Algebra-2E-Solution-by-Michael-Artin(1)

How many elements of order 2 does the symmetric group S4 tain?. Show by example that the product of elements of finite order in agroup need not have finite order.. Suppose a, b are our e

Selected Solutions to Artin’s Algebra, Second Ed.

Takumi Murayama July 22, 2014

These solutions are the result of taking MAT323 Algebra in the Spring of 2012, and also TA-ing for MAT346 Algebra II in the Spring of 2014, both at Princeton University This is not a complete set of solutions; see the List of Solved Exercises

at the end Please e-mail takumim@umich.edu with any corrections

Contents

2.1 Laws of Composition 4

2.2 Groups and Subgroups 4

2.4 Cyclic Groups 5

2.6 Isomorphisms 7

2.8 Cosets 8

2.10 The Correspondence Theorem 9

2.11 Product Groups 9

2.12 Quotient Groups 10

6 Symmetry 13 6.3 Isometries of the Plane 13

6.4 Finite Groups of Orthogonal Operators on the Plane 14

6.5 Discrete Groups of Isometries 16

6.6 Plane Crystallographic Groups 17

6.7 Abstract Symmetry: Group Operations 17

6.8 The Operation on Cosets 18

6.9 The Counting Formula 19

6.10 Operations on Subsets 19

6.11 Permutation Representations 20

6.12 Finite Subgroups of the Rotation Group 21

6.M Miscellaneous Problems 23

11 Rings 24 11.1 Definition of a Ring 24

11.2 Polynomial Rings 25

11.3 Homomorphisms and Ideals 27

11.4 Quotient Rings 32

11.5 Adjoining Elements 33

11.6 Product Rings 35

11.7 Fractions 37

11.8 Maximal Ideals 38

11.9 Algebraic Geometry 40

11.M Miscellaneous Problems 42

12 Factoring 45 12.1 Factoring Integers 45

12.2 Unique Factorization Domains 46

12.3 Gauss’s Lemma 50

12.4 Factoring Integer Polynomials 51

12.5 Gauss Primes 56

12.M Miscellaneous Problems 59

13 Quadratic Number Fields 61 13.1 Algebraic Integers 61

13.2 Factoring Algebraic Integers 62

13.3 Ideals in Z[√−5] 62

13.4 Ideal Multiplication 63

13.5 Factoring Ideals 63

14 Linear Algebra in a Ring 64 14.1 Modules 64

14.2 Free Modules 65

14.4 Diagonalizing Integer Matrices 66

14.5 Generators and Relations 66

14.7 Structure of Abelian Groups 67

14.8 Applications to Linear Operators 67

14.M Miscellaneous Problems 69

15 Fields 70

15.2 Algebraic and Transcendental Elements 70

15.3 The Degree of a Field Extension 71

15.4 Finding the Irreducible Polynomial 72

15.6 Adjoining Roots 73

15.7 Finite Fields 74

15.8 Primitive Elements 76

15.9 Function Fields 77

15.10 The Fundamental Theorem of Algebra 77

15.M Miscellaneous Problems 78

16 Galois Theory 81 16.1 Symmetric Functions 81

16.2 The Discriminant 83

16.3 Splitting Fields 86

16.4 Isomorphisms of Field Extensions 86

16.5 Fixed Fields 87

16.6 Galois Extensions 88

16.7 The Main Theorem 88

16.8 Cubic Equations 91

16.9 Quartic Equations 92

16.10 Roots of Unity 94

16.11 Kummer Extensions 95

16.12 Quintic Equations 95

16.M Miscellaneous Problems 96

2 Groups

Exercise 2.1.2 Prove the properties of inverses that are listed near the end of thesection

Remark The properties are listed on p 40 as the following:

(a) If an element a has both a left inverse l and a right inverse r, i.e., if la = 1and ar = 1, then l = r, a is invertible, r is its inverse

(b) If a is invertible, its inverse is unique

(c) Inverses multiply in the opposite order: If a and b are invertible, so is theproduct ab, and (ab)−1 = b−1a−1

(d) An element a may have a left inverse or a right inverse, though it is notinvertible

Proof of (a) We see l = lar = r

Proof of (b) Let b, b0 be inverses of a Then, b = bab0 = b0, by (a)

Proof of (c) Consider ab We see that b−1a−1is the inverse of ab since (b−1a−1)(ab) =

b−1a−1ab = b−1b = 1 by associativity Uniqueness follows by (b)

Proof of (d) Consider Exercise 2.1.3 below s is not invertible since it does not have

a two-sided inverse, but it does have a left inverse

Exercise 2.1.3 Let N denote the set {1, 2, 3, } of natural numbers, and let s : N →

N be the shift map, defined by s(n) = n + 1 Prove that s has no right inverse, butthat it has infinitely many left inverses

Proof s does not have a right inverse since s does not map any element of N back

to 1; however, we can define a left inverse rk(n) = n − 1 for n > 1, and rk(1) = k forsome k ∈ N; we see that this is a left inverse of s, i.e., that rk◦ s = idN Since k isarbitrary this implies that there is an infinite number of rk’s

Exercise 2.2.3 Let x, y, z, and w be elements of a group G

(a) Solve for y, given that xyz−1w = 1

(b) Suppose that xyz = 1 Does it follow that yzx = 1? Does it follow thatyxz = 1?

