What is a semigroup? What is a group? What is the difference between a semigroup and a group?

The second exam will be on Thursday, July 5, 2012. The syllabus will be Sections IV.5 (RSA Encryption), III.1, III.2, III.3, III.4 and III.8, III.9, plus the handout on Burnside coloring arguments. Of course, the material is cumulative, and the listed sections depend on earlier sections, which it is assumed that you still know. Following are some of the concepts and results you should know: What is a semigroup? What is a group? What is the difference between a semigroup and a group? Know examples of groups such as S n, D 4, Z, Z n, cyclic groups. In particular, know what elements of these groups are and what the group operation is in each case. Know the order of each of the groups which is finite. Know the cancelation rules in a group. For example, if ab = ac then b = c. Exponential rules in groups. For example, a m a n = a m+n, (a m ) n = a mn, and (ab) 1 = b 1 a 1. What does it mean to be a subgroup? Know the criterion to be a subgroup (Proposition 3.8) and how to use it to check that H is a subgroup of a group G. What is the center Z(G) of a group G? What is an abelian group? What is a cyclic group? What is a generator of a cyclic group? What is the condition for an element a Z n to have a multiplicative inverse? (Answer: a and n should be relatively prime integers. When a has a multiplicative inverse, know how to find it using the Euclidean algorithm. What is the order of a group (denoted G )? What is the order o(g) of an element g G? (o(g) is the smallest positive integer m such that g m = e.) If o(g) = m, then g n = e if and only if m n. Every subgroup of a cyclic group is cyclic (Proposition 3.19). The left cosets of a subgroup H are the sets ah = {ah : h H}. They are precisely the equivalence classes of the equivalence relation a H b a 1 b H. The set of left cosets of H in G is denoted G/H. Lagrange s Theorem: [G : H] = G / H. G / H. In particular, H G. That is, the number of cosets of H is exactly The order of every element of a group divides the order of the group (Corollary 4.9). If the order of G is prime, then G is cyclic (Corollary 4.11). S n denotes the set of all permutations of the set {1,..., n} of integers from 1 to n. cardinality of S n is S n = n!. The Know how to represent permutations in the two rowed notation, and how to multiply permutations using this notation. Know what a cycle of length r is. A cycle of length 2 is a transposition. 1

Know what it means to say that two cycles π and σ are disjoint. Disjoint cycles commute. (Lemma 8.4, Page 109) Know how to compute the cycle decomposition of permutations in S n. Know how to go back and forth between two rowed notation for permutations and cycle decompositions. Know how to multiply permutations given in either format and express the result in either two rowed or cycle notation. Know what is meant by the order of a permutation: o(π) is the smallest positive integer k such that π k = id. That is, o(π) is the order of π as an element of the group S n The order of an r-cycle is r. Know how to compute the order of a permutation from the cycle structure: If π = τ 1 τ 2 τ k is a product of disjoint cycles, then the order of π is the least common multiple of the lengths of the cycles τ 1,..., τ k. A transposition is a cycle of length 2. Every permutation is a product of transpositions. The number of transpositions in such a product for a permutation σ is always even or always odd. σ is even if it is a product of an even number of transpositions; σ is odd if it is a product of an odd number of transpositions. An r-cycle (j 1, j 2,..., j r ) is an even permutation if r is odd and it is an odd permutation if r is even. This follows from the factorization (j 1 j 2... j r ) = (j 1 j r )(j 1 j r 1 ) (j 1 j 2 ). The alternating group A n S n is the subgroup of all even permutation. The order of A n is A n = n!/2. Know what it means to say that a group G acts on a set X (Definition 9.2, Page 118). Namely, each element g of G determines a permutation of X, denoted by x gx, in such a way that: (a) ex = x for all x X, where e is the identity of G. That is the permutation of X determined by e is the identity permutation. (b) (gh)x = g(hx) for all g, h G and x X. That is, the multiplication of g and h in G corresponds to the permutation of X that is the composition of the permutation determined by h and that determined by g. If G is a group acting on a set X and a X, know what the orbit of a under the action of G is. Namely, Orb(a) = {ga : g G}. That is, the orbit of a consists of all of the elements of X that are the image of a under one of the permutations of X determined by an element of G. Know what the stabilizer of a, denoted G a, is: G a = {g G : ga = a}. That is the stabilizer of a consists of all the group elements g that do not move a under the permutation of X determined by g. 2

