Groups. Chapter 1. If ab = ba for all a, b G we call the group commutative.

Chapter 1 Groups A group G is a set of objects { a, b, c, } (not necessarily countable) together with a binary operation which associates with any ordered pair of elements a, b in G a third element ab in G (closure) The binary operation (called group multiplication) is subject to the following requirements: 1) There exists an element e in G called the identity element (also called neutral element) such that eg = ge = g for all g G ) For every g G there exists an inverse element g 1 in G such that gg 1 = g 1 g = e 3) Associative law The identity (ab)c = a(bc) is satisfied for all a, b, c G If ab = ba for all a, b G we call the group commutative If G has a finite number of elements it has finite order n(g), where n(g) is the number of elements Otherwise, G has infinite order Lagrange theorem tells us that the order of a subgroup of a finite group is a divisor of the order of the group If H is a subset of the group G closed under the group operation of G, and if H is itself a group under the induced operation, then H is a subgroup of G Let G be a group and S a subgroup If for all g G, the right coset Sg := { sg : s S } 1

Problems and Solutions is equal to the left coset gs := { gs : s S } then we say that the subgroup S is a normal or invariant subgroup of G A subgroup H of a group G is called a normal subgroup if gh = Hg for all g G This is denoted by H G We also define ghg 1 := { ghg 1 : h H } The center Z(G) of a group G is defined as the set of elements z G which commute with all elements of the group, ie Z(G) := { z G : zg = gz for all g G } Let G be a group For any subset X of G, we define its centralizer C(X) to be C(X) := { y G : xy = yx for all x X } If X Y, then C(Y ) C(X) A cyclic group G is a group containing an element g with the property that every other element of G can be written as a power of g, ie such that for all h G, for some n Z, h = g n We then say that G is the cyclic group generated by g Let (G 1, ) and (G, ) be groups A function f : G 1 G with f(a b) = f(a) f(b), for all a, b G 1 is called a homomorphism If f is invertible, then f is an isomorphism Groups have matrix representations with invertible n n matrices and matrix multiplication as group multiplication The identity element is the identity matrix The inverse element is the inverse matrix An important subgroup is the set of unitary matrices U, where U = U 1 Let GL(n, F) be the group of invertible n n matrices with entries in the field F, where F is R or C Let G be a group A matrix representation of G over the field F is a homomorphism ρ from G to GL(n, F) The degree of ρ is the integer n Let ρ : G GL(n, F) Then ρ is a representation if and only if ρ(g h) = ρ(g)ρ(h) for all g, h G

Groups 3 Problem 1 Let x, y R Does the composition x y := 3 x 3 + y 3 define a group? Solution 1 Yes We have 3 x 3 + y 3 R The neutral element is 0 The inverse element of x is x The associative law (x y) z = ( 3 x 3 + y 3 ) z = 3 ( 3 x 3 + y 3 ) 3 + z 3 = 3 x 3 + y 3 + z 3 = 3 x 3 + ( 3 y 3 + z 3 ) 3 = x (y z) also holds The group is commutative since x y = y x for all x, y R Problem Let x, y R \ {0} and denotes multiplication in R Does the composition x y := x y define a group? Solution Yes We have (x y)/ R \ {0} The neutral element is and 4/x is inverse to x The associative law also holds The group is commutative Problem 3 Let x, y R Is the composition x y := x + y associative? Here denotes the absolute value Solution 3 The answer is no We have 0 = ( 1 + ( 1) ) + 0 1 + ( ( 1) + 0 ) = Problem 4 Consider the set G = { (a, b) R : a 0 } We define the composition (a, b) (c, d) := (ac, ad + b) Show that this composition defines a group Is the group commutative?

4 Problems and Solutions Solution 4 The composition is associative The neutral element is (1, 0) The inverse element of (a, b) is (1/a, b/a) The group is not commutative since (a, b) (c, d) = (ac, ad + b) and (c, d) (a, b) = (ca, cb + d) Problem 5 Show that the set { +1, 1, +i, i } forms a group under multiplication Find all subgroups Solution 5 The multiplication table is 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 we find that the group is commutative We can also deduce this property from the commutativity of complex multiplication From the table we find that the inverse elements are 1 1 = 1, ( 1) 1 = 1, (i) 1 = i, ( i) 1 = i The associative law follows from the associative law for multiplication of complex numbers Thus the set {+1, 1, +i, i} forms a group under multiplication We classify subgroups by their orders There is only one subgroup of order 1, the trivial group - {1} Subgroups of order contain two elements - the identity element e = 1 and one additional element a There are two possibilities: a = e or a = a The first possibility provides one subgroup: {1, 1} which is commutative The element ( 1) is inverse to itself, it is in involution The second possibility reduce to the trivial group {1} There are no subgroups of order 3, because the order of a subgroup must divide the order of the group (Lagrange s theorem) There is only one subgroup of order 4 - this is the group itself Therefore the group has in total 1 + 1 + 1 = 3 subgroups with one proper subgroup Problem 6 Let i = 1 Let S be the set of complex numbers of the form q + pi 5, where p, q Q and are not both simultaneously 0 Show that this set forms a group under multiplication of complex numbers Solution 6 Consider the product of two numbers q 1 + ip 1 5 and q + ip 5 q 3 + ip 3 5 = (q1 + ip 1 5)(q + ip 5) = q1 q 5p 1 p + i 5(q 1 p + p 1 q )

Groups 5 Thus and q 3 = q 1 q 5p 1 p Q, p 3 = q 1 p + p 1 q Q q 3 + 5p 3 = (q 1 + 5p 1)(q + 5p ) > 0 Therefore the product of two numbers of the form q + ip 5 belongs to the same set The identity element is 1, ie q = 1 and p = 0 The inverse element is q ip 5 q + 5p = q p q + 5p + q + 5p i 5 The existence of the inverse element follows from the fact that p, q Q and are not both simultaneously 0 Thus the inverse element belongs to the same set The associative law follows from the same law of multiplication of complex numbers Thus the set of numbers of the form q + ip 5, where p, q Q forms an abelian group under multiplication Problem 7 Let p be a prime number with p 3 Let r and s be rational numbers (r, s Q) with r + s > 0 Show that the set given by the numbers r + s p form a commutative group Solution 7 Associativity and commutativity follow from the multiplication of real numbers We have (r 1 + s 1 p)(r + s p) = (r1 r + ps 1 s ) + (r 1 s + r s 1 ) p Since r 1 r + ps 1 s Q and r 1 s + r s 1 Q the operation is closed The neutral element is 1, ie r = 1 and s = 0 The inverse element is r s p (r + s p)(r s p) = r p p r s p + r s p Problem 8 Show that the set { e iα : α R } forms a group under multiplication Note that e iα = 1 Solution 8 We have e iα e iβ = e i(α+β) The neutral element is given by α = 0, ie e 0 = 1 The inverse element of e iα is e iα The associative and commutative laws follow from the same