Solution for (a) We claim that y = x−1w−1z This follows since

in GL2(R) shows that xyz = 1 does not imply yxz = 1

Exercise 2.2.4 In which of the following cases is H a subgroup of G?

(a) G = GLn(C) and H = GLn(R)

(b) G = R× and H = {1, −1}

(c) G = Z+ and H is the set of positive integers

(d) G = R× and H is the set of positive reals

(e) G = GL2(R) and H is the set of matrices a 00 0

, with a 6= 0

Solution for (a) H is a subset since R ⊂ C implies H ⊂ GLn(R) H is a subgroupsince GLn(R) is a group, hence contains an identity and is closed under multiplicationand inversion

Solution for (b) H is a subgroup since it is clearly a subset, contains the identity 1,and −1 × −1 = 1 implies H is closed under multiplication and inversion

Solution for (c) H is not a subgroup since −1 /∈ H, though it is the inverse of 1.Solution for (d) H is a subgroup since it is clearly a subset, contains 1, is closedunder multiplication since the product of two positive real numbers is a positive realnumber, and since x ∈ H has inverse 1/x ∈ H, which is still positive and real.Solution for (e) H is not a subgroup since it is not even a subset of G

Exercise 2.4.3 Let a and b be elements of a group G Prove that ab and ba havethe same order.

Proof Suppose (ab)n = 1 We note that b = b(ab)n = (ba)nb, but this implies(ba)n= 1, and so both have order n

Exercise 2.4.9 How many elements of order 2 does the symmetric group S4 tain?

con-Solution The order 2 elements of S4 consist of 42
= 6 two-cycles, and 4

2
× 1

2 = 3products of disjoint two-cycles, and so there are 9 elements of order 2

Exercise 2.4.10 Show by example that the product of elements of finite order in agroup need not have finite order What if the group is abelian?

Solution Consider GL2(R), and the following matrices in GL2(R):

We see that A2 = B2 = 1, and so they are of order 2, whereas

Now suppose the group is abelian Suppose a, b are our elements of finite order,

of order n, m respectively Then, (ab)nm = anmbnm = (an)m(bm)n = 1, and so ab isnecessarily of finite order

Thus, ϕ is fully determined by what 1 maps to By the above, we then have that ϕn:

z nz for n ∈ Z+ are all the homomorphisms of Z+ The injective homomorphismsconsist of those ϕnfor n 6= 0 The surjective homomorphisms consist of those ϕn for

n = ±1; these are also the isomorphisms of Z+ since they are injective

Exercise 2.6.3 Show that the functions f = 1/x, g = (x − 1)/x generate a group offunctions, the law of composition being composition of functions, that is isomorphic

to the symmetric group S3

This proves it is a homomorphism since all of the multiplications are accurate, and

is an isomorphism since every element in S3 is mapped to, with inverse defined bymatching entries This shows f2, f6 generate the group of functions since (12), (123)generate S3 as on p 42

Exercise 2.6.6 Are the matrices 1 1

1

,1

1 1

conjugate elements of the group

GL2(R)? Are they conjugate elements of SL2(R)?

Solution We explicitly calculate the conjugation for the conjugation matrix A:

we have that g = gr(a−b) = hr(i−j) ∈ H, a contradiction But this contradicts thecounting formula (2.8.8), since |G| = 35 6= 49 = 7 · 7 = |H|[G : H], and so G contains

an element of order 5

Now suppose G had no elements of order 7; then all non-identity elements haveorder 5 as before Letting h have order 5 and H, g as before, the same argument givesthat H, gH, g2H, , g4H are disjoint left cosets in G This contradicts the countingformula (2.8.8) again, since |G| = 35 6= 25 = 5 · 5 = |H|[G : H], hence G contains anelement of order 7

Exercise 2.8.8 Let G be a group of order 25 Prove that G has at least one subgroup

of order 5, and that if it contains only one subgroup of order 5, then it is a cyclicgroup

Proof Any element in G has order in {1, 5, 25} by Cor 2.8.10 If G had no elements

of order 5, it must have an element x of order 25, but then |x5| = 5, hence hx5i is asubgroup of order 5

Now suppose H is the only subgroup of order 5 in G, and pick x /∈ H x hasorder 5 or 25 by Cor 2.8.10 again, and since e ∈ H But |x| = 5 implies H = hxi,hence x ∈ H, and so we know |x| = 25, i.e., G = hxi is cyclic