If G acts on a set X, then the orbits of this action form a partition of X. If G acts on a set X, then the stabilizer G a of each element a X is a subgroup of G. Know the Orbit-Stabilizer theorem for group actions (Proposition 9.11, Page 121): If G acts on X and a X is any element of X, then Orb(a) = [G : G a ] = G G a = the number of cosets of G a in G. Another way to state the same thing is: Orb(a) G a = G. If G is a group acting on the set X and g G, know what the fixed set of g, denoted X g is: X g = {x X : gx = x}. That is, X g is the set of elements of X that are not moved by the permutation of X determined by g. Know the Burnside counting theorem (Theorem 9.15, Page 122: If G is a finite group acting on a finite set X, then the number N of orbits in X under this action of G is N = 1 G X g g G where X g is the number of elements in X that are fixed by g. A convenient way to think about this theorem is that it says that the number of orbits is the average (as g varies over G) of the number of elements of X fixed by an element g of G. Know how to use Burnside s theorem to compute the number of distinct patterns in the colorings of a given set of points in the plane, using the action of the symmetry group of the points. Review Exercises Be sure that you know how to do all assigned homework exercises. The following are a few supplemental exercises similar to those already assigned as homework. These exercises are listed randomly. That is, there is no attempt to give the exercises in the order of presentation of material in the text. 1. If a, b, and c are elements of a group G, then solve the equation axb = c for x G. Solution. Multiply on the left by a 1 and on the right by b 1 to get x = exe = a 1 axbb 1 = a 1 cb 1. 2. Let α = S 5. 5, β = 2 5 1 3 4 (a) Solve for γ: αγ = β. (b) Solve for δ: δα = β. (c) Solve for ν: ανα 1 = β. 5. Solve each of the following equations in 5 1 3 2 4 3

5 5 5 Solution. γ = α 1 β = =. 3 1 4 5 2 5 1 3 2 4 2 3 4 1 5 5 5 5 δ = βα 1 = =. 5 1 3 2 4 3 1 4 5 2 3 5 2 4 1 5 5 5 ν = α 1 βα = γα = =. 2 3 4 1 5 2 5 1 3 4 3 5 2 4 1 3. Let G = {1, 1, i, i} C. Recall that C is the multiplicative group of nonzero complex numbers. (a) Verify that G is a subgroup of C. (Constructing a Cayley table for G may be useful.) (b) Verify that S = {1, i} is not a subgroup of G. (c) Verify that G is cyclic and list all the generators of G. Solution. The following is the multiplication table for G: 1 1 i i 1 1 1 i i 1 1 1 i i i i i 1 1 i i i 1 1 From the table it is clear that G is closed under multiplication so G is a subgroup of C Since i 2 = 1 / S, S is not closed under multiplication, and hence S is not a subgroup of G. Since (1) = {1}, ( 1) = {1, 1}, (i) = {i, 1, i, 1} = G, and ( i) = { i, 1, i, 1} = G, it follows that i and i are generators for G, while 1 and 1 are not. 4. Suppose that G is the group defined by the following Cayley table. (a) Which element is the identity of G? (b) What is g 1? (c) Find C(g) = {x G : xg = gx}. (d) Find the order of d and the order of g. Solution. a b c d e f g h a a b c d e f g h b b a h g f e d c c c d e f g h a b d d c b a h g f e e e f g h a b c d f f e d c b a h g g g h a b c d e f h h g f e d c b a (a) The identity of G is a since (from the table) ax = x = xa for all x G. (b) g 1 = c since gc = cg = a and a is the identity of G. 4