6 Problems and Solutions laws of multiplication for complex numbers Thus the set {e iα : α R} forms a 1-dimensional group under multiplication of complex numbers Problem 9 Consider the additive group (Z, +) Give a proper subgroup Solution 9 An example of a proper subgroup would be all even numbers, since the sum of two even numbers is again an even number Problem 10 Let S := R \ { 1} We define the binary operation on S a b := a + b + ab Show that S, forms a group Is the group commutative? Solution 10 When a, b are elements of S, then a b is an element of S Suppose instead that a + b + ab = 1 Then a(1 + b) = (1 + b) Thus b = 1 or a = 1 Since a 1 and b 1 we have a + b + ab 1 The neutral element is e = 0 since a e = a + 0 + a0 = a, e b = 0 + b + 0b = b To find the inverse of an arbitrary element g we consider right multiplication by its inverse g 1 We obtain g g 1 = g + g 1 + gg 1 = 0 g 1 = g/(1 + g) All elements are invertible, because g 1 The associative law holds (a b) c = (a + b + ab) c = (a + b + ab) + c + (a + b + ab)c = a + b + c + (ab + bc + ca) + abc = a (b c) The group is commutative since a b = b a Problem 11 Let G be a group and x, y G Show that (xy) 1 = y 1 x 1 Solution 11 Let e be the neutral element of G We have xyy 1 x 1 = xex 1 = xx 1 = e and y 1 x 1 xy = y 1 ey = y 1 y = e

Groups 7 Problem 1 Let S be the set of all rational numbers Q in the interval 0 q < 1 Define the operation (q, p S) q p := { q + p if 0 q + p < 1 q + p 1 if q + p 1 Show that S with this operation is an abelian group Solution 1 The group neutral element is 0, ie p + 0 = 0 + p = p The inverse element is p 1 = 1 p since p + p 1 = 1 1 p p 1 = 1 1 = 0 The inverse element exists for all p, because p < 1 Associativity follows from the associative law for addition of rational numbers Commutativity follows from the commutativity of addition of rational numbers We can write each nonzero element of the group as the proper fraction p = a, a, b N, a < b b where the fraction is irreducible (a and b have no common divisors) Problem 13 Show that the finite set Z n := { 0, 1,, n 1 } for n 1 forms an abelian group under addition modulo n The group is referred to as the group of integers modulo n Solution 13 The neutral element is 0 The inverse of 0 is 0 For any j > 0 in Z n the inverse of j is n j Obviously the associative law holds The group law is a + b < n a b = a + b a + b n a b = a + b n, a, b Z n The expressions are symmetric in the two arguments a, b Thus the group is abelian Problem 14 Let n N Let U(n) be the set of all positive integers less than n and relatively prime to n Then U(n) is a commutative group under multiplication modulo n Find the group table for U(8) Solution 14 Obviously 1 and n 1 are elements of U(n), where 1 is the neutral element For n = 8 we have U(8) = { 1, 3, 5, 7 } The group table is

8 Problems and Solutions mod 8 1 3 5 7 1 1 3 5 7 3 3 1 7 5 5 5 7 1 3 7 7 5 3 1 Problem 15 Consider the subset of odd integers { 1, 3, 7, 9, 11, 13, 17, 19 } Show that this set forms an abelian group under multiplication modulo 0 Solution 15 The multiplication table modulo 0 reads 1 3 7 9 11 13 17 19 1 1 3 7 9 11 13 17 19 3 3 9 1 7 13 19 11 17 7 7 1 9 3 17 11 19 13 9 9 7 3 1 19 17 13 11 11 11 13 17 19 1 3 7 9 13 13 19 11 17 3 9 1 7 17 17 11 19 13 7 1 9 3 19 19 17 13 11 9 7 3 1 The table shows that products of each two elements of the set is again an element of the set The identity element is 1 - implied by the same property of modulo 0 multiplication for all integer numbers (or from the multiplication table) The inverse elements are 1 1 = 1, 3 1 = 7, 7 1 = 3, 9 1 = 9, 11 1 = 11, 13 1 = 17, 17 1 = 13, 19 1 = 19 The associative law for modulo 0 multiplication for all integer numbers implies the same law for the set in question Problem 16 Solution 16 group table is Give the group table for the cyclic group Z 6 of 6 elements The neutral element is 0 The group is commutative The

Groups 9 + 3 4 5 0 3 4 5 1 1 3 4 5 0 3 4 5 3 3 4 5 4 4 5 3 5 5 3 4 Problem 17 Consider the finite group Z Z 3 which has 3 = 6 elements given by (0, 0), (0, 1), (0, ), (1, 0), (1, 1), (1, ) The neutral element is (0, 0) Show that Z Z 3 is cyclic Solution 17 It is only necessary to find a generator We start with (1, 1) Then (1, 1) = (1, 1) (1, 1) = (1, 1) + (1, 1) = (0, ) 3(1, 1) = (1, 1) + (1, 1) = (1, 0) 4(1, 1) = 3(1, 1) + (1, 1) = (1, 0) + (1, 1) = (0, 1) 5(1, 1) = 4(1, 1) + (1, 1) = (0, 1) + (1, 1) = (1, ) 6(1, 1) = 5(1, 1) + (1, 1) = (1, ) + (1, 1) = (0, 0) Therefore the element (1, 1) generates all elements of the commutative group Z Z 3 Problem 18 Consider the functions defined on R \ { 0, 1 } f 1 (x) = x, f (x) = 1 x, f 3(x) = 1 x, f 4 (x) = x x 1, f 5(x) = 1 1 x, f 6(x) = 1 1 x Show that these functions form a group with the function composition f j f k, where (f j f k )(x) := f j (f k (x))

10 Problems and Solutions Solution 18 The neutral element is f 1 The group table is f 1 f f 3 f 4 f 5 f 6 f 1 f 1 f f 3 f 4 f 5 f 6 f f f 1 f 5 f 6 f 3 f 4 f 3 f 3 f 6 f 1 f 5 f 4 f f 4 f 4 f 5 f 6 f 1 f f 3 f 5 f 5 f 4 f f 3 f 6 f 1 f 6 f 6 f 3 f 4 f f 1 f 5 For example (f 3 f 4 )(x) = f 3 (f 4 (x)) = 1 x x 1 = 1 1 x = f 5(x) (f 4 f 3 )(x) = f 4 (f 3 (x)) = 1 x 1 x 1 = 1 1 x = f 6(x) Each element has an inverse The associativity law holds for function composition The group is not commutative Problem 19 An isomorphism of a group G with itself is an automorphism Show that for each g G the mapping i g : G G defined by xi g := g 1 xg is an automorphism of G, the inner automorphism of G under conjugation by the group element g We have to show that i g is an isomorphism of G with itself Thus we have to show it is one to one, onto, and that for all g G (xy)i g = (xi g )(yi g ) Solution 19 If xi g = yi g, then g 1 xg = g 1 yg Thus x = y For onto, if x G, then applying the associative law yields Now (xy)i g = g 1 xyg and (gxg 1 )i g = g 1 (gxg 1 )g = x (xi g )(yi g ) = (g 1 xg)(g 1 yg) = g 1 xyg since gg 1 = e Thus (xy)i g = (xi g )(yi g ) Problem 0 Let C n be the cyclic group Show that C 6 C 3 C

