Groups and Galois Theory

Course Notes

Alberto Elduque Departamento de Matem´aticas Universidad de Zaragoza 50009 Zaragoza, Spain c 2009-2019 Alberto Elduque Contents

Syllabus v

What is this course about? vii

1 Groups 1 § 1. Deﬁnitions and examples ...... 1 § 2. Cyclic groups. Generators ...... 3 § 3. Homomorphisms ...... 5 § 4. Symmetric group ...... 9 § 5. Group actions ...... 13 § 6. Conjugation. Translation ...... 17 6.1. Action by conjugation ...... 17 6.2. Action by translation ...... 19 § 7. Sylow’s Theorems ...... 20 § 8. Direct products ...... 22 § 9. Finitely generated abelian groups ...... 24 § 10. Solvable groups ...... 27 § 11. Simple groups ...... 30 Exercises ...... 32

2 Galois Theory 37 § 1. Algebraic extensions ...... 37 § 2. Splitting ﬁelds. Algebraic closure ...... 42 § 3. Separable extensions ...... 47 § 4. Galois group ...... 49 § 5. The Fundamental Theorem of Galois Theory ...... 55 § 6. Finite ﬁelds ...... 58 § 7. Primitive elements ...... 59 § 8. Ruler and compass constructions ...... 60 § 9. Galois groups of polynomials ...... 62 § 10. Solvability by radicals ...... 69 Exercises ...... 75

Previous exams 79

iii

Syllabus

This is a required course for Math Majors at the University of Zaragoza (Science School). It consists of 6 credits. It is a sequel of (and relies heavily on) the course Algebraic Structures.

Lecturer: Alberto Elduque Office: Math. Building. Second floor. Algebra Section. Office no. 2. e-mail: [email protected] http://www.unizar.es/matematicas/algebra/elduque Office hours: Tuesday and Friday, from 17:00 to 20:00.

Course description: The goal of the course is to understand the basic properties of the groups, which form the basic algebraic structure to understand and measure the symmetry, and then to proceed to study the symmetry of the algebraic equations (Galois Theory). The peak of the course will be the proof of the impossibility to solve by radicals the algebraic equations of degree ≥ 5.

Exam, exercises: There will be a ﬁnal exam, which will consist of some questions of a theoretical nature and some exercises. Students will be required to solve and explain to their classmates some of the exercises in the notes. This will constitute a ten percent of the ﬁnal mark.

References: There are many good references on Abstract Algebra, which include chapters devoted to Groups and to Galois Theory. Here are just a few of them: - D.S. Dummit and R.M. Foote: Abstract Algebra (2nd edition). John Wiley and Sons, 1999.

- J.J. Rotman: A First Course in Abstract Algebra (2nd edition), Pren- tice Hall, 2000.

- R.B. Ash: Abstract Algebra. The Basic Graduate Year http://www.math.uiuc.edu/ r-ash/Algebra.html

v vi SYLLABUS

Some more speciﬁc textbooks are the following:

- I. Stewart: Galois Theory. 4th ed., CRC Press, 2015.

- J.P. Tignol: Galois’ Theory of Algebraic Equations, World Scientiﬁc, 2001.

- J.S. Milne: Group Theory and Fields and Galois Theory, http://www.jmilne.org/math/CourseNotes/index.html What is this course about?

Symmetries of an equilateral triangle

Fix an equilateral triangle and consider the set

2 G = {isometries of R ﬁxing the triangle}

G consists of three reflections relative to the lines that pass through a vertex and the middle point of the opposite side (see Figure 1) together with the rotations of 0, 120 and 240 degrees. Thus G consists of 6 elements: |G| = 6. Each isometry fixing the triangle permutes the vertices. The three reflections correspond to the transposition of two vertices. The rotations of 120 and 240 degrees correspond to the two cyclic permutations of the vertices.

. . σ1 ...... σ3 σ2

Figure 1: Symmetries of an equilateral triangle.

For us, the important property of this set G is that its elements can be composed (they are bijective maps!), and the composition of any two of its elements again belongs to G. Thus, for example, σ1σ2 is the clockwise rotation of 120 degrees. These sets, endowed with a binary operation satisfying the usual properties of the composition of bijective maps, are called groups.

vii viii WHAT IS THIS COURSE ABOUT?

Solutions of equations by radicals We are quite used to ﬁnd the solutions of the degree 2 equations:

ax2 + bx + c = 0, which are given by the familiar formula: √ −b ± b2 − 4ac x = . 2a In the course Algebraic Structures we learned how to solve the equations of degree 3 and 4, which involve the use of square and cubic roots. The clue to deal with equations of degree ≥ 5 is to study the symmetries of the equation. These form a group, called the Galois group of the equation. Many properties of the equation, like its solvability by radicals, are determined by the structure of its Galois group. The theory devoted to the study of the algebraic equations and their Galois groups is called Galois Theory.

This course is devoted to study the basic properties of groups and of the Galois Theory of algebraic equations. Chapter 1

Groups

The purpose of this chapter is the study of those sets which, like the set of isometries preserving an equilateral triangle, are endowed with a binary operation satisfying the usual properties.

§ 1. Deﬁnitions and examples

1.1 Deﬁnition. A group is a set G endowed with a binary operation (usually called multiplication)

G × G −→ G (x, y) 7→ xy, which is associative ((xy)z = x(yz) for any x, y, z ∈ G), there is a neutral element (which will usually be denoted by e: ex = x = xe for any x ∈ G), and such that for any element x ∈ G there exists an element, denoted by x−1 and called the inverse of x, such that xx−1 = x−1x = e. Moreover, if the multiplication is commutative (xy = yx for any x, y ∈ G), then the group is said to be commutative or abelian.

1.2 Examples.

(i) Any ring R is an abelian group with the operation given by its addition. In particular we get the abelian groups (Z, +) or (Z/nZ, +). (ii) If R is a unital ring, the set of its units R× = {r ∈ R : ∃s ∈ R such that rs = sr = 1} is a group. The operation is given by the product of the ring.

(iii) Classical groups: For any natural number n ∈ N we have the following groups of matrices (the operation is the usual product):

1 2 CHAPTER 1. GROUPS

• GL(n, R) = {A ∈ Matn(R) : det A 6= 0} (general linear group), • SL(n, R) = {A ∈ Matn(R) : det A = 1} (special linear group), 0 • O(n, R) = {A ∈ Matn(R): AA = In} (orthogonal group), • SO(n, R) = O(n, R) ∩ SL(n, R) (special orthogonal group). (iv) Let Ω be an arbitrary set, then the set

S(Ω) = {f :Ω → Ω: f is a bijective map}

is a group under the composition of maps. It is called the group of permutations of Ω. The groups contained in S(Ω) are called groups of transformations of Ω.

The famous Erlangen Program (Felix Klein, 1872) proposed that Geome- try is the study of invariants under the action of a group. Thus the euclidean geometry becomes the study of invariants under the isometry group (angles, distances, ..., are invarians), the aﬃne (or projective) geometry becomes the study of invariants under the aﬃne (or projective) group (for instance, the barycenter is an invariant), the topology becomes the study of invariants under the action of the group of homeomorphisms (compactness, connect- edness, fundamental groups, ..., are invariants), ...

1.3 Remark. Thanks to the associativity of the product, there is no need to put parentheses in products x1x2 ··· xn (n > 2). Also, we may deﬁne recursively the powers of an element of a group G as follows: x0 = e, n+1 n −n n −1 x = x x, x = (x ) , for any n ∈ N. The associativity of the product n m n+m immediately gives the property x x = x for any x ∈ G and n, m ∈ Z. The associativity also implies the uniqueness of the inverse: If xy = e = zx, then z = ze = z(xy) = (zx)y = ey = y.

As (xy)(y−1x−1) = e, it follows that (xy)−1 = y−1x−1 .

1.4 Deﬁnition. Let G be a group and let H be a nonempty subset of G. H is said to be a subgroup of G if for any x, y ∈ H, both xy and x−1 belong to H. (It will be denoted by H ≤ G.)

1.5 ‘Silly’ properties. Let H be a nonempty subset of the group G.

• H ≤ G =⇒ e ∈ H.

• H ≤ G ⇐⇒ ∀x, y ∈ H, xy−1 ∈ H.

• H is a subgroup of G if and only if H is closed under the multiplication in G, and it is a group with this multiplication. § 2. CYCLIC GROUPS. GENERATORS 3

1.6 Examples.

• SO(n, R) ≤ SL(n, R) ≤ GL(n, R), O(n, R) ≤ GL(n, R). × • Let H be the ring of Hamilton quaternions, then H = {a+bi+cj+dk : a2 + b2 + c2 + d2 6= 0}. Consider the subset Q = {±1, ±i, ±j, ±k}. × Then Q is a subgroup of H consisting of 8 elements. It is called the quaternion group. × Another subgroup of H is given by the three-dimensional sphere: S3 = {a + bi + cj + dk : a2 + b2 + c2 + d2 = 1}.

• Any group G contains two trivial subgroups: {e} and G. Any other subgroup is said to be a proper subgroup of G.

By abuse of notation, the trivial subgroup {e} is usually denoted simply by 1 (or 0 if additive notation is being used). Notation: Given a group G, its cardinal will be denoted by |G|. Thus, if G is ﬁnite, |G| is the number of elements of G. This is called the order of G. In general, multiplicative notation as above is used, unless we are dealing with abelian groups, like (Z, +); in this case, additive notation is used:

Multiplicative notation Additive notation Neutral element e or 1 0 operation xy x + y inverse x−1 −x powers xn nx

§ 2. Cyclic groups. Generators

n 2.1 Deﬁnition. Let G be an arbitrary group and g ∈ G. Then {g : n ∈ Z} is a subgroup of G which is called the cyclic subgroup generated by g. It will be denoted by hgi. If there exists an element g ∈ G such that G = hgi, then G is said to be a cyclic group.

Note than any cyclic group is commutative, as gngm = gn+m = gmgn for any n, m ∈ Z.

2.2 Examples.

(i) (Z, +) is a cyclic group generated by 1 (and by −1). 4 CHAPTER 1. GROUPS

(ii) Let Pn be a regular polygon of n sides on the euclidean plane, and let

G = {rotations of the plane which leave Pn ﬁxed}.

2π Then G = hϕi, where ϕ is the rotation of angle n centered at the barycenter of the polygon. Note that G = {id, ϕ, ϕ2, . . . , ϕn−1} and (ϕi)−1 = ϕn−i for any i.

2.3 Proposition. Let G be a group and g ∈ G. Then: ( ∞, if gn 6= e for any n ≥ 1, |hgi| = n r = min{n ∈ N : g = e}, otherwise.

In the first case, we say that the order of g is infinite (and write |g| = ∞), while in the second case that g has finite order r (and write |g| = r).

n n m Proof. If g 6= e for any n ≥ 1 and there are n ≤ m ∈ Z such that g = g , then multiply by g−n to get gm−n = e, so n − m = 0. Therefore, if gn 6= e n for any n ≥ 1, then the elements g , n ∈ Z, are all diﬀerent and |hgi| = ∞. On the contrary, if there exists a natural number n such that gn = 1, n take r = min{n ∈ N : g = e}. As before, it is easy to see that the elements 2 r−1 e, g, g , . . . , g are all diﬀerent. Moreover, for any m ∈ Z, there are unique m r q s q s q, s ∈ Z with 0 ≤ s < r such that m = qr + s. Then g = (g ) g = e g = gs, so that hgi = {e, g, . . . , gr−1} has order r.

Note that the proof above shows that if the order of g ∈ G is r, then n g = e for some n ∈ Z if and only if r divides n. Also note that hgi is the smallest subgroup containing the element g. In the same way, given a subset S of a group G, the subgroup generated by S is, by deﬁnition, the smallest subgroup of G containing S, and it is denoted by hSi. In other words, hSi is the intersection of all the subgroups of G containing S (as the intersection of a family of subgroups is always a subgroup): \ hSi = {H : H ≤ G and S ⊆ H}.

2.4 Proposition. Given a subset S of a group G, the subgroup generated by S is the subset of G consisting of e and of all the possible products

t1t2 ··· tn

−1 with n ∈ N and such that for any 1 ≤ i ≤ n, either ti ∈ S or ti ∈ S.

2.5 Examples.

• hGi = G, h∅i = 1 (trivial subgroups). § 3. HOMOMORPHISMS 5

• Let Q be the quaternion group, then Q = hi, ji. 2.6 Definition. A group G is said to be finitely generated if there exists a finite subset S of G such that G = hSi. Note that, in particular, any cyclic group is finitely generated. 2.7 Proposition. Let G be a group, and let a, b be two elements of G such that ab = ba, |a| = s, |b| = t, with s, t ∈ N and gcd(s, t) = 1. Then ha, bi = habi is a cyclic group of order st. Proof. Obviously ab ∈ ha, bi, so we get habi ⊆ ha, bi. Let d be an element in hai ∩ hbi, then there are natural numbers i and j such that d = ai = bj. Thus, bsj = (bj)s = ds = (ai)s = ais = (as)i = ei = e. Therefore t divides sj, and since gcd(s, t) = 1, we conclude that t divides j, but then d = bj = e. Hence hai ∩ hbi = 1. Now, for any natural number n, we get (ab)n = e ⇐⇒ anbn = e since ab = ba, ⇐⇒ an = b−n ∈ hai ∩ hbi, ⇐⇒ an = bn = e, ⇐⇒ both s and t divide n, ⇐⇒ st divides n (since gcd(s, t) = 1). Therefore |ab| = st and habi is a cyclic group of order st. Moreover, ha, bi = {aibj : 0 ≤ i ≤ s − 1, 0 ≤ j ≤ t − 1}, and hence the order of ha, bi is at most st. Since habi is contained in ha, bi, and |ab| = st, we conclude that habi = ha, bi, as required.

§ 3. Homomorphisms

3.1 Deﬁnition. Let G and H be two groups and let ϕ : G → H be a map. Then: • ϕ is said to be a group homomorphism (or just a homomorphism) if for any x, y ∈ G, ϕ(xy) = ϕ(x)ϕ(y). • If ϕ is a group homomorphism, then its kernel is the subset ker ϕ = ϕ−1(e) of G, while its image is the set im ϕ = ϕ(G). • A group homomorphism is said to be a monomorphism if it is one- to-one, an epimorphism if it is surjective, and an isomorphism if it is a bijection. Moreover, the isomorphisms ψ : G → G are called automorphisms. 6 CHAPTER 1. GROUPS

3.2 Examples.

• Let Cn be a cyclic group of order n. If g is a generator of Cn, then m the map ϕ :(Z, +) → G given by ϕ(m) = g is an epimorphism with ker ϕ = nZ.

• Two cyclic groups are isomorphic if and only if they have the same order. To see this, note that if G is an inﬁnite cyclic group, then the example above shows that G is isomorphic to (Z, +), while the same arguments show that Cn is isomorphic to (Z/nZ, +).

+ x • The map ϕ :(R, +) → (R ,.), ϕ(x) = e , is an isomorphism. (Here + R denotes the set of positive real numbers.)

cos t − sin t • The map ϕ :(R, +) → SO(2, R), ϕ(t) = sin t cos t , is an epimorphism with ker ϕ = 2πZ.

• The determinant map det : GL(n, R) → (R \{0},.), A 7→ det A, is an epimorphism with ker det = SL(n, R).

• Let G be a group and let a ∈ G. The map Ia : G → G given by −1 Ia(x) = axa (conjugation by the element a) is an automorphism, which is called an inner automorphism. The set Aut G of automorphisms of G is a group under the composition of maps, and the map I : G → Aut G, given by I(a) = Ia is a homomorphism, because IaIb = Iab. Its image im I is the subgroup of inner automorphisms, denoted by Int G, while its kernel is ker I = {a ∈ G : Ia = id} = {a ∈ G : ax = xa ∀x ∈ G}, which is called the center of G (and denoted by Z(G)).

3.3 Properties. Let ϕ : G → H be a group homomorphism. Then:

(i) ϕ(e) = e and ϕ(x−1) = ϕ(x)−1 for any x ∈ G,

(ii) im ϕ is a subgroup of H,

(iii) ker ϕ is a subgroup of G which satisﬁes the following extra condition: ∀x ∈ ker ϕ, ∀a ∈ G, axa−1 ∈ ker ϕ.

This last property deserves a name:

3.4 Deﬁnition. Let G be a group and let N be a subgroup of G. Then N is said to be a normal subgroup of G if for any x ∈ N and a ∈ G, the −1 element axa belongs to N. (This will be denoted by N E G.) § 3. HOMOMORPHISMS 7

Therefore the kernels of group homomorphisms are normal subgroups. The trivial subgroups 1 and the whole G are always normal subgroups of a group G. If G is abelian, then any subgroup is a normal subgroup. Given a group G and a subgroup H, deﬁne the following relation among elements in G: a ∼ b if a−1b ∈ H.

This is clearly an equivalence relation. The equivalence class of the element a is the subset

{b ∈ G : a−1b ∈ H} = {b ∈ G : ∃h ∈ H such that b = ah} =: aH, which is called the left coset of a modulo H. In the same vein we may deﬁne the equivalence relation where a ∼ b if ab−1 ∈ H, whose equivalence classes are the right cosets modulo H (the sets Ha for a ∈ G). The set of left cosets modulo H (the quotient set by the equivalence relation) is denoted by G/H, and its cardinal by [G : H], which is called the index of the subgroup H in G. Note that the map H = eH → aH, h 7→ ah, is bijective, and hence all the left cosets modulo H have the same cardinal, equal to |H|. It follows that G is the disjoint union of the equivalence classes, and hence

|G| = [G : H] |H|.

3.5 Proposition. (Lagrange’s Theorem) The order of a ﬁnite group is a multiple of the order of any of its subgroups.

3.6 Corollary.

(i) The order of any element of a ﬁnite group divides the order of the group.

(ii) If the order of a group G is a prime number, then G is cyclic.

In general, given a divisor of the order of a ﬁnite group, there may be no subgroup whose order is this divisor. However, for cyclic groups we have the following result:

3.7 Theorem. Any subgroup of a cyclic group is cyclic. Moreover, the subgroups of (Z, +) are the subgroups (mZ, +) for m ∈ N ∪ {0}, and the subgroups of (Z/nZ, +) are in one-to-one correspondence with the divisors of n. 8 CHAPTER 1. GROUPS

Proof. Let H be a subgroup of either (Z, +) or (Z/nZ, +). Then for any m ∈ Z and a ∈ H, ma = a + ··· + a (m summands) if m > 0, while ma = (−a) + ··· + (−a)(−m summands) if m < 0. In both cases, ma ∈ H as H is closed under the operation (the sum) and the inverse (the opposite). It follows that H is an ideal of Z or of Z/nZ, and the result follows from the corresponding result in the course Algebraic Structures. (Anyway, it is very easy to provide a direct proof.)

Using the fact that any cyclic group is isomorphic to either (Z, +) or to (Z/nZ, +) for some n ∈ N, we get the following consequence:

3.8 Corollary. Let G = hai be a cyclic group of order q ∈ N and let d ∈ N be a divisor of q. Then the subgroup of G of order d is (with additive notation) q H = h d ai = {b ∈ G : db = 0}.

Let now N be a normal subgroup of a group G. Then the equivalence relation considered above (a ∼ b if a−1b ∈ N) satisﬁes the following property:

a ∼ b and c ∼ d =⇒ ac ∼ bd, because (ac)−1bd = c−1a−1bd = c−1(a−1b)c(c−1d), and both c−1d and c−1(a−1b)c are in N (the latter one because N is normal). Therefore we may deﬁne a multiplication on the quotient set G/N as follows: (aN)(bN) = (ab)N, for any a, b ∈ G. This multiplication is associative, the element eN is its neutral element, and the inverse of aN is given by a−1N. Hence, G/N is a group, called the quotient group of G modulo N. With the same sort of arguments used for rings, the following results can easily be proved:

3.9 Theorem.

1. First Isomorphism Theorem: Let ϕ : G → H be a group homomorphism, then the quotient group G/ ker ϕ is isomorphic to im ϕ through the isomorphism

ϕ¯ : G/ ker ϕ → im ϕ a ker ϕ 7→ ϕ(a).

2. Let N be a normal subgroup of a group G, then the map π : G → G/N, a 7→ aN, is an epimorphism, called the natural projection of G over G/N. Besides, ker π = N. In particular, this shows that any normal subgroup is the kernel of some homomorphism. § 4. SYMMETRIC GROUP 9

3. Let ϕ : G → H be a group homomorphism, then ϕ is a monomorphism if and only if ker ϕ = 1, while ϕ is an epimorphism if and only if im ϕ = H.

4. Second Isomorphism Theorem: Let H be a subgroup and N a normal subgroup of a group G. Then HN = {hn : h ∈ H, n ∈ N} is a subgroup of G, N is a normal subgroup of HN, H ∩ N is a normal subgroup of H and the map

H/H ∩ N → HN/N h(H ∩ N) 7→ hN,

is an isomorphism.

5. Third isomorphism theorem: Let H and K be two normal subgroups of a group G with H ⊆ K, then K/H is a normal subgroup of G/H and the quotient groups (G/H)/(K/H) and G/K are isomorphic.

6. Let N be a normal subgroup of a group G, then the map

{subgroups of G containing N} → {subgroups of G/N} H 7→ H/N,

is a bijection. The inverse map is given by H˜ ≤ G/N 7→ H = {g ∈ G : gN ∈ H˜ }. The same result is valid changing subgroups for normal subgroups.

§ 4. Symmetric group

We have already seen that given an arbitrary set Ω, the set S(Ω) = {f : Ω → Ω: f is a bijective map} is a group with the composition of maps. Take Ω = {1, 2, . . . , n}, then the group S(Ω) is denoted by Sn and called the symmetric group of degree n. The elements of Sn can be represented like this:

1 2 . . . n π = . π(1) π(2) . . . π(n)

For obvious reasons, the elements of Sn are called permutations.

4.1 Example. Consider the symmetric group S4 and its elements

1 2 3 4 1 2 3 4 σ = , τ = . 2 3 4 1 4 3 2 1 10 CHAPTER 1. GROUPS

Then,

1 2 3 4 ↓ ↓ ↓ ↓ 1 2 3 4 1 2 3 4 στ = 4 3 2 1 = , τσ = . 1 4 3 2 3 2 1 4 ↓ ↓ ↓ ↓ 1 4 3 2

4.2 Proposition. The order of the symmetric group of degree n is |Sn| = n!. 1 2 ... n Its neutral element is the identity map: ( 1 2 ... n ). Let us try to decompose any permutation into a product of simpler permutations: 1 2 3 4 5 6 7 8 4.3 Example. π = 2 4 8 1 3 5 7 6 Under the action of π, the elements are permuted as follows:

1 7→ 2 7→ 4 7→ 1, 3 7→ 8 7→ 6 7→ 5 7→ 3, 7 7→ 7.

There are thus three disjoint cycles. We will write then π = (1 2 4)(3 8 6 5)(7) or simply (omitting the cycles of length 1):

π = (1 2 4)(3 8 6 5).