(c) C(g) = {a, c, e, g}. (d) d = { d, a = d 2} so o(d) = d = 2, and g = { g, e = g 2, c = g 3, a = g 4} so o(g) = g = 4. 5. (a) If G is a group with G = 21, what are the possible orders of elements of G? (b) If G = 21, and H is a subgroup of G other than G itself, explain why H must be cyclic. (c) Let K be a subgroup of a group L and let L be a subgroup of a group M. What are the possible values of L if K = 6 and M = 72? Solution. (a) The order of an element of a group must divide the order of the group (Corollary 4.9, Page 78). Hence the possible orders of elements of G are the divisors of G = 21, i.e., 1, 3, 7, and 21. (b) The possible orders for H are 1, 3, and 7 by Lagrange s Theorem. H = 21 since H = G by assumption. Thus H is 1, in which case H = {e} which is clearly cyclic, or H = 3 or 7 which are both prime numbers and thus H is cyclic by Corollary 4.11, Page 78. (c) Since K is a subgroup of L, L is a multiple of K = 6 and since L is a subgroup of M, L divides M = 72. Thus L is a multiple of 6 which also divides 72, so the possibilities for L are 6, 12, 13, 24, 36, and 72. 6. Let α = ( 5 6 ) and β = ( 1 2 ) ( 4 5 6 ). (Note that we are using the cycle notation for α and β which we are assuming are in S 6. (a) Write α and β in two rowed notation. (b) Express α 1 and α 2 as products of disjoint cycles. (c) Write β as a product of 2 cycles. Is β even or odd? (d) Let θ = στ where σ and τ are disjoint cycles of length 9 and 6, respectively. Find the smallest positive integer s such that θ s is the identity permutation. Solution. (a) α = 5 6 and β = 2 3 4 5 6 1 (b) α 1 = ( 1 6 5 4 3 2 ) and α 2 = ( 1 3 5 ) ( 2 4 6 ). (c) β = ( 1 2 ) ( 4 6 ) ( 4 5 ). Hence β is an odd permutation. 5 6. 2 1 3 5 6 4 (d) The smallest possible s is the least common multiple of 6 and 9, i.e. 18. (See Lemma 8.7, Page 110.) 7. Write each of the following permutations as a product of disjoint cycles. (Remember that a single cycle qualifies.) 5 6 (a) 3 5 6 4 2 1 5

(f) ( 1 2 ) 5 6 (b) 1 6 4 5 3 2 (c) ( 1 2 ) ( 1 3 ) ( 1 4 ) (d) ( 1 3 ) 1 ( 2 4 ) ( 2 3 5 ) 1 (e) ( 1 4 5 ) ( ) ( 1 3 ) (f) ( 1 2 3 ) 1 ( 2 3 ) ( 1 2 3 ) Solution. (a) ( 1 3 6 ) ( 2 5 ) (b) ( 2 6 ) ( 3 4 5 ) (c) ( 1 4 3 2 ) (d) ( 1 3 4 2 5 ) (e) ( 1 5 ) ( 2 3 ) 8. Determine all the left cosets of H in G if: (a) G = Z 24, H = ( 4). (b) G = S 3, H = ( ( 2 3 ) ). (c) G = D 4, H = (β) where β is the reflection about the horizontal centerline of a square. (See Page 58). Solution. (a) H = { 0, 4, 8, 12, 16, 20} 1H = { 1, 5, 9, 13, 17, 21} 2H = { 2, 6, 10, 14, 18, 22} 3H = { 3, 7, 11, 15, 19, 23} (b) H = { ε, ( 2 3 )} ( 1 2 ) H = {( 1 2 ), ( 1 2 3 )} ( 1 3 ) H = {( 1 3 ), ( 1 3 2 )} (c) H = {ε, β} αh = {α, αβ} α 2 H = { α 2, α 2 β } α 3 H = { α 3, α 3 β } 6

(d) ( 1 4 ) 9. Let H = 5Z in Z. Then H is a subgroup of the additive group Z. Determine whether the following cosets of H are the same. (Remember the group operation is + so cosets will be written a + H, rather than ah.) (a) 12 + H and 27 + H (b) 13 + H and 2 + H (c) 126 + H and 1 + H Solution. (a) 27 12 = 15 = 5 3 H. Thus the two cosets are equal. (b) 2 13 = 15 H. Thus the two cosets are equal. (c) 1 126 = 127 / H. Thus the two cosets are not equal. 10. Assume that σ = and τ = are permutations in S 1 4 3 2 3 1 4 2 4. Compute each of the following elements of S 4 : (a) στ (b) τσ (c) σ 2 (d) τ 2 (e) τ 3 (f) τ 4 (g) σ 1 (h) τ 1 Solution. στ =, τσ =, σ 3 1 2 4 3 2 4 1 2 =, τ 2 =, τ 4 3 2 1 3 =, τ 2 4 1 3 4 =, σ 1 =, 1 4 3 2 τ 1 = 2 4 1 3 11. Write each of the following permutations as a single cycle or a product of disjoint cycles. 5 6 (a) (b) 1 4 1 5 1 2 3 5 1 6 4 5 3 2 1 (c) 1 2 3 2 3 1 2 3 (d) 2 4 5 1 3 5 4 1 2 3 Solution. (a) ( 2 6 ) ( 3 4 5 ) (b) ( 1 2 5 3 4 ) (c) ( 1 2 ) 12. In S 10, let α = (1, 3, 5, 7, 9), β = (1, 2, 6), γ = (1, 2, 5, 3), and let σ = αβγ. Write σ as a product of disjoint cycles, and use this to find its order and its inverse. Is σ even or odd? Solution. σ = (1, 6, 3, 2, 7, 9). Thus σ is a 6-cycle so the order of σ is 6, σ is odd, and σ 1 = (1, 9, 7, 2, 3, 6). 13. Show that S 10 has elements of order 10, 12, and 14, but not 11 or 13. 7