Exercise 2.8.10 Prove that every subgroup of index 2 is a normal subgroup, andshow by example that a subgroup of index 3 need not be normal

Proof If H 6 G with [G : H] = 2, we see that, picking a /∈ H, G = H q aH =

H q Ha, since this is the only we can form cosets of a in G Thus, aH = Ha, hence

H C G by Prop 2.8.17

Now we consider S3; the 2-cycle y = (12) generates a subgroup H of order 2, andtherefore of index 3 by the counting formula (2.8.8) However, by the multiplicationtable constructed last week, we see that, from (2.8.4) and (2.8.16),

xH = {x, xy} 6= {x, x2y} = Hx

Exercise 2.10.3 Let G and G0 be cyclic groups of orders 12 and 6, generated byelements x and y, respectively, and let ϕ : G → G0 be the map defined by ϕ(xi) = yi.Exhibit the correspondence referred to in the Correspondence Theorem explicitly.Solution We note that K = ker ϕ = {e, x6} The subgroups of G that contain theseare hxi, hx2i, hx3i, hx6i, and each correspond to hyi, hy2i, hy3i, e respectively, whichare all the subgroups of G0

Exercise 2.11.1 Let x be an element of order r of a group G, and let y be anelement of G0 of order s What is the order of (x, y) in the product group G × G0?Solution The order is lcm(r, s), since (x, y)n = (xn, yn) = (e, e) implies r, s | n

Exercise 2.11.3 Prove that the product of two infinite cyclic groups is not infinitecyclic.

Proof Recall that a cyclic group must be generated by a single element; since allinfinite cyclic groups are isomorphic to Z, we can consider Z × Z Suppose that (a, b)

is this single element But then, we see that (2a, b) cannot be obtained from adding(a, b) to itself, which implies that Z × Z is not infinite cyclic

where ∗ represents an arbitrary real number Show that H is a subgroup of GL3, that

K is a normal subgroup of H, and identify the quotient group H/K Determine thecenter of H

Proof H 6 GL3 since clearly I ∈ H,

The quotient group H/K is then represented by matrices of the form

ka to a coset with arbitrary parameters b, c

The center of H contains K, since K commutes with elements of H as shownabove; we claim the center is solely K If A, B ∈ H, then by the above AB = BA is

This implies a11b22= a22b11 But if A is fixed and B is an arbitrary matrix in H, then

a11b22 = a22b11 must hold for all matrices B, and so a11 = a22 = 0, hence A ∈ K.Thus, the center of H is K ≈ R+

Exercise 2.12.4 Let H = {±1, ±i} be the subgroup of G = C× of fourth roots ofunity Describe the cosets of H in G explicitly Is G/H isomorphic to G?

Solution By (2.8.5), cosets aH, bH are equal if and only if b = ah for some h ∈ H,and so if a = re2πiθ, b = se2πiη for r, s ∈ R>0, θ, η ∈ [0, 1), then aH = bH if andonly if r = s and θ − η ∈ {0, 1/4, 1/2, 3/4} Hence the cosets of H are {re2πθH | r ∈

R>0, θ ∈ [0, 1/4)}

Now consider the map ϕ : G → G, x x4 Then, this is trivially a phism of G; it is moreover surjective since any nonzero complex number has a fourthroot We see that ker ϕ = H, and so G/H ≈ G by the isomorphism theorem

homomor-Exercise 2.12.5 Let G be the group of upper triangular real matrices a b

0 d

, with

a and d different from zero For each of the following subsets, determine whether ornot S is a subgroup, and whether or not S is a normal subgroup If S is a normalsubgroup, identify the quotient group G/S

(i) S is the subset defined by b = 0

(ii) S is the subset defined by d = 1

(iii) S is the subset defined by a = d

Solution for (i) S 6 G since the diagonal entries would multiply with each otherand not affect the other entries of the matrices, I ∈ S, and the inverse can be found

by letting the inverse have entries a−1, d−1 S is not a normal subgroup since

implies it is closed under multiplication; the identity is trivially in S

We now see that it is a normal subgroup:

and since each matrix of the form above gives a different coset because multiplying

by elements in S keep d constant

Solution for (iii) This is a subgroup since it satisfies closure:

and clearly the identity is in S.

