Group Actions and Burnside’s Lemma I. Introduction In the text, the Orbit–Stabilizer Theorem (Theorem 7.3, p. 145) is stated and proved for a subgroup G ≤ SYM(S); in this case, the group G moves the elements of set S around—giving rise to the orbits— because G is given to be a group of permutations of the set S. The first purpose of this handout is to prove Orbit–Stabilizer Theorem in a more general context by extending slightly the notion of how a group can permute the elements of a set. (This generalization, though slight, is important; it brings several important applications within the scope of the theory. One of these will be treated in some detail in this handout; you will encounter two others in the spring.) The second purpose will be the statement and proof of the theorem known as “Burnside’s Lemma,” which gives a method for counting the orbits induced by a group action. The central definition is this: Definition 1. Let G be a group, and let X be a nonempty set. An action of G on X is a function from the cartesian product G × X to the set X, written (g, x) 7−→ g · x, satisfying the following two conditions: (1) For all x ∈ X, e · x = x; and (2) For all g, h ∈ G and all x ∈ X, g · (h · x) = (gh) · x. Example 1. If G ≤ SYM(X), then putting g · x equal to g(x) makes an action of G on X. (This is the context of Theorem 7.3 in Gallian. Examples 2 and 3 need the more general setting.) Example 2. Let X be the underlying set of the group G. Define g · x to be the element gxg −1 of X. This gives an action of G on X. Exercise 1. Prove that the functions defined in examples 1 and 2 are indeed group actions—that is, that they satisfy the clauses of the definition. (There is a worked example similar to this exercise at the bottom of page 2 of this handout.) Exercises 3 and 4 below show that the existence of an action of G on X is equivalent to the existence of a group homomorphism from G into SYM(X). Exercise 2. Show that if g · x = y, then g −1 · y = x. Exercise 3. Let G be a group which acts on set X. (a): For each g ∈ G, Show that the function φg : X −→ X given by φg (x) = g · x is in SYM(X); that is, show that φg is one–to–one and onto. (b): Show that the map G −→ SYM(X) given by g 7−→ φg is a group homomorphism. Exercise 4. Conversely: If ψ: G −→ SYM(X) is a group homomorphism, one defines g · x to be ψ(g) (x). Show that this definition makes the function (g, x) 7−→ g · x an action of G on X. 1 Example 3. Let G be D4 , the group of symmetries of the square ———— |2 1| | | | | |3 4| ———— (Think of G as a subgroup of S4 , with each element permuting the four vertices of the square.) The set X will be the set of all functions f : {1, 2, 3, 4} −→ {R, Y, G, B, O}; each function f can be thought of as a coloring of the four vertices, using the colors Red, Y ellow, Green, Blue, and Orange. (To make sure you understand the set X, convince yourself that there are 5 × 5 × 5 × 5 = 625 such functions.) Now, the action of G on X is the following. For g ∈ D4 and function f ∈ X, g · f is the function given by (g · f )(i) = f g −1 (i) , 1 ≤ i ≤ 4. (∗) If you think of f as a coloring of the vertices, then you can think of g · f as another coloring, the one you get by applying the symmetry g to the colors of f while keeping the square fixed. Thus, for example, if f is the coloring O ———— Y |2 1| | | | | |3 4| B ———— G, and if g is the counterclockwise rotation R90◦ , then g · f is the coloring you get by rotating the colors counterclockwise by 90◦ : Y ———— G |2 1| | | | | |3 4| O ———— B. Let us check that the correspondence (g, f ) 7−→ g · f given in (∗) does indeed define an action of D4 on the set X of colorings—that it satisfies the two clauses of Definition 1. Clause (1). Let f be any coloring. For all 1 ≤ i ≤ 4, · f (i) = f −1 (i) = f (i) = f (i), so f and · f are the same function: f ≡ · f . Clause (2). Let f be any coloring, and let g and h be any group elements. Then for all 1 ≤ i ≤ 4, g · (h · f ) (i) = h · f g −1 (i) = f h−1 g −1 (i) = f (gh)−1 (i) = (gh) · f ) (i). Thus g · (h · f ) and (gh) · f are the same function: g · (h · f ) ≡ (gh) · f . 