In general, given any permutation π ∈ Sn and any i ∈ Ω = {1, . . . , n}, the orbit of the element i under the action of π is the set {i, π(i), π2(i),...} ⊆ Ω. Sooner or later, there appear natural numbers h < k with πh(i) = πk(i), so that πk−h(i) = i. Hence, if l is the lowest natural number such that πl(i) = i, then the orbit of i is precisely {i, π(i), . . . , πl−1(i)}. This number l is then called the length of the orbit. The diﬀerent orbits give a partition of Ω: Ω = Ω1 ∪ · · · ∪ Ωp (disjoint union). l −1 If |Ωk| = lk, then for any i ∈ Ωk,Ωk = {i, π(i), . . . , π k (i)}. The permutation π acts independently on each orbit and hence π is the product π = π1 ··· πp, where for each k, ( π(i) if i ∈ Ωk, πk(i) = i otherwise.

The permutations πk are cycles of length lk. In case lk = 1, then πk = id. Therefore we get the following result:

4.4 Theorem. Any permutation in Sn diﬀerent from the identity can be expressed uniquely as a product of “disjoint” cycles of length ≥ 2, up to the order of these cycles. § 4. SYMMETRIC GROUP 11

4.5 Corollary. Let π be an element of Sn expressed as a product π = π1 ··· πp of disjoint cycles of length ≥ 2. Let the length of πk be lk for k = 1, . . . , p. Then the order of π is the lowest common multiple of the lk’s: |π| = lcm(l1, . . . , lp).

Proof. Since the πk’s are disjoint, they commute, so for any n ∈ N,

n n n π = π1 ··· πp ,

n n and π = 1 if and only if πk = 1 for any k, if and only if lk divides n for any k, as required. (Note that the symbol 1 is used for the identity map.)

4.6 Example. What are the possible orders of the elements of S6? Because of the previous Corollary, it is enough to see what are the different ways to decompose {1, 2, 3, 4, 5, 6} into disjoint orbits. That is, the diﬀerent ways of decomposing 6 as a sum of natural numbers:

Decomposition example order 1+1+1+1+1+1 id 1 2+1+1+1+1 (1 2) 2 2+2+1+1 (1 2)(3 4) 2 2+2+2 (1 2)(3 4)(5 6) 2 3+1+1+1 (1 2 3) 3 3+2+1 (1 2 3)(4 5) 6 3+3 (1 2 3)(4 5 6) 3 4+1+1 (1 2 3 4) 4 4+2 (1 2 3 4)(5 6) 4 5+1 (1 2 3 4 5) 5 6 (1 2 3 4 5 6) 6

4.7 Deﬁnition. A cycle of length 2 is called a transposition.

4.8 Proposition. Any permutation can be expressed as a product of transpositions.

Proof. It is enough to check that any cycle is a product of transpositions. For this just note the following:

(1 2 ··· l) = (1 l)(1 l − 1)(1 l − 2) ··· (1 3)(1 2).

Note that the expression of a permutation as a product of transpositions is not unique in general. For instance, in S4 we have:

(1 2 3) = (1 3)(1 2) = (2 3)(1 3) = (1 3)(2 4)(1 2)(1 4). 12 CHAPTER 1. GROUPS

4.9 Proposition. Given a natural number n, consider the map:

ϕ : Sn −→ GL(n, R)  

σ 7→ Eσ(1) ··· Eσ(n)

where Ei denotes the column with a 1 in the i-th position, and 0’s elsewhere. In other words, ϕ(σ) is the matrix aij with aij = 1 if i = σ(j) and aij = 0 otherwise. Then ϕ is a group homomorphism. Proof. For any σ, τ ∈ Sn, consider ϕ(σ) = aij , ϕ(τ) = bij and ϕ(στ) = cij . Then, for any i, j: ( 1 if i = στ(j), ai1b1j + ai2b2j + ··· + ainbnj = aiτ(j)bτ(j)j = 0 otherwise.

Therefore cij = ai1b1j + ai2b2j + ··· + ainbnj and ϕ(στ) = ϕ(σ)ϕ(τ).

× Since the determinant map det : GL(n, R) → R is a homomorphism too, so is the composition det ◦ϕ. Moreover, det ◦ϕ(σ) = ±1, as the determinant of the matrix associated to any transposition is −1 (the corresponding matrix is the identity matrix with two columns permuted), and any σ is a product of transpositions.

4.10 Deﬁnition. The epimorphism sgn : Sn → C2 = {1, −1} (C2 is the cyclic group of two elements) given by sgn(σ) = detϕ(σ) is called the signature homomorphism. The value sgn(σ) is said to be the signature of the permutation σ. Its kernel is denoted by An and called the alternating group of degree n. The permutations in An are said to be even, while the permutations in Sn \ An are said to be odd.

Note that An is a normal subgroup of Sn of index 2, as Sn/An is isomorphic to C2.

4.11 Theorem. Let π be a permutation in the symmetric group Sn.

(i) If π = τ1τ2 ··· τp for some transpositions τi, i = 1, . . . , p, then sgn π = (−1)p.

(ii) If π = π1π2 ··· πq for some disjoint cycles πi of length li, then sgn π = Pq (l −1) (−1) i=1 i . Qp p Proof. For (i) just note that sgn π = i=1 sgn(τi) = (−1) , as the signature of any transposition is −1. For (ii) note that the signature of a cycle of length l is (−1)l−1, as (1 2 ··· l) = (1 l)(1 l − 1)(1 l − 2) ··· (1 3)(1 2) (a product of l − 1 transpositions). Now (i) can be applied. § 5. GROUP ACTIONS 13

Symmetric groups are quite important, partly because of the next result: 4.12 Cayley’s Theorem. Let G be a group of order n. Then G is isomorphic to a subgroup of Sn.

Proof. We may identify Sn with S(G) = {f : G → G : f is a bijection}. Consider the “left action”: L : G −→ S(G)

a 7→ La : G → G x 7→ ax

For any a ∈ G, La is bijective with inverse La−1 , and Lab = La ◦ Lb for any a, b ∈ G. Thus, L is a group homomorphism. Moreover, if a ∈ ker L, then La = id, that is ax = x for any x ∈ G. In particular a = ae = e. Hence ker L = 1, and G ' im G ≤ S(G).

§ 5. Group actions

Groups made their appearance as “groups of symmetries”; that is, their elements are symmetry maps on some object. 5.1 Deﬁnition. Let G be a group and Ω a set. An action of G on Ω (on the left) is a map Φ: G × Ω −→ Ω (g, x) 7→ gx satisfying the following two properties: (i) ex = x for any x ∈ Ω, (ii) (gh)x = g(hx) for any g, h ∈ G and x ∈ Ω. Note that if Φ is a group action, then the map (which will be again denoted by Φ): Φ: G −→ S(Ω)

g 7→ Φg :Ω → Ω x 7→ gx is a group homomorphism. And conversely, given any group homomorphism Φ: G → S(Ω), g 7→ Φg, the map G × Ω → Ω, given by (g, x) 7→ Φg(x), is an action of the group G on Ω. Therefore, {Actions of G on Ω} =∼ {Homomorphisms G → S(Ω)}, and an action of a group can be considered too as a homomorphism Φ : G → S(Ω). 14 CHAPTER 1. GROUPS

5.2 Examples.

(i) Any subgroup of Sn acts on {1, 2, . . . , n} in a natural way. (ii) In Cayley’s Theorem we considered the action of a group on itself by left multiplication: L : G −→ S(G)

g 7→ Lg : G → G x 7→ gx which corresponds to the map G × G → G,(g, x) 7→ gx. (iii) Recall the deﬁnition of similar matrices in Linear Algebra. This corresponds to the action

GL(n, R) × Matn(R) −→ Matn(R) (A, B) 7→ ABA−1.

5.3 Deﬁnition. Let Φ : G → S(Ω) be an action of a group G on a set Ω. Then: • ker Φ is said to be the kernel of the action. If ker Φ = 1, then the action is said to be faithful, and then G can be identiﬁed to a subgroup of S(Ω) through Φ. • There appears the following equivalence relation on Ω: x ∼ y ⇐⇒ ∃g ∈ G such that gx = y. The equivalence classes are called the orbits of the action. The orbit of an element x ∈ Ω is denoted by orb(x) or simply by Gx. If there is just one orbit, then the action is called transitive. (This means that for any x, y ∈ Ω, there is an element g ∈ G such that gx = y.) • For any x ∈ Ω, the stabilizer of x is the subgroup

Gx = {g ∈ G : gx = x}.

5.4 Proposition. Let Φ: G → S(Ω) be an action of a group G on a set Ω. Then:

(i) For any x ∈ Ω, |Gx| = [G : Gx]. In particular, if G is ﬁnite, the length (cardinal) of any orbit divides the order of G. Therefore, if {Ωi : i ∈ I} is the set of orbits of the action, and {xi : i ∈ I} is a set of representatives of the orbits (that is, xi ∈ Ωi for any i ∈ I) then X X |Ω| = |Ωi| = [G : Gxi ] i∈I i∈I § 5. GROUP ACTIONS 15

−1 (ii) Let g ∈ G and x, y ∈ Ω be such that gx = y, then Gy = gGxg = Ig(Gx).

(iii) Cauchy-Frobenius: If G is ﬁnite and for any g ∈ G we consider F (g) = {x ∈ Ω: gx = x} (the set of ﬁxed elements by g), then

1 X number of orbits = |F (g)| |G| g∈G

0 −1 0 Proof. For (i) just note that gx = g x if and only if g g ∈ Gx, if and only 0 if g ∈ gGx. Hence the map Gx → G/Gx, gx 7→ gGx is well deﬁned and bijective, so |Gx| = [G : Gx]. Also note that if gx = y and hy = y, then hgx = gx or g−1hgx = x, −1 and conversely. Then h ∈ Gy if and only if g hg ∈ Gx, if and only if −1 h ∈ gGxg , thus obtaining the result in (ii). Therefore, if x and y belong to the same orbit, then |Gx| = |Gy| and [G : Gx] = [G : Gy]. From here, the last part of (i) follows. Consider now the set X = {(g, x) ∈ G × Ω: gx = x}. We may count the elements of X in two ways: X X |X| = |F (g)| and |X| = |Gx|. g∈G x∈Ω

But because of (ii) we get: X X |X| = |Gx| = |Gxi||Gxi | x∈Ω i∈I X = |G| = |G|(number of orbits). i∈I

Hence X |G|(number of orbits) = |X| = |F (g)|, g∈G as required.

5.5 Example. A bracelet is made by stringing together six beads, two of them are green and the other four beads are red. How many different patterns are possible? Let P6 be the regular hexagon. We put the beads on the vertices of P6. 6 6 There are 2 = 4 = 15 ways of doing this. However, two of these ways give the same pattern if there is a symmetry of P6 which takes one to the other. Therefore, the number of different patterns is the number of orbits of the action of the group of symmetries of P6 acting on the set of the 15 different possibilities. 16 CHAPTER 1. GROUPS

6...... 1 ...... 5 ...... 2 ...... 34

The group of symmetries of the regular hexagon is the dihedral group D6 of degree 6:

2 5 45 56 61 D6 = {id, ϕ, ϕ , . . . , ϕ , s14, s25, s36, s12, s23, s34}

2π where ϕ is the clockwise rotation of angle 6 , s14 is the reflection relative 45 to the axis which goes through the vertices 1 and 4, s12 is the reflection relative to the axis which goes through the middle points of the sides which join the vertices 1 and 2 and the vertices 4 and 5, and similarly for the other elements. Now we compute the fixed elements in the set Ω of our 15 possibilities by each element of the group D6, in order to apply Cauchy-Frobenius Theorem: • |F (id)| = 15,

• |F (ϕ)| = |F (ϕ2)| = |F (ϕ4)| = |F (ϕ5)| = 0,

• |F (ϕ3)| = 3 (which correspond to the green beads in vertices 1 and 4, in 2 and 5, and in 3 and 6),

• |F (s14)| = 3 (which correspond to the green beads in 1 and 4, in 2 and 6 and in 3 and 5), and in the same vein |F (s25)| = |F (s36)| = 3,

45 • |F (s12)| = 3 (which correspond to the green beads in 1 and 2, in 3 and 56 61 6 and in 4 and 5), and in the same vein |F (s23)| = |F (s34)| = 3. Therefore: 1 36 number of patterns = 15 + 3 + 3 × 3 + 3 × 3 = = 3. 12 12 Note that the result is clear: the green beads may be in adjacent vertices, or separated by one vertex, or in opposite vertices. 5.6 Example. How many diﬀerent chemical compounds can be obtained adding radicals CH3 or H to a benzene ring? Here we have two possible colors (CH3 or H) for each vertex of the hexagon. Hence there are 26 = 64 diﬀerent possibilities. A similar computation to the one performed before gives: § 6. CONJUGATION. TRANSLATION 17

...... CC ...... C C ...... CC......

• |F (id)| = 64,

• |F (ϕ)| = |F (ϕ5)| = 2,

• |F (ϕ2)| = |F (ϕ4)| = 22,

• |F (ϕ3)| = 23,

4 • |F (s14)| = |F (s25)| = |F (s36)| = 2 ,

45 56 61 3 • |F (s12)| = |F (s23)| = |F (s34| = 2 . And hence: 1 number of compounds = 64 + 2 × 2 + 2 × 22 + 23 + 3 × 24 + 3 × 23 = 13. 12

§ 6. Conjugation. Translation

6.1. Action by conjugation The action by conjugation of a group on itself is given by the homomorphism Φ: G −→ Aut G ≤ S(G)

g 7→ Ig : G → G x 7→ gxg−1 We already know that its kernel is the center of G:

ker Φ = Z(G) = {g ∈ G : gx = xg ∀x ∈ G}.

The orbits of this action are called conjugacy classes:

orb(x) = {y ∈ G : ∃g ∈ G such that y = gxg−1}.

The stabilizer of an element x ∈ G is called the centralizer of x:

CG(x) = {g ∈ G : gx = xg}. 18 CHAPTER 1. GROUPS

Note that the orbit of x restricts to {x} if and only if CG(x) = G, if and only if x ∈ Z(G). Therefore, if G is a ﬁnite group and x1, . . . , xr are representatives of the orbits containing at least two elements, then

r X |G| = |Z(G)| + [G : CG(xi)] , i=1 which is called the class equation of the group G. 6.1 Corollary. Let p be a prime number and let G be a ﬁnite p-group (that is, the order of G is a power of p). Then Z(G) 6= 1. Moreover, if |G| = p2, then G is abelian (G = Z(G)).

Proof. Consider the class equation of G, and note that since [G : CG(xi)] ≥ 2, and it divides the order of G,[G : CG(xi)] is a power of p for any i. It follows that p divides |Z(G)|, and hence Z(G) 6= 1. Moreover, assume that G is a non abelian group with |G| = p2. Take an element x ∈ G \ Z(G). By the above Z(G) is a subgroup of order p, and hence it is cyclic: Z(G) = hyi. Then hx, yi is a subgroup, whose order is greater than p and divides the order of G. Since |G| = p2, it follows that G = hx, yi. But x and y commute as y ∈ Z(G), so G is abelian, a contradiction.

6.2 Example. (Conjugation in Sn) Let π, σ be two permutations in Sn and assume that π(i) = j. Then

σπσ−1σ(i) = σπ(i) = σ(j).

−1 That is, if we have i −→π j, then σ(i) −−−−→σπσ σ(j). Therefore, if π is expressed as a product of disjoint cycles:

π = (i1 . . . ir)(j1 . . . js) ··· , then −1 σπσ = σ(i1) . . . σ(ir) σ(j1) . . . σ(jr) ···

Therefore, two elements of Sn are conjugate if and only if they have the same cycle structure (that is, they can be expressed as a product of disjoint cycles of the same length). The type of the permutation π is the sequence (α1, . . . , αn), where αi is the number of cycles of length i (the number of orbits of length i in the action of π on {1, 2, . . . , n}). Note that α1 + 2α2 + ··· + nαn = n. Thus, for example, for n = 5, the elements

1, (1 2), (1 2)(3 4), (1 2 3), (1 2 3)(4 5), (1 2 3 4), (1 2 3 4 5), form a representative system of the conjugacy classes in S5. § 6. CONJUGATION. TRANSLATION 19

The action by conjugation can be extended to an action of G on Ω = {subsets of G}:

Φ:˜ G −→ S(Ω)

g 7→ Ig :Ω → Ω S 7→ gSg−1

Two subsets S and T are said to be conjugate if there is an element g ∈ G such that T = gSg−1. Given a subgroup H of G, its stabilizer under this action is called the normalizer of H:

−1 NG(H) = {g ∈ G : gHg = H}.

Note that H is a normal subgroup of its normalizer, and that H is normal if and only if NG(H) = G. The length of the orbit of H is then [G : NG(H)].

6.2. Action by translation The action by (left) translation of a group on itself is given by the homomorphism

L : G −→ S(G)

g 7→ Lg : G → G x 7→ gx

This action has already been used in Cayley’s Theorem (4.12). If H is a subgroup of G, this action L induces an action

LH : G −→ S(G/H) H g 7→ Lg : G/H → G/H xH 7→ gxH which is well deﬁned, as xH = yH if and only if x−1y ∈ H, if and only if (x−1g−1)(gy) ∈ H, if and only if gxH = gyH. The kernel of this action is given by:

ker LH = {g ∈ G : gxH = xH ∀x ∈ G} = {g ∈ G : x−1gx ∈ H ∀x ∈ G} = {g ∈ G : g ∈ xHx−1 ∀x ∈ G} −1 = ∩x∈GxHx , which is the largest normal subgroup of G contained in H. 20 CHAPTER 1. GROUPS

6.3 Corollary. (i) If H is a subgroup of G of index n ([G : H] = n) and such that H does not contain any nontrivial normal subgroup of G, then |G| divides n!. (ii) If p is the lowest prime dividing the order of a ﬁnite group G, and if H is a subgroup of G with [G : H] = p, then H is a normal subgroup. H ∼ Proof. For (i) note that L : G → S(G/H) = Sn is a monomorphism. For (ii) consider again the action LH and its kernel K = ker LH . K is a normal subgroup of G contained in H, and [G : K] divides p! = |S(G/H)|. But [G : K] = [G : H][H : K] = p[H : K]. Hence [H : K] divides (p − 1)!. However, the prime factors of [H : K] are greater or equal than p. The only possibility is [H : K] = 1, or H = K, which is normal.

§ 7. Sylow’s Theorems

7.1 Definition. Let G be a finite group and let p be a prime number with n n |G| = p m and p - m. The subgroups of G of order p are called the Sylow p-subgroups of G. The objective of this section is the proof of the following result: 7.2 Theorem. (Sylow, 1872) Let G be a finite group and let p be a n prime number with |G| = p m and p-m. Then: (i) G contains Sylow p-subgroups, (ii) Any two Sylow p-subgroups are conjugate, (iii) The number of Sylow p-subgroups of G is congruent to 1 modulo p. A previous Lemma, due to Cauchy, is needed: 7.3 Lemma. Let G be a finite group and let p be a prime number dividing the order of G. Then there are elements of G of order p. Proof. Assume first that G is abelian, and take a nontrivial element g ∈ G. r If p | |g|, then |g| = pr for some r ∈ N, and g is an order p element. Otherwise, H = hgi is a normal subgroup of G (as G is abelian), and it can be assumed by an inductive argument that there is an element xH ∈ G/H of order p. If |x| = m, then we have (xH)m = eH, so p = |xH| divides m, and again xm/p has order p. Pr If G is not abelian, consider its class equation |G| = |Z(G)| + i=1[G : CG(xi)]. If p | |Z(G)|, then there are elements of order p in the abelian subgroup Z(G). Otherwise there is an i such that p-[G : CG(xi)]. But then p divides |CG(xi)|, and an inductive argument finishes the proof. § 7. SYLOW’S THEOREMS 21

Proof of the Theorem. For part (i), consider again the class equation Pr |G| = |Z(G)| + i=1[G : CG(xi)]. If p - |Z(G)|, then again there is an n 0 0 i such that p - [G : CG(xi)], so that |CG(xi)| = p m for some m (since |G| = [G : CG(xi)]|CG(xi)|). An inductive argument shows that there is a n subgroup of CG(xi) of order p , and this is a Sylow subgroup of G. Otherwise p | |Z(G)|, so by Cauchy’s Lemma, there is an element z ∈ Z(G) of order p. Since z is central, N = hzi is a normal subgroup of G and |G/N| = pn−1m. By an inductive argument there is a subgroup P/N ≤ G/N of order pn−1, and hence |P | = pn, so it is a Sylow p-subgroup. We will prove an assertion which is stronger than (ii): Let P be a Sylow p-subgroup of G, and let Q be any p-subgroup of G, then we will check that Q is contained in a conjugate of P . Actually, consider the action Q −→ S(G/P ) P g 7→ Lg : G/P → G/P xP 7→ gxP.

Write Ω = G/P , and let Ω1,..., Ωr be the orbits of this action. Since |Ωi| Pr divides |Q| (a power of p) for any i and m = |G/P | = |Ω| = i=1|Ωi|, and p - m, it follows that there is an i such that |Ωi| = 1. That is, there is an element x ∈ G such that gxP = xP for any g ∈ Q. This means that x−1gx ∈ P for any g ∈ Q, or Q ≤ xP x−1. As for (iii), let P be a Sylow p-subgroup, and let Ω be the set of Sylow p-subgroups of G, which is the set of subgroups conjugate to P . Consider the action by conjugation: P −→ S(Ω) g 7→ Ω → Ω Q 7→ gQg−1.

If an orbit of this action contains just one element Q, then gQg−1 = Q for any g ∈ P , so that P ≤ NG(Q), but then P and Q are Sylow p-subgroups of NG(Q), and hence they are conjugate in NG(Q). Since Q is a normal subgroup of NG(Q), it follows that Q = P . Therefore there is just one orbit consisting of one element, namely {P }. The length of any other orbit is a nontrivial divisor of |P |, and hence it is a power of p. We conclude that |Ω| = 1 + P(powers of p), and this is congruent to 1 modulo p. 7.4 Remark. If P is a Sylow p-subgroup of a ﬁnite group G, then the number of Sylow p-subgroups is the number of subgroups conjugate to P , which is [G : NG(P )], and this is a divisor of [G : P ] and of |G|. 7.5 Corollary. Let P be a Sylow p-subgroup of a ﬁnite group G, then P is the only Sylow p-subgroup of G if and only if P is a normal subgroup of G, if and only if P contains all p-subgroups of G. 22 CHAPTER 1. GROUPS

7.6 Example. Let p be a prime number, consider the following classical group:

a b G = SL(2, ) = : a, b, c, d ∈ , ad − bc = 1 . Zp c d Zp

We are going to compute the number of Sylow p-subgroups of G. × First, since det : GL(2, Zp) → Zp is a group epimorphism, and SL(2, Zp) equals ker det, it follows that

|GL(2, )| (p2 − 1)(p2 − p) |G| = Zp = = p(p2 − 1). p − 1 p − 1

On the other hand, it is clear that

1 a 1 0 P = : a ∈ and P = : a ∈ 1 0 1 Zp 2 a 1 Zp are two Sylow p-subgroups. Hence P1 is not a normal subgroup, so that NG(P1) 6= G. Also, for any 0 6= b ∈ Zp,

b 0 1 a b−1 0 1 b2a = , 0 b−1 0 1 0 b 0 1

b a × so that H = 0 b−1 : b ∈ Zp , a ∈ Zp is a subgroup contained in NG(P1). But |H| = p(p − 1), NG(P1) 6= G, and [G : NG(P1)] = 1 + kp for some k ≥ 1 (because of the “Third Sylow’s Theorem”). The only possibility is [G : NG(P1)] = p + 1 and H = NG(P1). Therefore the number of Sylow p-subgroups of G is p + 1.