We now check this subgroup is normal:

for d ∈ R× since this would cover all of G:

and since each matrix of the form above gives a different coset since multiplying byelements in S cannot give another matrix of the same form unless d is the same

6 Symmetry

Exercise 6.3.2 Let m be an orientation-reversing isometry Prove algebraicallythat m2 is a translation

6.4 Finite Groups of Orthogonal Operators on the Plane

Exercise 6.4.2

(a) List all subgroups of the dihedral group D4, and decide which ones are normal.(b) List the proper normal subgroups N of the dihedral group D15, and identify thequotient groups D15/N

(c) List the subgroups of D6 that do not contain x3

Solution for (a) We claim the subgroups of D4 form the lattice diagram

4 subgroups are either generated by an order 4 element or by two order 2 elements;these give the third row Finally, D4 is the unique order 8 subgroup

We claim the normal subgroups of D4 are

{e}, hxi, hx2i, hx2, yi, hx2, xyi, D4

hx2i is normal since x2 commutes with all other elements of D4; the other propersubgroups are normal because they are of index 2 by the counting formula (2.8.8),and by Exercise 2.8.10 Lastly, the other four subgroups in the second row of thediagram above are not normal since they are not closed under conjugation by x.Solution for (b) Any proper subgroup H 6 D15 must have |H| ∈ {2, 3, 5, 15} byLagrange’s theorem (Thm 2.8.9) The order 2 subgroups are of the form hxiyi; theseare not normal since they are not closed under conjugation by x Since 2 - 3, 5, 15,

we know no other subgroup contains an element of the form xiy Thus, every normalsubgroup of D15 must be of the form hxii for i = 1, 3, 5, since any other i gives asubgroup equal to one of these three These are all in fact normal subgroups sincethey are kernels of homomorphisms D15 → Di for i = 1, 3, 5 mapping x x, y y

By the first isomorphism theorem (Thm 2.12.10), this implies the quotient groupsfor each nontrivial N are isomorphic to

D15/hxi ≈ D1, D15/hx3i ≈ D3, D15/hx5i ≈ D5.Solution for (c) By Lagrange’s theorem (Thm 2.8.9), any subgroup H 6 D6 musthave |H| ∈ {1, 2, 3, 4, 6, 12} {e} is the unique order 1 subgroup, and does not contain

x3 The order 2 subgroups are hx3i and hxiyi for 0 ≤ i ≤ 5; only the first contains

x3 The unique order 3 subgroup is hx2i, which does not contain x3 The order 4subgroups not containing x3 are of the form hxiy, xjyi for i 6= j since D6 contains noelements of order 4, but this subgroup is of order 4 if and only if xj−i = x3 = xi−j,hence there are no order 4 subgroups not containing x3 The order 6 subgroups must

be generated by an order 2 and an order 3 element, hence those not containing x3are of the form hx2, xiyi But different choices i, j for i give the same subgroup ifand only if i ≡ j mod 2, hence the only subgroups of order 6 not containing x3 are

hx2, yi and hx2, xyi The unique order 12 subgroup D16 contains x3 In summary,the subgroups that do not contain x3 are

{e}, hyi, hxyi, hx2yi, hx3yi, hx4yi, hx5yi, hx2i, hx2, yi, hx2, xyi

x5 Thus H C D10

Now the surjective homomorphism D10 → D5 sending x x, y y has kernel

H, hence D10/H ≈ D5 by the first isomorphism theorem (Thm 2.12.10)

Solution for (c) Note D5 ,→ D10 by having x x2, hence to show D10 ≈ D5× H,

it suffices by Prop 2.11.4(d) to show HT D5 = {e}, HD5 = G, and H, D5C G withthis embedding of D5 The normality of both groups follows by (b) and since D5 is

of index 2 in G (Exercise 2.8.10) Their intersection is e since x5 ∈ D/ 5 HD5 = Gsince x5 commutes with every element of D5 ⊂ D10 by (b), and so we can get anyelement of the form xi, xiy for 0 ≤ i ≤ 9

6.5 Discrete Groups of Isometries

Exercise 6.5.5 Prove that the group of symmetries of the frieze pattern C C C C

C C C is isomorphic to the direct product C2× C∞ of a cyclic group of order 2 and

an infinite cyclic group

Proof Let G 6 M be the group of symmetries By Thm 6.3.2, any symmetry arises

as tvρθ or tvρθr where tv are translations, ρθ are rotations, and r are reflections acrossthe x-axis In both cases, θ must be zero to be a symmetry, and v must be an integermultiple of the length of C Hence the group C2 of reflections across the x-axis andthe group C∞ of translations by integer multiples of lengths of C generate G

We claim C2× C∞ ≈ G Since clearly C2∩ C∞ = {e}, by Prop 2.11.4(d) it onlyremains to show C2, C∞C G C∞C G since it has index 2 in G by the classification

of symmetries above, and then by Exercise 2.8.10 C2C G since our translations alllie on the x-axis, and then by using (6.3.3)

Exercise 6.5.9 Let G be a discrete subgroup of M whose translation group is nottrivial Prove that there is a point p0 in the plane that is not fixed by any element of

G except the identity

Proof We first claim that the set of points fixed by any nontrivial isometry m ∈ Ghas Lebesgue measure zero We proceed by considering each kind of isometry inThm 6.3.4 If m is a nontrivial translation, then there are no fixed points If m is

a nontrivial rotation around a point p, then p is a fixed point If m is a nontrivialreflection about a line `, then any point on ` is a fixed point If m is a nontrivial glide,then m has no fixed points Thus, in all cases the set of fixed points has Lebesguemeasure zero in R2