Groups 11 Solution 0 The cyclic group C 3 contains three elements and the cyclic group C contains two elements Let b = c, generating the cyclic group C 3 = {e, b, b } = B where e is the neutral element Let a = c 3 generating the cyclic group C = {e, a} = A Now b and a commute and every element of the cyclic group C 6 can be written as a r b s Thus c = ab, c = b, c 3 = a, c 4 = b and c 5 = ab Therefore C 6 = A B = C 3 C Problem 1 (i) Let Z(G) be the center of the group G Show that Z(G) is a commutative subgroup of G (ii) Let a Z(G) and g G Then ag = ga Show that a 1 g = ga 1 Solution 1 have (i) The neutral element e is in Z(G) If z 1, z Z(G) we z 1 z g = z 1 gz = gz 1 z Thus Z(G) is closed under the group operation It is therefore a subgroup Furthermore it is commutative since we can take g = z in the defining relation z 1 g = gz 1 We multiply both sides of identity gz = zg by z 1 from left and right This gives z 1 g = gz 1 z 1 Z(G) The associative law follows from the same law for the group G Thus the set Z(G) forms a subgroup of group G (ii) Let e be the neutral element From ag = ga we obtain using the associative law a 1 (ag)a 1 = a 1 (ga)a 1 (a 1 a)(ga 1 ) = (a 1 g)(aa 1 ) ega 1 = a 1 ge ga 1 = a 1 g Problem Show that the center Z(G) of a group G is a normal subgroup of the group G Solution g G we have Let z 1 and z be two elements of the center Z(G) For each gz 1 z = z 1 gz = z 1 z g z 1 z Z(G) Multiplying both sides of gz = zg by z 1 from left and right yields z 1 g = gz 1

1 Problems and Solutions Thus z 1 Z(G) The associative law follows from the same law for group G Hence the set Z(G) forms a subgroup of group G The commutative law follows from zg = gz for all g G zg = gz for all g Z(G) Thus Z(G) is a commutative subgroup of the group G Let H = Z(G) We must show that H = ghg 1 for all g G Let z be in H Then we know that gz = zg Thus gzg 1 = zgg 1 = z Hence z is in ghg 1 On the other hand, suppose z is in ghg 1 Then z = zgg 1 = gzg 1 so that zg = gz and hence z is in H Hence the center Z(G) is a normal subgroup of G Problem 3 Let H be a subgroup of a group G The normalizer N(H) of H in G is defined by N(H) := { g G : ghg 1 = H } (i) Show that N(H) is a subgroup of G (ii) Show that H is a normal subgroup of N(H) Solution 3 (i) First we show that the normalizer is closed under multiplication Let g, k be two elements of N(H) Then for all h H we have (ghg 1 H) (khk 1 H) Note that for any h H (gk)h(gk) 1 = (gk)h(k 1 g 1 ) = g(khk 1 )g 1 Since k N(H) we have khk 1 = h for some h H Thus (gk)h(gk) 1 = gh g 1 Since g N(H), gh g 1 is also in H Therefore, we conclude that (gk)h(gk) 1 is in H for all h H This implies that the element gk is also in N(H) Let g be in N(H) Now for any h H ghg 1 = h g 1 h g = h Therefore when h runs over all elements of H, h also runs over all elements of H Thus for all h H g 1 h g = h H However g 1 h g = (g 1 )h (g 1 ) 1

Groups 13 Therefore, g 1 is also in N(H) Thus N(H) is a subgroup of G (ii) Let h be arbitrary element of H and n be arbitrary element of N(H) Conjugation of h by n is c = nhn 1 H by definition of the normalizer Thus H is a normal subgroup of N(H) Problem 4 Consider a group G We define the group commutator (x, y G) [x, y] := xyx 1 y 1 Consider the two matrices with determinant 1 t 1 1 b x =, y = 1 0 c 1 + bc where t, b, c R This means x and y are elements of SL(, R) Find the group commutator Solution 4 We have x 1 = ( 0 1 1 t ), y 1 = 1 + bc b c 1 Obviously these matrices are also elements of SL(, R) Thus xyx 1 y 1 = b c(c+t) b(c t+c(t ) t)+c +1 b (c + t)+b(ct + t 1) c b c + b(ct 1) b bt + 1 This matrix also has determinant equal to 1 Problem 5 Let G be a group Given two elements g 1, g G One defines the group commutator of g 1 and g to be the element g 1 g g 1 1 g 1 Consider the compact Lie group SO(, R) with cos α sin α cos β sin β g 1 (α) =, g sin α cos α 1 (β) = sin β cos β Find the group commutator Solution 5 Since the group is commutative we find g 1 (α)g (β)g1 1 1 0 (α)g 1 (β) =

14 Problems and Solutions Problem 6 (i) Show that the set of all group commutators aba 1 b 1 of a group G generates a normal subgroup G (the so-called commutator subgroup) of G (ii) Show that G/G is abelian Solution 6 (i) Obviously the commutators generate a subgroup G of G The inverse of a commutator is again a commutator since (aba 1 b 1 ) 1 = bab 1 a 1 Furthermore the identity element e is a commutator since e = eee 1 e 1 Then G consists of all of all finite products of commutators Next we show that for x G we have g 1 xg G for all g G We have to show that if x is a product of commutators, so is g 1 xg for all g G By inserting the neutral element e = gg 1 between each product of commutators occurring in x, we find that it is sufficient to show that for each commutator cdc 1 d 1 that g 1 (cdc 1 d 1 )g is in the subgroup G We have g 1 (cdc 1 d 1 )g = (g 1 cdc 1 )(e)(d 1 g) which is in G Thus G is normal in G (ii) From we find that G/G is abelian = (g 1 cdc 1 )(gd 1 dg 1 )(d 1 g) = ((g 1 c)d(g 1 c) 1 d 1 )(dg 1 d 1 g) (ag)(bg ) = abg = ab(b 1 a 1 ba)g = (abb 1 a 1 )bag = bag = (bg )(ag ) Problem 7 the relation Let H and K be two subgroups of a group G Show that g g if g = hgk for some h H, k K is an equivalence relation, partitioning G into double cosets Hg j K for g j G Solution 7 The relation is reflexive Choose h = k = e in the definition g g if g = hgk This is possible, because the subgroup H is a group Thus e H and the same for K It follows that g g The relation is symmetric Let

Groups 15 h = h 1, k = k 1 This is possible, because the subgroup H is a group, so h H h H and the same for K It follows that g g g = hgk g = h 1 g k 1 = h g k g g The relation is transitive Suppose that (for appropriate h 1, h H and k 1, k K) a b a = h 1 bk 1, b c b = h ck Then using h = h 1 h and k = k 1 k we find a = h 1 bk 1 = h 1 (h ck )k 1 = (h 1 h )c(k k 1 ) = h ck a c Here we used the associative law and the group property and the same for K h 1, h H h = h 1 h H Problem 8 Let G be a group with the composition Let H be a subgroup of G We define the relation a b a b 1 H for a, b G Is this relation an equivalence relation? Solution 8 Yes We have a a a a 1 = e H where e is the identity element of the group If a b we have a b 1 H Thus (b a 1 ) 1 H Therefore b a 1 H b a Let a b and b c Then a b 1 H and b c 1 H It follows that Hence a c (a b 1 ) (b c 1 ) = a c 1 H Problem 9 Let G be a group Let G 1 and G be two subgroups of G Show that the intersection of G 1 and G is itself a subgroup Solution 9 Obviously the identity e lies in both G 1 and G Thus G 1 G is non-empty If x, y are elements of the intersection G 1 G, they are both elements of G 1 and both elements of G Since G 1 and G are subgroups the product xy 1 lies in G 1 and G Therefore xy 1 G 1 G

