 3 & 27 & 27 & 9 & 9 \cr}$$

For example, $m_x$ , reflection about the x-axis, fixes paintings where the top and bottom sides have the same color. Each side can be a different color, so there are $3\cdot 3\cdot 3 = 27$ paintings fixed by $m_x$ .

The total number of fixed points is 168. Since $|D_4| = 8$ , there are $168/8 = 21$ distinct paintings.


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