Now for each m 6= e, let Fm be the set of fixed points of m; by the above, ithas Lebesgue measure zero We claim there are only countably many m ∈ G ByThm 6.3.2, any m is of the form tvρθ or tvρθ By Thm 6.5.5, there are only countablymany tv By Prop 6.5.10, there are only finitely many ρθ and ρθr Thus, there areonly countably many m ∈ G

Finally, the set of all points fixed by some nontrivial element of G has Lebesguemeasure

6.6 Plane Crystallographic Groups

Exercise 6.6.2 Let G be the group of symmetries of an equilateral triangular lattice

L Determine the index in G of the subgroup of translations in G

Proof The index of translations is given by [G : T ] By Theorem 6.3.2, we see thatevery isometry should be given by tvρθr; we therefore want the cardinality of the set

of ρθr’s But this is exactly the set D3, and so [G : T ] = |D3| = 6

Exercise 6.7.1 Let G = D4 be the dihedral group of symmetries of the square.(a) What is the stabilizer of a vertex? of an edge?

(b) G operates on the set of two elements consisting of the diagonal lines What

is the stabilizer of a diagonal?

Solution for (a) If the vertex lies along the axis of reflection for y, {e, y} is thestabilizer If it does not, {e, x2y} is the stabilizer If the given edge lies immediately

to the +θ direction of the line of reflection, the stabilizer is {e, xy} If it does not,{e, x3y} is the stabilizer

Solution for (b) The stabilizer of the diagonal is {e, x2, y, x2y}

Exercise 6.7.2 The group M of isometries of the plane operates on the set of lines

in the plane Determine the stabilizer of a line

Solution It suffices to consider the classes of isometries in Thm 6.3.4 The stabilizer

of a line ` consists of translations along `, rotations of an angle π about a point p ∈ `,reflections about `, and glide reflections about `

Exercise 6.7.8 Decompose the set C2×2of 2×2 matrices into orbits for the followingoperations of GL2(C):

(a) left multiplication,

(b) conjugation

Solution (a) Left multiplication by elements of GL2(C) corresponds to products ofelementary row operations on matrices in C2×2 By row reduction, we know thatapplying elementary row operations on a given matrix in C2×2gives a unique matrix

of one of the following forms:

Solution (b) Since every matrix is conjugate (similar) to a unique Jordan form (up

to ordering of Jordan blocks), we have that the orbits are generated by the differentJordan forms, i.e., C2×2 decomposes as

Exercise 6.7.11 Prove that the only subgroup of order 12 of the symmetric group

S4 is the alternating group A4

Proof We first prove a lemma: If H is a subgroup of G, then H is normal if andonly if H is the union of conjugacy classes in G But H is normal if and only if it isclosed under conjugation if and only if H =S

h∈HG ∗ h

Now suppose H 6 S4 has order 12 Then, by the counting formula (2.8.8),

|S4| = 24 = 12 · [S4 : H] = |H|[S4 : H] implies [S4 : H] = 2, and so H C S4 byExercise 2.8.10 So, we use the classification of conjugacy classes in S4 from p 201,following Prop 7.5.1:

Partition Element No in Conj Class

Exercise 6.8.4 Let H be the stabilizer of the index 1 for the operation of the metric group G = Sn on the set of indices {1, , n} Describe the left cosets of H

sym-in G and the map (6.8.4) sym-in this case

Solution We see that H consists of all cycles that hold 1 fixed, i.e., permutations ofthe remaining n − 1 elements, hence is isomorphic to Sn−1 We claim

G/H = {(1i)H | i ∈ {1, , n}}

Each (1i) gives a different coset since no (1i)H contains (1j) for i 6= j, and then by(2.8.5) These are all the cosets since by Pop 6.8.4, |G/H| = |O1| = n Thus, themap  : G/H → O1 is the map (1i) i

Exercise 6.9.4 Identify the group T0 of all symmetries of a regular tetrahedron,including orientation-reversing symmetries

Solution T0 6 O3(R) operates transitively on the set F of faces of order 4, hence isgiven by a permutation of the set {f1, f2, f3, f4}, and so there is a homomorphism

T0 → S4 by Prop 6.11.2 This is injective since if t ∈ T0 fixes three faces, then it fixesthe three vectors defining the centers of each face, hence is the identity matrix in

O3(R) We claim this homomorphism T0 → S4 is surjective, hence an isomorphism;

it suffices to show |T0| = 24 But the stabilizer Gf of a given face f is the group

D3 generated by a rotation by 2π/3 about the center of f and a reflection about anaxis in f ; |D3| = 6, and so the counting formula (6.9.2) gives |T0| = 6 · 4 = 24, hence