16 Problems and Solutions Problem 30 The set {1, 1} forms a group G 1 under multiplication The set {0, 1} forms a group G under the XOR-operation, ie 0 0 = 0, 0 1 = 1, 1 0 = 1, 1 1 = 0 Show that the two groups are isomorphic Solution 30 Since the neutral element of the group with the XORoperation is 0, we have the isomorphism φ : G 1 G with φ(1) = 0, φ( 1) = 1 Note that all groups with elements are isomorphic Problem 31 Show that any infinite cyclic group G is isomorphic to the group Z of integers under addition Solution 31 Let g be the generator of G Then G = { g n : n Z } We define the map φ : G Z by φ(g n ) = n for all g n G If φ(g n ) = φ(g m ), then n = m and g n = g m Thus the map φ is one to one if G is not finite For any n Z, the element g n G is mapped onto n by the map φ Thus φ is onto Z Now and Thus φ(g n g m ) = φ(g n+m ) = n + m φ(g n ) + φ(g m ) = n + m φ(g n g m ) = φ(g n+m ) = n + m = φ(g n ) + φ(g m ) Problem 3 Let φ 1 : G 1 G and φ : G G 1 be homomorphisms such that φ 1 φ = φ φ 1 = i, where i is the identity map This means φ 1 φ : G 1 G 1 and φ φ 1 : G G are both the identity map The First Isomorphism Theorem states that the image of a group homomorphism, im(φ 1 ) is isomorphic to the quotient group G 1 / ker(φ 1 ) (i) Show that both φ 1 and φ are isomorphisms of G 1 with G (ii) Show that φ 1 = (φ ) 1 Solution 3 (i) One can deduce that φ 1 maps the identity element e 1 of G 1 to the identity element e of G, and it also maps inverses to inverses as φ 1 (u 1 ) = φ 1 (u) 1

Groups 17 Therefore φ 1 is compatible with the group structure We define the kernel of φ 1 to be the set of elements in G 1 which are mapped to the identity in G ker(φ 1 ) = {u G : φ 1 (u) = e } and the image of φ 1 to be im(φ 1 ) = {φ 1 (u) : u G 1 } Consider the condition φ 1 φ = φ φ 1 = i It follows that ker(φ 1 ) = e 1, im(φ 1 ) = G This means that G is isomorphic to G 1 (ii) It follows that both φ 1 and φ are isomorphisms Then φ 1 = (φ ) 1 Problem 33 Show that the finite groups G 1 := {+1, 1 : } and G = { 1 0, } : 1 0 are isomorphic, where denotes matrix multiplication Solution 33 have The identity of G 1 must map to the identity in G We φ(+1) = 1 0, φ( 1) = 1 0 Problem 34 G = Consider the matrix groups with matrix multiplication { } a b 0 a 1 : a, b R, a > 0, N = { } 1 b : b R (i) Show that N is a normal subgroup of G (ii) Show that the factor group G/N is isomorphic to the additive group (R, +) Solution 34 ( a1 b 1 0 a 1 1 (i) Matrix multiplication yields ) a b 0 a 1 = The inverse of an element of G is given by 1 a b 0 a 1 = ( a1 a a 1 b + b 1 a 1 0 a 1 1 a 1 a 1 b 0 a )

18 Problems and Solutions Setting a = 1 we see that N is a subgroup Now let 1 β N It follows that ( a b 0 a 1 ) 1 1 β a b 0 a 1 = 1 βa N Thus the subgroup N is normal (ii) Let R + be the group of real numbers under multiplication The map from G onto R + given by a b 0 a 1 a is a homomorphism whose kernel is N Consequently G/N is isomorphic to R + which is isomorphic to the additive group R Problem 35 Let G be a group Let H be a subgroup of G For any fixed g G one defines ghg 1 := { ghg 1 : h H } The group ghg 1 is called a conjugate of the subgroup H Show that if H is cyclic, then ghg 1 is cyclic Solution 35 Let H be cyclic and d H be its generator Then all elements h of H are equal to d p for some p Z Let a = gdg 1, then a ghg 1 Consider powers of the element a a p = (gdg 1 )(gdg 1 )(gdg 1 ) (gdg 1 ) = gd(g 1 g)d(g 1 g)d(g 1 g)dg 1 = gd dg 1 = gd p g 1 = ghg 1 Thus all powers of the element a are in ghg 1 Since all elements of H are of the form d p we find that all elements of ghg 1 are of the form gd p g 1 = ghg 1 Thus ghg 1 is cyclic and its generator is gdg 1 Problem 36 Let G 1, G,, G n be groups For (g 1, g,, g n ) G 1 G G n, (h 1, h,, h n ) G 1 G G n

Groups 19 we define (g 1, g,, g n ) (h 1, h,, h n ) to be (g 1 h 1, g h,, g n h n ) Show that G 1 G G n is a group under this composition (called the external direct product of the groups G j ) Solution 36 Since g j, h j G j we have g j h j G j Thus the definition given above provides that G 1 G G n is closed under the composition If e j is the identity element in G j, then (e 1, e,, e n ) is the identity element of G 1 G G n The inverse of (g 1, g,, g n ) is (g1 1, g 1,, g 1 n ) The associative law (g 1,, g n ) ((h 1,, h n ) (k 1,, k n )) is also satisfied = ((g 1,, g n ) (h 1,, h n )) (k 1,, k n ) Problem 37 Let G be a finite group A conjugacy class for a given element g of G is the set defined by C(g) := { aga 1 : a G } Show that the number of elements in a conjugacy class divides the order of G Solution 37 Let G be a finite group Then for any group element x, the elements in the conjugacy class of x are in one-to-one correspondence with cosets of the centralizer C G (x) The centralizer C G (x) of each element x is a subgroup of group G Thus the left and right cosets exist (for each x) When any two elements b and c belong to the same left coset, then b = cz for some z in the centralizer C G (x) This implies that two conjugations of x by b and c are equal bxb 1 = (cz)x(cz) 1 = cxc 1 Therefore the number of elements in the conjugacy class of x is the index [G : C G (x)] of the centralizer C G (x) in G Lagrange s theorem then implies that the size of each conjugacy class is a divisor of the size of the group Problem 38 The 3 3 permutation matrices 1 0 0 0, 0 0, 0 1 0 0 0 1 0 0 0 form a group under matrix multiplication Find the conjugacy classes

0 Problems and Solutions Solution 38 Since the group is commutative the conjugacy classes are 1 0 0 0, 0 0, 0 1 0 0 0 1 0 0 0 Problem 39 Let D n be the dihedral group with n 3 This is the finite nonabelian group of rigid motions of a regular n-gon The order of the group is n Find the center Z(D n ) := { c D n : cx = xc for all x D n } Solution 39 Let a, b be the generators of the group D n, ie D n = a, b : a n = b = e, ba = a 1 b where a is a rotation by π/n, b is a flip and e is the neutral element It suffices to find those elements which commute with the generators a and b Since n 3 it follows that a 1 a Suppose a s b Z(D n ) Hence a s+1 b = a(a s b) = (a s b)a = a s 1 b Thus a = e This is a contradiction Hence no element of the form a s b is in the center Z(D n ) Analogously, if for 1 r < n a r b = ba r = a r b then a r = e, which is possible only if r = n Thus a r commutes with b if and only if n = r If n = r, the center of D n is { e, a r } If n is odd the center is given by the neutral element { e } Problem 40 The number of elements of a group (finite or infinite) is called its order The order of G is denoted by G The order of an element g in the group G is the smallest positive integer k such that g k = e, where e is the identity element of the group It is denoted by g If no such element exists, one says that g has infinite order Consider the 3 3 permutation matrix P = 0 0 1 0 0 and matrix multiplication What is the order of P in the group of permutation matrices? What is order of the group of all 3 3 permutation matrices?

