groups and galois theorypersonal.unizar.es/elduque/files/groupsgalois2019.pdfvi syllabus some more...

Groups and

Galois Theory

Course Notes

Alberto Elduque

Departamento de MatematicasUniversidad de Zaragoza50009 Zaragoza, Spain

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 . . . . . . . . . . . . . . . . . . 176.2. 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 . . . . . . . . . . . . . . . . . . . . . . 69Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 heavilyon) the course Algebraic Structures.

Lecturer: Alberto ElduqueOffice: Math. Building. Second floor. Algebra Section. Office no. 2.e-mail: [email protected]

http://www.unizar.es/matematicas/algebra/elduqueOffice hours: Tuesday and Friday, from 17:00 to 20:00.

Course description: The goal of the course is to understand the basicproperties of the groups, which form the basic algebraic structure to under-stand and measure the symmetry, and then to proceed to study the symme-try of the algebraic equations (Galois Theory). The peak of the course willbe the proof of the impossibility to solve by radicals the algebraic equationsof degree ≥ 5.

Exam, exercises: There will be a final exam, which will consist of somequestions of a theoretical nature and some exercises. Students will be re-quired to solve and explain to their classmates some of the exercises in thenotes. 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 ofthem:

- D.S. Dummit and R.M. Foote: Abstract Algebra (2nd edition). JohnWiley 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 Yearhttp://www.math.uiuc.edu/ r-ash/Algebra.html

v

http://www.unizar.es/matematicas/algebra/elduque

http://www.math.uiuc.edu/~r-ash/Algebra.html

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

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

G = isometries of R2 fixing the triangle

G consists of three reflections relative to the lines that pass through a vertexand the middle point of the opposite side (see Figure 1) together with therotations of 0, 120 and 240 degrees. Thus G consists of 6 elements: |G| = 6.

Each isometry fixing the triangle permutes the vertices. The three reflec-tions correspond to the transposition of two vertices. The rotations of 120and 240 degrees correspond to the two cyclic permutations of the vertices.

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σ3 σ2

σ1

Figure 1: Symmetries of an equilateral triangle.

For us, the important property of this set G is that its elements canbe composed (they are bijective maps!), and the composition of any two ofits elements again belongs to G. Thus, for example, σ1σ2 is the clockwiserotation of 120 degrees.

These sets, endowed with a binary operation satisfying the usual prop-erties 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:

x =−b±

√b2 − 4ac

2a.

In the course Algebraic Structures we learned how to solve the equationsof 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 symme-tries of the equation. These form a group, called the Galois group of theequation. 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 theirGalois groups is called Galois Theory.

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

Chapter 1

Groups

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

§ 1. Definitions and examples

1.1 Definition. A group is a setG endowed with a binary operation (usuallycalled multiplication)

G×G −→ G

(x, y) 7→ xy,

which is associative ((xy)z = x(yz) for any x, y, z ∈ G), there is a neutralelement (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 byx−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 ringR 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 bythe product of the ring.

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

1

2 CHAPTER 1. GROUPS

• GL(n,R) = A ∈ Matn(R) : detA 6= 0 (general linear group),

• SL(n,R) = A ∈ Matn(R) : detA = 1 (special linear group),

• 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 ofpermutations of Ω.

The groups contained in S(Ω) are called groups of transformations ofΩ.

The famous Erlangen Program (Felix Klein, 1872) proposed that Geome-try is the study of invariants under the action of a group. Thus the euclideangeometry becomes the study of invariants under the isometry group (angles,distances, ..., are invarians), the affine (or projective) geometry becomes thestudy of invariants under the affine (or projective) group (for instance, thebarycenter is an invariant), the topology becomes the study of invariantsunder 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 needto put parentheses in products x1x2 · · ·xn (n > 2). Also, we may definerecursively the powers of an element of a group G as follows: x0 = e,xn+1 = xnx, x−n = (xn)−1, for any n ∈ N. The associativity of the productimmediately gives the property xnxm = xn+m 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 belongto 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 multiplicationin 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 thequaternion 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 othersubgroup is said to be a proper subgroup of G.

By abuse of notation, the trivial subgroup e is usually denoted simplyby 1 (or 0 if additive notation is being used).

Notation: Given a group G, its cardinal will be denoted by |G|. Thus, ifG is finite, |G| is the number of elements of G. This is called the order ofG.

In general, multiplicative notation as above is used, unless we are dealingwith 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 −xpowers xn nx

§ 2. Cyclic groups. Generators

2.1 Definition. Let G be an arbitrary group and g ∈ G. Then gn : n ∈ Zis a subgroup of G which is called the cyclic subgroup generated by g. Itwill be denoted by 〈g〉.

