Suppose that we wish to color the vertices of a square with two different colors, say black and white. We might suspect that there would be24= 16different colorings. However, some of
these colorings are equivalent. If we color the first vertex black and the remaining vertices white, it is the same as coloring the second vertex black and the remaining ones white since we could obtain the second coloring simply by rotating the square90◦ (Figure14.17).
B W W W W B W W W W B W W W W B
14.3. BURNSIDE’S COUNTING THEOREM 163 Burnside’s Counting Theorem offers a method of computing the number of distinguish- able ways in which something can be done. In addition to its geometric applications, the theorem has interesting applications to areas in switching theory and chemistry. The proof of Burnside’s Counting Theorem depends on the following lemma.
Lemma 14.18. Let X be aG-set and suppose that x ∼y. Then Gx is isomorphic to Gy.
In particular, |Gx|=|Gy|.
Proof. Let G act on X by (g, x) 7→ g·x. Since x ∼ y, there exists a g ∈ G such that g·x=y. Leta∈Gx. Since
gag−1·y=ga·g−1y=ga·x=g·x=y,
we can define a mapϕ:Gx→Gy byϕ(a) =gag−1. The map ϕis a homomorphism since
ϕ(ab) =gabg−1=gag−1gbg−1=ϕ(a)ϕ(b).
Suppose that ϕ(a) = ϕ(b). Then gag−1 =gbg−1 or a=b; hence, the map is injective. To show thatϕis onto, let bbe inGy; theng−1bg is in Gx since
g−1bg·x=g−1b·gx=g−1b·y=g−1·y=x;
and ϕ(g−1bg) =b.
Theorem 14.19 Burnside. Let G be a finite group acting on a set X and let k denote the number of orbits of X. Then
k= 1
|G| ∑
g∈G
|Xg|.
Proof. We look at all the fixed pointsx of all the elements in g∈G; that is, we look at all g’s and allx’s such thatgx =x. If viewed in terms of fixed point sets, the number of
all g’s fixing x’s is ∑
g∈G
|Xg|.
However, if viewed in terms of the stabilizer subgroups, this number is ∑ x∈X |Gx|; hence,∑g∈G|Xg|= ∑ x∈X|Gx|. By Lemma 14.18, ∑ y∈Ox |Gy|=|Ox| · |Gx|.
By Theorem14.11 and Lagrange’s Theorem, this expression is equal to|G|. Summing over all of the kdistinct orbits, we conclude that
∑ g∈G |Xg|= ∑ x∈X |Gx|=k· |G|.
Example 14.20. Let X = {1,2,3,4,5} and suppose that G is the permutation group G={(1),(13),(13)(25),(25)}. The orbits of X are{1,3},{2,5}, and {4}. The fixed point sets are
X(13)={2,4,5}
X(13)(25)={4}
X(25)={1,3,4}. Burnside’s Theorem says that
k= 1 |G| ∑ g∈G |Xg|= 1 4(5 + 3 + 1 + 3) = 3. A Geometric Example
Before we apply Burnside’s Theorem to switching-theory problems, let us examine the number of ways in which the vertices of a square can be colored black or white. Notice that we can sometimes obtain equivalent colorings by simply applying a rigid motion to the square. For instance, as we have pointed out, if we color one of the vertices black and the remaining three white, it does not matter which vertex was colored black since a rotation will give an equivalent coloring.
The symmetry group of a square, D4, is given by the following permutations:
(1) (13) (24) (1432)
(1234) (12)(34) (14)(23) (13)(24)
The group G acts on the set of vertices {1,2,3,4} in the usual manner. We can describe the different colorings by mappings from X into Y = {B, W} where B and W represent the colors black and white, respectively. Each map f : X → Y describes a way to color the corners of the square. Every σ ∈D4 induces a permutation eσ of the possible colorings
given byσe(f) =f◦σ forf :X→Y. For example, suppose that f is defined by f(1) =B
f(2) =W f(3) =W f(4) =W
and σ = (12)(34). Then eσ(f) = f ◦σ sends vertex 2 to B and the remaining vertices to W. The set of all such eσ is a permutation group Ge on the set of possible colorings. LetXe denote the set of all possible colorings; that is, Xe is the set of all possible maps fromX to Y. Now we must compute the number of G-equivalence classes.e
1. Xe(1) =Xe since the identity fixes every possible coloring. |Xe|= 24 = 16.
2. Xe(1234) consists of allf ∈Xe such thatf is unchanged by the permutation (1234). In
this casef(1) =f(2) =f(3) =f(4), so that all values off must be the same; that is, eitherf(x) =B orf(x) =W for every vertexx of the square. So |Xe(1234)|= 2. 3. |Xe(1432)|= 2.