T0 ≈ S4

Exercise 6.10.1 Determine the orders of the orbits for left multiplication on theset of subsets of order 3 of D3

Solution We know that there are 63
= 20 subsets of order 3 of D3; we also know

by the counting formula (6.9.2) that there can only be orbits of order 1, 2, 3, 6 sinceonly these divide |D3| We see that only the subset {e, x, x2} is a subgroup, whichproduces the orbit {e, x, x2}, {y, yx, yx2} The other 18 subsets form 3 orbits of order

6 of D3 Letting H1 = {e, x, y}, we have

H1 = {e, x, y}, xH1 = {x, x2, yx2}, x2H1 = {x2, e, yx},

yH1 = {y, yx, e}, yxH1 = {yx, yx2, x2}, yx2H1 = {y, yx2, x}.Letting H2 = {e, x, yx}, we have

H2 = {e, x, yx}, xH2 = {x, x2, y}, x2H2 = {x2, e, yx2},

yH2 = {y, yx, x}, yxH2 = {yx, yx2, e}, yx2H2 = {y, yx2, x2}

Letting H3 = {e, x, yx2}, we have

H3 = {e, x, yx2}, xH3 = {x, x2, yx}, x2H3 = {x2, e, y},

yH3 = {y, yx, x2}, yxH3 = {yx, yx2, x}, yx2H3 = {y, yx2, e}

We therefore have one orbit of order 2 and three orbits of order 6; this means wehave 1 × 2 + 3 × 6 = 20 subsets in these orbits, and so we have found all of them

If f (x) = e, then there is no restriction on f (y), thus there are ten phisms f of this kind

homomor-If f (x) = (abc), then since yx has order 2, f (yx)2 = f (y)2f (x)2 = e Thus

f (y) 6= e Now suppose f (y) acts nontrivially on the fourth element d unaffected by

f (x); then without loss of generality, f (y) interchanges a, d, and so f (yx) · d = a.Hence f (yx) must map a d as well, and so f (y) · b = d But then f (y)2 · b =

a 6= b, contradicting that f (y) has order 2 Hence f (y) is a permutation of order

2 of the subset {a, b, c}, which are 2-cycles; assume without loss of generality that

f (y) = (ab) In the argument above, there are 4 choices for the fixed point d of f (x),

3 choices for the fixed point of f (y), and two choices for a = f (x) · c Each choicegives a homomorphism f by just renaming a, b, c as 1, 2, 3, hence realizing f as thecanonical operation of S3 on the subset {1, 2, 3} ⊂ {1, 2, 3, 4} Thus, there are 24homomorphisms f such that f (x) 6= e

Exercise 6.11.5 A group G operates faithfully on a set S of five elements, and thereare two orbits, one of order 3 and one of order 2 What are the possible groups?Hint: Map G to a product of symmetric groups

Solution Let S = {1, 2, 3, 4, 5} such that the two orbits are O3 = {1, 2, 3}, O2 ={4, 5}, respectively G operates on O3, O2 separately, hence the action of G on

O3, O2 respectively correspond to group homomorphisms f3: G → S3 and f2: G →

S2 This defines a group homomorphism f = (f3, f2) : G → S3 × S2 defined by

g (f3(g), f2(g)) Since G acts faithfully on S, g is injective, hence G ≈ f (G) byCor 2.12.11 Now the only group that acts transitively on O2 is S2 itself, and so

f2(G) ≈ S2 On the other hand, there are two groups that act transitively on O3:

C3 and S3 Hence G ≈ C3 × S2 or S3 × S2

Exercise 6.11.6 Let F = F3 There are four one-dimensional subspaces of the space

of column vectors F2 List them Left multiplication by an invertible matrix permutesthese subspaces Prove that this operation defines a homomorphism ϕ : GL2(F ) → S4.Determine the kernel and the image of this homomorphism

Proof We have the following one-dimensional subspaces, denoting F = {0, 1, 2}:



, V3 =1

1

,22



, V4 =2

1

,12



Call eithe first listed vector in Vi; note that Vi = {ei, 2ei} These are all the subspaces

in F2 since there are only 3 · 3 − 1 = 8 nontrivial vectors in F2

Left multiplication by an invertible matrix defines an action on the vectors in F2

since left multiplication by matrices in F is associative, and since the identity matrixdefines the trivial action; this descends to a well-defined action on the subspaces Visince the span of Aei is equal to the span of 2Aei for any A ∈ GL2(F ) Hence, byCor 6.11.3, this action defines a homomorphism ϕ : GL2(F ) → S4

Now ker ϕ = {I, 2I} since ⊃ clearly holds, and for A ∈ GL2(F ), Ae1 ∈ V1 impliesthe first column of A is either e1 or 2e1, Ae2 ∈ V2 implies the second column of A iseither e2 or 2e2, and Ae3 ∈ V3 implies that the coefficients on ei in each column have

to be the same, hence A = I or 2I

We claim im ϕ = S4 By Cor 2.8.13, we have |GL2(F )| = |ker ϕ||im ϕ|, and since

|S4| = 24 and |ker ϕ| = 2 from above, it suffices to show |GL2(F )| = 48 Now everymatrix in GL2(F ) consists of a pair of linearly independent vectors in F2; there are