If there exists an element g ∈ G such that G = 〈g〉, then G is said to bea 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.

Then G = 〈ϕ〉, where ϕ is the rotation of angle 2πn 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:

|〈g〉| =

∞, if gn 6= e for any n ≥ 1,

r = minn ∈ N : gn = 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).

Proof. If gn 6= e for any n ≥ 1 and there are n ≤ m ∈ Z such that gn = gm,then multiply by g−n to get gm−n = e, so n −m = 0. Therefore, if gn 6= efor any n ≥ 1, then the elements gn, n ∈ Z, are all different and |〈g〉| =∞.

On the contrary, if there exists a natural number n such that gn = 1,take r = minn ∈ N : gn = e. As before, it is easy to see that the elementse, g, g2, . . . , gr−1 are all different. Moreover, for any m ∈ Z, there are uniqueq, s ∈ Z with 0 ≤ s < r such that m = qr + s. Then gm = (gr)qgs = eqgs =gs, so that 〈g〉 = e, g, . . . , gr−1 has order r.

Note that the proof above shows that if the order of g ∈ G is r, thengn = e for some n ∈ Z if and only if r divides n.

Also note that 〈g〉 is the smallest subgroup containing the element g.In the same way, given a subset S of a group G, the subgroup generatedby S is, by definition, the smallest subgroup of G containing S, and it isdenoted by 〈S〉. In other words, 〈S〉 is the intersection of all the subgroupsof G containing S (as the intersection of a family of subgroups is always asubgroup):

〈S〉 =⋂H : H ≤ G and S ⊆ H.

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

t1t2 · · · tn

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

2.5 Examples.

• 〈G〉 = G, 〈∅〉 = 1 (trivial subgroups).

§ 3. HOMOMORPHISMS 5

• Let Q be the quaternion group, then Q = 〈i, j〉.

2.6 Definition. A group G is said to be finitely generated if there exists afinite subset S of G such that G = 〈S〉.

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 suchthat ab = ba, |a| = s, |b| = t, with s, t ∈ N and gcd(s, t) = 1. Then〈a, b〉 = 〈ab〉 is a cyclic group of order st.

Proof. Obviously ab ∈ 〈a, b〉, so we get 〈ab〉 ⊆ 〈a, b〉.Let d be an element in 〈a〉 ∩ 〈b〉, 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 〈a〉 ∩ 〈b〉 = 1.

Now, for any natural number n, we get

(ab)n = e⇐⇒ anbn = e since ab = ba,

⇐⇒ an = b−n ∈ 〈a〉 ∩ 〈b〉,⇐⇒ an = bn = e,

⇐⇒ both s and t divide n,

⇐⇒ st divides n (since gcd(s, t) = 1).

Therefore |ab| = st and 〈ab〉 is a cyclic group of order st.Moreover, 〈a, b〉 = aibj : 0 ≤ i ≤ s − 1, 0 ≤ j ≤ t − 1, and hence the

order of 〈a, b〉 is at most st. Since 〈ab〉 is contained in 〈a, b〉, and |ab| = st,we conclude that 〈ab〉 = 〈a, b〉, 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) iffor any x, y ∈ G, ϕ(xy) = ϕ(x)ϕ(y).

• If ϕ is a group homomorphism, then its kernel is the subset kerϕ =ϕ−1(e) of G, while its image is the set imϕ = 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 itis a bijection. Moreover, the isomorphisms ψ : G → G are calledautomorphisms.

6 CHAPTER 1. GROUPS

3.2 Examples.

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

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

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

• The map ϕ : (R,+) → SO(2,R), ϕ(t) =(

cos t − sin tsin t cos t

), is an epimor-

phism with kerϕ = 2πZ.

• The determinant map det : GL(n,R)→ (R \ 0, .), A 7→ detA, is anepimorphism with ker det = SL(n,R).

• Let G be a group and let a ∈ G. The map Ia : G → G given byIa(x) = axa−1 (conjugation by the element a) is an automorphism,which is called an inner automorphism.

The set AutG of automorphisms of G is a group under the compo-sition of maps, and the map I : G → AutG, given by I(a) = Iais a homomorphism, because IaIb = Iab. Its image im I is the sub-group of inner automorphisms, denoted by IntG, while its kernel isker I = a ∈ G : Ia = id = a ∈ G : ax = xa ∀x ∈ G, which is calledthe 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) imϕ is a subgroup of H,

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

This last property deserves a name:

3.4 Definition. Let G be a group and let N be a subgroup of G. ThenN is said to be a normal subgroup of G if for any x ∈ N and a ∈ G, theelement axa−1 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 agroup G. If G is abelian, then any subgroup is a normal subgroup.