4. ForXe(13)(24),f(1) =f(3)and f(2) =f(4). Thus, |Xe(13)(24)|= 22= 4.
5. |Xe(12)(34)|= 4. 6. |Xe(14)(23)|= 4.
14.3. BURNSIDE’S COUNTING THEOREM 165 7. For Xe(13), f(1) = f(3) and the other corners can be of any color; hence, |Xe(13)| =
23 = 8.
8. |Xe(24)|= 8.
By Burnside’s Theorem, we can conclude that there are exactly
1 8(2
4+ 21+ 22+ 21+ 22+ 22+ 23+ 23) = 6
ways to color the vertices of the square.
Proposition 14.21. Let G be a permutation group of X and Xe the set of functions from X to Y. Then there exists a permutation group Ge acting onX, wheree σe∈Ge is defined by e
σ(f) =f◦σ for σ ∈G and f ∈X. Furthermore, ife n is the number of cycles in the cycle decomposition of σ, then |Xeσ|=|Y|n.
Proof. Let σ ∈ G and f ∈ X. Clearly,e f ◦σ is also in X. Suppose thate g is another function from X toY such thateσ(f) =σe(g). Then for each x∈X,
f(σ(x)) =eσ(f)(x) =eσ(g)(x) =g(σ(x)).
Sinceσ is a permutation ofX, every elementx′ inX is the image of somex inX under σ; hence, f and g agree on all elements of X. Therefore, f =g and σe is injective. The map σ7→σe is onto, since the two sets are the same size.
Suppose that σ is a permutation of X with cycle decomposition σ =σ1σ2· · ·σn. Any
f in Xeσ must have the same value on each cycle of σ. Since there are n cycles and |Y|
possible values for each cycle,|Xeσ|=|Y|n.
Example 14.22. Let X = {1,2, . . . ,7} and suppose that Y = {A, B, C}. If g is the permutation of X given by (13)(245) = (13)(245)(6)(7), then n = 4. Any f ∈ Xeg must
have the same value on each cycle in g. There are |Y|= 3 such choices for any value, so |Xeg|= 34 = 81.
Example 14.23. Suppose that we wish to color the vertices of a square using four different colors. By Proposition 14.21, we can immediately decide that there are
1 8(4
4+ 41+ 42+ 41+ 42+ 42+ 43+ 43) = 55
possible ways.
Switching Functions
In switching theory we are concerned with the design of electronic circuits with binary inputs and outputs. The simplest of these circuits is a switching function that hasninputs and a single output (Figure 14.24). Large electronic circuits can often be constructed by combining smaller modules of this kind. The inherent problem here is that even for a simple circuit a large number of different switching functions can be constructed. With only four inputs and a single output, we can construct 65,536 different switching functions. However, we can often replace one switching function with another merely by permuting the input leads to the circuit (Figure14.25).
f f(x1, x2, . . . , xn) xn x2 x1 .. .
Figure 14.24: A switching function of nvariables
We define a switching or Boolean function of n variables to be a function from Zn
2
to Z2. Since any switching function can have two possible values for each binary n-tuple and there are 2n binary n-tuples, 22n switching functions are possible for n variables. In
general, allowing permutations of the inputs greatly reduces the number of different kinds of modules that are needed to build a large circuit.