§ 8. Direct products

8.1 Deﬁnition. Let G1,...,Gn be groups, then the cartesian product G = G1 × · · · × Gn is a group with the operation deﬁned componentwise. It is called the direct product of the groups G1,...,Gn.

Note that if G = G1 × · · · × Gn is a direct product, for each i the subset Hi = {(e, . . . , x, . . . , e): x ∈ Gi} (n-tuples with an element of Gi in the ith position, and the neutral element of the group Gj in the jth position for any j 6= i) is a normal subgroup of G isomorphic to Gi. If each of the groups Gi is abelian, so is G. Moreover, if additive notation is used for the abelian groups, then the direct product is denoted by G1 ⊕ · · · ⊕ Gn. § 8. DIRECT PRODUCTS 23

8.2 Theorem. Let G be a group, and let H1,...,Hn be normal subgroups of G satisfying that G = hH1,...,Hni and that for any i = 1, . . . , n, Hi ∩ hH1,...,Hi−1,Hi+1,...,Hni = 1. Then the map

ϕ : H1 × · · · × Hn −→ G

(h1, . . . , hn) 7→ h1h2 ··· hn is a group isomorphism.

Proof. For any hi ∈ Hi and hj ∈ Hj, i 6= j,

( −1 −1 −1 −1 (hihjhi )hj ∈ Hj since Hj E G, hihjhi hj = −1 −1 hi(hjhi hj ) ∈ Hi since Hi E G,

−1 −1 so that, as Hi ∩ Hj = 1, it follows that hihjhi hj = e, or hihj = hjhi. This shows that ϕ is a group homomorphism. Moreover, H1H2 ··· Hn = {h1h2 ··· hn : hi ∈ Hi ∀i} is then a normal subgroup of G which contains all the Hi’s. Hence G = hH1,...,Hni = H1H2 ··· Hn, and hence ϕ is an epimorphism. Finally, if (h1, h2, . . . , hn) ∈ ker ϕ, then h1h2 ··· hn = e, so that for any −1 −1 i, hi = h1 ···ˆı ··· hn ∈ Hi ∩ hH1,...,ˆı, . . . , Hni = 1, and this shows that ϕ is a monomorphism.

In the situation of the theorem above, G is said to be the inner direct product of the subgroups H1,...,Hn. Any element of G can be written in a unique way as a product h1h2 ··· hn, with hi ∈ Hi, i = 1, . . . , n.

The case n = 2 of the previous theorem asserts that if H and K are two normal subgroups of a group G with HK = G and H ∩ K = 1, then G is isomorphic, in a natural way, to the direct product H × K.

If we only assume that H is a normal subgroup of G, but K is just a subgroup, and again HK = G and H∩K = 1, then G is called the semidirect product of the normal subgroup H by the subgroup K. In this case note that for h1, h2 ∈ H and k1, k2 ∈ K:

−1 (h1k1)(h2k2) = h1(k1h2k1 ) k1k2 = h1Ik1 (h2) k1k2 , and note that we have a natural homomorphism K → Aut H, given by k 7→ Ik|H . Conversely, assume that H and K are two groups, and that ϕ : K → Aut H, x 7→ ϕx, is a group homomorphism. On the cartesian product H ×K deﬁne a product by: (a, x)(b, y) = aϕx(b), xy , 24 CHAPTER 1. GROUPS for any a, b ∈ H and x, y ∈ K. With this product H ×K is a group (CHECK THIS!), denoted by H oϕ K, and called the outer semidirect product of H and K with respect to ϕ. This semidirect product has two distinguished subgroups, the normal subgroup {(h, e): h ∈ H} isomorphic to H, and {(e, k): k ∈ K} isomorphic to K.

§ 9. Finitely generated abelian groups

Throughout this section, additive notation will be used.

9.1 Deﬁnition. Let A be an abelian group. The subgroup

tor A = {x ∈ A : ∃n ∈ N such that nx = 0} = {x ∈ A : |x| < ∞} is called the torsion subgroup of A. The group A is said to be torsion free in case tor A = 0.

9.2 Remark. Let A be any abelian group. Then A/ tor A is a torsion free group. (Why?)

9.3 Theorem. Let A be a nonzero ﬁnitely generated abelian group. Then there are a natural number n ∈ N, an integer r ∈ Z with 0 ≤ r ≤ n, natural numbers m1, . . . , mr ≥ 2, and a system of generators {a1, . . . , an} of A such that:

• A = ha1i ⊕ · · · ⊕ hani is the inner direct sum of the haii’s.

• |ai| = mi for any i = 1, . . . , r, |ai| = ∞ for r + 1 ≤ i ≤ n, and m1 |m2 |· · ·|mr.

Moreover, the numbers n, r, m1, . . . , mr are uniquely determined by A, n − r is called the Betti number of A, and m1, . . . , mr the invariant factors of A. 9.4 Remark. Note that the Theorem implies that A is isomorphic to the n−r group Zm1 ⊕ · · · ⊕ Zmr ⊕ Z ⊕ · · · ⊕ Z. Proof. Let n be the minimal number of generators of A. First we will show the existence of the decomposition in the Theorem by induction on n, and we will check also that there appear exactly n summands. If n = 1 A is cyclic, so it is either isomorphic to Z or to Zm for some m and we are done. Assume the result is true for n − 1. If there is a generating system {a1, . . . , an} such that q1a1 + ··· + qnan = 0 for q1, . . . , qn ∈ Z implies q1 = ··· = qn = 0, then it is clear that |ai| = ∞ and A = ha1i ⊕ · · · ⊕ hani ' n Z⊕· · ·⊕Z (because for any i haii∩ ha1i+···+hai−1i+hai+1i+···+hani = § 9. FINITELY GENERATED ABELIAN GROUPS 25

0). Otherwise the set Ω of those natural numbers x ∈ N such that there exists a generating system {b1, . . . , bn} of A and integers x2, . . . , xn ∈ Z such that xb1 + x2b2 + ··· + xnbn = 0 is not empty. Let m1 be the minimum of Ω and let {b1, . . . , bn} be a generating system and s2, . . . , sn ∈ Z integers such that m1b1 + s2b2 + ··· + snbn = 0. For any i = 2, . . . , n, let qi, ri ∈ Z, with 0 ≤ ri < m1 be the unique integers satisfying si = qim1 + ri. Then

m1(b1 + q2b2 + ··· + qnbn) + r2b2 + ··· + rnbn = 0.

Since {a1 = b1 + q2b2 + ··· + qnbn, b2, . . . , bn} is a generating system of A, and m1 is the minimum of Ω, it follows that r2 = ··· = rn = 0. Therefore m1a1 = 0 and, by minimality, |a1| = m1. Moreover, A is the inner direct sum A = ha1i⊕hb2, . . . , bni, as otherwise there would exist a nonzero element ra1 ∈ ha1i ∩ hb2, . . . , bni with 0 < r < m1. But then r ∈ Ω and r < m1, a contradiction. By the induction argument, hb2, . . . , bni = ha2i ⊕ · · · ⊕ hani, for some elements a2, . . . , an such that |ai| = mi < ∞ for i = 2, . . . , r and |ai| = ∞ for r ≤ i ≤ n, for some r and mi’s, with m2 |· · ·|mr. Finally, the division algorithm gives us integers q, s with m2 = qm1 + s and 0 ≤ s < m1. Then

0 = m1a1 + m2a2 + 0a3 + ··· + 0an

= m1(a1 + qa2) + sa2 + 0a3 + ··· + 0an, and since {a1 + qa2, a2, . . . , an} is a generating system, it follows that s ∈ Ω unless s = 0. Since s < m1, the only possibility left is s = 0, so that m1 |m2. Let us prove the uniqueness now. Note that

tor A = ha1i ⊕ · · · ⊕ hari, and ∼ ∼ n−r B = A/ tor A = har+1i ⊕ · · · ⊕ hani = Z ⊕ · · · ⊕ Z. Hence ∼ n−r B/2B = Z2 ⊕ · · · ⊕ Z2, which contains 2n−r elements. Therefore we obtain:

n − r = log2 (|(A/ tor A)/2(A/ tor A)|) .

Now tor A = ha1i⊕· · ·⊕hari and hence mr is the maximum of the orders of the elements in tor(A):

mr = max{|x| : x ∈ tor(A)}.

Also, for any j = 1, . . . , r − 1 and d ∈ N with d|mj+1, we have r d tor(A) = ⊕i=1hdaii 26 CHAPTER 1. GROUPS and r j !  r  Y Y Y mi |d tor(A)| = |da | = |da | · . i i  d  i=1 i=1 i=j+1 1 Hence |d tor(A)| = dr−j mj+1 ····· mr if and only if dai = 0 for any i = 1, . . . , j, if and only if mj divides d. Therefore we get: 1 m = min{d ∈ : d|m and |d tor(A)| = m ····· m }. j N j+1 dr−j j+1 r We conclude that m1, . . . , mr are uniquely determined by tor(A), and hence by A. 9.5 Corollary. Let 0 6= A be a finite abelian group. Then there are natural numbers r, m1, . . . , mr with m1 ≥ 2 and m1 | m2 | · · · | mr, uniquely ∼ determined by A, such that A = Zm1 ⊕ · · · ⊕ Zmr . s1 sr 9.6 Lemma. Let m = p1 ··· pr be the prime factorization of a natural number m ≥ 2 (p1, . . . , pr are different prime numbers and s1, . . . , sr ∈ N). Then the abelian group m is isomorphic to s1 ⊕ · · · ⊕ sr . Z Zp1 Zpr Proof. Actually, the Chinese Remainder Theorem shows us that there is a ∼ ring isomorphism m = s1 ⊕· · ·⊕ sr . This isomorphism is, in particular, Z Zp1 Zpr a group isomorphism. 9.7 Example. If A is a finite abelian group with invariant factors 6 and 18. Then we get: ∼ ∼ ∼ A = Z6 ⊕ Z18 = Z2 ⊕ Z3 ⊕ Z2 ⊕ Z9 = Z2 ⊕ Z2 ⊕ Z3 ⊕ Z9.

On the other hand, if an abelian group B is isomorphic to, say, Z2 ⊕Z4 ⊕Z3 ⊕ Z3 ⊕Z9, then again by the previous Lemma B is isomorphic to Z3 ⊕Z6 ⊕Z36. By the uniqueness of the Theorem above, we conclude that its invariant factors are 3, 6 and 36. The arguments in this example give the following result: 9.8 Corollary. Any nonzero ﬁnite abelian group is isomorphic to a direct sum of cyclic groups whose orders and powers of prime numbers. These powers are uniquely determined by the group (and are called the elementary divisors of the group). 9.9 Example. How many abelian groups of order 180 do exist, up to isomorphism? Since 180 = 22 · 32 · 5, there are the following possibilities for the elementary divisors: ∼ 2, 2, 3, 3, 5 : Z2 ⊕ Z2 ⊕ Z3 ⊕ Z3 ⊕ Z5 = Z6 ⊕ Z30, ∼ 4, 3, 3, 5 : Z4 ⊕ Z3 ⊕ Z3 ⊕ Z5 = Z3 ⊕ Z60, ∼ 2, 2, 9, 5 : Z2 ⊕ Z2 ⊕ Z9 ⊕ Z5 = Z2 ⊕ Z90, ∼ 4, 9, 5 : Z4 ⊕ Z9 ⊕ Z5 = Z180. § 10. SOLVABLE GROUPS 27

§ 10. Solvable groups

The solvable groups are the ones that will appear as groups of symmetries of the algebraic equations which are solvable by radicals.

10.1 Deﬁnition. Let G be a group and let x, y ∈ G.

• The element [x, y] = xyx−1y−1 is called the commutator of x and y.

• The subgroup generated by the commutators: G0 = h[a, b]: a, b ∈ Gi, is called the derived subgroup of G.

Note that [x, y] = e if and only if xy = yx, and hence a group G is abelian if and only if G0 = 1. In a way, G0 measures how far the group G is from being abelian.

10.2 Example. For any x, y in the symmetric group Sn,

sgn[x, y] = sgn x sgn y(sgn x)−1(sgn y)−1 = 1, so that the derived subgroup of Sn is contained in the alternating group An. Moreover, [(i j), (i k)] = (i j)(i k)(i j)(i k) = (i j k), 0 and hence all the 3-cycles are in Sn. But, for diﬀerent i, j, h, k:

(i j)(i k) = (i k j), (i j)(h k) = (i j)(i h)(i h)(h k) = (i h j)(h k i), and hence any even product of transpositions is a product of 3-cycles. There- fore we get 0 An ⊆ hcycles of length 3i ⊆ Sn ⊆ An, and we conclude that the alternating group An is the derived subgroup of the symmetric group Sn, and it is generated by the cycles of length 3. 10.3 Proposition. Let G be a group.

(i) If K is a normal subgroup of G, then so is K0. In particular G0 is a normal subgroup of G.

(ii) The quotient group G/G0 is abelian, and if K is a normal subgroup of G with G/K abelian, then G0 is contained in K.

(iii) Any subgroup of G containing G0 is normal.

Proof. For (i) note that ∀g ∈ G and ∀x, y ∈ K, Ig([x, y]) = [Ig(x),Ig(y)] ∈ 0 0 0 K , since Ig is an automorphism. Therefore Ig(K ) ⊆ K for any g ∈ G and K0 is a normal subgroup. 28 CHAPTER 1. GROUPS

For (ii) it is enough to note the following equation in the quotient group G/G0 valid for any x, y ∈ G: (xG0)(yG0) = (xy)G0 = (xy)[y−1, x−1]G0 = yxG0 = (yG0)(xG0). Hence G/G0 is abelian. Besides, if G/K is abelian, then xyK = yxK for any x, y ∈ G, and this is equivalent to [x, y] ∈ K. Thus G0 ⊆ K. Finally, if G0 ⊆ K ⊆ G, and x ∈ K, g ∈ G, gxg−1 = [g, x]x ∈ K, because [g, x] ∈ G0 ⊆ K, so K is a normal subgroup.

We already have two subgroups related to the commutativity of a group G: G is abelian ⇐⇒ G0 = 1 ⇐⇒ Z(G) = G. Deﬁne recursively the following normal subgroups of a group G: 0 • G(0) = G, G(n+1) = G(n) , • Z0 = 1, Zn+1/Zn = Z G/Zn .

10.4 Deﬁnition. Let G be a group. • The descending chain of normal subgroups G = G(0) ⊇ G0 = G(1) ⊇ G(2) ⊇ · · · is called the derived series of G.

• The ascending chain of normal subgroups 1 = Z0 ⊆ Z(G) = Z1 ⊆ Z2 ⊆ · · · is called the upper central series of G.

• The group G is said to be solvable if there is an n ∈ N such that G(n) = 1.

• The group G is said to be nilpotent if there is an n ∈ N such that Zn = G.

10.5 Example. The symmetric group S2 is the cyclic group of order 2, which is abelian, and hence solvable. The symmetric group S3 satisﬁes that its derived subgroup is A3, which is cyclic of order 3. Hence its derived chain is S3 D A3 D 1 and S3 is solvable too. The symmetric group S4 satisﬁes that its derived subgroup is A4 of order 12. Consider the so called Klein four group V = {1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}.

It is straightforward to check that this is a subgroup, isomorphic to Z2 ×Z2, and it is normal as any conjugate of a product of two disjoint transpositions is a product of two disjoint transpositions. Note that V is contained in 0 A4 and A4/V has order 3, and hence it is cyclic. Hence (A4) ⊆ V . But [(1 2 3), (1 2 4)] = (1 2)(3 4). It follows that the derived series of S4 is S4 D A4 D V D 1. Again S4 is solvable. § 10. SOLVABLE GROUPS 29

10.6 Proposition.

(i) Any p-group (p a prime number) is nilpotent.

(ii) A group G is solvable if and only if there is chain of subgroups G D G1 D ··· D Gm = 1, with each factor Gi/Gi+1 being abelian. (iii) Any nilpotent group is solvable.

(iv) If N is a normal subgroup of a group G, then G is solvable if and only if so are both N and G/N.

Proof. (i) If G is a p-group: |G| = pn, we know (see 6.1 Corollary) that Z1 = Z(G) is not trivial (1 & Z1). Now, if Z1 6= G, then G/Z1 is a p-group too, so 1 6= Z2/Z1 = Z(G/Z1) and hence Z1 & Z2. Continuing in this way we get 1 & Z1 & Z2 & ··· . Since G is ﬁnite, it turns out that G = Zn for some n. (ii) One implication is clear, because if G is solvable, its derived series is a chain with abelian consecutive factors. Conversely, if G D G1 D ··· D Gm = 1 is a descending chain with abelian consecutive factors, then we may prove (i) inductively that G ⊆ Gi for any i = 0, 1, 2,... (with G0 = G). Indeed, (i) this is trivial for i = 0, and if G ⊆ Gi, then since Gi/Gi+1 is abelian, (i+1) (i)0 0 G = G ⊆ Gi ⊆ Gi+1, as desired. (iii) If G is nilpotent, the chain G = Zn D Zn−1 D ··· D Z0 = 0 is a descending chain with abelian consecutive factors, and hence G is solvable. (iv) Note that if N is a normal subgroup of a solvable group G with G(n) = 1, then1 N (n) ⊆ G(n) = 1, and also G/N(n) ⊆ G(n)N/N = N/N = 1. Hence both N and G/N are solvable. Conversely, if both N and G/N are solvable, we may join any descending chains with consecutive abelian factors of both N and G/N to get such a chain for G: N = N0 D N1 D ··· D Nr = 1 and G/N = G0/N D G1/N D ··· D Gs/N = N/N = 1 give the descending chain G = G0 D G1 D ··· D Gs = N = N0 D N1 D ··· D Nr = 1.

10.7 Proposition. Let p, q and r be three diﬀerent prime numbers. Then the groups of order pn (n ≥ 1), pq, p2q, p2q2 and pqr are all solvable.

Proof. We have already proved in the previous proposition that any p-group is nilpotent, and hence solvable. pq : We may assume p > q. If |G| = pq, the number of Sylow p-subgroups of G is Np, and Np | q, since Np = [G : NG(P )] | [G : P ] for a ﬁxed Sylow subgroup P of G, while Np ≡ 1 (mod p). The only possibility is Np = 1,

1This is valid even for a non normal subgroup and shows that any subgroup of a solvable group is solvable 30 CHAPTER 1. GROUPS and hence there is a unique (normal) Sylow p-subgroup P . Then |P | = p, so P is cyclic (and hence abelian and solvable) and |G/P | = q and hence G/P is solvable too. Thus, G is solvable. p2q : If p > q the same argument as before gives that there is a unique Sylow p-subgroup P which, as any p-group, is nilpotent and hence solvable, while G/P is cyclic, and the result follows. If p < q, then the number of 2 Sylow q-subgroups Nq divides p and is ≡ 1 (mod q). Then either Nq = 1 2 and, as before, G is solvable, or Nq = p . In this case, there are exactly p2 Sylow q-subgroups, and each of them is cyclic of order q, so there are p2(q − 1) elements of order q in G. But |G| = p2q = p2(q − 1) + p2, so there is room for a unique Sylow p-subgroup, and this implies again that G is solvable. 2 2 2 p q : We may assume that p > q and, as before, either Np = 1 or Np = q . In the ﬁrst case we are done. In the second case, as Np ≡ 1 (mod p), p | q2 − 1 = (q − 1)(q + 1), so p | q + 1 and p = q + 1 as p > q. The only possibility is q = 2 and p = 3. Thus we have N3 = 4. Let P be a Sylow 3-subgroup of G, then G acts by left multiplication on the quotient G/P : Φ: G → S(G/P ) ' S4. But |S4| = 24 and |G| = 36. Thus ker Φ, which is the largest normal subgroup contained in P is not trivial. We conclude that ker Φ has order 3, and hence it is solvable, while G/ ker Φ has order 12 and hence it is solvable too. pqr : We may assume p > q > r and Np,Nq,Nr > 1. Since Np | qr, Np ≡ 1 (mod p), and p > q, r, we conclude that Np = qr, and hence there are (p − 1)qr elements of order p in G. Also Nq | pr and Nq ≡ 1 (mod q) so Nq ≥ p and there are at least p(q − 1) elements of order q. But (p − 1)qr + p(q − 1) = pqr + (p(q − 1) − qr) > pqr = |G|, a contradiction.

There are two deep results of Group Theory which we mention here without proofs: 10.8 Burnside’s Theorem (1904). Let p and q be two diﬀerent prime n m numbers and let n, m ∈ N. Then any group of order p q is solvable. 10.9 Feit-Thompson’s Theorem (1963). Any ﬁnite group of odd order is solvable.

§ 11. Simple groups

11.1 Deﬁnition. A nontrivial group G is said to be simple if it contains no proper normal subgroup. 11.2 Proposition. Let G be a solvable group. Then G is simple if and only if it is cyclic of prime order. § 11. SIMPLE GROUPS 31

Proof. It is clear that any cyclic group of prime order is simple. Conversely, 0 0 given a simple solvable group G, as G E G, and G 6= G, it follows that 0 G = 1, so G is abelian. Then for any element 1 6= x ∈ G, we have hxi E G, so that G = hxi is cyclic. But if |x| = nm, with n, m > 1, then hxmi is a proper normal subgroup of G. Hence G is cyclic of prime order.

Our purpose now is to show that the alternating groups An for n ≥ 5 are simple.

11.3 Lemma. Assume n ≥ 3.

(i) The alternating group An is generated by the cycles of length 3.

(ii) If a normal subgroup of An contains a cycle of length 3, then it is the whole An.

Proof. Part (i) is already known (see 10.2). Now, if N E An and (a b c) ∈ N, then (a b d)(a b c)2(a b d)−1 = (b c d) = (d b c) ∈ N. In this way we may show that any cycle of length 3 lies in N, and hence N = An.

11.4 Theorem. The alternating group An is simple for any n ≥ 5.

Proof. Let N be a normal subgroup of An, N 6= 1. It is enough to check that N contains a cycle of length 3. Let 1 6= σ ∈ N and let r be the greatest length of the cycles involved in the cycle decomposition of σ. If r ≥ 4, then σ = (a1 a2 . . . ar)τ, where τ is “disjoint” to (a1 a2 . . . ar). −1 −1 Then, as [σ, γ] = σ γσ γ ∈ N for any γ ∈ An, we get

−1 −1 [σ, (a1 a2 a3)] = σ(a1 a2 a3)σ (a1 a2 a3) = σ(a1) σ(a2) σ(a3) (a1 a3 a2)

= (a2 a3 a4)(a1 a3 a2) = (a1 a4 a2) ∈ N.