Groups 1 Solution 40 We have P = 0 1 0 0 = P T 0 and P 3 = I 3, where T denotes transpose Thus the order of the group element P is 3 The order of the group of all 3 3 permutation matrices is 3! = 6 Problem 41 Consider the set of Pauli spin matrices σ x, σ y, σ z 0 i 1 0 σ x :=, σ 1 0 y :=, σ i 0 z := 0 1 and the identity matrix I Can we extend this set so that we obtain a group under matrix multiplication? Solution 41 We have σ x = σ y = σ z = I and σ x σ y = iσ z, σ y σ x = iσ z σ y σ z = iσ x, σ z σ y = iσ x σ z σ x = iσ y, σ x σ z = iσ y Furthermore (iσ z )(iσ z ) = I, σ z (iσ z ) = ii, σ z ( iσ z ) = ii etc Thus we have to extend the set to the 16 elements { I, σ x, σ y, σ z, I, σ x, σ y, σ z, ii, iσ x, iσ y, iσ z, ii, iσ x, iσ y, iσ z } to obtain a group under matrix multiplication Note that all matrices are unitary A generating set is { σ x, σ z } Problem 4 The group SL(, R) consists of all matrices over R with determinant equal to 1 Let 0 1 0 1 A =, B = 1 0 1 1 be elements of SL(, R) What is the order of A, B and AB? Solution 4 The order of the group element A is equal to 4 We have A 1 0 = = I 0 1

Problems and Solutions and A 3 = A Thus A 4 = I The order of the matrix B is equal to 3 We have B 1 1 = 1 0 and therefore B 3 = I We have AB = 1 1 0 1 By induction we can show that (AB) n ( 1) n ( 1) = n+1 n 0 ( 1) n Thus AB has infinite order Problem 43 Consider the set N = { 1,,, n } The set of all permutations of N is called the symmetric group of degree n and is denoted by S n The elements of S n have the form 1 n s = s(1) s() s(n) The order of the group is n! (i) Consider S 3 Give all the elements of this group (ii) Is the group commutative? (iii) How many subgroups does S 3 have? Solution 43 e = r = (ii) We have and (i) We have the six elements (e is the identity) ) ( 1 3, t =, t 1 3 = 3 1 3 1 ( 1 3 1 3 1 3, tr = 1 3 tr = rt = 1 3 3 1 1 3 1 3 1 3, t r = 1 3 1 3 = 1 3 1 3 = 3 1 ), 1 3 3 1 ( 1 ) 3 1 3 1 3 3 1 Thus tr rt Therefore the group is non-commutative

Groups 3 (iii) The symmetric group S 3 has 6 subgroups There is obviously one subgroup of order 1 There are three subgroups of order, namely {e, r}, {e, tr}, {e, t r} There is one subgroup of order 3, namely {e, t, t } This is the cyclic permutation subgroup There is one subgroup of order 6, namely S 3 itself Problem 44 Consider the square where the corners are numbered counter-clockwise 1,, 3, 4 Show that the symmetry group D 4 of the square is a subgroup of the symmetric group S 4 The group S 4 is defined as the set of all permutations of {1,, 3, 4} Each element of the symmetry group D 4 transforms every corner of the square to another corner Thus it transforms every element of {1,, 3, 4} (the number of the corner) to another element Solution 44 We demonstrate the bijection between all elements of the group D 4 group and some subset of S 4 We prove that the product of every two elements of this subset belongs again to the subset Then the subset is a subgroup of S 4 The characteristic property of the elements D 4 is that they preserve some geometric shapes - edges and diagonals Two vertices having a common edge before transformation, have a common edge after transformation also, and the same for diagonals Perform two such transformations one after another and let v 1, v be two vertices, which have a common edge before the transformations This implies that after the first transformation they also have a common edge Therefore after the second transformation they possess the same property This proves that the product of every two elements of D 4 belongs again to D 4 Thus it is a subgroup of the group S 4 The relation between shapes before and after transformation is a bijection Therefore the inverse transformation in the sense of S 4 also preserves the geometric shapes Thus when some transformation belongs to D 4, its inverse (in the sense of S 4 ) transformation also belongs to D 4 Therefore the symmetry group D 4 of the square is a subgroup of the symmetric group S 4 The group is generated by the two elements s = 1 3 4, t = 3 4 1 1 3 4 1 4 3 The order of the group D 4 is 8 Problem 45 The group of all even permutations of n objects is known as the alternating group of degree n and is denoted by A n For n > 1, the alternating group has order n!/ Find all elements of the group A 3

4 Problems and Solutions Solution 45 e = Since n = 3 we have 3!/ = 3 elements The elements are 1 3, s = 1 3 1 3, t = 3 1 1 3 3 1 Expressed with permutation matrices (which permutes row vectors) we have e 1 0 0 0, 0 s 0 1 0 0, 0 t 0 0 1 0 0 Problem 46 If G is a group and g G, then H := { g n : n Z } is a subgroup of G This group is the cyclic subgroup of G generated by g Consider the 3 3 permutation matrix A = 0 0 1 0 0 Show that A, A, A 3 form a group under matrix multiplication Solution 46 From A 3 = I 3 we obtain A 1 = A = A T Thus we have a group This is the cyclic group C 3 Problem 47 Consider the 4 4 permutation matrix 0 0 0 0 A = 0 0 Let I 4 be the 4 4 identity matrix Find A, A 3, A 4 and thus show that we have a cyclic group Solution 47 We obtain 0 0 0 0 A 0 0 =, A 3 = 0 0 0 0 0 0 and A 4 = I 4 Note that A 1 = A 3, (A ) 1 = A, (A 3 ) 1 = A

Groups 5 Problem 48 (i) Consider the 4 4 matrix 0 0 0 0 A = 0 1 0 0 Obviously rank(a) = 4 and therefore the inverse exists Let I 4 be the 4 4 identity matrix Find the inverse of A (ii) Does the set { A, A 1, I 4 } form a group under matrix multiplication? If not can we find a finite extension to the set to obtain a group? (iii) Calculate the determinants of the matrices found in (ii) Show that these numbers form a group under multiplication Solution 48 (i) We find 0 0 1 0 A 1 0 0 0 1 = = A 0 0 (ii) Since A = (A 1 ) = I 4 and A 1 = A we find that the set { A, A, I 4, I 4 } forms a group under matrix multiplication (iii) We obtain +1 for all the determinants of the matrices Thus we have the trivial group Problem 49 Show that the following matrices form a group under matrix multiplication s 1 = 1 0 0 0, s = 1 0 0 0, s 3 = 0 1 0 0, 0 0 0 s 4 = 0 0, s 5 = 0 1 0 0, s 6 = 0 0 1 0 0 0 1 0 0 These matrices are the six 3 3 permutation matrices