f f(a, b) a b f f(b, a) =g(a, b) a b
Figure 14.25: A switching function of two variables
The possible switching functions with two input variables a and b are listed in Ta- ble 14.26. Two switching functions f and g are equivalent if g can be obtained from f by a permutation of the input variables. For example, g(a, b, c) = f(b, c, a). In this case g ∼ f via the permutation (acb). In the case of switching functions of two variables, the permutation (ab) reduces 16 possible switching functions to 12 equivalent functions since
f2∼f4 f3∼f5 f10∼f12 f11∼f13. Inputs Outputs f0 f1 f2 f3 f4 f5 f6 f7 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 1 0 1 0 1 0 1 0 1 Inputs Outputs f8 f9 f10 f11 f12 f13 f14 f15 0 0 1 1 1 1 1 1 1 1 0 1 0 0 0 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 1 0 1 0 1 0 1 0 1
Table 14.26: Switching functions in two variables
For three input variables there are 223 = 256 possible switching functions; in the case of four variables there are 224 = 65,536. The number of equivalence classes is too large to
14.3. BURNSIDE’S COUNTING THEOREM 167 reasonably calculate directly. It is necessary to employ Burnside’s Theorem.
Consider a switching function with three possible inputs, a, b, and c. As we have mentioned, two switching functions f and g are equivalent if a permutation of the input variables off givesg. It is important to notice that a permutation of the switching functions is not simply a permutation of the input values {a, b, c}. A switching function is a set of output values for the inputsa,b, andc, so when we consider equivalent switching functions, we are permuting23possible outputs, not just three input values. For example, each binary triple (a, b, c) has a specific output associated with it. The permutation (acb) changes outputs as follows: (0,0,0)7→(0,0,0) (0,0,1)7→(0,1,0) (0,1,0)7→(1,0,0) .. . (1,1,0)7→(1,0,1) (1,1,1)7→(1,1,1).
LetXbe the set of output values for a switching function innvariables. Then|X|= 2n. We can enumerate these values as follows:
(0, . . . ,0,1)7→0 (0, . . . ,1,0)7→1 (0, . . . ,1,1)7→2 .. . (1, . . . ,1,1)7→2n−1.
Now let us consider a circuit with four input variables and a single output. Suppose that we can permute the leads of any circuit according to the following permutation group:
(a), (ac), (bd), (adcb),
(abcd), (ab)(cd), (ad)(bc), (ac)(bd).
The permutations of the four possible input variables induce the permutations of the output values in Table 14.27.
Hence, there are
1 8(2
16+ 2·212+ 2·26+ 3·210) = 9616
possible switching functions of four variables under this group of permutations. This number will be even smaller if we consider the full symmetric group on four letters.
Group Number Permutation Switching Function Permutation of Cycles
(a) (0) 16 (ac) (2,8)(3,9)(6,12)(7,13) 12 (bd) (1,4)(3,6)(9,12)(11,14) 12 (adcb) (1,2,4,8)(3,6.12,9)(5,10)(7,14,13,11) 6 (abcd) (1,8,4,2)(3,9,12,6)(5,10)(7,11,13,14) 6 (ab)(cd) (1,2)(4,8)(5,10)(6,9)(7,11)(13,14) 10 (ad)(bc) (1,8)(2,4)(3,12)(5,10)(7,14)(11,13) 10 (ac)(bd) (1,4)(2,8)(3,12)(6,9)(7,13)(11,14) 10
Table 14.27: Permutations of switching functions in four variables
Sage Sage has many commands related to conjugacy, which is a group action. It also has commands for orbits and stabilizers of permutation groups. In the supplement, we illustrate the automorphism group of a (combinatorial) graph as another example of a group action on the vertex set of the graph.
Historical Note
William Burnside was born in London in 1852. He attended Cambridge University from 1871 to 1875 and won the Smith’s Prize in his last year. After his graduation he lectured at Cambridge. He was made a member of the Royal Society in 1893. Burnside wrote approximately 150 papers on topics in applied mathematics, differential geometry, and probability, but his most famous contributions were in group theory. Several of Burnside’s conjectures have stimulated research to this day. One such conjecture was that every group of odd order is solvable; that is, for a group G of odd order, there exists a sequence of subgroups
G=Hn⊃Hn−1 ⊃ · · · ⊃H1 ⊃H0 ={e}
such thatHi is normal inHi+1 and Hi+1/Hi is abelian. This conjecture was finally proven
by W. Feit and J. Thompson in 1963. Burnside’s The Theory of Groups of Finite Order, published in 1897, was one of the first books to treat groups in a modern context as opposed to permutation groups. The second edition, published in 1911, is still a classic.