If r = 3 and σ contains at least two disjoint cycles of length 3, then σ = (a b c)(d e f)τ, then

[σ, (a b d)] = σ(a) σ(b) σ(d)(a d b) = (b c e)(a d b) = (a d c e b) ∈ N, and the argument for r ≥ 4 works. If r = 3 and σ contains exactly one cycle of length 3: σ = (a b c)τ and τ 2 = 1, then σ2 = (a c b) ∈ N. Finally, if r = 2, then σ = (a b)(c d)τ, where τ is a product of an even number (possibly 0) of transpositions. Then

[σ, (a b c)] = σ(a) σ(b) σ(c)(a c b) = (b a d)(a c b) = (a c)(b d) ∈ N, and since n ≥ 5,

[(a c e), (a c)(b d)] = (c e)(b d)(a c)(b d) = (a e c) ∈ N. 32 CHAPTER 1. GROUPS

11.5 Corollary. The symmetric group Sn, for n ≥ 5, is not solvable.

0 Proof. We know that Sn = An. Since An is simple and it is not abelian, its 0 derived subgroup is An = An. And hence the derived series of Sn is just Sn D An.

Exercises

1. Let G be a semigroup, that is, a nonempty set with a binary operation which is associative. Assume that the following two conditions hold: (a) There exists an element e ∈ G such that ex = x for any x ∈ G (a left neutral element), (b) For any x ∈ G there is an element y ∈ G such that yx = e (a left inverse). Prove that G is a group.

2. Let G be a semigroup containing a left neutral element e and such that any element has a right inverse (relative to e). Is G a group?

3. Let G be a group such that g2 = 1 for any g ∈ G. Prove that G is abelian.

4. Let S be the subgroup of the cyclic group of order 154: C154 = hxi, generated by {x28, x88}. Find a natural number n such that S = hxni.

5. Let G = hxi be an infinite cyclic group. Prove that any nontrivial subgroup of G is an infinite cyclic group of finite index generated by n x for some n ∈ N. 6. Let G 6= 1 be a group containing no subgroup different from 1 and G. Prove that G is cyclic of prime order.

7. Let G be a cyclic group of order n. Compute the number of generators of G (that is, compute |{g ∈ G : hgi = G}|).

× 8. Consider the multiplicative group Zn of the units of the ring Zn. Prove × that Zn is cyclic if n is prime.

9. Let Cn be a cyclic group of order n and let m ∈ N. Prove that the map m αm : Cn → Cn such that x 7→ x for any x ∈ Cn is an automorphism if and only if gcd(m, n) = 1. Prove that Aut Cn is an abelian group of order φ(n). Compute Int Cn.

9 Recall that φ(n) denotes the Euler map: φ(n) = |{1 ≤ m ≤ n : gcd(m, n) = 1}|. EXERCISES 33

0 −1 0 −1 10. Consider the elements A = and B = in GL ( ). 1 0 1 −1 2 Z Prove that A has order 4, B has order 3, while the order of AB is inﬁnite.

11. Give an example of a nonabelian ﬁnite group such that all its proper subgroups are cyclic (respectively normal).

12. Consider the additive group Q of the rational numbers. (a) Is it cyclic? (b) Take any two elements x, y ∈ Q, is hx, yi cyclic? (c) What is the answer to the two previous questions if we substitute × Q by the multiplicative group Q = Q \{0}? 13. Let x, y, z be elements of a group G. Are the following assertions true? Why? (a) The elements x, x−1, xy = yxy−1 have the same order. (b) xy and yx have the same order. (c) xyz and zyx have the same order.

14. Prove that, up to isomorphism, the only groups of order 4 are C4 and C2 × C2. 15. Let G be a group such that G/Z(G) is cyclic. Prove that G is abelian.

16. Let G be an abelian group. Prove that the set of ﬁnite order elements of G is a subgroup of G.

17. Let H and K be two subgroups of a group G. (a) Prove that H ∪ K is a subgroup of G if and only if either H ⊆ K or K ⊆ H. (b) Let HK = {xy : x ∈ H, y ∈ K}. Give an example showing that, in general, HK is not a subgroup of G. (c) Prove that HK is a subgroup of G if and only if HK = KH.

n λ −µ o 18. Prove that the set of matrices µ λ : λ, µ ∈ R, (λ, µ) 6= (0, 0) is × a multiplicative group isomorphic to C = C \{0}. 19. Prove, without the use of group actions, that any index two subgroup of a group G is normal.

20. Let Φ : G1 → G2 be a group homomorphism, and let N be a normal subgroup of G1 and K a subgroup of G2.

(a) Is Φ(N) a normal subgroup of G2? 34 CHAPTER 1. GROUPS

−1 (b) Is Φ (K) = {x ∈ G1 : Φ(x) ∈ K} a subgroup of G1? −1 (c) If K is normal, is Φ (K) a normal subgroup of G1? 21. Find the order of H, where:

(a) H is the subgroup of S3 generated by {(1 2), (1 3)}.

(b) H is the subgroup of S5 generated by {(1 2), (1 3)(4 5)}

22. Prove that Sn is generated by (1 2) and (1 2 . . . n).

23. Find abelian and nonabelian order 6 subgroups of S6.

24. Prove that for any n ≥ 3, the center of Sn is trivial.

25. Compute the centralizer of the element (1 2 3 4) in S4. How many conjugates does this element have in S4? 26. Compute the center of GL2(Z) = {A ∈ Mat2(Z): ∃B ∈ Mat2(Z) such that AB = I2 = BA} .

27. Consider the dihedral group D6 of isometries of the euclidean plane which ﬁx a regular hexagon. (a) Prove that this group is generated by the clockwise rotation x of π angle 3 and by the reﬂection y through two opposite vertices. (b) Show that x6 = 1 = y2 and yx = x−1y. n m (c) Show that D6 = {x y : 0 ≤ n < 6, 0 ≤ m < 2}.

28. Let G be a ﬁnite group generated by two diﬀerent elements of order 2. Prove that G is isomorphic to C2 × C2 or to Dn for some n ≥ 3.

29. Compute Aut Dn (n ≥ 3).

±1 λ 30. Prove that the multiplicative group Gn = 0 1 : λ ∈ Zn is isomorphic to Dn.

31. Prove that the only groups of order 6 are, up to isomorphism, C6 and D3. 27 It can be checked that any equation satisﬁed by x and y is a consequence of the equations in (ii). This allows us to write

6 2 −1 D6 = hx, y : x = y = 1, yx = x yi.

This is a presentation by generators and relations of D6. These presentations constitute a very useful way to define groups. The formal definition of these presentations require the definition of free groups, which will not be given in this course. 28 Let u and v be the two order 2 generators. If uv = vu, then show that G =∼ C2 × C2. Otherwise, show that G is generated by x = uv and y = v, and that these elements satisfy the relations of the generators of a dihedral group. EXERCISES 35

32. Prove that the only nonabelian groups of order 8 are, up to isomorphism, D4 and the quaternion group.

33. Prove that A4 has no order 6 subgroups.

34. How many diﬀerent bracelets can be done with 3 red beads and 4 green beads?

35. Let the order of a group G be a product of two prime numbers p > q. Prove that either G is abelian or q |p − 1.

36. Let G be a group with order 2p for an odd prime number p. Prove that either G is cyclic or isomorphic to the dihedral group Dp.

37. Compute, up to isomorphism, the groups of order 21.

38. Compute, up to isomorphism, the groups of order 12.

39. Let G be a nonabelian group of order p3 for a prime number p. Prove that G0 equals Z(G).

n 2 −1 40. For n ≥ 3, let Dn = hx, y : x = y = 1, yx = x yi be the dihedral group of degree n.

n (a) Prove that Z(Dn) = 1 for odd n, but Z(Dn) = hx 2 i for even n. 0 2 (b) Prove that Dn = hx i (which is equal to hxi for odd n).

41. Let H be a subgroup of a simple nonabelian group G of index m ≥ 5. Prove that G is isomorphic to a subgroup of Am.

42. Let G be a ﬁnite group of order 2m for an odd m > 1. Prove that G is not simple.

43. Prove that there are no simple groups of order 56, 80, 84, 132, 300 or 1000.

44. Prove that, up to isomorphism, the only simple nonabelian group of order ≤ 60 is A5.

45. Prove the following relations in any group G:

[x, y]−1 = [y, x], [x, yz] = [x, y](y[x, z]y−1), [xy, z] = (x[y, z]x−1)[x, z].

42 Otherwise G would be isomorphic, through the action by left translation, to a subgroup of A2m. Then argue with the possible expressions as a product of disjoint cycles of an order 2 element x of G, taking into account that the left multiplication by x does not ﬁx any element of G. 36 CHAPTER 1. GROUPS

46. Let H and K be subgroups of a group G. Deﬁne the commutator subgroup [H,K] as

[H,K] = h[h, k]: h ∈ H, k ∈ Ki.

Prove that [H,K] = [K,H] and that H is a normal subgroup of G if and only if [G, H] ≤ H.

47. The lower central series of a group G is deﬁned by means of L0 = G, Ln+1 = [G, Ln] for n ≥ 0.

(a) Prove that Ln is a normal subgroup of G for any n. (b) Show that we have indeed a descending series: L0 D L1 D L2 D ....

(d) Prove that the group G is nilpotent if and only if Ln = 1 for some n.

48. Prove that subgroups, quotients and ﬁnite direct products of solvable (respectively nilpotent) groups are solvable (respectively nilpotent).

49. Give an example of a nonnilpotent group G with a normal subgroup N such that both N and G/N are nilpotent.

50. Give examples of nilpotent nonabelian groups and of solvable nonnilpotent groups.

51. Consider the following two groups:

5 4 3 G1 = hx, y : x = 1 = y , yx = x yi, 11 3 −1 −1 G2 = hx, y, z : x = 1 = z , [x, y] = 1, zx = yz, zy = x y zi.

(a) Compute their orders and the number of its Sylow subgroups. (b) Describe these groups as semidirect products. (c) Compute the derived subgroups, the centers, and the corresponding quotient groups. (d) How many conjugacy classes do they contain? (e) Study the solvability and nilpotency of these groups.

51 0 0 2 The groups G1 and G1/G1 are cyclic. The subgroup of G1 generated by x and y is a dihedral group. These facts, together with the number of Sylow subgroups, allows you to conclude that G1 contains 5 elements of order 2 and that its center is trivial. For G2 consider the subgroup H generated by x and y, which is normal and abelian. 0 0 Compute [x, z] and [y, z] and deduce from here G2 and G2/G2. Argue that Z(G2) cannot contain order 3 or 11 elements. Using Sylow subgroups, check that G can only contain elements of order 1, 3 or 11. Chapter 2

Galois Theory

We will start this chapter by recalling in the ﬁrst section some concepts you should already know.

§ 1. Algebraic extensions

1.1. Let F be a ﬁeld with unity 1F = 1. Consider the map:

ϕ : Z −→ F n n 7→ n1 = 1+ ··· +1 (n ∈ N) 0 7→ 0 n −n 7→ −n1 = (−1)+ ··· +(−1) (n ∈ N)

Then ϕ is a ring homomorphism and there are two possibilities:

1. ker ϕ = 0 (that is, ϕ is a monomorphism). Then ϕ extends to a monomorphism

ψ : Q −→ F m 7→ ϕ(m)ϕ(n)−1 n

since Q is the field of fractions of Z. In this case, it is said that the characteristic of F is 0 and Q is identified with its image under ψ, which is the smallest subfield of F and it is called the prime subfield of F .

2. ker ϕ = pZ for some p ∈ N, so that ϕ induces a monomorphism

ϕ¯ : Z/pZ −→ F n + pZ 7→ ϕ(n) = n1.

37 38 CHAPTER 2. GALOIS THEORY

Since F has no zero divisors, neither does Z/pZ, so that p is a prime number, which is called the characteristic of F . In this case the field Fp = Z/pZ is identified with its image underϕ ¯, which again is the smallest subfield of F and called the prime subfield of F . Notice that p is the smallest natural number such that p1 = 0 and that for any α ∈ F

p p pα = α+ ··· +α = (1+ ··· +1)α = 0α = 0

1.2 Examples.

• Q, R and C are ﬁelds of characteristic 0.

• Fp = Z/pZ, for a prime number p, is a ﬁeld of characteristic p, and so is Fp(X) (the fraction ﬁeld of Fp[X]).

1.3. Let K/F be a field extension, then 1K = 1F and the characteristics of F and K coincide. Moreover, K is a vector space over F in a natural way. Its dimension dimF K is called the degree of the extension and it is denoted by [K : F ]. If it is finite, then K/F is said to be a finite extension.

The most important example. Let F be a field and f(X) ∈ F [X] be an irreducible polynomial. Then f(X) is a maximal ideal, so that K = F [X]/f(X) is a field. We may view F as a subfield of K by means of the map F,→ K, a 7→ a + f(X). Take the element θ = X + f(X) ∈ K and n assume that f(X) = a0 + a1X + ··· + anX with an 6= 0. Then θ is trivially a root of f(X) in K, because

n f(θ) = a0 + a1θ + ··· + anθ n = a0 + a1(X + f(X) ) + ··· + an(X + f(X) ) n = a0 + a1X + ··· + anX + f(X) = f(X) + f(X) = 0.

Moreover, {1, θ, . . . , θn−1} is a basis of K as a vector space over F . In particular, [K : F ] = deg f(X).

Proof. Let us show ﬁrst that {1, θ, . . . , θn−1} is free. Assume that there are n−1 scalars b0, . . . , bn−1 ∈ F such that b0 + b1θ + ··· + bn−1θ = 0. As before,

n−1 n−1 b0 + b1θ + ··· + bn−1θ = (b0 + b1X + ··· + bn−1X ) + f(X) ,

n−1 so that we conclude that b0 + b1X + ··· + bn−1X ∈ f(X) . That is, n−1 f(X) divides b0 + b1X + ··· + bn−1X and this is only possible if b0 = ··· = bn−1 = 0. § 1. ALGEBRAIC EXTENSIONS 39

To show that {1, θ, . . . , θn−1} is a spanning set, take an arbitrary element g(X) + f(X) ∈ K. Since F [X] is an euclidean domain, there are polynomials c(X), r(X) ∈ F [X] with deg r(X) ≤ n − 1 such that n−1 g(X) = c(X)f(X) + r(X). If r(X) = r0 + r1X + ··· + rn−1X , then

n−1 g(X) + f(X) = r(X) + f(X) = r01 + r1θ + ··· + rn−1θ is a linear span with coeﬃcients in F of the elements in {1, θ, . . . , θn−1}, as required.

Therefore,

K = F [X]/f(X) = F 1 + ··· + F θn−1 = {r(θ): r(X) ∈ F [X] and deg r(X) < n}, and the operations in K are given by: ( r1(θ) + r2(θ) = (r1 + r2)(θ),

r1(θ)r2(θ) = r(θ), for r1(X), r2(X) ∈ F [X] of degree at most n − 1, and where r(X) is the remainder of the division of r1(X)r2(X) by f(X).

Let us consider a couple of examples:

∼ 2 • C = R[X]/(X + 1).

3 • The polynomial X − 2 ∈ Q[X] is irreducible (Eisenstein’s criterion). 3 3 2 Let K = Q[X]/(X −2) and θ = X+(X −2). Then K = {a+bθ+cθ : 3 4 a, b, c ∈ Q}, θ = 2, θ = 2θ. Therefore, the multiplication in K is given by:

(a + bθ + cθ2)(a0 + b0θ + c0θ2) = (aa0 + 2bc0 + 2cb0) + (ab0 + ba0 + 2cc0)θ + (ac0 + ca0 + bb0)θ2.

If we want to compute (1 + θ)−1, we proceed as follows: since X3 − 2 is irreducible, gcd(X3 − 2,X + 1) = 1 and we compute the coeﬃcients of Bezout’s identity to get:

1 1 1 = − (X3 − 2) + (X2 − X + 1)(X + 1), 3 3

θ2 − θ + 1 θ2 − θ + 1 so that 1 = (θ + 1) . Therefore, (1 + θ)−1 = . 3 3 40 CHAPTER 2. GALOIS THEORY

1.4. Let K/F be a ﬁeld extension and let α ∈ K. The evaluation map:

ψ : F [X] −→ K f(X) 7→ f(α) is a ring homomorphism and there are two possibilities: 1. ker ψ = 0. In this case there is no nonzero polynomial in F [X] for which α is a root, and ψ extends to a ring monomorphism F (X) ,→ K, whose image is F (α). In particular, [F (α): F ] = [F (X): F ] ≥ dimF F [X] = ∞. Then α is said to be transcendental over F . (Here, and in what follows, given a subset S of K, F (S) denotes the smallest subﬁeld of K containing both F and S.)

2. ker ψ 6= 0. Since F [X] is a principal ideal domain and im ψ is an integral domain (it is a subring of the ﬁeld K), ker ψ = m(X) for a unique monic irreducible polynomial m(X) ∈ F [X]. Hence any polynomial f(X) ∈ F [X] with f(α) = 0 is a multiple of m(X). Hence m(X) is the monic polynomial of lowest degree that annihilates α. The polynomial m(X) is called the minimal polynomial of α over F and it is denoted by mα,F (X). In this case, α is said to be algebraic over F . Note that F (α) is isomorphic to F [X]/ mα,F (X) and hence [F (α): F ] = deg mα,F (X). The extension ﬁeld K/F is said to be algebraic if for any α ∈ K, α is algebraic over F .

Examples.

• C/R is algebraic, because for any α = a + bi ∈ C (a, b ∈ R), α is a root 2 2 2 of (X − α)(X − α¯) = X − 2aX + (a + b ) ∈ R[X].

• Q(X)/Q is not algebraic.

1.5 Example. (Quadratic extensions) Let K/F be a ﬁeld extension with [K : F ] = 2 and assume that the characteristic of F is 6= 2. Then any α ∈ K \ F is a root of a polynomial X2 + bX + c ∈ F [X]. But

b 2 b2 − 4c X2 + bX + c = X − − , 2 4 so that β = 2α−b (∈ K \F ) satisﬁes β2 = b2 −4c ∈ F . With D = b2 −4c this 2 shows that the minimal polynomial√ of β is X − D. Since [F (β): F ] = 2, necessarily K = F (β) = F ( D). § 1. ALGEBRAIC EXTENSIONS 41

1.6 Corollary. Let K/F be a ﬁeld extension and α ∈ K. Then α is algebraic over F if and only if [F (α): F ] < ∞. 1.7 Corollary. Any ﬁeld extension K/F with [K : F ] < ∞ is algebraic.

1.8 Proposition. Let K/F and L/K be two ﬁeld extensions. Then [L : F ] = [L : K][K : F ].

Proof. Let {βi : i ∈ I} be a basis of L over K and let {αj : j ∈ J} be a basis of K over F . Then, any γ ∈ L can be written uniquely as

γ = b1βi1 + ··· + brβir , with b1, . . . , br ∈ K, and each bh can be written uniquely as

bh = ah,1αj1 + ··· + ah,sαjs , with ah,1, . . . , ah,s ∈ F . Therefore, γ can be written uniquely as γ = P ah,kαjk βih . Hence {αjβi : I ∈ I, j ∈ J} is a basis of L over F . Thus, [L : F ] = |I||J| = [L : K][K : F ]. √ √ 1.9 Example. Consider the ﬁeld extension Q( 2, 3)/Q. Then √ √ √ √ √ Q $ Q( 2) ⊆ Q( 2)( 3) = Q( 2, 3) (⊆ R). √ ( 26∈Q) √ Now, [ ( 2) : ] = 2, since m√ (X) = X2 − 2 (irreducible by Eisen- Q Q √ 2,Q √ 2 stein’s criterion). Also, 3 is a root√ of √X − 3 √∈ Q[X] ⊆ Q( √2)[X√], so m√ √ (X) | X2−3, and either [ ( 2, 3) : ( 2)] = 2, or ( 2, 3) = √3,Q( 2) √ √ Q Q Q Q( 2), that is, √3 ∈ Q( 2).√ But in the latter case,√ there would√ exist 2 2 a, b ∈ Q such that 3 = a + b 2, so 3 = (a + 2b ) + 2ab 2. Since 2 6∈ Q, √this last equation forces either a = 0 or b = 0. But√ b = 0 would imply 3 ∈ Q, which is not true,√ while√ a = 0√ implies 6 = 2b ∈ Q, again a contradiction. Therefore, [Q( 2, 3) : Q( 2)] = 2 and √ √ √ √ √ √ [Q( 2, 3) : Q] = [Q( 2, 3) : Q( 2)][Q( 2) : Q] = 2 × 2 = 4. √ √ √ Moreover,√ √ {1, 2} √is a basis of Q( 2) over Q and {1, 3} is a basis of ( 2, 3) over ( 2) (because deg m√ √ (X) = 2). By the argument Q Q √3,Q(√2) √ √ √ in the previous proof, it follows that {1, 2, 3, 6} is a basis of Q( 2, 3) over Q. 1.10 Theorem. Let K/F be a ﬁeld extension. Then [K : F ] < ∞ if and only if there are r ∈ N and α1, . . . , αr ∈ K, all of them algebraic over F , such that K = F (α1, . . . , αr).

In this case, if deg mαi,F (X) = ni for any i = 1, . . . , r, then [K : F ] ≤ n1 ····· nr. 42 CHAPTER 2. GALOIS THEORY

Proof. If [K : F ] < ∞ and {α1, . . . , αr} is a basis of K over F , then for any i = 1, . . . , r,[F (αi): F ] ≤ [K : F ] < ∞, so that αi is algebraic over F and, evidently, K = F (α1, . . . , αr). Now, assume that K = F (α1, . . . , αr) with αi algebraic over F for any i, and let deg mαi,F (X) = ni. Thus, mαr,F (X) is a polynomial in

F (α1, . . . , αr−1)[X] which ‘kills’ αr, and therefore mαr,F (α1,...,αr−1)(X) divides mα,F (X) in F (α1, . . . , αr−1)[X]. Hence [K : F (α1, . . . , αr−1)] ≤ nr. By an easy induction argument on r we conclude that

[K : F ] = [K : F (α1, . . . , αr−1)][F (α1, . . . , αr−1): F ]

≤ nr · (n1 ····· nr−1) = n1 ····· nr < ∞. 1.11 Corollary. Let K/F be a field extension and let α, β ∈ K be two algebraic elements over F . Then α + β, α − β, αβ and αβ−1 (if β 6= 0) are algebraic over F too. In particular, the set {γ ∈ K : γ is algebraic over F } is a subfield of K which contains F . Proof. From the Theorem above we conclude that [F (α, β): F ] < ∞, and therefore the field extension F (α, β)/F is algebraic. Now everything follows since the elements α+β, α−β, αβ and αβ−1 (if β 6= 0) are all in F (α, β).