8 · 6 = 48 of these pairs by the above decomposition of F2 into subspaces, hence

|GL2(F )| = 48

Exercise 6.12.3 Let O be the group of rotations of a cube, and let S be the set offour diagonal lines connecting opposite vertices Determine the stabilizer of one ofthe diagonals

Solution O acts on S, hence there is a homomorphism O → S4 by Prop 6.11.2 Wefirst show this homomorphism is injective If s ∈ O fixes all four diagonals, then

s either fixes or interchanges the two endpoints of each diagonal If s 6= e, i.e., itacts nontrivially on the vertexes, then we can pick three pairs of opposite vertexessuch that s interchanges one of the pairs of vertexes, or such that s interchanges allthree pairs of vertexes In either case, give R3 a basis that points along the threediagonals connecting these pairs of vertexes; then the matrix for s in this basis has

an odd number of −1’s on the diagonal, and so has determinant −1, contradictingthat O 6 SO3(R) The homomorphism O → S4 is surjective, since looking at thefaces of the cube, O acts transitively on the set of faces, and each face has stabilizer

C4, hence |O| = |Of||O · f | = 4 · 6 = 24 by the counting formula (6.9.2)

Now we know O acts like S4 on the set of diagonals, hence the stabilizer of one

of the diagonals is the same as fixing one index in the set of indexes {1, 2, 3, 4} as

in Exercise 6.8.4, hence is equal to S3

Exercise 6.12.7 The 12 points (±1, ±α, 0)t, (0, ±1, ±α)t, (±α, 0, ±1)t form thevertices of a regular icosahedron if α > 1 is chosen suitably Verify this, and deter-mine α

Proof We have that the 12 points form three categories of the form above By thedistance formula, if they are in the same category, we have three possibilities forsquare distances:

Vertex Change sign of 1 Change sign of α Change sign of both

If they are in different categories, then we have two possibilities for square distances

If our coordinates are (x1, x2, x3) and (y1, y2, y3), then there is only one i such that

xi, yi 6= 0, and xi, yi must have different absolute values for each i, giving the table

Vertex xi, yi have same sign xi, yi have differing sign

by choosing the root greater than 1 This shows each vertex has exactly five neighbors

of distance 2 away from it; we claim that the polyhedron formed by connectingvertexes of distance 2 away from each other forms an icosahedron

Now we show each face formed by the edges is a congruent equilateral triangle.This is true since any face is formed by neighboring vertexes, which are distance 2away from each other by the above Each vertex moreover has the same number offaces meeting there since every vertexes has exactly five neighbors by the above.Finally, suppose we have two neighboring vertexes v, v0 forming an edge; we claimthere are only two faces intersecting at that edge It suffices to show there are onlytwo vertexes w that are of distance 2 from both Suppose vi = v0i = 0; in thefollowing, we consider subscripts mod 3 Then vi+1 = −v0i+1 with absolute value

1, so wi+1 = 0, |wi+2| = 1, |wi| = 3 by the table above Next vi+2 = vi+20 , hence

wi+2 must have the same sign as vi+2, vi+20 Finally, wi can have either sign since

vi = vi0 = 0, hence there are only two vertexes that are of distance 2 from v, v0

Exercise 6.M.7 Let G be a finite group operating on a finite set S For each element

g of G, let Sg denote the subset of elements of S fixed by g: Sg = {s ∈ S | gs = s},and let Gs be the stabilizer of s

(a) We may imagine a true-false table for the assertion that gs = s, say with rowsindexed by elements of G and columns indexed by elements of S Constructsuch a table for the action of the dihedral group D3 on the vertices of a triangle.(b) Prove the formula P

x2 False False False

yx2 False False Truewhere S = {s1, s2, s3} is the set of the vertices of the triangle, x = (s1s2s3) rotations,and y is the reflection with preferred vertex s1

Proof of (b) By summing over different sets, we have

Exercise 11.1.2 Prove that, for n 6= 0, cos(2π/n) is an algebraic number

Proof Suppose n > 0 Recall that the Chebyshev polynomials Tn(x) are defined bythe recurrence relations T0(x) = 1, T1(x) = x, and Tn(x) = 2x Tn−1(x) − Tn−2(x)

We claim Tn(cos θ) = cos nθ for any θ This is clear for n = 0, 1 For arbitrary n,

Tn(cos θ) = 2 cos θ Tn−1(cos θ) − Tn−2(cos θ)

= 2 cos θ cos ((n − 1)θ) − cos ((n − 2)θ)