6 Problems and Solutions Solution 49 The multiplication table is s 1 s s 3 s 4 s 5 s 6 s 1 s 1 s s 3 s 4 s 5 s 6 s s s 1 s 4 s 3 s 6 s 5 s 3 s 3 s 5 s 1 s 6 s s 4 s 4 s 4 s 6 s s 5 s 1 s 3 s 5 s 5 s 3 s 6 s 1 s 4 s s 6 s 6 s 4 s 5 s s 3 s 1 From the table we see that the element s 1 = e commutes with all other elements and only s 4 and s 5 commute: s 4 s 5 = s 5 s 4 = s 1 = e All other elements do not commute This means that the group of 3 3 permutation matrices is not commutative The inverse elements can be found from the multiplication table s 1 1 = s 1, s 1 = s, s 1 3 = s 3, s 1 4 = s 5, s 1 5 = s 4, s 1 6 = s 6 The associative law follows the from associative law for matrix multiplication Thus the set of the 3 3 permutation matrices forms a noncommutative group under matrix multiplication Problem 50 Given two manifolds M and N, a bijective map φ from M to N is called a diffeomorphism if both φ : M N and its inverse φ 1 are differentiable Let f : R R be given by the analytic function f(x) = 4x(1 x) and the analytic function g : R R be given by g(x) = 1 x (i) Can one find a diffeomorphism φ : R R such that g = φ f φ 1? (ii) Consider the diffeomorphism ψ : R R Calculate ψ g ψ 1 ψ(x) = sinh(x) Solution 50 (i) We find φ(x) = x 1 Thus the inverse is φ 1 (x) = (x + 1)/ Therefore (f φ 1 )(x) = 4 x + 1 ( 1 x + 1 ) = 1 x

Groups 7 It follows that (φ f φ 1 )(x) = ( x + 1) 1 = 1 x (ii) For the left composition we obtain (ψ g)(x) = sinh(1 x ) Then (ψ g ψ 1 )(x) = sinh ( 1 (Arsinh(x)) ) Problem 51 Let G be the dihedral group defined by D 8 := a, b : a 4 = b = e, b 1 ab = a 1 where e is the identity element in the group Define the invertible matrices 1 0 A =, B = 1 0 0 1 Let I be the identity matrix Show that A 4 = B = I, B 1 AB = A 1 and thus show that we have a representation of the dihedral group Solution 51 We have A = I Thus A 4 = I Furthermore B = I The inverse of B is given by B 1 = B Thus B 1 1 0 1 0 AB = BAB = = A 1 0 1 1 0 0 1 It follows that ρ : a j b k A j B k, j = 0, 1,, 3, k = 0, 1 is a representation of D 8 over R Problem 5 Consider the matrices ( ) 1/ 3/ C 3 = 1 0, σ 3/ 1/ 3 = 0 1 together with the identity matrix I Do these matrices form a group under matrix multiplication? If not find the smallest set of matrices with C 3, σ 3 and I as elements which forms a group under matrix multiplication Solution 5 We have σ3 = I and ( ) C3 1/ 3/ =, C3 3 = I 3/ 1/

8 Problems and Solutions Furthermore σ 3 C 3 = ( ) ( ) 1/ 3/ 1/ 3/, C 3/ 1/ 3 σ 3 = 3/ 1/ Thus σ 3 C 3 C 3 σ 3 Since σ 3 C 3 σ 3 = C 3 = C 1 3 it follows that Thus the elements (σ 3 C 3 ) = I, (C 3 σ 3 ) = I { I, σ 3, C 3, C 3, σ 3 C 3, C 3 σ 3 } form a non-commutative group under matrix multiplication Problem 53 Let c R and c 0 Show that the matrices c c A(c) = c c form a group under matrix multiplication Notice that the matrices A(c) are not invertible in the usual sense, ie det(a(c)) = 0 Find exp(a(c)) Solution 53 Multiplication of such two matrices yields c1 c 1 c c c1 c = c 1 c c 1 c 1 c c c 1 c c 1 c Thus the right-hand side is an element of the set again The neutral element is the matrix 1/ 1/ 1/ 1/ The inverse element is the matrix 1/(4c) 1/(4c) 1/(4c) 1/(4c) We obtain exp(a(c)) = I + 1 1 c j 1 1 j! j 1 j=1 Problem 54 (i) Find all matrices A over R with det(a) = a 11 a a 1 a 1 = 1 (1)

Groups 9 and A 1 1 = 1 1 () 1 1 (ii) Do these matrices form a group under matrix multiplication? Solution 54 (i) From () we find the system of linear equations a 11 + a 1 = 1, a 1 + a = 1 From det(a) = 1 we have a 11 a a 1 a 1 = 1 Eliminating a we obtain a 11 + a 1 = 1, a 11 (1 a 1 ) a 1 a 1 = 1 Eliminating a 1 we obtain a 11 a 1 = 1 with a 1 = 1 a 11 and a = a 11 Thus a11 1 a A = 11 a 11 1 a 11 with a 11 arbitrary (ii) Consider the matrices α 1 α α 1 α where α R For α = 1 one has the identity matrix I Then α 1 α β 1 β γ 1 γ = α 1 α β 1 β γ 1 γ where γ = α + β 1 The inverse of the matrix exists and is given by α α 1 1 α α Thus we have a group under matrix multiplication Problem 55 Let G be a group If a subset H of G itself is a group under the operation of G, then H is called a subgroup of G Consider the group of the 3 3 permutation matrices and matrix multiplication Find a subgroup of order 3 Solution 55 We find the abelian group 1 0 0 0, 0 0, 0 1 0 0 0 1 0 0 0

30 Problems and Solutions Problem 56 Consider finite groups Then Lagrange s theorem tells us that the order of a subgroup of a finite group is always a divisor of the order of the group Show that we cannot conclude that if G is a finite group and if m is a divisor of the order of G that G will contain a subgroup of order m Consider the subgroup A 4 with determinant +1 of the group of all 4 4 permutation matrices The order of this group is 4!/ = 1 Solution 56 We show that there is no subgroup of order 6 of the group A 4 The group A 4 has order 1 According to Lagrange s theorem the order of a subgroup of a finite group is always a divisor of the order of the group Thus a subgroup of A 4 can have order 6 However the group A 4 has no subgroup of order 6 A 4 is the smallest group demonstrating that a group does not need to not have a subgroup of every order that divides the group s order We list all elements of the group A 4 in lexicographical order 0 0 0 0 0 0 g 1 =, g 0 0 =, g 0 0 3 =, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 g 4 =, g 0 0 5 =, g 6 =, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 g 7 =, g 0 0 8 =, g 0 0 9 =, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 g 10 =, g 0 1 =, g 1 0 0 = 0 0 0 0 0 0 The multiplication table of the group A 4 is given by