1.12 Corollary. If K/F and L/K are algebraic ﬁeld extensions, so is L/F . n−1 n Proof. Let α ∈ L and assume mα,K (X) = b0 + b1X + ··· + bn−1X + X , with b0, . . . , bn−1 ∈ K. Thus α is algebraic over the subﬁeld F (b0, . . . , bn−1) of K, and hence [F (b0, . . . , bn−1, α): F (b0, . . . , bn−1)] < ∞. But b0, . . . , bn−1 are all algebraic over F , since they all belong to K and K/F is algebraic. Thus, by the Theorem above [F (b0, . . . , bn−1): F ] < ∞. Hence,

[F (b0, . . . , bn−1, α): F ]

= [F (b0, . . . , bn−1, α): F (b0, . . . , bn−1)][F (b0, . . . , bn−1): F ] < ∞, so that F (b0, . . . , bn−1, α)/F is an algebraic ﬁeld extension and, in particular, α is algebraic over F .

§ 2. Splitting ﬁelds. Algebraic closure

The definition and first properties of splitting fields were considered too in Algebraic Structures. 2.1 Definition. Let F be a field and 0 6= f(X) ∈ F [X]. A field extension K of F is said to be a splitting field of f(X) over F if, as a polynomial in K[X], it splits, that is:

f(X) = a(X − α1) ··· (X − αn) § 2. SPLITTING FIELDS. ALGEBRAIC CLOSURE 43 where a ∈ F (since it is the leading coeﬃcient of f(X)), α1, . . . , αn ∈ K, and K = F (α1, . . . , αn).

2.2 Examples.

2 • C is a splitting field of X + 1 ∈ R[X]. √ 2 • Q( 2) is a splitting field of X − 2 over Q. √ √ 2 2 • Q( 2, 3) is a splitting field of (X − 2)(X − 3) over Q. √ • The splitting field of X3 − 2 over is not ( 3 2), because the roots √ √ √Q Q √ (in ) of X3 −2 are 3 2, ω 3 2, ω2 3 2, where ω = −1+i 3 is a cubic root C √ 2 3 of√ 1. Note√ that if 2 and ω√are√ in a field K containing Q, then so is 3 3 i 3 = −3. Then√ K = Q( 2, −3)√ is a splitting field of X − 2 over 3 2 Q.√ Since [K : Q( 2)] = 2, because −√3 is a root√ of X + 3 and K 6= 3 3 3 Q( 2) (⊆ R), we get [K : Q] = [K : Q( 2)][Q( 2) : Q] = 2 × 3 = 6.

2.3 Proposition. Let K be a splitting ﬁeld of a polynomial f(X) of degree n over a ﬁeld F . Then [K : F ] ≤ n!.

Proof. In case deg f(X) = 1, then K = F and [K : F ] = 1 = 1!. Now, if deg f(X) = n > 1 and α ∈ K is a root of f(X), then f(X) = (X −α)g(X) for some g(X) ∈ F (α)[X] of degree n−1. Then K is a splitting ﬁeld too of g(X) over F (α), so by an inductive argument we get

[K : F ] = [K : F (α)][F (α): F ] ≤ (n − 1)! × n = n!

Our next goal is to recall that there always exist splitting ﬁelds of polynomials and that, up to isomorphism, they are unique.

2.4 Lemma. Let ϕ : F → Fˆ be a field isomorphism, f(X) = a0 + a1X + n ··· + anX ∈ F [X] an irreducible polynomial, fˆ(X) = ϕ(a0) + ϕ(a1)X + n ··· + ϕ(an)X ∈ Fˆ[X] (which is irreducible too), α a root of f(X) in some field extension of F and β a root of fˆ(X) in some field extension of Fˆ. Then there exists a field isomorphism σ : F (α) → Fˆ(β) such that σ|F = ϕ and σ(α) = β.

Proof. It is enough to concatenate the following isomorphisms:

F (α) =∼ F [X]/f(X) =∼ Fˆ[X]/fˆ(X) =∼ Fˆ(β) ˆ a ∈ F a + f(X) 7→ ϕ(a) + f(X) ϕ(a) ˆ α X + f(X) 7→ X + f(X) β

2.5 Theorem. Let F be a ﬁeld and let 0 6= f(X) ∈ F [X]. Then: 44 CHAPTER 2. GALOIS THEORY

(i) There exist splitting ﬁelds of f(X) over F .

(ii) All of them are isomorphic. More precisely, if ϕ : F → Fˆ is a field isomorphism, fˆ(X) is the polynomial which results of applying ϕ to the coefficients of f(X), E is a splitting field of f(X) over F , and Eˆ a splitting field of fˆ(X) over Fˆ, then there is a field isomorphism σ : E → Eˆ such that σ|F = ϕ. Proof. For (i), we use induction of deg f(X). If deg f(X) ≤ 1, then F itself is a splitting field. Now, if deg f(X) = n > 1, we know that there exists a field extension of F which contains a root α1 of some irreducible factor of f(X). Let K1 = F (α1). Then, in K1[X], f(X) = (X −α1)g(X), and deg g(X) = n−1. By induction hypothesis, there exists a splitting field E of g(X) over K1. Hence, in E[X], g(X) = a(X − α2) ··· (X − αn) for some α2, . . . , αn ∈ E and E = K1(α2, . . . , αn). But then, in E[X], f(X) = (X − α1)g(X) = a(X − α1) ··· (X − αn) and E = K1(α2, . . . , αn) = F (α1, . . . , αn); so that E is a splitting field of f(X) over F . Induction on n = deg f(X) will be used too for (ii). Again, if n = 1, then E = F and Eˆ = Fˆ because the roots (if any) of f(X) (respectively fˆ(X)) are in F (respectively in Fˆ). Hence σ = ϕ. Now, if n > 1, let f1(X) be an irreducible factor of f(X) and let fˆ1(X) be the corresponding irreducible factor of fˆ(X). Let α ∈ E be a root of f1(X) and β ∈ Eˆ a root of fˆ1(X). Thus F ≤ F (α) ≤ E and Fˆ ≤ Fˆ(β) ≤ Eˆ and, by the previous Lemma, there exists σ1 : F (α) → Fˆ(β), a field isomorphism, extending ϕ and such that σ1(α) = β. In F (α)[X], f(X) = (X−α)g(X), and in Fˆ(β)[X], fˆ(X) = (X−β)ˆg(X), whereg ˆ(X) is the quotient of fˆ(X), which is the polynomial obtained from f(X) by applying σ1 to all its coefficients, by (X−β), which is the polynomial obtained from (X − α) by applying σ1 to all its coefficients. Thereforeg ˆ(X) is the polynomial obtained from g(X) by applying σ1 to all its coefficients. Moreover, E is a splitting field of g(X) over F (α) and Eˆ is a splitting field ofg ˆ(X) over Fˆ(β) so, by the induction hypothesis, there exists a field ˆ isomorphism σ : E → E, such that σ|F (α) = σ1, which implies that σ|F = ϕ, as required.

2.6 Definition. A finite field extension K/F is said to be normal if there is a polynomial f(X) ∈ F [X] such that K is the splitting field of f(X) over F .

2.7 Theorem. Let K/F be a ﬁnite ﬁeld extension, then the following are equivalent:

• K/F is normal, § 2. SPLITTING FIELDS. ALGEBRAIC CLOSURE 45

• For any α ∈ K, mα,F (X) splits in K[X]. (In other words, K contains all the roots of mα,F (X).)

Proof. Assume first that mα,F (X) splits for any α ∈ K, and write K = Qr F (α1, . . . , αr). Then K is the splitting field of the polynomial i=1 mαi,F (X) over F , and hence K/F is normal. Conversely, if K is the splitting field of a polynomial f(X) ∈ F [X], for an arbitrary α ∈ K take a splitting field L of mα,F (X) over K. Hence L is a splitting field of the polynomial mα,F (X)f(X) over F and F ⊆ K ⊆ L. Let β ∈ L be a root of mα,F (X). Then the isomorphism τ : F (α) → F (β), such that τ|F = 1 and τ(α) = β, extends to an automorphism σ of L. But σ permutes the roots of f(X) ∈ F [X], and hence σ(K) = K. It follows that β = σ(α) is in K, and hence K contains all the roots of mα,F (X).

2.8 Definition. Let K/F be a field extension. Then K is said to be an algebraic closure of F if K/F is an algebraic field extension and any polynomial f(X) ∈ F [X] with deg f(X) ≥ 1 splits in K[X]. 2.9 Definition. A field F is said to be algebraically closed if any polynomial of degree ≥ 1 over F has a root in F . (That is to say, the irreducible polynomials in F [X] are the degree 1 polynomials.)

2.10 Example. C is algebraically closed by the Fundamental Theorem of Algebra (proved in Algebraic Structures) and hence it is an algebraic closure of R. 2.11 Theorem. Let F be a ﬁeld. (i) If F¯ is an algebraic closure of F , then F¯ is algebraically closed.

(ii) There exists an algebraically closed ﬁeld extension K of F .

(iii) If K is an algebraically closed ﬁeld extension of F , then F¯ = {α ∈ K : α is algebraic over F } is an algebraic closure of F .

(iv) Up to isomorphisms of field extensions, there is a unique algebraic closure of F . Proof. For the first part let f(X) ∈ F¯[X] be a monic polynomial, deg f(X) ≥ 1, and let α be a root of f(X) in some field extension L of F¯. Then α is algebraic over F¯ and F¯ /F is algebraic, so that α is algebraic over F . But mα,F (X) ∈ F [X] factors over F¯ as a product of degree 1 polynomials: mα,F (X) = (X − α1) ··· (X − αr), with α1, . . . , αr ∈ F¯. Since Qr 0 = mα,F (α) = i=1(α − αi) in L, it follows that there is an i such that α = αi, and thus α ∈ F¯, as required. For the second part we will follow the proof by Artin. For each monic polynomial f(X) ∈ F [X] of degree ≥ 1, let Xf be a variable, and consider 46 CHAPTER 2. GALOIS THEORY

the ring R of polynomials in the variables Xf with coeﬃcients in F . Let I be the ideal generated by the set {f(Xf ): f(X) ∈ F [X], monic of degree ≥ 1}. In case I = R, then 1 ∈ I, so that there are monic degree ≥ 1 polynomials f1(X), . . . , fr(X) and polynomials g1, . . . , gr in in the variables Xf ’s such that 1 = g1f1(Xf1 ) + ··· + grfr(Xfr ). Rename the variables Xf1 =

X1,...,Xfr = Xr, and denote by Xr+1,...,Xn the remaining variables ap- pearing in the gi’s. Then we have:

(2.12) 1 = g1(X1,...,Xn)f1(X1) + ··· + gr(X1,...,Xn)fr(Xr).

0 0 Let F be a splitting field of f1(X) ··· fn(X) over F . In F there are elements αi such that fi(αi) = 0 for i = 1, . . . , r. Then evaluate equation (2.12) with Xi 7→ αi, for i = 1, . . . , r, and Xi 7→ 0, for i = r + 1, . . . , n, to get 1 = 0, a contradiction. Therefore the ideal I is not the whole R. Let M be a maximal ideal of R containing I, and consider the field K1 = R/M. K1 is a field extension of F through the embedding ι : F → R/M, a 7→ a + M, and if xf = Xf + M for any monic polynomial of degree ≥ 1, xf is a root of f(X) over K1. Thus K1 contains a root for each monic polynomial in F [X] of degree ≥ 1. We repeat the process with K1 to get a field extension K2 of K1 such that any degree ≥ 1 monic polynomial in K1[X] has a root in K2, and then we repeat the process with K2, ... Eventually we get a chain

F = K0 ⊆ K1 ⊆ K2 ⊆ · · · ⊆ Kn ⊆ Kn+1 ⊆ · · · where any degree ≥ 1 monic polynomial in Kn[X] has a root in Kn+1 for any ∞ n ≥ 0. Consider K = ∪n=0Kn, which is a field extension of F . Also, given any degree ≥ 1 monic polynomial h(X) ∈ K[X], there is an n ∈ N such that h(X) ∈ Kn[X], and hence h(X) has a root in Kn+1 ⊆ K. Therefore, K is algebraically closed. Now the third part of the Theorem is easy. F¯ is a subfield of K containing F , F¯ /F is an algebraic field extension and if f(X) ∈ F [X] is a polynomial of degree ≥ 1, then f(X) = a(X −α1) ··· (X −αr) with a ∈ F and α1, . . . , αn ∈ K. But then α1, . . . , αr ∈ F¯, as they are roots of f(X), and hence algebraic over F . Finally, because of items (ii) and (iii) there are algebraic closures of F . If K1 and K2 are two such algebraic closures, consider the set S consisting of all the triples (L1,L2, ϕ), where L1 is a field extension of F contained in K1, L2 is a field extension of F contained in K2 and ϕ : L1 → L2 is an F -isomorphism (that is, an isomorphism which restricts to the identity 0 0 0 map of F ). The set S is partially ordered with (L1,L2, ϕ) ≤ (L1,L2, ϕ ) 0 0 0 if L1 ⊆ L1, L2 ⊆ L2 and ϕ = ϕ |L1 . Besides, any chain has an upper bound (obtained as the union of the first and second components, and the natural isomorphism among these unions). By Zorn’s Lemma, there exists a maximal element (L1,L2, ϕ) in S. If L1 6= K1, take an element α ∈ K1 \ L1. § 3. SEPARABLE EXTENSIONS 47

Since α is algebraic over L1, ϕ can be extended (as in the proof of Theorem 2.5) to an isomorphism from L1(α) onto a subﬁeld of K2 containing L2, a contradiction. Hence L1 = K1, and similarly we prove that L2 = K2. Therefore, K1 and K2 are isomorphic.

§ 3. Separable extensions

3.1 Definition. Let f(X) be a degree ≥ 1 polynomial over a field F . Then f(X) is said to be separable if it does not have multiple roots (in a splitting field). As proved in Algebraic Structures, under these circumstances: f(X) is separable ⇐⇒ gcdf(X), f 0(X) = 1 ⇐⇒ Df(X) 6= 0, where Df(X) is the discriminant of f(X). n Recall from Algebraic Structures that if f(X) = a0 + a1X + ··· + anX m and g(X) = b0 + b1X + ··· + bmX are polynomials over a field F with n, m ≥ 1 (although we admit that an or bm may be 0), then the resultant of f(X) and g(X) is the determinant of the Sylvester matrix:

 an an−1 an−2 ...... a0 0 ...... 0  0 an an−1 ...... a0 0 ... 0  ......  ...... m rows  0 ... 0 an ...... a1 a0 0   0 ...... 0 an ...... a2 a1 a0  Resn,m(f, g) = b b b ...... b 0 ...... 0  m m−1 m−2 0  0 b b ...... b 0 ... 0  m m−1 0  ......  ...... n rows  0 ... 0 bm ...... b1 b0 0   0 ...... 0 bm ...... b2 b1 b0  | {z } n + m columns

If deg f(X) = n and deg g(X) = m (that is, if an 6= 0 6= bm), then we write simply Res(f, g). n The discriminant of the polynomial f(X) = a0 + a1X + ··· + anX ∈ F [X] of degree n ≥ 2 (an 6= 0) is the scalar (in F )

n (2) −1 0 D(f) = (−1) an Resn,n−1(f, f ).

In case f(X) = (X − α1) ··· (X − αn) in some ﬁeld extension of F , then Q 2 Q D(f) = 1≤i

3.2 Examples.

pn • X − X ∈ Fp[X] is separable, because its derivative is −1.

• Over a ﬁeld of characteristic 0 any irreducible polynomial is separable, as gcdf(X), f 0(X) = 1, since f(X) is irreducible, and 0 6= f 0(X) is not a multiple of f(X).

• Assume that F is a ﬁeld of characteristic p > 0. The polynomial f(X) = Xn − 1 satisﬁes ( 0 if p|n, f 0(X) = nXn−1 = 6= 0 if p-n,

and hence, f(X) is separable if and only if p-n.

• Over a finite field (and hence of prime characteristic) any irreducible polynomial is separable. Actually, let F be a finite field of characteristic p, and let f(X) = n a0 + a1X + ··· + anX be an irreducible polynomial over F (an 6= 0). If f 0(X) 6= 0, we conclude as before that f(X) is separable. Otherwise f 0(X) = 0 and hence the powers of X which appear on f(X) with a nonzero coefficient are of the form Xps for some s. But ϕ : F → F , a 7→ ap is one-to-one and, since F is finite, it is a bijection. Hence, for any a ∈ F there is a unique b ∈ F with bp = a. Hence

p rp p p p p rp f(X) = a0 + apX + ··· + arpX = b0 + b1X + ··· + brX r p = (b0 + b1X + ··· + brX ) ,

for some b0, b1, . . . , br ∈ F . But this is a contradiction with the irreducibility of f(X).

3.3 Deﬁnition.

• Given a ﬁeld extension K/F , an element α ∈ K is said to be separable over F if it is algebraic and mα,F (X) is separable.

• A ﬁnite ﬁeld extension K/F is said to be separable if any element of K is separable over F .

• A field F is said to be perfect if all its finite field extensions are separable. § 4. GALOIS GROUP 49

3.4 Properties.

(i) A ﬁeld F is perfect if and only if any irreducible polynomial in F [X] is separable.

(ii) The fields of characteristic 0 and the finite fields are perfect.

(iii) Let F be a field of characteristic p and let F p = {αp : α ∈ F }. (This is a subfield of F !) Then F is perfect if and only if F = F p. Actually, if F p = F , the same arguments as in the last example above show that F is perfect. Conversely, if F is perfect and α ∈ F , take K a splitting field of Xp − α over F , and take a root β in K, so that βp = α and Xp − α = Xp − βp = (X − β)p in K[X]. Since F is p perfect, mβ,F (X) is separable, but mβ,F (X) | X − α. We conclude p that mβ,F (X) = X − β and β ∈ F . Hence F = F . p 3.5 Example. Fp(X) is not a perfect field, because X 6∈ Fp(X) .

§ 4. Galois group

Let K be a field and let σ : K → K be an automorphism. The element a ∈ K is fixed by σ if σ(a) = a. Note that we have σ(1) = 1, and hence σ(m1) = m1 for any m ∈ Z. Hence any automorphism fixes the prime subfield of K.

4.1 Deﬁnition.

(i) Let K/F be a ﬁeld extension. The Galois group of K/F is the group

Gal(K/F ) = {σ ∈ Aut K : σ(a) = a ∀a ∈ F } ≤ Aut K.

(ii) Let F be a ﬁeld and let 0 6= f(X) ∈ F [X] be a nonzero polynomial. Let E be the splitting ﬁeld of f(X) over F . Then Gal(E/F ) is said to be the Galois group of the polynomial f(X) over F .

4.2 Proposition. Let K/F be a ﬁeld extension, let α ∈ K be algebraic over F , and let σ be an element in Gal(K/F ). Then σ(α) is a root of mα,F (X). (That is, σ induces a permutation of the roots of mα,F (X) which lie in K.)

n−1 n Proof. Assume mα,F (X) = a0+a1X+···+an−1X +X , ai ∈ F for all i = n−1 n 0, . . . , n−1. Then 0 = mα,F (X) = a0 +a1α+···+an−1α +α , and hence, n−1 n since σ(ai) = ai for any i we get 0 = a0 + a1σ(α) + ··· an−1σ(α) + σ(α) , so that σ(α) is a root too of mα,F (X). 50 CHAPTER 2. GALOIS THEORY

4.3 Examples. √ √ √ (i) Take K√= Q( 2). Then any τ ∈ Gal(K/Q) satisfies τ( 2) = ± 2, 2 since τ( 2) is a root of X − 2. Therefore Gal(K/Q) = {1, σ},√ where 1 denotes√ the identity automorphism and σ is given by σ(a + b 2) = a − b 2. Therefore Gal(K/Q) is the cyclic group of two elements. √ 3 3 (ii) Now let K = ( 2). Since m √3 (X) = X − 2 and the other two Q 2,Q √ √ roots of this polynomial are not in K, we conclude that τ( 3 2) = 3 2 for any τ ∈ Gal(K/Q), and hence Gal(K/Q) = 1. The proofs of the following two propositions are straightforward. 4.4 Proposition. Let K be a field and let H be a subgroup of Aut K. Then KH = {a ∈ K : σ(a) = a ∀σ ∈ H} is a subfield of K, which is called the fixed field by H. 4.5 Proposition. Let K be a field.

(i) If F1 and F2 are two subﬁelds of K with F1 ⊆ F2, then Gal(K/F2) is a subgroup of Gal(K/F1). (Thus Gal(K/F2) ≤ Gal(K/F1) ≤ Aut K.)

H (ii) Let H1 and H2 be two subgroups of Aut K with H1 ≤ H2. Then K 2 is a subﬁeld of KH1 . Therefore the two maps:

{subﬁelds of K} → {subgroups of Aut K} F 7→ Gal(K/F ), and {subgroups of Aut K} → {subﬁelds of K} H 7→ KH , reverse containments.

4.6 Dedekind’s Lemma. Let G be a group, K a field, and σ1, . . . , σn : × G → K different group homomorphisms. Then σ1, . . . , σn are “linearly Pn independent”. That is, if a1, . . . , an ∈ K satisfy i=1 aiσi(g) = 0 for any g ∈ G, then a1 = ··· = an = 0. Proof. The proof is done by induction on n, the result being clear if n = 1. Hence assume the result to be true for n − 1 homomorphisms and that × there are different group homomorphisms σ1, . . . , σn : G → K and scalars a1, . . . , an ∈ K, not all of them different from 0, such that

a1σ1(g) + ··· + anσn(g) = 0, § 4. GALOIS GROUP 51 for any g ∈ G. If one of these scalars is 0, then we get a contradiction with the n − 1 case. Hence ai 6= 0 for any i. Now, as σ1 6= σn, there is an element h ∈ G such that σ1(h) 6= σn(h). Then we have:

n X aiσi(hg) = a1σ1(h)σ1(g) + ··· + anσn(h)σn(g) = 0 i=1 n X σn(h) aiσi(g) = a1σn(h)σ1(g) + ··· + anσn(h)σn(g) = 0. i=1 If we subtract the two equations we get: a1 σ1(h) − σn(h) σ1(g) + ··· + an−1 σn−1(h) − σn(h) σn−1(g) = 0. But since a1 σ1(h) − σn(h) is not 0, we get a contradiction with the induction hypothesis.

The next theorem is usually attributed to Artin, but it is already present in the work of Dedekind:

4.7 Theorem. Let K be a ﬁeld and let G be a subgroup of Aut K of order G n ∈ N, then [K : K ] = n.

G Proof. Let F = K and G = {σ1 = 1, . . . , σn}. Assume ﬁrst [K : F ] < n and let {ω1, . . . , ωm} be a basis of K over F (m < n). Consider the following homogeneous system of m linear equations with n unknowns over K:  σ1(ω1)x1 + ··· + σn(ω1)xn = 0,   . .   σ1(ωm)x1 + ··· + σn(ωm)xn = 0.