Solution. If α = (1, 2)(3, 4, 5, 6, 7), β = (1, 2, 3)(4, 5, 6, 7), γ = (1, 2)(3, 4, 5, 6, 7, 8, 9), then the order of α is 10, the order of β is 12 and the order of γ is 14 (see Theorem 4.2.6). Since 11 is prime, any permutation of order 11 must be an 11-cycle or a product of disjoint 11-cycles. In S(10) there are no cycles of length 11. Hence there can be no elements of S(10) of order 11. The same argument applies to 13 since 13 is also a prime bigger than 10. 14. Write each of the following permutations as a product of disjoint cycles. (a) (1, 2, 3)(1, 4, 5) (b) (1, 2, 3, 4)(1, 5, 6, 7) (c) (1, 2, 3, 4, 5)(1, 6, 7, 8, 9) (d) (1, 4)(2, 4)(3, 4)(1, 2)(2, 4)(2, 3) 15. Let α = (1, 3)(1, 5)(1, 6)(2, 1)(2, 4)(2, 6) and β = (1, 2, 5)(3, 2, 6)(1, 4). Find the disjoint cycle factorizations of α, α 1, β, and β 99 and give the order of each of these four permutations. Solution. α = (1, 2, 5, 3)(4, 6), α 1 = (1, 3, 5, 2)(4, 6), and β = (1, 4, 2, 6, 3, 5). Since β is a 6-cycle, it has order 6. Since 99 = 6 16 + 3, β 99 = β 3 = (1, 6)(2, 5)(4, 3). The order of α and α 1 is lcm {4, 2} = 4. Since β 99 is a product of disjoint 2-cycles, the order is 2. 16. Count the number of ways that a 3 3 grid of squares can be colored red or black (adjacent squares are allowed to have the same color). We will assume that two colorings are the same when one can be obtained from the other by clockwise rotation by 90, 180, or 270. It may be helpful to use the following two grids for visualization purposes. 1 2 3 4 5 6 7 8 9 7 4 1 8 5 2 9 6 3 The grid on the right is obtained from the one on the left by rotation ρ by 90, so that ρ can be identified with the following permutation of the nine small squares: ρ = ( 1 3 9 7 ) ( 2 6 8 4 ) ( 5 ) The following represents a particular coloring of the grid and the coloring obtained by 90 clockwise rotation. Thus these two colorings will be considered the same. R B B R R B B R R B R R R R B R B B Solution. This is an application of Burnside s theorem (Theorem III (9.15)). Let X be the set of all possible ways to color the 3 3 grid, without taking into account any symmetry. Thus, each of the 9 small squares can be colored Red or Black, independent of any other square, so that X = 2 9. The identification of colorings that we are using means that the group G of rotations by 90, 180, 270, and 360 acts on X, and the number of distinct colorings is the number N of orbits of this group action. Since G = { ε, ρ, ρ 2, ρ 3}, we need to 8