14.4
Exercises
1. Examples 14.1–14.5 in the first section each describe an action of a group G on a set X, which will give rise to the equivalence relation defined by G-equivalence. For each example, compute the equivalence classes of the equivalence relation, the G-equivalence classes.
2. Compute allXg and all Gx for each of the following permutation groups.
(a) X={1,2,3},G=S3={(1),(12),(13),(23),(123),(132)}
(b) X={1,2,3,4,5,6},G={(1),(12),(345),(354),(12)(345),(12)(354)}
3. Compute the G-equivalence classes ofX for each of the G-sets in Exercise14.4.2. For each x∈X verify that|G|=|Ox| · |Gx|.
14.4. EXERCISES 169 4. LetGbe the additive group of real numbers. Let the action of θ∈Gon the real plane
R2 be given by rotating the plane counterclockwise about the origin throughθradians. Let
P be a point on the plane other than the origin. (a) Show that R2 is aG-set.
(b) Describe geometrically the orbit containingP. (c) Find the groupGP.
5. LetG=A4 and suppose thatGacts on itself by conjugation; that is,(g, h) 7→ ghg−1.
(a) Determine the conjugacy classes (orbits) of each element ofG. (b) Determine all of the isotropy subgroups for each element of G.
6. Find the conjugacy classes and the class equation for each of the following groups.
(a) S4 (b) D5 (c) Z9 (d) Q8
7. Write the class equation forS5 and for A5.
8. If a square remains fixed in the plane, how many different ways can the corners of the square be colored if three colors are used?
9. How many ways can the vertices of an equilateral triangle be colored using three different colors?
10. Find the number of ways a six-sided die can be constructed if each side is marked differently with1, . . . ,6dots.
11. Up to a rotation, how many ways can the faces of a cube be colored with three different colors?
12. Consider12straight wires of equal lengths with their ends soldered together to form the edges of a cube. Either silver or copper wire can be used for each edge. How many different ways can the cube be constructed?
13. Suppose that we color each of the eight corners of a cube. Using three different colors, how many ways can the corners be colored up to a rotation of the cube?
14. Each of the faces of a regular tetrahedron can be painted either red or white. Up to a rotation, how many different ways can the tetrahedron be painted?
15. Suppose that the vertices of a regular hexagon are to be colored either red or white. How many ways can this be done up to a symmetry of the hexagon?
16. A molecule of benzene is made up of six carbon atoms and six hydrogen atoms, linked together in a hexagonal shape as in Figure14.28.
(a) How many different compounds can be formed by replacing one or more of the hy- drogen atoms with a chlorine atom?
(b) Find the number of different chemical compounds that can be formed by replacing three of the six hydrogen atoms in a benzene ring with aCH3 radical.
H
H
H H
H H
Figure 14.28: A benzene ring
17. How many equivalence classes of switching functions are there if the input variables x1,x2, andx3 can be permuted by any permutation inS3? What if the input variablesx1,
x2,x3, and x4 can be permuted by any permutation inS4?
18. How many equivalence classes of switching functions are there if the input variables x1, x2, x3, and x4 can be permuted by any permutation in the subgroup of S4 generated
by the permutation (x1x2x3x4)?
19. A striped necktie has 12 bands of color. Each band can be colored by one of four possible colors. How many possible different-colored neckties are there?
20. A group acts faithfully on a G-setX if the identity is the only element of G that leaves every element of X fixed. Show that G acts faithfully on X if and only if no two distinct elements of Ghave the same action on each element of X.
21. Letpbe prime. Show that the number of different abelian groups of order pn (up to isomorphism) is the same as the number of conjugacy classes in Sn.
22. Leta∈G. Show that for anyg∈G,gC(a)g−1 =C(gag−1).
23. Let|G|=pn be a nonabelian group for pprime. Prove that |Z(G)|< pn−1. 24. LetGbe a group with orderpnwherepis prime and X a finiteG-set. IfX
G={x∈
X:gx=x for all g∈G} is the set of elements inX fixed by the group action, then prove that|X| ≡ |XG| (mod p).
25. If G is a group of order pn, where p is prime and n≥2, show thatG must have a proper subgroup of orderp. Ifn≥3, is it true thatGwill have a proper subgroup of order p2?