As m < n, there is a nontrivial solution α1, . . . , αn. Now, for any 0 6= α ∈ K, there are scalars λ1, . . . , λm ∈ F such that α = λ1ω1 + ··· + λmωm. Note G that σi(λj) = λj for any i, j, because F = K , so if we multiply in the system above, with the xi’s substituted by the αi’s, the ﬁrst row by λ1, the second row by λ2, ..., and we add the resulting equations we get

σ1(α)α1 + ··· + σn(α)αn = 0 for any 0 6= α ∈ K, and this contradicts Dedekind’s Lemma. (Here we have × × to consider σ1, . . . , σn as group homomorphisms K → K .) Note that we have not used here that G is a group, only that σ1, . . . , σn are diﬀerent automorphisms of K.

Assume now [K : F ] > n, and let ω1, . . . , ωn+1 ∈ K be a family of linearly independent elements (over F ). Again there is a nontrivial solution 52 CHAPTER 2. GALOIS THEORY

α1, . . . , αn+1 of the homogeneous linear system of equations:  σ1(ω1)x1 + ··· + σ1(ωn+1)xn+1 = 0,   . .   σn(ω1)x1 + ··· + σn(ωn−1)xn+1 = 0.

As σ1 = 1 and the ωi’s are linearly independent over F , not all the αi’s are in F (look at the ﬁrst row of the system). Among all the possible nontrivial solutions we take one (α1, . . . , αn+1) with the lowest possible number r of nonzero elements. After reordering the ωi’s, we may assume that this solution is of the form (β1, . . . , βr, 0,..., 0) with βr = 1 and β1 6∈ F . Therefore, for any j = 1, . . . , n we have

σj(ω1)β1 + ··· + σj(ωr−1)βr−1 + σj(ωr) = 0.

Now, since β1 6∈ F , there is an index k such that σk(β1) 6= β1. Apply σk to the last equation to get for any j:

σkσj(ω1)σk(β1) + ··· + σkσj(ωr−1)σk(βr−1) + σkσj(ωr) = 0.

But since G is a group, we have G = {σkσ1, . . . , σkσn}, and hence the tuple (σk(β1), . . . , σk(βr), 0,..., 0) is another solution of the homogeneous linear system. Therefore we get the new solution β1 − σk(β1), . . . , βr−1 − σk(βr−1), 1 − 1, 0,..., 0) with at most r − 1 nonzero components, and with the ﬁrst component diﬀerent from 0. This contradicts our assumption on the minimality of r.

4.8 Corollary. Let K/F be a finite field extension. Then |Gal(K/F )| ≤ [K : F ], and the equality holds if and only if F is the fixed field by Gal(K/F ).

Proof. Write G = Gal(K/F ). The ﬁrst part of the proof above shows that |G| is at most [K : KG], but [K : F ] = [K : KG][KG : F ], so that |G| ≤ [K : KG] ≤ [K : F ]. In particular G is ﬁnite. Now, the previous theorem shows that |G| = [K : KG], and hence |G| = [K : F ] if and only if [KG : F ] = 1, if and only if F = KG.

4.9 Corollary. Let K be a ﬁeld and let G and H be two ﬁnite subgroups of Aut K. Then KG = KH if and only if G = H.

Proof. If KG = KH and σ ∈ H \G, then the fixed subfield by G∪{σ} is KG, but the first part of the proof of Theorem 4.7 shows that |G| + 1 ≤ [K : KG] and also |G| = [K : KG], thus getting a contradiction. The converse is trivial.

4.10 Corollary. Let K be a ﬁeld and let G be a ﬁnite subgroup of Aut K. Then Gal(K/KG) = G. § 4. GALOIS GROUP 53

Proof. The containment G ≤ Gal(K/KG) is clear, and both subgroups have the same ﬁxed subﬁeld, so the previous corollary applies.

4.11 Theorem. Let K/F be a ﬁnite ﬁeld extension, and let G = Gal(K/F ) be its Galois group. The following conditions are equivalent:

(i) F = KG,

(ii) |Gal(K/F )| = [K : F ],

(iii) Any irreducible polynomial h(X) ∈ F [X] which has a root in K is separable and splits over K.

(iv) K is the splitting ﬁeld of a separable polynomial f(X) ∈ F [X].

Proof. The previous corollaries prove the equivalence of conditions (i) and (ii). (i) ⇒ (iii): Let f(X) ∈ F [X] be an irreducible polynomial and let α ∈ K be a root of f(X). Consider the different elements α1, . . . , αr in the set {σ(α): σ ∈ G} (which is contained in the set of roots of f(X)) and take the polynomial g(X) = (X − α1) ··· (X − αr) ∈ K[X]. Each σ ∈ G permutes the αi’s, and hence the coefficients of g(X) are fixed by σ. By (i) we get g(X) ∈ F [X]. But g(X) | f(X) and f(X) is irreducible, so we get f(X) = ag(X) for some 0 6= a ∈ F , which is separable with all its roots in K.

(iii) ⇒ (iv): Let {ω1, . . . , ωn} be a basis of K over F . Item (iii) implies that mωi,F (X) is separable for any i and all its roots are in K. Hence K is the splitting field of the least common multiple of these polynomials mωi,F (X), which is separable. (iv) ⇒ (ii): We know that if ϕ : F → F ] is a field isomorphism, K is a splitting field of a polynomial f(X) ∈ F [X] over F , and K] is a splitting field of the polynomial f ](X) = ϕf(X) over F ] (this is the polynomial obtained from f(X) by applying ϕ to all its coefficients), then there is a field isomorphism ψ : K → K] which extends ϕ. We will prove by induction on [K : F ] that the number of ψ’s extending ϕ is at most [K : F ], and that it is exactly [K : F ] if f(X) is separable. The particular case of F = F ], K = K] and ϕ = 1 proves (iv) ⇒ (ii). If [K : F ] = 1, then K = F , K] = F ] and ψ = ϕ is the only possibility. Assume then that [K : F ] > 1 and that the result is valid for lower degree extensions. Let h(X) be an irreducible factor of f(X) with a root α ∈ K \F , and let h](X) be its “image” under ϕ. If ψ extends ϕ then α] = ψ(α) is a ] ] ] root of h (X) and τ = ψ|F (α) : F (α) → F (α ) is an isomorphism extending 54 CHAPTER 2. GALOIS THEORY

ϕ. We get the diagram:

ψ : K −→∼ K] ↑ ↑ ∪ ∪ τ : F (α) −→∼ F ](α]) ↑ ↑ ∪ ∪ ϕ : F −→∼ F ]

Conversely, if α] is a root of h](X), there is an isomorphism τ : F (α) → F ](α]) extending ϕ and such that τ(α) = α], and since K (respectively K]) is a splitting ﬁeld of f(X) (respectively f ](X)) over F (respectively F ]), there is an isomorphism ψ : K → K] extending τ. Therefore the number of extensions ψ : K → K] of ϕ equals the product of the number of extensions τ : F (α) → F ](α]) of ϕ by the number of extensions of each such τ to an isomorphism ψ : K → K]. This last number is, by the induction hypothesis, at most [K : F (α)], with equality if f(X) is separable, while the ﬁrst number (of extensions τ) is the number of roots of h(X), which is at most deg h(X) = [F (α): F ], again with equality if f(X) (and hence h(X)) is separable. Therefore the number of extensions ψ of ϕ is at most [K : F (α)][F (α): F ] = [K : F ], with equality if f(X) is separable.

4.12 Definition. A finite field extension K/F satisfying the equivalent conditions in the previous theorem is said to be a Galois extension.

4.13 Corollary. Let K/F be a ﬁnite ﬁeld extension. Then K/F is a Galois extension if and only if it is separable and normal.

Proof. If K/F is a Galois extension, then for any α ∈ K, mα,F (X) is separable by item (iii) in the previous theorem, and hence K/F is separable. By item (iv) K/F is normal too. Conversely, if K/F is a finite normal field extension, then K is the splitting field of a monic polynomial f(X) ∈ F [X]. Consider its factorization m ms 0 into irreducible factors: f(X) = f1(X) 1 ··· fs(X) (the fi(X) s are irreducible and coprime). Then K is also the splitting field over F of the polynomial f1(X) ··· fs(X). Now, if K/F is separable, each fi(X) is separable, and hence so is its product. Hence item (iv) of the previous theorem is fulfilled. § 5. THE FUNDAMENTAL THEOREM OF GALOIS THEORY 55

4.14 Consequences.

(i) Let K/F be a ﬁnite ﬁeld extension, and assume that K = F (α1, . . . , αr) with αi separable over F for any i = 1, . . . , r. Then K/F is separable.

(ii) Let K/F be a finite field extension. Consider the set S = {α ∈ K : α is separable over F }. Then S is a subfield of K containing F . More- over, if the characteristic of F is 0, then S = K, while if the characteristic of F is p > 0, then for any α ∈ K there is an n ∈ N ∪ {0} such that αpn ∈ S.

(iii) If L/K and K/F are two ﬁnite ﬁeld extensions, then L/F is separable if and only if so are L/K and K/F .

Proof. For (i) note that if L is a splitting field over F , and containing K, of the least common multiple of the mαi,F (X)’s, then L/F is a Galois extension. Hence L/F is separable and so is K/F . For (ii) note that F is trivially contained in S and that, because of (i), for any two elements α, β ∈ S, the subfield F (α, β) is contained in S, in particular, α + β, αβ, α−1 (if α 6= 0) are in S and S is a subfield. If the characteristic is 0 any field extension is separable, so the last assertion follows. Assume now that the characteristic is p, and take α ∈ K and f(X) = mα,F (X). Then there is an n ∈ N ∪ {0} and a polynomial g(X) ∈ F [X] such that f(X) = g(Xpn ) and g0(X) 6= 0. Since f(X) is irreducible, so is g(X), and hence g(X) is separable. It follows that αpn ∈ S. For (iii) note that if L/F is separable, then so is K/F . Also for any α ∈ L, mα,K (X) divides mα,F (X), which is separable, and hence mα,K (X) is separable too. This shows that L/K is separable. Conversely, if L/K and K/F are separable, then if the characteristic is 0 obviously L/F is separable, while if the characteristic is p, by (ii) the subfield S = {α ∈ L : α is separable} contains K, and for any α ∈ L, there is an n such that pn pn pn pn α ∈ S. Then mα,S(X) | X − α = (X − α) . But L/K is separable, and hence so is L/S. Therefore mα,S(X) = X − α and α ∈ S. Thus L = S is separable over F .

§ 5. The Fundamental Theorem of Galois Theory √ √ 5.1 Example. Let K = Q( 2, 3) and F = Q. Then K is a splitting ﬁeld 2 2 of (X − 2)(X − 3) over F , and hence K/F is a Galois extension.√ For any ξ ∈ Gal(K/F ),√ξ is determined by the valued ξ( 2) (which is a root of X2 − 2) and ξ( 3) (a root of X2 − 3). Hence there are four 56 CHAPTER 2. GALOIS THEORY possibilities: ( √ √ ( √ √ ( √ √ ( √ √ 2 7→ 2, 2 7→ − 2, 2 7→ 2, 2 7→ − 2, √ √ √ √ √ √ √ √ 3 7→ 3, 3 7→ 3, 3 7→ − 3, 3 7→ − 3, 1 σ τ στ = τσ

Since Gal(K/F ) has order 4, it follows that Gal(K/F ) = {1, σ, τ, στ} = hσi × hτi ' C2 × C2. Let us have a look at the subgroups of Gal(K/F ) and the corresponding fixed subfields: Subgroups of Gal(K/F ) Fixed subfields √ √ 1 K = Q(√ 2, 3) hσi Q(√3) hτi Q(√2) hστi Q( 6) hσ, τi = Gal(K/F ) Q

Here is the diagram of subgroups:

1 ...... 22 ...... 2 ...... hσi hστi hτi ...... 2. 2...... 2 ...... hσ, τi and here the diagram of ﬁxed subﬁelds:

K ...... 22 ...... 2 ...... √..... √...... √ ( 3) ( 6) ( 2) Q . Q . Q...... 2. 2...... 2 ...... Q

The Fundamental Theorem of Galois Theory establishes a bijection between the set of intermediate subfields in a Galois extension K/F and the set of subgroups of the group Gal(K/F ). The first set seems, in principle, quite difficult to deal with (its elements are usually quite complicated infinite objects), while the second set is a finite discrete set, more suitable for direct computations (with or without a computer). § 5. THE FUNDAMENTAL THEOREM OF GALOIS THEORY 57

5.2 Fundamental Theorem of Galois Theory. Let K/F be a Galois extension with Galois group G = Gal(K/F ). Then the maps

ϕ {Subﬁelds of K containing F } −−→{Subgroups of G} E 7→ Gal(K/E), and

ψ {Subgroups of G} −−→{Subﬁelds of K containing F } H 7→ KH ,

(both of which reverse containments) are bijective and mutually inverse. Moreover, for H ≤ G and E = KH , we have [E : F ] = [G : H] and [K : E] = |H|. Besides K/E is a Galois extension, while the extension E/F is a Galois extension if and only if H is a normal subgroup of G. In this case, Gal(E/F ) is isomorphic to G/H.

H Proof. We already know that if H1 and H2 are subgroups of G and K 1 = H K 2 , then H1 = H2. Hence ψ is one-to-one. If E is a subfield with F ⊆ E ⊆ K, then since K/F is a Galois extension, K is a splitting field over F of a separable polynomial f(X) ∈ F [X], and hence K is a splitting field too of f(X) over E. Therefore, K/E is a Galois extension. On the other hand, the extension K/E is a Galois extension if and only if E is the fixed field by Gal(K/E), that is: E = KGal(K/E) = ψ(ϕ(E)). This shows that ψϕ = 1, and hence ψ is onto too. We conclude that ψ is a bijection and ϕ = ψ−1. Moreover, since K/E is a Galois extension, we get [K : E] = |H|, with H = Gal(K/E). But ( [G : H]|H|, |G| = [K : F ] = [K : E][E : F ], and we obtain [E : F ] = [G : H]. Finally, let H be a subgroup of G and let E = KH . Let us prove the equivalence of the following condtions: (i) E/F is a Galois extension, (ii) σ(E) = E for any σ ∈ G, (iii) H is a normal subgroup of G. (i)⇒(ii): If E/F is a Galois extension, then E is the splitting field over F of a separable polynomial f(X) ∈ F [X]: f(X) = (X − α1) ··· (X − αr), where α1, . . . , αr are all different and E = F (α1, . . . , αr). Now for any σ ∈ G and any i, 1 ≤ i ≤ r, σ(αi) is a root of f(X). Thus σ permutes the αi’s, and hence σ(E) = E. (iii)⇒(ii): If H is normal, for any σ ∈ G, τ ∈ H, and γ ∈ E, τσ(γ) = σ(σ−1τσ)(γ) = σ(γ), because σ−1τσ is in H. Hence σ(γ) ∈ KH = E and σ(E) = E. 58 CHAPTER 2. GALOIS THEORY

(ii) ⇒ (i) and (iii): If σ(E) = E for any σ ∈ G, we have the group homomorphism:

G = Gal(K/F ) −→π Gal(E/F )

σ 7→ σ|E, with kernel Gal(K/E) = H, so that H is normal. Besides π is onto, since any automorphism τ : E → E can be extended to K because K is a splitting ﬁeld over F . By the First Isomorphism Theorem G/H is isomorphic to Gal(E/F ) and |Gal(E/F )| = [G : H] = [E : F ]. We conclude that E/F is a Galois extension. (Alternatively, for any α ∈ EGal(E/F ) and σ ∈ G, G Gal(E/F ) σ(α) = σ|E(α) = α, so α ∈ K = F . Hence F = E , and E/F is a Galois extension.)

5.3 Corollary. Let K/F be a Galois extension. Then there is a ﬁnite number of intermediate subﬁelds between F and K.

§ 6. Finite ﬁelds

6.1 Lemma. Let F be a ﬁeld and let G be a ﬁnite subgroup of the multiplicative group F ×. Then G is cyclic.

Proof. G is a ﬁnite abelian group, and hence G is isomorphic to a direct product of cyclic groups G ' Cm1 × · · · × Cmr , with m1 | m2 | · · · | mr. Therefore any x ∈ G is a root of the polynomial Xmr − 1. But F contains at most mr roots of this polynomial. We conclude that r = 1, G is cyclic, and G is the set of roots of the separable polynomial Xmr − 1 in F .

If F is a finite field, then its characteristic is a prime number p. If n denotes the dimension of F over its prime subfield Fp: n = dimFp F , then |F | = pn. The following result is due to Galois:

6.2 Theorem. For each prime number p and each natural number n there exists a unique (up to isomorphism) ﬁeld F with pn elements. Moreover, the extension F/Fp is a Galois extension and Gal(F/Fp) is the cyclic group of order n. (Notation: F = GF (pn).)

Proof. If F is a field with pn elements, then F × is a cyclic group of order pn−1 whose elements are the roots of the separable polynomial Xpn−1 − 1. Therefore F is the splitting field of the separable polynomial Xpn − X ∈ Fp[X]. This proves the uniqueness. On the other hand, let E be a splitting field of the separable polynomial pn f(X) = X − X over Fp (note that the derivative of this polynomial is § 7. PRIMITIVE ELEMENTS 59

−1, and hence gcd(f(X), f 0(X)) = 1 and it contains no multiple roots). Let F = {x ∈ E : f(x) = 0}. Then since f(X) has no multiple roots, |F | = pn. But F is a ﬁeld, because (a + b)pn = apn + bpn ,(ab)pn = apn bpn and, if b 6= 0, (bpn )−1 = (b−1)pn for any a, b ∈ E. Hence F = E and |E| = pn. This shows the existence. Finally, for F = E as above, the map σ : F → F , x 7→ xp is an automorphism (called the Frobenius automorphism). Since F × is cyclic, F × = hαi for some 0 6= α ∈ F , with αpn−1 = 1 and αi 6= 1 for any i < pn − 1. Then σn(α) = αpn = α, while σi(α) 6= α for i < pn − 1. Hence the order of σ is n = |Gal(F/Fp)|, and hence Gal(F/Fp) = hσi is cyclic.

6.3 Remark.

(i) By the Fundamental Theorem of Galois Theory, the subfields of GF (pn) are in one-to-one correspondence with the subgroups of Gal(F/Fp) = hσi, and hence with the set of divisors of n. For any divisor d of n, d d hσ i is a subgroup of hσi = Gal(F/Fp) and the fixed subfield by σ is {β ∈ F : βpd = β} = GF (pd). Thus the set of subfields of GF (pn) is {GF (pd): d|n}.

n × n (ii) If α is a generator of the cyclic group GF (p ) , then GF (p ) = Fp(α) and deg mα,Fp (X) = n. In particular, for any n ∈ N there is an irreducible polynomial in Fp[X] of degree n.

§ 7. Primitive elements

7.1 Deﬁnition. A ﬁeld extension K/F is said to be simple if there is an element α ∈ K such that K = F (α). In this case, the element α is said to be a primitive element of the extension.

The following theorem is due to Steinitz:

7.2 Theorem. Let K/F be a finite field extension. Then K/F is simple if and only if the number of intermediate subfields between F and K is finite.

Proof. Assume first that K = F (α) is a simple extension, and let f(X) = mα,F (X). Let E be an intermediate subfield: F ⊆ E ⊆ K. Then the r r+1 polynomial mα,E(X) = a0 +a1X +···+arX +X divides f(X) (in E[X], 0 and hence in K[X]). Consider the subfield E = F (a0, a1, . . . , ar). Then F ⊆ 0 0 E ⊆ E ⊆ K and mα,E(X) | mα,E0 (X) in E[X], but mα,E(X) ∈ E [X], and 0 it is irreducible over E, and hence over E , so equality follows: mα,E(X) = mα,E0 (X). But then we have [K : E] = [E(α): E] = deg mα,E(X) = [K : E0] and E = E0. 60 CHAPTER 2. GALOIS THEORY

Therefore the intermediate subfields are generated over F by the coefficients of monic factors of f(X), and there are only a finite number of such factors. For the converse, if F is finite, then K is finite too and hence K = Fp(α) = F (α) for some α. If F is infinite, it is enough to show that for any α, β ∈ K, there is an element c ∈ F such that F (α, β) = F (α +cβ). Since F is infinite and there is only a finite number of intermediate subfields, there are elements c1 6= c2 in F such that F (α + c1β) = F (α + c2β), but then α + c1β, α + c2β ∈ F (α + c2β), and it follows that F (α + c2β) contains both α and β, so F (α + c2β) = F (α, β).

7.3 Primitive Element Theorem. Let K/F be a finite separable field extension. Then K/F is simple. Proof. Let K = F (α1, . . . , αr) and f(X) = lcm mα1,F (X), . . . , mαr,F (X) . Then f(X) is a least common multiple of separable polynomials, and hence it is separable. Let E be a splitting field of f(X) over F with F ⊆ K ⊆ E. Then E/F is a Galois extension so the number of intermediate fields between F and E is finite. Hence so is the number of intermediate fields between F and K and Steinitz’s Theorem applies.

§ 8. Ruler and compass constructions

Recall from Algebraic Structures that an element x ∈ R≥0 is said to be constructible if, starting with the unit segment, it is possible to construct, with ruler and compass only, a segment of length x. An element x ∈ R<0 is constructible if so is −x. It was shown that the set C = {x ∈ R : x is constructible} is a subﬁeld of R containing Q. The main result on C proved in that course is the following:

8.1 Theorem. Let a ∈ R, then a ∈ C if and only if there are n ∈ N ∪ {0} and subﬁelds F0,F1,...,Fn of R such that F0 = Q ≤ F1 ≤ · · · ≤ Fn, [Fi : Fi−1] = 2 for any i = 1, . . . , n, and a ∈ Fn. In particular, if a ∈ C, then [Q(a): Q] is a power of 2.

Now we can improve a little bit on this.

8.2 Lemma. Let K/F be a Galois extension of characteristic 6= 2 with [K : F ] = 2r for some r ≥ 0, and let E/K be a degree 2 ﬁeld extension. Then there is a ﬁeld extension L/E such that L/F is a Galois extension whose degree is again a power of 2. § 8. RULER AND COMPASS CONSTRUCTIONS 61

Proof. Because of the restriction on the characteristic, E = K(α) for some 2 α ∈ E with mα,K (X) = X − µ for some µ ∈ K. Write G = Gal(K/F ). The Primitive Element Theorem shows that there is an element β ∈ K such that K = F (β). Write mβ,F (X) = f(X). Let µ1, . . . , µm be the different elements in the set {σ(µ): σ ∈ G}, and consider the polynomial Qm 2 g(X) = j=1 X − µj . This is a polynomial in F [X], because its coefficients are fixed by the elements in G (which permute the µj’s). Let L be a splitting field over F of the separable polynomial lcmf(X), g(X) containing E. Hence L/F is a Galois extension. Let α1 = α, α2, . . . , αr be the roots of g(X) in L and note that none of them is in K (as γ ∈ K and γ2 = σ(µ) implies σ−1(γ)2 = µ, and hence X2 − µ would have a root in K). Then L = K(α1, . . . , αr) and the degree of each extension K(α1, . . . , αi)/K(α1, . . . , αi−1) is ≤ 2. Hence [L : K] is a power of 2, and so is [L : F ] = [L : K][K : F ].