2 II. Orbits and Stabilizers. Definitions 2 and 3. Let G be a group with an action on a set X, and let x ∈ X. The orbit of x under G is the set orb G (x) = {g · x, g ∈ G}; and the stabilizer of x in G is the set stab G (x) = {g ∈ G g · x = x}. Note that orb G (x) is a subset of X, while stab G (x) is a subset of G. Exercise 5. Let us define a binary relation R on X by x1 R x2 ⇐⇒ (∃g ∈ G)(g · x1 = x2 ). Show that R is an equivalence relation on X and the equivalence classes it generates are precisely the orbits of X under G. Exercise 6. Show that stab G (x) ≤ G. Theorem 1 (The Orbit–Stabilizer Theorem). Let a finite group G act on a finite set X. For each x ∈ X, stab G (x) · orb G (x) = G. (1) Proof . Say orb G (x) = t; let orb G (x) = {x1 , x2 , . . . , xt }, where x = x1 . To each xi (1 ≤ i ≤ t) associate the set G i = {g ∈ G : g · x1 = xi }. That is, x1 ←→ G1 = {g ∈ G : g · x1 = x1 } x2 ←→ G2 = {g ∈ G : g · x1 = x2 }; x3 ←→ G3 = {g ∈ G : g · x1 = x3 }; ··· ··· ··· ··· xt ←→ Gt = {g ∈ G : g · x1 = xt }. = stab G (x1 ); The collection {G1 , . . . , Gi , . . . , Gt } is obviously a partition of G into t equivalence classes. CLAIM: For each 1 ≤ i ≤ t, the equivalence class Gi is the left coset gi stab G (x1 ), where gi is any fixed element of G such that gi · x1 = xi . Proof of claim. First, observe that for any h ∈ stab G (x1 ), (gi h) · x1 = gi · (h · x1 ) = gi · x1 = xi , so that gi h ∈ Gi . It follows that stab G (x1 ) ⊆ Gi . Conversely, given any k ∈ Gi , we have (gi−1 k) · x1 = gi−1 · (k · x1 ) = gi−1 · xi = ↑ x1 , (Ex. 2) so that gi−1 k ∈ stab G (x1 ), which implies that k ∈ gi stab G (x1 ). It follows that Gi ⊆ stab G (x1 ). (CLAIM) The theorem now follows from Lagrange’s Theorem: G stab G (x1 ) × # cosets of stab G (x1 ) = ↑ (Lagrange) 3 = ↑ claim stab G (x1 ) × t = stab G (x1 ) · orb G (x1 ). The following two corollaries are immediate from equation (1): Corollary 1.1. For any x ∈ X, orb G (x) divides G. Corollary 1.2. If x and y are in the same orbit, then stab G (x) = stab G (y). When x and y are in the same orbit, one can say more than this. The stabilizer subgroups are not only equal in size, they are also what is known as “conjugate” (Gallian p.90). This is what exercise 7 says. Exercise 7. Let x and y be in the same orbit. Choose and fix any g ∈ G such that g · x = y. Show that stab G (y) = g stab G (x) g −1 . III. Burnside’s Lemma. Throughout this section, G will be a finite group acting on a finite set X. The question to be addressed is: How many orbits will there be? For many group actions, this is the main question—the real reason for introducing the group to begin with. For instance, consider example 3 again. The set X contains 54 = 625 different functions, but these do not really constitute 625 different colorings of the vertices. Here is the reason. If g ∈ D4 and f is a function in this set X, then f and g · f are not really different as colorings, since they differ by a symmetry of the square: you can make one look exactly like the other just by moving the square from one position to another. Two functions are “really different” as colorings precisely if they are in different orbits, so that you cannot get from one to the other via a symmetry. In other words, the number of different ways to color the vertices of the square is the number of different orbits of X under the action of D4 . It turns out that this counting problem is harder than that of counting cosets. One cannot simply compute the number of orbits by performing a division, since in most examples, the orbits turn out to be of different