= cos nθ + cos ((n − 2)θ) − cos ((n − 2)θ) = cos nθ,and so Tn(cos θ) = cos nθ for any θ as desired Letting θ = 2π/n, we have that

Tn(cos(2π/n)) = cos 2π = 1, and so cos(2π/n) satisfies the polynomial equation

Tn(x) − 1 = 0

If n < 0, then using the fact that cos(2π/n) = cos(−2π/n), we see cos(2π/n)satisfies the polynomial equation T−n(x) − 1 = 0 by the above

Exercise 11.1.3 Let Q[α, β] denote the smallest subring of C containing the rationalnumbers Q and the elements α =√2 and β =√

c + dαβ for c odd, d even when k evenThis is clear for k = 0, k = 1 For arbitrary n, consider first when k is even Then,

by inductive hypothesis

γk= γk−1γ = (aα + bβ)(α + β) = 2a + 3b + (a + b)αβ,for a, b odd, hence 2a + 3b is odd and a + b is even Likewise, when k is odd,

γk= γk−1γ = (c + dαβ)(α + β) = (3d + c)α + (c + 2d)β,for c odd, d even, hence 3d + c, c + 2d are odd, and we are done

and so x4 + 3x3 + x2 + 7x + 5 = (x2 + 2x − 2)(x2 + x + 1) + (7x + 7), where theremainder 7x + 7 ≡ 0 mod n if and only if n ∈ {1, 7} Hence x2 + x + 1 divides

x4+ 3x3+ x2+ 7x + 5 in [Z/(n)][x] if and only if n ∈ {1, 7}

Exercise 11.2.2 Let F be a field The set of all formal power series p(t) = a0 +

a1t + a2t2+ · · · , with ai in F , forms a ring that is often denoted by F [[t]] By formalpower series we mean that the coefficients form an arbitrary sequence of elements of

F There is no requirement of convergence Prove that F [[t]] is a ring, and determinethe units in this ring

Proof We denote p(t) = P

iaiti, q(t) = P

ibiti, r(t) = P

iciti Let + defined asp(t) + q(t) =P

k(ak+ bk)tk and × as p(t) × q(t) =P

k

P

i+j=kaibjtk.+ makes F [[t]] an abelian group because associativity and commutativity followsince + is defined termwise, and because having ai = 0 for all i defines an identityand letting q(t) such that bi = −ai for all i defines an inverse for p(t)

The identity is p(t) such that a0 = 1, ai = 0 for all i > 0

It remains to show the distributive property:

Exercise 11.3.3 Find generators for the kernels of the following maps:

Remark We will denote each map as ϕ

Solution for (a) We claim that (x, y) = ker ϕ By the division algorithm, any nomial f ∈ R[x, y] can be written g + a0 for g ∈ (x, y), and so ϕ(f ) = a0 = 0 if andonly if a0 = 0 if and only if f = g ∈ (x, y)

poly-Solution for (b) We claim that (x2− 4x + 5) = ker ϕ x2− 4x + 5 = (x − (2 + i))(x −(2 − i)), hence (x2 − 4x + 5) ⊂ ker ϕ Conversely, let f ∈ ker ϕ By the divisionalgorithm, we can write f = g + r for g ∈ (x2 − 4x + 5), where r = a1x + a0 for

ai ∈ R has degree less than 2 Then, ϕ(f) = ϕ(g) + ϕ(r) = a1(2 + i) + a0, which iszero only if a1 = a0 = 0, i.e., only if f = g ∈ (x2− 4x + 5)

Solution for (c) We claim that (x2−2x−1) = ker ϕ x2−2x−1 = (x−(1+√2))(x−(1 −√

2)), hence (x2 − 2x − 1) ⊂ ker ϕ Conversely, let f ∈ ker ϕ By the divisionalgorithm, we can write f = g + r for g ∈ (x2 − 4x + 5), where r = a1x + a0 for

ai ∈ Z has degree less than 2, since x2− 2x − 1 is monic Then, ϕ(f ) = ϕ(g) + ϕ(r) =

a1(1 +√

2) + a0, which is zero only if a1 = a0 = 0 since 1,√

2 are linearly independentover Z, i.e., only if f = g ∈ (x2− 2x − 1)

Solution for (d) We claim that (x4− 10x2+ 1) = ker ϕ We have (x4− 10x2+ 1) ⊂ker ϕ, since ϕ(x4− 10x2 + 1) = (√

2 +√3)4 − 10(√2 +√

3)2+ 1 = 0 Conversely,

let f ∈ ker ϕ By the division algorithm, we can write f = g + r, where r =

a3x3+ a2x2+ a1x + a0 for ai ∈ Z has degree less than 4 Then,

ϕ(f ) = ϕ(g + r) = r(√

2 +√3)

= a3(√

2 +√3)3 + a2(√

2 +√3)2+ a1(√

2 +√3) + a0







,

= −2

0 1 11