8.3 Theorem. Let α be a real number algebraic over Q, and let K be a splitting ﬁeld of mα,Q(X) over Q. Then α is constructible if and only if [K : Q] is a power of 2.

Proof. If α is constructible, then there is a tower of field extensions F0 = Q ≤ F1 ≤ · · · ≤ Fn, such that Fi is a subfield of R and [Fi : Fi−1] = 2 for any i = 1, . . . , n, and α ∈ Fn. An inductive argument using the previous lemma shows that there is a Galois extension L/Q with Fn ⊆ L ⊆ C and [L : Q] a power of 2. Since α ∈ L and L/Q is a Galois extension, there is a splitting field of mα,Q(X) contained in L. The uniqueness of splitting fields give the result. Conversely, assume that α belongs to K, which may be assumed to be contained in C, and K/Q is a Galois extension of degree a power of 2. Then G = Gal(K/Q) is a 2-group and a chain of subgroups can be obtained: 1 = G0 ≤ G1 ≤ · · · ≤ Gr = G, such that [Gi : Gi−1] = 2 for any i = 1, . . . , r. Considering the corresponding fixed fields, we get a tower of subfields of C:

Gr Gr−1 G0 Q = K0 = K ≤ K1 = K ≤ · · · ≤ Kr = K = K , such that [Kj : Kj−1] = 2 for any j = 1, . . . , r. The problem is that for any j, Kj is contained in C, and not necessarily in R. Write Kj = Kj−1(aj +bji), 2 with (aj + bji) ∈ Kj−1. For each j consider the ﬁeld Fj generated over Q by the real and imaginary parts of the elements of Kj. It is easy to check that F0 = Q and Fj = Fj−1(aj, bj) for any j ≥ 1. Note that α ∈ R ∩ K, so that α ∈ Fr. 2 2 2 But (aj + bji) ∈ Kj−1, which implies aj − bj , ajbj ∈ Fj−1, and hence we have two extensions

2 2 Fj−1 ≤ Fj−1(aj + bj ) ≤ Fj−1(aj, bj), 62 CHAPTER 2. GALOIS THEORY

2 2 2 2 2 2 2 whose degrees are either 1 or 2, because (aj +bj ) = (aj −bj ) +(2ajbj) is in Fj−1, and if, for instance, aj 6= 0, then bj ∈ Fj−1(aj) (as ajbj ∈ Fj−1), and 2 1 2 2 1 2 2 2 2 aj = 2 (aj + bj ) + 2 (aj − bj ) ∈ Fj−1(aj + bj ). (The same argument applies if bj 6= 0.) Therefore the tower of subfields Q = F0 ≤ · · · ≤ Fr can be refined to a tower of subfields with degree 2 consecutive field extensions.

§ 9. Galois groups of polynomials

Recall that if F is a field and f(X) ∈ F [X] is a degree ≥ 1 polynomial over F , then the Galois group of f(X) over F is the Galois group Gal(E/F ), where E is a splitting field of f(X) over F . If f(X) is separable, then E/F is a Galois extension. 9.1 Proposition. Let F be a field and let f(X) ∈ F [X] be a polynomial of degree ≥ 1. Let E be a splitting field of f(X) over F . Consider the n nr complete factorization f(X) = af1(X) 1 ··· fr(X) , where 0 6= a is the leading coefficient and f1(X), . . . , fr(X) are the different monic irreducible factors of f(X). Denote by Ωi the set of roots of fi(X) in E, for any i = 1, . . . , r, and by Ω = Ω1 ∪ · · · ∪ Ωr (disjoint union) the set of roots of f(X) in E. Then:

(i) Any σ ∈ Gal(E/F ) permutes the elements of Ωi for any i = 1, . . . , r. Hence there is an associated one-to-one group homomorphism: Gal(E/F ) → S(Ω1) × · · · × S(Ωr) ⊆ S(Ω) .

(ii) For any α ∈ Ωi (i = 1, . . . , r), the orbit of α by the action of Gal(E/F ) is the whole Ωi. (That is, Gal(E/F ) acts transitively on each Ωi.)

Proof. If α is in Ωi, it is a root of fi(X) ∈ F [X], and so is σ(α) for any σ ∈ Gal(E/F ). Hence σ(α) ∈ Ωi too. On the other hand, if an element σ ∈ Gal(E/F ) satisfies σ(α) = α for any α ∈ Ω then, as E = F (Ω), σ = 1, so the above group homomorphism is one-to-one. For any i = 1, . . . , r and α, β ∈ Ωi there is a field isomorphism τ : F (α) → F (β) which acts as the identity on F and takes α to β. But E is a splitting field of f(X) over F , and hence this τ extends to an automorphism σ of E. Hence there is an element σ ∈ Gal(E/F ) such that σ(α) = β.

2 2 9.2 Example. For the polynomial√ √ f(X) = (X − 2)(X − 3) ∈ Q[X], the splitting ﬁeld√ is E√= Q( 2, 3), and√ the√ orbits in the proposition above are Ω1 = { 2, − 2} and Ω2 = { 3, − 3}. In this case Gal(E/F ) ' S(Ω1) × S(Ω2) ' C2 × C2.

Let us check now that the symmetric groups Sn appear as Galois groups of certain polynomials. § 9. GALOIS GROUPS OF POLYNOMIALS 63

9.3 Definition. Let F be a field and for n ∈ N let x1, . . . , xn be n different unknowns. The elements s1, . . . , sn ∈ F (x1, . . . , xn) defined by:

 s1 = x1 + ··· + xn,   X  s2 = xixj,   1≤i

9.4 Remark. Note that

pn(X) = (X − x1)(X − x2) ··· (X − xn) n n−1 n−2 n = X − s1X + s2X − · · · + (−1) sn ∈ F (s1, s2, . . . , sn)[X], and that F (x1, . . . , xn) is a splitting ﬁeld of the separable polynomial pn(X) over F (s1, . . . , sn). Hence F (x1, . . . , xn)/F (s1, . . . , sn) is a Galois extension. 9.5 Theorem. The Galois group Gal F (x1, . . . , xn)/F (s1, . . . , sn) is isomorphic to the symmetric group Sn. Proof. By the previous proposition the map Gal F (x1, . . . , xn)/F (s1, . . . , sn) −→ Sn σ 7→ {1, . . . , n} → {1, . . . , n}

i 7→ j if σ(xi) = xj, is a one-to-one group homomorphism. But for any τ ∈ Sn we may deﬁne the automorphism:

τ˜ : F (x1, . . . , xn) → F (x1, . . . , xn)

f(x1, . . . , xn) 7→ f(xτ(1), . . . , xτ(n)). Thenτ ˜(si) = si for any i, so we haveτ ˜ ∈ Gal F (x1, . . . , xn)/F (s1, . . . , sn) and the monomorphism above is an isomorphism.

9.6 Corollary. The general polynomial of degree n: pn(X), is irreducible over F (s1, . . . , sn).

Proof. Just notice that Sn acts transitively on {1, . . . , n}. 64 CHAPTER 2. GALOIS THEORY

9.7 Deﬁnition. A rational function f(x1, . . . , xn) ∈ F (x1, . . . , xn) is said to be symmetric if for any τ ∈ Sn, f(x1, . . . , xn) = f(xτ(1), . . . , xτ(n)). 2 9.8 Example. (x1 − x2)(x1 − x3)(x2 − x3) is symmetric in F (x1, x2, x3).

9.9 Proposition. A rational function f(x1, . . . , xn) ∈ F (x1, . . . , xn) is symmetric if and only if it belongs to F (s1, . . . , sn).

Proof. Note that f(x1, . . . , xn) is symmetric if and only if it belongs to the ﬁxed ﬁeld by Gal F (x1, . . . , xn)/F (s1, . . . , sn) ' Sn, and this is F (s1, . . . , sn).

For the rest of this section, the characteristic of the ﬁelds involved will be assumed to be 6= 2. Note that the discriminant of the general polynomial is

Y 2 D = (xi − xj) . 1≤i

We will identify any σ ∈ Sn with the corresponding element in the Galois Q group Gal F (x1, . . . , xn)/F (s1, . . . , sn) . Write ∆ = 1≤i

σ(∆) = ∆ ⇐⇒ σ ∈ An.

9.10 Proposition. Let f(X) ∈ F [X] be a monic separable polynomial of degree ≥ 1, let E be a splitting ﬁeld of f(X) over F , let α1, . . . , αn be the diﬀerent roots of f(X) in E (so that f(X) = (X − α1) ··· (X − αn), Q 2 ∆ = 1≤i

Galois groups of degree 2 polynomials. 2 If f(X) = X + bX + c is a monic degree√ 2 polynomial over a ﬁeld F , then D = Df(X) = b2 − 4c, and ∆ = b2 − 4c. Let E be a splitting ﬁeld of f(X) over F , and G = Gal(E/F ) the associated Galois group. Then G . S2 ' C2, so either G = 1 and ∆ ∈ F , or G is the cyclic group of order 2, ∆ 6∈ F and E = F (∆). § 9. GALOIS GROUPS OF POLYNOMIALS 65

Galois groups of degree 3 polynomials. (Characteristic 6= 2, 3) Let f(X) = X3 + aX2 + bX + c be a monic degree 3 polynomial over a field F of characteristic 6= 2, 3. Let E be a splitting field of f(X) over F and let G be the associated Galois group. If f(X) is reducible, then either it is a product of polynomials of degree 1 over F , and hence E = F and G = 1, or f(X) is a product of a polynomial of degree 1 and an irreducible polynomial of degree 2. The considerations above show that in this case G is the cyclic group of order 2. Hence we will assume from now on that f(X) is irreducible. The characteristic being 6= 3 assures that f(X) is separable, so there are three different roots α1, α2, α3 ∈ E with f(X) = (X − α1)(X − α2)(X − α3). The Galois group G is, up to isomorphism, a subgroup of S3 which acts transitively on these roots. Therefore either G = A3 or G = S3 and by the previous propo- q sition G = A3 if and only if ∆ = D f(X) = (α1 − α2)(α1 − α3)(α2 − α3) is in F . Note that the discriminant is computed as follows:

1 a b c 0 1 a b c 0

0 1 a b c 0 1 a b c D f(X) = − 3 2a b 0 0 = − 0 −a −2b −3c 0

0 3 2a b 0 0 0 −a −2b −3c

0 0 3 2a b 0 0 3 2a b

1 a b c 2 a − 2b ab − 3c ac 0 a2 − 2b ab − 3c ac = = a 2b 3c 0 a 2b 3c 3 2a b 0 3 2a b = a2b2 − 4b3 − 4a3c − 27c2 + 18abc. Recall from Algebraic Structures that in order to get the roots, we may 1 proceed as follows. By means of the change of variables given by Y = X+ 3 a, we are left with a degree 3 equation of the form

Y 3 + pY + q = 0,

1 2 1 3 with p = 3 (3b − a ) and q = 27 (2a − 9ab + 27c). Write Y = u + v to get u3 + v3 + (3uv + p)(u + v) + q = 0,

p which, with uv = − 3 , gives  3 3  u + v = −q, p uv = − .  3 And this allows us to compute u3 and v3, and hence u and v, by solving a degree 2 equation. Let ω be a primitive cubic root of 1 in an extension ﬁeld 66 CHAPTER 2. GALOIS THEORY of E. Hence 1+ω +ω2 = 0 and if u, v is a solution of the previous equations, then the possible solutions are: u and v, ωu and ω2v, and ω2u and ωv. We thus get the roots of our original equation: u + v, ωu + ω2v, ω2u + ωv.

Galois groups of degree 4 polynomials. (Characteristic 6= 2, 3)

Let f(X) = X4 + aX3 + bX2 + cX + d be a monic degree 4 polynomial over a field F of characteristic 6= 2, 3. Let E be a splitting field of f(X) over F and let G be the associated Galois group. To simplify later computations, we can always perform the change of 1 variables Y = X + 4 a, and hence assume that our polynomial is f(X) = X4 + pX2 + qX + r. If f(X) is reducible, the computation of G reduces to the computation for lower degree polynomials. Hence we will assume that f(X) is irreducible. Since the characteristic is 6= 2, it is separable and hence f(X) = (X − α1)(X − α2)(X − α3)(X − α4) for different elements α1, α2, α3, α4 ∈ E. We know that G acts transitively on these roots and, as usual, G will be identified, through its action on the roots, with a subgroup of S4. Consider Klein’s four group

V = {1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)},

which is a normal subgroup of S4 such that S4/V is isomorphic to S3. Let H = G ∩ V . Consider now the following elements of EH :

 θ1 = (α1 + α2)(α3 + α4),  θ2 = (α1 + α3)(α2 + α4),   θ3 = (α1 + α4)(α2 + α3).

The elements of G permute the θi’s, and hence the elementary symmetric G polynomials on the θi’s are in E = F . Our hypotheses imply that α1 + α2 + α3 + α4 = 0 and a few easy computations yield:

X θ1 + θ2 + θ3 = 2 αiαj = 2p, i

θ1θ2+θ1θ3 + θ2θ3 2 = (α1 + α2)(α1 + α3) 2 2 + (α1 + α3)(α1 + α4) + (α1 + α4)(α1 + α2) 2 2 = α1 + α1(α2 + α3) + α2α3 2 2 2 2 + α1 + α1(α3 + α4) + α3α4 + α1 + α1(α2 + α4) + α2α4 2 2 2 = α2α3 − α1α4 + α3α4 − α1α2 + α2α4 − α1α3 X 2 X 2 = (αiαj) − 6α1α2α3α4 = (αiαj) − 6r 1≤i

Hence θ1, θ2, θ3 are the roots of the polynomial

3 2 2 2 h(X) = (X − θ1)(X − θ2)(X − θ3) = X − 2pX + (p − 4r)X + q , which is called the cubic resolvent of f(X). Moreover, we have θ1 −θ2 = −(α1 −α4)(α2 −α3) and similarly for θ1 −θ3 and θ2 − θ3, so we get: Y 2 Y 2 D f(X) = (αi − αj) = (θi − θj) = D h(X) , 1≤i

G∩A G∩V Therefore we have F (∆) = E 4 and F (θ1, θ2, θ3) = E :

We are left with several possibilities:

• If h(X) is irreducible, then |G| = [E : F ] is a multiple of both 3 since E contains the subﬁeld F (θ1) and [F (θ1): F ] = deg h(X) = 3, and 4 since [F (α1): F ] = 4. Hence |G| is a multiple of 12, and G is a subgroup of S4, whose order is 24. The only subgroup of S4 of order 12 is A4, and hence either G = A4 (and ∆ ∈ F ), or G = S4 (and ∆ 6∈ F ).

• If h(X) is reducible, and θ1, θ2, θ3 ∈ F , then necessarily |G| = 4, and G = V . (Note that in this case ∆ ∈ F .)

• If h(X) is reducible and, without loss of generality, θ1 ∈ F , but θ2, θ3 6∈ F , then any σ ∈ G satisﬁes σ(θ1) = θ1, so that

σ ∈ V ∪ (1 2)V = V ∪ {(1 2), (3 4), (1 3 2 4), (1 4 2 3)} = D4,

the dihedral group of degree 4. Also there is a σ ∈ G with σ(θ2) 6= θ2, so that G is not contained in V . Hence G is not contained in A4 (as D4 ∩ A4 = V ), so we get ∆ 6∈ F . Therefore, G is a group whose order is a multiple of 4, contained in D4, diﬀerent from V and which acts transitively on {1, 2, 3, 4}. There are only two possibilities: either G = D4 or G = h(1 3 2 4)i = C4. In the ﬁrst case G ∩ A4 = G ∩ V = V and the action of GalE/F (∆) = V on the roots is transitive, so that f(X) is irreducible over F (∆). In the second case the action of Gal E/F (∆) = C2 is not transitive, and hence f(X) is reducible over F (∆).

We summarize our ﬁndings in the next table: h(X) irreducible h(X) reducible ∆ ∈ F A4 V ∆ 6∈ F S4 D4 or C4

4 2 To obtain the roots of the equation X + pX + qX + r = (X − α1)(X − α2)(X − α3)(X − α4) we proceed as follows. First we compute the roots of § 10. SOLVABILITY BY RADICALS 69 the cubic resolvent h(X), which are θ1, θ2 and θ3. But:

(α1 + α2) + (α3 + α4) = 0,

(α1 + α2)(α3 + α4) = θ1, √ √ √so we get α1+α2 = −θ√1 and α3+α4 = − −θ√1. In a similar vein, α1+√α3 = −θ2 and α2 + α4 = − −θ2, and α1 + α4 = −θ3 and α2 + α3 = − −θ3. √ √ √ 3 2 Besides, −θ1 −θ2 −θ3 = (α1 +α2)(α1 +α3)(α1 +α4) = α1 +α1(α2 +α3 + α ) + P α α α = −q (recall α + α + α + α = 0), and hence 4 1≤i

9.11 Remark. The group S5 is not solvable, so we do not have a chain of normal subgroups like the one used for S4: S4 ⊇ A4 ⊇ V ⊇ 1.

§ 10. Solvability by radicals

Under what conditions can the roots of a polynomial in one variable be obtained by just using the operations of addition, subtraction, multiplication, division and extraction of roots?

10.1 Definition. (i) An extension K/F is said to be purely radical if there is an element n α ∈ K and a natural number n ∈ N such that K = F (α) and α ∈ F . (Hence α is an nth root of an element in F .) (ii) Let f(X) ∈ F [X] be a degree ≥ 1 polynomial over a field F . Then f(X) is said to be solvable by radicals if there exists a ‘tower’ of field extensions: F = K0 ⊆ K1 ⊆ K2 ⊆ · · · ⊆ Kr = K

such that each Ki/Ki−1 is purely radical (i = 1, . . . , r) and there are α1, . . . , αs ∈ K with f(X) = a(X − α1) ··· (X − αs). The extension K/F is said to be a radical extension. 70 CHAPTER 2. GALOIS THEORY

10.2 Remark. A ﬁeld extension K/F is then radical if K = F (γ1, . . . , γr) ni and for each i = 1, . . . , r there is a natural number ni ∈ N such that γi ∈ F (γ1, . . . , γi−1).

th 10.3 Lemma. Let F be a ﬁeld and let µn be the set of n roots of unity in n F : µn = {α ∈ F : α = 1}. Then µn is a cyclic group whose order divides n.

× Proof. It is clear that µn is a ﬁnite subgroup of F , hence it is cyclic (Lemma n 6.1). If γ is a generator of µn, then γ = 1, so that the order of µn, which equals the order of γ, divides n.

In the event that |µn| = n, the generators of µn are called the primitive th i th n roots of 1. Note that if µn = hζi with |ζ| = n, then ζ is a primitive n root of 1 if and only if gcd(i, n) = 1.

10.4 Lemma. Let F be a field and n a natural number. The polynomial Xn − 1 ∈ F [X] has n different roots (in a splitting field over F ) if and only if the characteristic of F does not divide n.

Proof. Take f(X) = Xn − 1. Then f 0(X) = nXn−1 is 6= 0 if and only if the characteristic of F does not divide n, and in this case gcd(f(X), f 0(X)) = 1, and f(X) has no multiple roots. On the contrary, if f 0(X) = 0, then f(X) contains multiple roots.

th 10.5 Proposition. Let F be a ﬁeld, n ∈ N, and let ζ be a primitive n root of 1 in a ﬁeld extension of F (so that the characteristic does not divide n). Then F (ζ)/F is a Galois extension and GalF (ζ)/F is an abelian group.

Proof. Since ζ ∈ F (ζ) and the roots of Xn − 1 are the powers of ζ, it follows that F (ζ) is the splitting ﬁeld of the separable polynomial Xn − 1 over F , and hence F (ζ)/F is a Galois extension. Any σ ∈ GalF (ζ)/F is determined by the value σ(ζ). Hence there is a one-to-one group homomorphism

× Gal F (ζ)/F −→ Z/nZ σ 7→ ¯ı such that σ(ζ) = ζi.

× But Z/nZ) is an abelian group, and hence so is Gal F (ζ)/F .

10.6 Proposition. Let F be a field and let n ∈ N such that the characteristic of F does not divide n and such that F contains all the nth roots of unity (that is, |µn| = n). Let K/F be a finite field extension. Then K/F is a Galois extension with Gal(K/F ) a cyclic group whose order is a divisor of n if and only if there is an element α ∈ K such that K = F (α) and αn ∈ F . § 10. SOLVABILITY BY RADICALS 71

Proof. If K = F (α) and αn = a ∈ F , then either a = 0 and hence K = F and Gal(K/F ) is trivial, or K/F is a splitting ﬁeld over F of the separable polynomial Xn − a (whose roots are α, ζα, . . . , ζn−1α for a primitive nth root of unity ζ, and they all lie in F ). Hence K/F is a Galois extension. Moreover, the map

Gal(K/F ) −→ Z/nZ σ 7→ ¯ı such that σ(α) = ζiα, is a one-to-one group homomorphism. Therefore Gal(K/F ) is isomorphic to a subgroup of the cyclic group of order n, and it is thus cyclic of order a divisor of n. Conversely, assume that K/F is a Galois ﬁeld extension whose Galois group is cyclic of order a divisor of n. Let m = |Gal(K/F )| and let ξ = ζn/m, which is a primitive mth root of 1, ξ ∈ F . Let σ be a generator of Gal(K/F ). For any α ∈ K consider the so called Lagrange resolvent of α:

(α, ξ) = α + ξσ(α) + ··· + ξm−1σm−1(α).

Note that for any i:

σi(α, ξ) = σi(α) + ξσi+1(α) + ··· = ξ−i(α, ξ).

As 1, σ, . . . , σm−1 are linearly independent (by Dedekind’s Lemma 4.6), there is an element α ∈ K such that (α, ξ) 6= 0. Write β = (α, ξ). Then for any i, σi(β) = ξ−iβ, so that β is not ﬁxed under any nontrivial element in Gal(K/F ). Therefore β does not belong to any proper subﬁeld of K containing F . Hence K = F (β). Besides, σ(βm) = σ(β)m = (ξ−1β)m = βm, and hence βm ∈ F . It follows that βn ∈ F too, as required.

10.7 Theorem. Let K/F be a radical extension and let E be an intermediate subﬁeld: F ⊆ E ⊆ K. Then the Galois group Gal(E/F ) is solvable.

Proof. Some reductions will be made: 1) The extension E/F can be assumed to be a Galois extension: Actually, Gal(E/F ) = Gal(E/F0) where F0 is the fixed field by Gal(E/F ), and hence E/F0 is a Galois extension. Besides, we have F ⊆ F0 ⊆ E ⊆ K and K/F0 is radical too. 2) We may assume that K is a splitting field of some polynomial over F (that is, K/F is a normal extension): mi To prove this note that K = F (u1, . . . , ur) with ui ∈ F (u1, . . . , ui−1) for each i = 1, . . . , r, for some elements ui’s in K and natural numbers mi’s.