Given a group G and a subgroup H, define the following relation amongelements in G:

a ∼ b if a−1b ∈ H.

This is clearly an equivalence relation. The equivalence class of the elementa 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 ifab−1 ∈ H, whose equivalence classes are the right cosets modulo H (the setsHa for a ∈ G).

The set of left cosets modulo H (the quotient set by the equivalencerelation) is denoted by G/H, and its cardinal by [G : H], which is called theindex of the subgroup H in G.

Note that the map H = eH → aH, h 7→ ah, is bijective, and hence allthe left cosets modulo H have the same cardinal, equal to |H|. It followsthat 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 isa 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 thegroup.

(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 beno subgroup whose order is this divisor. However, for cyclic groups we havethe following result:

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

8 CHAPTER 1. GROUPS

Proof. Let H be a subgroup of either (Z,+) or (Z/nZ,+). Then for anym ∈ Z and a ∈ H, ma = a + · · · + a (m summands) if m > 0, whilema = (−a) + · · ·+ (−a) (−m summands) if m < 0. In both cases, ma ∈ Has 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 fromthe corresponding result in the course Algebraic Structures. (Anyway, it isvery 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 = 〈a〉 be a cyclic group of order q ∈ N and let d ∈ N bea divisor of q. Then the subgroup of G of order d is (with additive notation)H = 〈 qda〉 = b ∈ G : db = 0.

Let now N be a normal subgroup of a group G. Then the equivalencerelation 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 itsneutral element, and the inverse of aN is given by a−1N . Hence, G/N is agroup, called the quotient group of G modulo N .

With the same sort of arguments used for rings, the following results caneasily be proved:

3.9 Theorem.

1. First Isomorphism Theorem: Let ϕ : G → H be a group ho-momorphism, then the quotient group G/ kerϕ is isomorphic to imϕthrough the isomorphism

ϕ : G/ kerϕ→ imϕ

a kerϕ 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 overG/N . Besides, kerπ = N . In particular, this shows that any normalsubgroup is the kernel of some homomorphism.

§ 4. SYMMETRIC GROUP 9

3. Let ϕ : G→ H be a group homomorphism, then ϕ is a monomorphismif and only if kerϕ = 1, while ϕ is an epimorphism if and only ifimϕ = H.

4. Second Isomorphism Theorem: Let H be a subgroup and N anormal subgroup of a group G. Then HN = hn : h ∈ H,n ∈ N isa subgroup of G, N is a normal subgroup of HN , H ∩N is a normalsubgroup 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 sub-groups of a group G with H ⊆ K, then K/H is a normal subgroup ofG/H and the quotient groups (G/H)/(K/H) and G/K are isomor-phic.

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

subgroups of G containing N → subgroups of G/NH 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 normalsubgroups.

§ 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 calledthe symmetric group of degree n.

The elements of Sn can be represented like this:

π =

(1 2 . . . n

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

σ =

(1 2 3 42 3 4 1

), τ =

(1 2 3 44 3 2 1

).

10 CHAPTER 1. GROUPS

Then,

στ =

1 2 3 4↓ ↓ ↓ ↓4 3 2 1↓ ↓ ↓ ↓1 4 3 2

=

(1 2 3 41 4 3 2

), τσ =

(1 2 3 43 2 1 4

).

4.2 Proposition. The order of the symmetric group of degree n is |Sn| = n!.Its neutral element is the identity map: ( 1 2 ... n

1 2 ... n ).

Let us try to decompose any permutation into a product of simplerpermutations:

4.3 Example. π =

(1 2 3 4 5 6 7 82 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, theorbit 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 thatπl(i) = i, then the orbit of i is precisely i, π(i), . . . , πl−1(i). This numberl is then called the length of the orbit. The different orbits give a partitionof Ω:

Ω = Ω1 ∪ · · · ∪ Ωp (disjoint union).