G, by Corollary 1.1; but they need not all be the same divisor sizes. (Each orbit size must be a divisor of of G .) There is a way to count the orbits, though: Burnside’s Lemma2 supplies a very elegant formula to do precisely this. Roughly stated, this lemma says that the number of orbits is the “average number of fixed points” of an element of G. Let me make precise what it is that will be averaged: Definition 4. Let G be a finite group acting on a finite set X. For g ∈ G, let fix(g) be the number of elements x of X such that g · x = x. In symbols, fix(g) = {x ∈ X | g · x = x}. A computation will clarify this definition. I will compute fix(g) for each group element g of Example 3. → fix R0◦ = 54 because each of the functions of X is fixed by the identity. → R90◦ · f will equal f precisely if f gives the same color to each vertex. The same is true for R270◦ . Thus fix R90◦ = fix R270◦ = 5. → Now consider R180◦ : For any f ∈ X, R180◦ · f = f ⇐⇒ 2 f (1) = f (3) ; f (2) = f (4) Gallian (p. 490) credits this theorem to Frobenius, and some scholars even attribute it to Cauchy. 4 and there are clearly 5 × 5 ways to make such a function f . Thus fix R180◦ = 52 . → Now let H be the reflection through the horizontal bisector, and let V be the reflection through the vertical bisector. Since f (1) = f (4) H · f = f ⇐⇒ f (2) = f (3) and V · f = f ⇐⇒ f (1) = f (2) , f (3) = f (4) we have fix(H) = fix(V ) = 52 . → Finally, let D and D0 be, respectively, the reflections in the two diagonals. Since D · f = f ⇐⇒ f (1) = f (3) and D0 · f = f ⇐⇒ f (2) = f (4), we have fix(D) = fix(D0 ) = 53 . Before stating and proving Burnside’s Lemma, I will establish one step of the proof in advance. Lemma. Let G be a finite group acting on a finite set X. Then X X stabG (x) = fix(g). x∈X g∈G Proof . Consider the X × G matrix A, with rows indexed by elements of X and columns indexed by elements of G, and with entries ( 1 if g · x = x, and Axg = 0 otherwise. A moment’s consideration shows that for any x ∈ X, the sum across row x equals stabG (x), while for any g ∈ G, the sum down column g equals fix(g). Thus X X stabG (x) = row sum of row x = total # of 1’s in the matrix, x∈X x∈X and X g∈G fix(g) = X column sum of column g = total # of 1’s in the matrix. g∈G I am now ready to state and prove the main result. Theorem 2 (Burnside’s Lemma). Let a finite group G act on a finite set X. The number of orbits that G induces is given by: 1 X number of orbits = fix(g). G g∈G X X stabG (x) = Proof . By the lemma, we have fix(g); so x∈X 1 X fix(g) G g∈G g∈G = ↑ (Lemma) X stabG (x) 1 X stabG (x) = . G G x∈X x∈X Now, equation (1) of Theorem 1 can be rearranged to read stabG (x) 1 , = G orb G (x) 5 (2) (3) and substituting (3) into (2) gives X stabG (x) X 1 X 1 fix(g) = · · · = = G G orb G (x) . g∈G x∈X x∈X The next step is to group this last sum by orbits, summing all of the elements of a given orbit together. This gives, up to this point: X X 1 X 1 = fix(g) = · · · = G orb G (x) g∈G x∈X orbits X 1 x∈ orbit ! orb G (x) . (4) Now, consider the inner sums in the last expression in (4). A moment’s thought shows that if a given orbit consists of {x1 , . . . , xt }, then t X X 1 1 = = 1. orb G (x) t i=1 x∈ orbit This addition to (4) completes the proof: X 1 X fix(g) = · · · = G g∈G orbits X x∈ orbit 1 orb G (x) ! = X 1 ! = number of orbits. orbits Application: Example 3, concluded. Burnside’s Lemma, together with the computations on page 5, allow us to compute the number of “really different” colorings of the vertices of a square using colors {R, Y, G, B, O}: 1 X 1 4 3 2 5 + 2 · 5 + 3 · 5 + 2 · 5 = 120. number of orbits = fix(G) = 8 G g∈G Exercise 8. Determine how many “really different” colorings of the square there are if you are allowed to use k colors (k = 1, 2, 3, · · ·). 6
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