compute the representation of each element of G as a permutation of the small squares in the grid: ρ = ( 1 3 9 7 ) ( 2 6 8 4 ) ( 5 ) ρ 2 = ( 1 9 ) ( 3 7 ) ( 2 8 ) ( 6 4 ) ( 5 ) ρ 3 = ( 1 7 9 3 ) ( 2 4 8 6 ) ( 5 ) Now we have to compute the number of elements in each fixed point set X g (g G). Here we observe that for a coloring to be fixed by g, each of the squares labeled with an index from a cycle in the disjoint cycle factorization of g must be assigned the same color. Thus, X g = 2 r where r is the number of disjoint cycles in the factorization of g. Hence we have the following table: g X g ε 2 9 ρ 2 3 ρ 2 2 5 ρ 3 2 3 Now apply Burnside s theorem to compute N: N = 1 G g G X g = 1 4 (29 + 2 3 + 2 5 + 2 3 ) = 560 4 = 140. 17. Find the number of different regular pentagons with vertices colored red, white, or blue. Solution. The group of symmetries of a regular pentagon is D 5 = { id, ρ,, ρ 2, ρ 3, ρ 4, β, ρβ, ρ 2 β, ρ 3 β, ρ 4 β }, which consists of the identity, four nontrivial rotations (the powers of ρ) and five reflections. Thus, D 5 = 10. The identity fixes all 3 5 = 243 possible 3-colored pentagons. Each nontrivial rotation will fix only those pentagons that have all vertices the same color, hence the fixed set of each nontrivial rotation contains 3 elements. Each reflection will fix a pentagon provided the corresponding vertices are colored the same. Thus there are 3 3 = 27 3-colored pentagons fixed by each reflection. By Burnside s Theorem, the number of distinct orbits is (243 + 4 3 + 5 27)/10 = 39. 18. A wheel is divided evenly into four compartments. Each compartment can be painted red, white, or blue. The back of the wheel is black. How many different such color wheels are there? Solution. The group of symmetries is a cyclic group of order 4, G = { id, α, α 2, α 3} where α denotes a rotation by 90. We have X id = 3 4, X α = 3 = X α 3, and X α 2 = 3 2. By Burnside s theorem, the number of different wheels is (81 + 6 + 9)/4 = 24. 19. A rectangular necktie is divided evenly into five bands and each band may be colored green, red, or blue. How many different neckties are there? 9

Solution. The symmetry group of the tie is the cyclic group G = {id, τ} of order 2 where τ is the rotation by 180. Then X id = 3 5 = 243, and X τ = 3 3 = 27 since the first and last bands must be colored the same and the second and fourth bands must be colored the same, while the center band can be any color. Thus, by Burnside s theorem the number of different neckties is (243 + 27)/2 = 135. 20. A rectangular electrical relay box appears as shown below: If each wire may be red or black, how many such boxes, taking symmetry into account, are there? If each wire may be red, black, or white, how many such boxes are there? Solution. The symmetry group of the rectangle is G = {id, α, β, γ} where α is reflection about the vertical axis through the middle of the box, β is reflection about the horizontal axis through the middle of the box, and γ = αβ is rotation through 180 through the center of the box. Then X id = 2 6 = 64, X α = 2 3 = 8, X β = 2 4 = 16 (since the single wires on each end can be either color), and X γ = 2 3 = 8. Thus, there are (64 + 8 + 16 + 8)/4 = 24 different boxes. If we allow 3 possible colors for each wire, then the number of different boxes is (3 6 + 3 3 + 3 4 + 3 3 )/4 = 216. 21. Think of the set of all six-digit binary words that is, strings of length 6 composed of 0 s and 1 s. For example: 010110. In some applications, two six-digit words are considered equivalent if one can be obtained from the other by applying the cyclic permutation σ: a 1 a 2 a 3 a 4 a 5 a 6 a 6 a 1 a 2 a 3 a 4 a 5 a number of times. For example, 010110 is equivalent to 110010 since the second string is obtained from the first by applying σ 3 times. Find the number of such non-equivalent words. Solution. This problem is the same type of application of Burnside s theorem as the previous exercise. In this case X is the set of all possible binary words of length 6, and G is the cyclic group generated by the 6-cycle σ = ( 5 6 ). Hence the non-identity elements of G are: σ = ( 5 6 ) σ 2 = ( 1 3 5 ) ( 2 4 6 ) σ 3 = ( 1 4 ) ( 2 5 ) ( 3 6 ) σ 4 = ( 1 5 3 ) ( 2 6 4 ) σ 5 = ( 1 6 5 4 3 2 ) 10

The following table then lists all the sizes of fixed point sets for G acting on X. g X g ε 2 6 σ 2 σ 2 2 2 σ 3 2 3 σ 4 2 2 σ 5 2 The number of equivalent words is then the number N of orbits under of action of G on X. By Burnside s theorem, this is N = 1 G X g = 1 6 (26 + 2 + 2 2 + 2 3 + 2 2 + 2) = 84 6 = 14. g G 11