If |Ωk| = lk, then for any i ∈ Ωk, Ωk = i, π(i), . . . , πlk−1(i). The per-mutation π acts independently on each orbit and hence π is the productπ = π1 · · ·πp, where for each k,

πk(i) =

π(i) if i ∈ Ωk,

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 beexpressed uniquely as a product of “disjoint” cycles of length ≥ 2, up to theorder 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 fork = 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,

πn = πn1 · · ·πnp ,

and πn = 1 if and only if πnk = 1 for any k, if and only if lk divides n for anyk, 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 dif-ferent ways to decompose 1, 2, 3, 4, 5, 6 into disjoint orbits. That is, thedifferent ways of decomposing 6 as a sum of natural numbers:

Decomposition example order

1+1+1+1+1+1 id 12+1+1+1+1 (1 2) 2

2+2+1+1 (1 2)(3 4) 22+2+2 (1 2)(3 4)(5 6) 2

3+1+1+1 (1 2 3) 33+2+1 (1 2 3)(4 5) 6

3+3 (1 2 3)(4 5 6) 34+1+1 (1 2 3 4) 4

4+2 (1 2 3 4)(5 6) 45+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 trans-positions.

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 transpositionsis 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).


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 = 0otherwise.

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:

ai1b1j + ai2b2j + · · ·+ ainbnj = aiτ(j)bτ(j)j =

1 if i = στ(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 homomorphismtoo, so is the composition det ϕ. Moreover, det ϕ(σ) = ±1, as the determi-nant of the matrix associated to any transposition is −1 (the correspondingmatrix is the identity matrix with two columns permuted), and any σ is aproduct of transpositions.

4.10 Definition. The epimorphism sgn : Sn → C2 = 1,−1 (C2 is thecyclic group of two elements) given by sgn(σ) = det

(ϕ(σ)

)is called the

signature homomorphism. The value sgn(σ) is said to be the signature ofthe permutation σ. Its kernel is denoted by An and called the alternatinggroup of degree n. The permutations in An are said to be even, while thepermutations in Sn \An are said to be odd.

Note that An is a normal subgroup of Sn of index 2, as Sn/An is isomor-phic 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 sgnπ =(−1)p.

(ii) If π = π1π2 · · ·πq for some disjoint cycles πi of length li, then sgnπ =

(−1)∑q

i=1(li−1).

Proof. For (i) just note that sgnπ =∏pi=1 sgn(τi) = (−1)p, 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 transposi-tions). 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 isomor-phic 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 anya, b ∈ G. Thus, L is a group homomorphism. Moreover, if a ∈ kerL, thenLa = id, that is ax = x for any x ∈ G. In particular a = ae = e. HencekerL = 1, and G ' imG ≤ S(G).

§ 5. Group actions

Groups made their appearance as “groups of symmetries”; that is, theirelements are symmetry maps on some object.

5.1 Definition. Let G be a group and Ω a set. An action of G on Ω (onthe 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 againdenoted by Φ):

Φ : G −→ S(Ω)

g 7→ Φg : Ω→ Ω

x 7→ gx

is a group homomorphism. And conversely, given any group homomorphismΦ : G→ S(Ω), g 7→ Φg, the map G×Ω→ Ω, given by (g, x) 7→ Φg(x), is anaction 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).


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 byleft 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 cor-responds 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 actionis said to be faithful, and then G can be identified to a subgroup ofS(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. (Thismeans that for any x, y ∈ Ω, there is an element g ∈ G such thatgx = 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, thelength (cardinal) of any orbit divides the order of G. Therefore, ifΩi : i ∈ I is the set of orbits of the action, and xi : i ∈ I is a setof representatives of the orbits (that is, xi ∈ Ωi for any i ∈ I) then

|Ω| =∑i∈I|Ωi| =

∑i∈I

[G : Gxi ]

§ 5. GROUP ACTIONS 15

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

Ig(Gx).

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

number of orbits =1

|G|∑g∈G|F (g)|

Proof. For (i) just note that gx = g′x if and only if g−1g′ ∈ Gx, if and onlyif g′ ∈ gGx. Hence the map Gx → G/Gx, gx 7→ gGx is well defined andbijective, so |Gx| = [G : Gx].

Also note that if gx = y and hy = y, then hgx = gx or g−1hgx = x,and conversely. Then h ∈ Gy if and only if g−1hg ∈ Gx, if and only ifh ∈ gGxg−1, thus obtaining the result in (ii). Therefore, if x and y belongto the same orbit, then |Gx| = |Gy| and [G : Gx] = [G : Gy]. From here, thelast part of (i) follows.

Consider now the set X = (g, x) ∈ G×Ω : gx = x. We may count theelements of X in two ways:

|X| =∑g∈G|F (g)| and |X| =

∑x∈Ω

|Gx|.