Groups 31 g 1 g g 3 g 4 g 5 g 6 g 7 g 8 g 9 g 10 g 11 g 1 g 1 g 1 g g 3 g 4 g 5 g 6 g 7 g 8 g 9 g 10 g 11 g 1 g g g 3 g 1 g 6 g 4 g 5 g 8 g 9 g 7 g 1 g 10 g 11 g 3 g 3 g 1 g g 5 g 6 g 4 g 9 g 7 g 8 g 11 g 1 g 10 g 4 g 4 g 7 g 10 g 1 g 8 g 11 g g 5 g 1 g 3 g 6 g 9 g 5 g 5 g 9 g 11 g 3 g 7 g 1 g 1 g 6 g 10 g g 4 g 8 g 6 g 6 g 8 g 1 g g 9 g 10 g 3 g 4 g 11 g 1 g 5 g 7 g 7 g 7 g 10 g 4 g 11 g 1 g 8 g 5 g 1 g g 9 g 3 g 6 g 8 g 8 g 1 g 6 g 10 g g 9 g 4 g 11 g 3 g 7 g 1 g 5 g 9 g 9 g 11 g 5 g 1 g 3 g 7 g 6 g 10 g 1 g 8 g g 4 g 10 g 10 g 4 g 7 g 8 g 11 g 1 g 1 g g 5 g 6 g 9 g 3 g 11 g 11 g 5 g 9 g 7 g 1 g 3 g 10 g 1 g 6 g 4 g 8 g g 1 g 1 g 6 g 8 g 9 g 10 g g 11 g 3 g 4 g 5 g 7 g 1 We calculate orders of all elements The result is given in the table below The first column is the group element, which we use to generate the cyclic group The second column is the order of the generator In the following columns sequential powers of the generator are presented Generator Order g 1 g g 3 g 1 1 g 1 g 3 g g 3 g 1 g 3 3 g 3 g g 1 g 4 g 4 g 1 g 5 3 g 5 g 7 g 1 g 6 3 g 6 g 10 g 1 g 7 3 g 7 g 5 g 1 g 8 3 g 8 g 11 g 1 g 9 g 9 g 1 g 10 3 g 10 g 6 g 1 g 11 3 g 11 g 8 g 1 g 1 g 1 g 1 From this table we see that A 4 has cyclic subgroups of order and 3 Suppose that some element g of a group G is in its subgroup H Then all powers of g are also in H All cyclic subgroups generated by G are in H and the orders of element g in G and H are equal and 3 are prime divisors of 6 Thus the subgroup of A 4 of order 6 according to Cauchy s theorem can have cyclic subgroups of orders and 3 These cyclic subgroups must exist in A 4 The subgroups share the identity element g 1 A subgroup containing a cyclic subgroup of order m and a different cyclic subgroup of order n has order of at least 1 + (m 1) + (n 1) = m + n 1 Thus to find subgroup of order 6 we use the following algorithm: - we choose an arbitrary cyclic subgroup of order and an arbitrary cyclic subgroup of order 3;

3 Problems and Solutions - we calculate all products of elements of the first cyclic subgroup by elements of the second cyclic subgroup; - we calculate all products of elements of the second cyclic subgroup by elements of the first cyclic subgroup; - we count all distinct elements in the result If the sum is more then 6 there is not a subgroup of order 6, containing both cyclic subgroups If the sum is equal to 6 additional considerations are necessary There are no such cyclic subgroups The results can be summarized in the following table e 1 e 1 e e e 3 Count g 4 g 1 g g 3 g 1 8 g 4 g 1 g 5 g 7 g 1 8 g 4 g 1 g 6 g 10 g 1 8 g 4 g 1 g 8 g 11 g 1 8 g 9 g 1 g g 3 g 1 8 g 9 g 1 g 5 g 7 g 1 8 g 9 g 1 g 6 g 10 g 1 8 g 9 g 1 g 8 g 11 g 1 8 g 1 g 1 g g 3 g 1 8 g 1 g 1 g 5 g 7 g 1 8 g 1 g 1 g 6 g 10 g 1 8 g 1 g 1 g 8 g 11 g 1 8 In the first two columns the elements of the first cyclic subgroup of order are presented In the next three columns the elements of the second cyclic subgroup of order 3 are presented The last column contains the count of distinct products of elements of the first cyclic subgroup by elements of the second cyclic subgroup In all the cases the count is more than 6 Thus the group A 4 of order 1 does not have a subgroup of order 6 Problem 57 Do the twelve orthogonal matrices 1 0 1 0 1 0,,,, 0 1 0 1, 1 0 1 1 1, 1 1, 1 0 ( 1 1 1 1 1 ), 0 1, 1 0 ( 1 1 1 1 1 0 1, 1 0 ) ( 1 1 1, 1 1 form a group under matrix multiplication? If not add the matrices so that one has a group )

Groups 33 Solution 57 ( 1 1 1 ) 1, 1 We have to add the orthogonal matrices 1 1 1, 1 1 1 1 1, 1 1 1 1 1 1 1 to the set Then we have a non-commutative group of order 16 Problem 58 Consider the group GL(n, R) of all invertible n n matrices over R Show that the n n matrices with integer entries and determinant +1 or 1 form a subgroup of GL(n, R) Solution 58 The set of all integers is a subset of R All n n matrices with integer entries and determinant +1 or 1 are a subset of GL(n, R) of all invertible n n matrices over R If the determinant of the matrix is equal to ±1, the matrix is invertible The identity element of the group GL(n, R) is the identity matrix I n All its entries are integers and the determinant is +1 Consider two n n matrices with integer entries and determinant +1 or 1 Their product is another n n matrix with integer entries and determinant +1 or 1 The determinant of the product is equal to product of the determinants The associative law follows from the same law for matrix multiplication The inverse of n n matrix with integer entries and determinant +1 or 1 is also n n matrix with integer entries and the same determinant The formula for components of the inverse is (A 1 ) ij = ( 1) i+j det(m ji )/ det(a) where det(m ji ) is the determinant of the (n 1) (n 1) matrix that results from deleting row j and column i of the matrix A When all entries of A are integers and det(a) = ±1, all entries of (A 1 ) ij are also integers The determinant of the inverse matrix is equal to 1/ det(a) This implies det(a) = ±1 det(a 1 ) = ±1 Thus the n n matrices with integer entries and determinant +1 or 1 form a subgroup of the group GL(n, R) Problem 59 Consider the group G = { a, b, c } with group operation : G G G defined by the table a b c a a b c b b c a c c a b

34 Problems and Solutions Let M(n) denote the vector space of n n matrices over C (i) Show that f : G M(3) is a faithful representation for (G, ) where f(a) := 1 0 0 0 0 3 i 1 0 0 f(b) := 0 1 3 0 3 1 3 i 1 0 0 f(c) := 0 1 3 0 3 1 (ii) Is the representation in (i) reducible? Prove or disprove Solution 59 (i) We have f(a)f(a) = 1 0 0 0 = f(a) 0 3 i 1 0 0 f(a)f(b) = 0 1 3 = f(b) 0 3 1 3 i 1 0 0 f(a)f(c) = 0 1 3 3 1 0 3 i 1 0 0 f(b)f(a) = 0 1 3 0 3 1 f(b)f(b) = 3 i 1 0 0 0 1 3 3 1 0 f(b)f(c) = 1 0 0 0 = f(a) 0 3 i 1 0 0 f(c)f(a) = 0 1 3 0 3 1 = f(c) = f(b) = f(c) = f(c)

f(c)f(b) = 1 0 0 0 = f(a) 0 3 i 1 0 0 f(c)f(c) = 0 1 3 = f(b) 0 3 1 Thus f is a representation of the group (ii) Let be the direct sum Since f(a) = 1 0 0 0 1 0 = (1) = (1) (1) (1) 0 3 i 1 0 0 ( ) ( f(b) = 0 1 3 3 = 0 i 1 3 1 ( ( = (e πi 3 cos πi ) ) 3 sin ) πi 3 sin πi 3 cos πi 3 3 i 1 0 0 ( ) ( f(c) = 0 1 3 3 = i 1 3 1 0 ( ( = (e 4πi 3 cos 4πi ) ) 3 sin 4πi 3 sin ( 4πi 3 cos ( 4πi 3 ) ) ) we find that the representation is reducible 1 Groups 35 3 3 1 ) 1 3 3 1 ) Problem 60 The direct sum of the matrices A, B is defined as A B := ( A ) 0 0 B where 0 is the zero matrix The matrices I = 1 0, N = 1 0 form a group under matrix multiplication Show that the 4 4 matrices I I, I N, N I, N N form a group under matrix multiplication