Consider the polynomial f(X) = mu1,F (X)mu2,F (X) ··· mur,F (X) ∈ F [X] and let K˜ be a splitting ﬁeld of f(X) over K (and hence also over F ). We get the containments F ⊆ E ⊆ K ⊆ K˜ . 72 CHAPTER 2. GALOIS THEORY

˜ If v is a root of muj ,F (X) in K, there is a unique isomorphism τ : F (uj) → F (v) such that τ|F = 1 and τ(uj) = v, and this τ extends to an isomorphism σ : K˜ → K˜ . Then F ⊆ σ(K) ⊆ K˜ and v ∈ σ(K). Moreover, m m σ being an isomorphism, we have v j = σ(u j ) ∈ F σ(u1), . . . , σ(uj−1) . Then with Gal(K/F˜ ) = {σ1 = 1, σ2, . . . , σm}, the ﬁeld K˜ is generated over F by the roots of f(X), and hence it equals K˜ = F u1, . . . , ur, σ2(u1), . . . , σ2(ur), . . . , σm(u1), . . . , σm(ur) , which is clearly a radical extension of F . Besides, F ⊆ E ⊆ K˜ . 3) We may assume that K/F is a Galois extension and that E = K. To prove this, and since E/F is assumed to be a Galois extension (and hence E is the splitting ﬁeld of some polynomial over F ), we have the restriction map

Gal(K/F ) −→ Gal(E/F )

σ 7→ σ|E : E → E, which is a group homomorphism. Also, since K is being assumed to be a splitting field, any τ ∈ Gal(E/F ) extends to a σ ∈ Gal(K/F ). Hence this map is onto. If Gal(K/F ) is solvable, so is Gal(E/F ), so it is enough to prove that Gal(K/F ) is solvable. Applying again the reduction 1), we may assume that K/F is a radical Galois extension. 4) Assume then that K/F is a radical Galois extension. Without loss of generality we may assume too that there are elements r ∈ N, u1, . . . , ur ∈ K mi and prime numbers m1, . . . , mr such that K = F (u1, . . . , ur) and ui ∈ F (u1, . . . , ui−1) for any i = 1 . . . , r. (Note that if, for instance, K = F (u) with u12 ∈ F , then (u6)2 ∈ F ,(u2)3 ∈ F (u6) and u2 ∈ F (u6, u2) = F (u2), so that K = F (u6, u2, u).) mi Also note that if some mi equals the characteristic of F , then ui ∈ mi mi Ki−1 = F (u1, . . . , ui−1), so that mui,Ki−1 (X) divides X − ui = (X − m ui) i . But ui is separable over F (since K/F is a Galois field extension) and hence also over Ki−1. Thus mui,Ki−1 (X) = X − ui, so ui ∈ Ki−1 and we can get rid of ui in the expression K = F (u1, . . . , ur). Hence we may assume that none of the prime numbers mi equals the characteristic of F . th Take now m = m1 ··· mr and ζ a primitive m root of 1 in a field extension of K. Take K˜ = K(ζ) = F (ζ, u1, . . . , ur). The extension K/F˜ is again radical. Moreover, K is the splitting field of a separable polynomial f(X) ∈ F [X] over F , and hence K˜ is a splitting field of the separable polynomial lcmXm − 1, f(X) over F , so K/F˜ is a Galois extension too. Since, as in reduction 3), Gal(K/F ) is a quotient of Gal(K/F˜ ), we may change K to K˜ and assume ζ ∈ K.

mi Thus, we assume K = F (ζ, u1, . . . , ur), with ui ∈ F (ζ, u1, . . . , ui−1), the mi’s being prime numbers diﬀerent from the characteristic. Take K0 = § 10. SOLVABILITY BY RADICALS 73

F (ζ), and Ki = F (ζ, u1, . . . , ui) for any i = 1, . . . , r:

Kr = K 1 S| T| . . . . S| Galois T|

K1 = F (ζ, u1) −→ G1 = Gal(K/K1) S| correspondence T|

K0 = F (ζ) G0 = Gal(K/K0) S| T| F G = Gal(K/F )

But K0/F is a Galois extension, and hence G0 is a normal subgroup of G with G/G0 ' Gal(K0/F ), which is abelian by Proposition 10.5. Since ζ ∈ Ki−1 for any i ≥ 1, the extension Ki/Ki−1 is a cyclic Galois extension (Proposition 10.6), and hence Gi is a normal subgroup of Gi−1 and Gi−1/Gi ' Gal(Ki/Ki−1) is a cyclic group. We then have the descending series of subgroups G D G0 D G1 D ··· D 1, with abelian quotients, and hence G is solvable.

10.8 Corollary. Let f(X) ∈ F [X] be a degree ≥ 1 polynomial over a field F which is solvable by radicals. Then the Galois group of f(X) is solvable. 10.9 Corollary. There are polynomials of degree ≥ 5 which are not solvable by radicals. Proof. Recall that Gal F (x1, . . . , xn)/F (s1, . . . , sn) is isomorphic to the symmetric group Sn, which is not solvable for n ≥ 5. Therefore the general polynomial of degree n ≥ 5 is not solvable by radicals. 5 For a more concrete example, take f(X) = X − 4X + 2 ∈ Q[X]. Then f(X) is irreducible (Eisenstein’s Criterion). Let E be the splitting field of f(X) with Q ⊆ E ⊆ C, and let G = Gal(E/Q). Let us check that G is not solvable. The derivative f 0(X) = 5X4 − 4, considered as a function of a real variable, grows for x ≥ 0 and decreases for x ≤ 0. It follows then easily that f(X) has at most three real roots. But f(−1) > 0 and f(1) < 0, while limx→−∞ f(x) = −∞, limx→∞ = ∞. It follows that f(X) has exactly three different real roots: α1, α2, α3, and two non real complex roots: α4, α5 =α ¯4. By irreducibility of f(X), 5 = [Q(α1): Q] divides |G|, and hence G contains an element of order 5. Identify G with a subgroup of S5 through its action on the roots. This means that G contains a cycle of length 5. On the other hand, the restriction σ of the complex conjugation to E gives a transposition in G. But an arbitrary cycle of length 5 and an arbitrary transposition generate S5 (you have essentially proved this as an exercise in Chapter 1). Hence G is the whole S5, which is not solvable. 74 CHAPTER 2. GALOIS THEORY

Finally, a partial converse to the previous theorem will be proved.

10.10 Proposition. Let E/F be a Galois extension with solvable Gal(E/F ) and such that the characteristic of F does not divide [E : F ]. Then there is a ﬁeld K with F ⊆ E ⊆ K such that the extension K/F is radical.

Proof. Again we will start with a reduction: 1) It can be assumed that F contains the roots of unity of order [E : F ]. Actually, let ζ be a primitive root of 1 of order [E : F ] (which is not a multiple of the characteristic) in some extension of E. Now E is a splitting ﬁeld of some separable polynomial f(X) ∈ F [X] over F , so that E(ζ) is a splitting ﬁeld of the separable polynomial lcmf(X),X[E:F ] −1 over F , and hence the extension E(ζ)/F is a Galois extension, and so is the extension E(ζ)/F (ζ). Note also that the map

GalE(ζ)/F (ζ) −→ Gal(E/F )

σ 7→ σ|E, is a one-to-one group homomorphism, which shows that GalE(ζ)/F (ζ) is solvable and [E(ζ): F (ζ)] divides [E : F ], so it is not a multiple of the characteristic. Finally, if K is an extension of E(ζ) such that K/F (ζ) is radical, then K/F is radical too. (Note that ζ[E:F ]/[E(ζ):F (ζ)] is a primitive root of unity of order [E(ζ): F (ζ)].) 2) Assuming that F contains the roots of unity of order [E : F ], take G = Gal(E/F ). Since G is solvable, there is a normal subgroup H of G such that [G : H] is a prime number p (just take for H any maximal subgroup containing G0). Using an inductive argument, since E/EH is a Galois extension, we may assume that there is a radical extension K/EH with E contained in K: F ⊆ EH ⊆ E ⊆ K. But |Gal(EH /F )| = p, which implies that Gal(EH /F ) is cyclic, and hence EH = F (α) with αp ∈ F (Proposition 10.6). In particular, EH /F is radical, and hence so is K/F .

10.11 Lemma. Let f(X) ∈ F [X] be a polynomial of degree n ≥ 1 over a ﬁeld F , and let E be a splitting ﬁeld of f(X) over F . Then if p is a prime factor of [E : F ], then p is lower than or equal to n (p|n!).

Proof. We proceed by induction on n, the case n = 1 being trivial. For n > 1, let α be a root of f(X) in E: f(X) = (X − α)g(X) in F (α)[X]. Hence [E : F ] = [E : F (α)][F (α): F ]. If p is a factor of [F (α): F ] the result is clear since [F (α): F ] ≤ deg f(X) = n. Otherwise p divides [E : F (α)] and by the inductive hypothesis (E is a splitting ﬁeld of g(X) over F (α)) we conclude p ≤ n − 1 = deg g(X).

Our last result is due to Galois: EXERCISES 75

10.12 Theorem. Let f(X) ∈ F [X] be a degree n ≥ 1 polynomial over a field F whose characteristic does not divide n! (that is, either the characteristic is 0 or greater than n). Then f(X) is solvable by radicals if and only if its Galois group is solvable. Proof. If f(X) is solvable by radicals, then we already know that its Galois group is solvable (Corollary 10.8). Conversely, assume that the Galois group of f(X) is solvable, and let E be a splitting field of f(X) over F . Take a complete factorization of r rm f(X): f(X) = ag1(X) 1 ··· gm(X) , with 0 6= a ∈ F and the gi(X)’s different monic irreducible polynomials. Because of our restrictions on the 0 characteristic gi(X) 6= 0 for any i, so that the gi(X)’s are separable, and E is a splitting field of the separable polynomial g1(X) ··· gm(X) over F . Therefore E/F is a Galois extension. Now the previous proposition gives the result.

Exercises

1. Prove that any degree 2 ﬁeld extension is normal.

2. Let F be a field of characteristic p > 0 and let t ∈ F . Prove that the polynomial f(X) = Xp −X −t has no multiple roots, and that if α is a root in some field extension, then the set of its roots is {α+γ : γ ∈ Fp}. Conclude that F (α) is a splitting field of f(X) over F .

3. Consider a chain of subﬁelds F ≤ E ≤ K. Check whether the following assertions are valid:

(a) If K/F is normal, then K/E is normal. (b) If K/F is normal, then E/F is normal. (c) If E/F and K/E are normal, then K/F is normal.

4. Check whether the following polynomials are separable over the field F : X2 + 2X − 1, X3 + 1, and X6 + X5 + X4 + X3 + X2 + X + 1, where F is either Q or Fp with p = 2, 3, 5. 5. Let K/F be a finite field extension whose degree is not a multiple of the characteristic. Prove that K/F is separable. Is the converse true?

6. Compute the Galois group Gal(F/Q), where F is a splitting ﬁeld of each of these polynomials:

• X2 − 2, • X3 − 2, 76 CHAPTER 2. GALOIS THEORY

• X5 − 2, • (X2 − 2)(X3 − 2), • X6 − 9, • (X3 − 2)(X4 − 2).

7. This exercise is designed to provide a proof of the Fundamental The- orem of Algebra, diﬀerent from the one given in Algebraic Structures. This proof will use the following facts.

• Any positive real number has a positive square root. • Any real polynomial of odd degree has a real root.

Prove the following assertions:

(a) C does not have extensions of degree 2. (b) R does not have proper extensions of odd degree.

Now let f(X) ∈ R[X] be an irreducible real polynomial, and let K be a splitting ﬁeld of f(X) over R.

(c) Prove that K(i)/R is a Galois extension, where i denotes a root of X2 + 1 in some extension of K. n (d) Write [K(i): R] = 2 m for n ≥ 0 and odd m. Prove that m = 1 by considering the fixed field of a Sylow 2-subgroup of the Galois group. (e) Prove that n = 1 by arguing with index 2 subgroups of the Galois group. (f) Conclude that f(X) splits over C. 8. Prove that an arbitrary finite field has irreducible polynomials of arbitrary degree.

9. Prove that the polynomial Xpn −X is the product of all the irreducible monic polynomials over Fp whose degree divides n.

10. Let F be a subﬁeld of R, and let f(X) ∈ F [X] be a degree 3 irreducible polynomial. Then prove the following assertions:

(a) Df(X)) > 0 if and only if the three roots of f(X) are real. (b) Df(X) < 0 if and only if f(X) has only one real root.

11. Let F be a ﬁeld of characteristic 6= 2, and let f(X) ∈ F [X] be a degree 3 polynomial whose discriminant is a square in F . Prove that either f(X) is irreducible or f(X) splits in F . EXERCISES 77

12. Prove that over any ﬁeld F either the cubic polynomial X3 − 3X + 1 is irreducible or it splits over F .

13. Let F be a ﬁeld of characteristic 6= 2, and let f(X) = X4 + bX2 + c ∈ F [X] be an irreducible polynomial. Let G be its Galois group. Prove the following assertions:

(a) If c is a square in F then G ' C2 × C2. (b) If c is not a square in F but c(b2 − 4c) is a square in F , then G ' C4. (c) If neither c nor c(b2 − 4c) are squares in F , then G is isomorphic to the dihedral group D4.

14. Compute the Galois group over Q of the following polynomials: (a) X3 − 3X + 1, (b) X3 + 3X2 − X − 1, (c) X4 − X + 1, (d) X4 − 6X2 + 8X + 28, (e) X4 − 2, (f) X4 + 4X2 + 2, (g) X4 + 1, (h) (X4 − 2)(X4 − 3).

15. For any n ∈ N consider the polynomial Φn(X) = (X − ζ1) ··· (X − ζr) th where ζ1, . . . , ζr are the primitive n roots of unity in C.Φn(X) is called the nth cyclotomic polynomial. Prove the following assertions:

n Q (a) X − 1 = d|n Φd(X). (b) Φn(X) ∈ Z[X]. (c) Φn(X) is irreducible (as a polynomial in Q[X]). (d) [Q(ζ): Q] = φ(n), where φ denotes the Euler map, and ζ a th × primitive n root of unity, and Gal Q(ζ)/Q ' Z/nZ . 16. Prove that if n ≥ 3, the regular polygon of n vertices is constructible m with ruler and compass if and only if n = 2 p1 ··· pr, where m, r ≥ 0 2m and p1, . . . , pr are Fermat primes (that is, of the form 2 + 1).

15 To show that Φn(X) is irreducible assume the contrary. Then Φn(X) = f(X)g(X) for two monic degree ≥ 1 polynomials in Z[X], with f(X) irreducible. Then show that there is a primitive root ζ and a prime number p which does not divide n such that f(ζ) = 0 6= f(ζp). Conclude that f(X) divides g(Xp). Now reduce modulo p (and use bars to denote the reductions) and show that Φ¯ n(X) has multiple roots. Get from here a contradiction.

Previous exams

February 2010

1. (a) List all the abelian groups of order 150 (up to isomorphism). (b) Give an example of a solvable nonabelian group of order 150. (c) Prove that any group of order 150 is solvable.

2. (a) Let H be a subgroup of a group G. Prove that H is a normal subgroup of G if and only if [G, H] ≤ H. (3 points) (b) Let G be a group of order pn, where p is a prime number and n ∈ N. Prove that there exists a chain of subgroups 1 = G0 ≤ G1 ≤ · · · ≤ Gn = G, such that for each i = 1, . . . , n, Gi is a i normal subgroup of G and |Gi| = p . (7 points)

4 2 3. Consider the polynomial f(X) = X − 2X − 4 ∈ Q[X].

(a) Compute its roots (in C). √ (b) Show that f(X) is not irreducible as a polynomial over Q( 5), but it is irreducible over Q. √ (c) Compute the Galois group of f(X) as a polynomial over Q( 5). (d) Compute the Galois group of f(X) as a polynomial over Q.

4. Given a natural number n ≥ 2, consider the primitive nth root of unity 2πi/n −1 ζ = e inside C, and the element θ = ζ + ζ .

(a) Prove that [Q(ζ): Q(θ)] equals 2. ∼ × (b) Which subgroup of Gal Q(ζ)/Q = (Zn) is Gal Q(ζ)/Q(θ) ?

2 Recall that [G, H] is the subgroup generated by the commutators [g, h], g ∈ G, h ∈ H.

79 80 PREVIOUS EXAMS

September 2010

1. Consider an odd number n 6= 1, the primitive nth root of unity ζ = 2πi/n e in C, and the subgroup D of the general linear group of degree 2 over C generated by the elements ζ 0 0 1 x = , y = . 0 ζ−1 −1 0

(a) Show that hxi is a normal subgroup of D. (b) Compute the order of D. (c) Compute the center of D. (d) Prove that D has a quotient isomorphic to the dihedral group Dn.

2. (a) Prove that any group of order 400 is solvable. (b) Give an example of a solvable but not nilpotent group of order 400.

4 3. Consider the polynomial f(X) = X + X + 1 ∈ Q[X]. (a) Prove that it is irreducible. (b) Compute its Galois group.

4. Let GF (27) be the ﬁeld of 27 elements.

(a) Give explicitly a basis of GF (27) as a vector space over F3. (b) Find a generator of the cyclic group GF (27)×. (c) Take a generator of the Galois group Gal GF (27)/F3 and compute its coordinate matrix in the basis you have got before.

3 The cubic resolvent of X4 + pX2 + qX + r is X3 − 2pX2 + (p2 − 4r)X + q2. PREVIOUS EXAMS 81

February 2011

1. (a) Let G be a nonempty set and let

G × G → G (x, y) 7→ xy,

be an associative binary operation such that

• for any x ∈ G the left multiplication by x (Lx : G → G, y 7→ xy) is bijective, • there exists an element e ∈ G such that xe = x for any x ∈ G. Prove that G is a group with this operation. (b) Give an example showing that the ﬁrst condition above is not suﬃcient for G to be a group.

2. (a) Prove that a group of order 180 is never simple. (b) Give examples of groups of order 180 which are: • solvable but not nilpotent, • not solvable.

5 3. Consider the polynomial f(X) = X − 3 ∈ Q[X]. (a) Prove that f(X) is irreducible. (3 points) (b) Compute its Galois group. (7 points)

4. (a) Find an irreducible polynomial of degree 4 over F2. (b) Give explicitly a basis of GF (16) (the ﬁeld of 16 elements) over F2. (c) Take a generator of Gal GF (16)/F2 and compute its coordinate matrix in the above basis. 82 PREVIOUS EXAMS

September 2011

1. Let G be a ﬁnite group and let N be a normal subgroup of G.

(a) Prove that if |N| = 2, then N is central (N ≤ Z(G)). Give an example to show that this fails for |N| = 3. (b) Assume that |G| is odd, and that |N| = 3. Prove that N is central too in this case.

2. (a) Prove that if p and q are diﬀerent prime numbers, any group of order pq or p2q is solvable. (b) How many groups of order 62 are there, up to isomorphism?

4 3. Consider the polynomial f(X) = X − 3 ∈ Q[X]. (a) Prove that f(X) is irreducible. (3 points) (b) Compute its Galois group. (7 points)

3 4. (a) Prove that the roots of the polynomial X − 3X + 1 ∈ Q[X] are all real. (b) Compute explicitly these roots. PREVIOUS EXAMS 83

February 2017

1. Describe explicitly all the Sylow subgroups of the symmetric group S4.

2. Consider the group, given by generators and relations: G = hx, y, z | x2 = y2 = z2 = 1, xy = yx, xz = zx, yz = zxyi. (a) What is the order of G? (b) Give an explicit isomorphism from G into a well-known group. √ 4 3. How many subﬁelds does Q( 3) contain? Why?

15 4. Let K be a splitting ﬁeld of X − 1 over F2. Compute [K : F2] and Gal(K/F2).

September 2017 1. Let G be a group of order 90.

(a) Show that if P1 and P2 are two diﬀerent Sylow 3-subgroups of G, then hP1,P2i = G and P1 ∩ P2 ≤ Z(G). (b) Prove that G is solvable.

2. Let O(n) be the orthogonal group: t O(n) = A ∈ Matn(R): A A = I , n and consider its natural action on R (identiﬁed with the set of n × 1 matrices). (a) What are the orbits of this action? 1 ! 0 (b) What is the stabilizer of . ? . 0

8 3. Let K be the splitting ﬁeld of X − 2 over Q. (a) Determine the degree [K : Q]. (b) Which group is, up to isomorphism, Gal(K/Q)? Why?

4. Find an irreducible polynomial f(X) ∈ F2[X] such that, if α is a root of f(X) in a ﬁeld extension K of F2, then the cyclic group generated by α in K× has order 7. 84 PREVIOUS EXAMS

January 2018

1. Consider the symmetric group S4 of degree 4.

(a) Prove that there are exactly 4 subgroups of S4 of order 3 and give a precise description of them.

(b) Show that the normalizer of any subgroup of S4 of order 3 is isomorphic to S3.

(d) Deduce that S4 contains exactly 4 subgroups of order 6 and give a precise description of them.

2. Let G be a group of order 8 × pn, where p is an odd prime number and n ∈ N.

(a) Prove that if the number Np of Sylow p-subgroups of G is > 1, then either p = 3 or p = 7.

(b) Show that if p = 3 and N3 > 1, then there is a nontrivial homomorphism G → S4 with solvable kernel. (c) If p = 7, prove that either G contains a unique Sylow 7-subgroup or G contains a normal subgroup of index 56. (d) Conclude that any group of order 8 × pn, with p and n as above, is solvable.

3. Compute the Galois group of the polynomial X4−4X3+6X2−4X−1 ∈ Q[X].

4. Consider the ﬁeld F7 of 7 elements.

(a) Compute the cubes of the elements in F7. 3 (b) Find a monic polynomial f(X) ∈ F7[X] such that GF (7 ) = F7[X]/ f(x) . 3 (c) Describe explicitly the elements in Gal GF (7 )/F7 in terms of the description of GF (73) in the previous item. PREVIOUS EXAMS 85

January 2019

1. Let G be a p-group for a prime number p, let N be a normal subgroup of G, and let x ∈ G be an element of order p. Prove the following assertions:

(a) N ∩ Z(G) 6= 1. (b) hxi is a normal subgroup of G if and only if x ∈ Z(G).

2. (a) Prove that any group of order 80 is solvable. (b) Give examples of both nilpotent and nonnilpotent groups of order 80.

4 2 3. Let K be a splitting ﬁeld of f(X) = X + X − 1 over Q.

(a) Compute the degree [K : Q] and give a basis of K as a vector space over Q. (b) What is the Galois group of f(X) over Q? Describe its elements.

4 4. (a) Show that the polynomial X + 1 is irreducible over Q. 4 (b) Show that, on the other hand, X + 1 is reducible over F2, F3, and F5. (c) Show that, in general, X4 + 1 splits in GF (p2) for any prime 4 number p, and deduce that X + 1 is reducible over Fp.