36 Problems and Solutions Solution 60 For the direct sum we find the permutation matrices 0 0 0 0 I I = = I 0 0 4, A = I N =, 0 0 0 0 0 0 0 0 0 0 B = N I =, C = N N = 0 0 0 0 0 0 0 0 The multiplication table is I 4 A B C I 4 I 4 A B C A A I 4 C B B B C I 4 A C C B A I 4 For the inverse elements we find A 1 = A, B 1 = B, C 1 = C Thus the set of direct sums of matrices forms a group under matrix multiplication Problem 61 Let Let A, B be matrices We define the composition a 11 0 0 a 1 0 b A B := 11 b 1 0 0 b 1 b 0 a 1 0 0 a E = 1 0, N = 1 0 Show that the 4 4 matrices E E, E N, N E, N N form a group under matrix multiplication Solution 61 We obtain the permutation matrices 0 0 0 0 E E =, E N =, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 N E =, N N = 0 0 0 0

Groups 37 Matrix multiplication shows that these four matrices form a group Problem 6 The Kronecker product of the matrices A, B is defined as the 4 4 matrix a11 B a A B := 1 B a 1 B a B Let N = 1 0 Show that the 4 4 matrices I I, I N, N I, N N form a group under matrix multiplication Solution 6 For the Kronecker product we obtain the permutation matrices 0 0 0 0 g 1 = I I = = I 0 0 4, g = I N =, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 g 3 = N I =, g 4 = N N = 0 0 0 0 The multiplication table is g 1 g g 3 g 4 g 1 g 1 g g 3 g 4 g g g 1 g 4 g 3 g 3 g 3 g 4 g 1 g g 4 g 4 g 3 g g 1 The inverse elements are given by g 1 = g, g 1 3 = g 3, g 1 4 = g 4 Thus the set { g 1, g, g 3, g 4 } forms a group under matrix multiplication Problem 63 Which of the 4 4 permutation matrices can be written as Kronecker products of permutation matrices? Solution 63 There are only two permutation matrices, namely 1 0 I =, N = 1 0

38 Problems and Solutions Thus we find four 4 4 permutation matrices 0 0 0 0 I I =, I 0 0 N =, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 N I =, N N = 0 0 0 0 Problem 64 ( 1 0 A = Show that the four matrices ( 1 0, C = 0 1 1 0 ), B = ), D = form a group under matrix multiplication Is the group abelian? ( 0 ) 1 1 0 Solution 64 The group table is A B C D A A B C D B B A D C C C D A B D D C B A From the table we find that the matrix product of any two matrices is again an element of the set {A, B, C, D} The inverse elements are A 1 = A, B 1 = B, C 1 = C, D 1 = D Thus the set of four matrices forms a group under matrix multiplication From the group table we see that the group is abelian Problem 65 Consider the set of all matrices over the set of integers Z with determinant equal to 1 Show that these matrices form a group under matrix multiplication This group is called SL(, Z) Solution 65 Let A, B SL(, Z) We have a11 a AB = 1 b11 b 1 a11 b = 11 + a 1 b 1 a 11 b 1 + a 1 b a 1 a b 1 b a 1 b 11 + a b 1 a 1 b 1 + a b

Groups 39 Since a ij b kl Z, and det A = a 11 a a 1 a 1 = 1, det B = b 11 b b 1 b 1 = 1 with det(ab) = det(a) det(b) = 1 we obtain that AB SL(, Z) The neutral element (identity) is the identity matrix The inverse of A is given by A 1 a a = 1 a 1 a 11 with det(a 1 ) = a 11 a a 1 a 1 = 1 and a 1, a 1 Z Matrix multiplication is associative Thus we have a group Problem 66 Consider the group of all 4 4 permutation matrices 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 A =, B =, C =, 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 D =, E =, F =, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G =, H =, I =, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 J =, K =, L =, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 M =, N =, O =, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 P =, Q =, R =, 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 S =, T =, U =, 0 0 0 0 0 0 0 0 0 0

40 Problems and Solutions 0 0 0 0 0 0 0 0 0 0 0 0 V =, W =, X = 0 0 0 0 0 0 Find all subgroups and discuss whether they are commutative or not Solution 66 Note that B = C = F = G = H = O = Q = V = X = A Applying the Lagrange s theorem we find that the proper subgroups must have order 1, 8, 6, 4, 3,, 1 The subgroup with order 1 is given by the set { A, D, E, H, I, L, M, P, Q, T, U, X } The group is not commutative since, for example, BX XB There are three subgroups with order 8 given by the sets { A, B, G, H, Q, R, W, X }, { A, C, H, K, N, Q, V, X }, { A, F, H, J, O, Q, S, X } There are four subgroups with order 6 given by the sets { A, B, C, D, E, F }, { A, B, O, P, U, V }, { A, C, G, I, M, O }, { A, F, G, L, T, V } There are seven subgroups with order 4 given by the sets { A, B, G, H }, { A, C, V, X }, { A, F, O, Q }, { A, H, Q, X }, { A, H, R, W }, { A, J, Q, S }, { A, K, N, X } There are four subgroups with order 3 given by the sets { A, D, E }, { A, I, M }, { A, L, T }, { A, P, U } There are nine subgroups with order given by the sets { A, B }, { A, C }, { A, F }, { A, G }, { A, H }, { A, O }, { A, Q }, { A, V }, { A, X } The subgroup with order 1 is obviously given by the set { A } Problem 67 The Heisenberg group is the group of 3 3 upper triangular matrices of the form 1 a 1 a 13 a 3 0

Groups 41 where a jk F with F an arbitrary field In the following we consider F = C (i) Is the Heisenberg group commutative? (ii) Let A, B be elements of the Heisenberg group Does the braid like relation ABBA = BAAB hold? Solution 67 (i) The Heisenberg group is not commutative We have AB = 1 a 1 + b 1 a 13 + a 1 b 3 + b 13 a 3 + b 3 0 and BA = 1 a 1 + b 1 a 13 + a 3 b 1 + b 13 a 3 + b 3 0 (ii) From (i) we find that ABBA = BAAB Thus the elements of the Heisenberg group satisfy the braid like relation Problem 68 The group SL(, Z) is generated by T = 1 1, S = 1 0 Find A SL(, Z) in terms of T and S with 1 A = 1 0 Solution 68 A = T T S = We obtain ( 1 1 ) ( 1 1 ) = 1 0 1 1 0 Problem 69 Consider the symmetric group S 3 of the 3 3 permutation matrices Do the group elements which satisfy the condition Pj = I 3 form a subgroup of S 3? Solution 69 No, they do not Consider the matrices P 1 = 1 0 0 0, P = 0 1 0 0 0 0