But because of (ii) we get:

|X| =∑x∈Ω

|Gx| =∑i∈I|Gxi||Gxi |

=∑i∈I|G| = |G|(number of orbits).

Hence

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

as required.

5.5 Example. A bracelet is made by stringing together six beads, twoof them are green and the other four beads are red. How many differentpatterns are possible?

Let P6 be the regular hexagon. We put the beads on the vertices of P6.

There are(

62

)=(

64

)= 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 theother. Therefore, the number of different patterns is the number of orbitsof the action of the group of symmetries of P6 acting on the set of the 15different possibilities.


..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

1

2

34

5

6

The group of symmetries of the regular hexagon is the dihedral groupD6 of degree 6:

D6 = id, ϕ, ϕ2, . . . , ϕ5, s14, s25, s36, s4512, s

5623, s

6134

where ϕ is the clockwise rotation of angle 2π6 , s14 is the reflection relative

to the axis which goes through the vertices 1 and 4, s4512 is the reflection

relative to the axis which goes through the middle points of the sides whichjoin the vertices 1 and 2 and the vertices 4 and 5, and similarly for the otherelements.

Now we compute the fixed elements in the set Ω of our 15 possibilities byeach 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 and6 and in 3 and 5), and in the same vein |F (s25)| = |F (s36)| = 3,

• |F (s4512)| = 3 (which correspond to the green beads in 1 and 2, in 3 and

6 and in 4 and 5), and in the same vein |F (s5623)| = |F (s61

34)| = 3.

Therefore:

number of patterns =1

12

(15 + 3 + 3× 3 + 3× 3

)=

36

12= 3.

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 obtainedadding radicals CH3 or H to a benzene ring?

Here we have two possible colors (CH3 or H) for each vertex of thehexagon. Hence there are 26 = 64 different possibilities. A similar compu-tation to the one performed before gives:

§ 6. CONJUGATION. TRANSLATION 17

.............................................................

..........................................................................................................................

.............................................................

.............................................................

.............................................................

.............................................................

.............................................................

.............................................................

.....................

.....................

.....................

.....................

.......................................... C

CC

C

C C

• |F (id)| = 64,

• |F (ϕ)| = |F (ϕ5)| = 2,

• |F (ϕ2)| = |F (ϕ4)| = 22,

• |F (ϕ3)| = 23,

• |F (s14)| = |F (s25)| = |F (s36)| = 24,

• |F (s4512)| = |F (s56

23)| = |F (s6134| = 23.

And hence:

number of compounds =1

12

(64 + 2×2 + 2×22 + 23 + 3×24 + 3×23

)= 13.

§ 6. Conjugation. Translation

6.1. Action by conjugation

The action by conjugation of a group on itself is given by the homomorphism

Φ : G −→ AutG ≤ 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.


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

|G| = |Z(G)|+r∑i=1

[G : CG(xi)] ,

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 (thatis, 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. Itfollows that p divides |Z(G)|, and hence Z(G) 6= 1.

Moreover, assume that G is a non abelian group with |G| = p2. Takean element x ∈ G \ Z(G). By the above Z(G) is a subgroup of order p,and hence it is cyclic: Z(G) = 〈y〉. Then 〈x, y〉 is a subgroup, whose orderis greater than p and divides the order of G. Since |G| = p2, it followsthat G = 〈x, y〉. But x and y commute as y ∈ Z(G), so G is abelian, acontradiction.

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

σπσ−1(σ(i)

)= σπ(i) = σ(j).

That is, if we have iπ−→ j, then σ(i)

σπσ−1

−−−−→ σ(j). Therefore, if π is expressedas a product of disjoint cycles:

π = (i1 . . . ir)(j1 . . . js) · · · ,

thenσπσ−1 =

(σ(i1) . . . σ(ir)

)(σ(j1) . . . σ(jr)

)· · ·

Therefore, two elements of Sn are conjugate if and only if they have thesame cycle structure (that is, they can be expressed as a product of disjointcycles 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 oforbits 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 ∈ Gsuch that T = gSg−1.

Given a subgroup H of G, its stabilizer under this action is called thenormalizer of H:

NG(H) = g ∈ G : gHg−1 = H.

Note that H is a normal subgroup of its normalizer, and that H is normal ifand 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 homo-morphism

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)

g 7→ LHg : 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:

kerLH = g ∈ G : gxH = xH ∀x ∈ G= g ∈ G : x−1gx ∈ H ∀x ∈ G= g ∈ G : g ∈ xHx−1 ∀x ∈ G= ∩x∈GxHx−1,

which is the largest normal subgroup of G contained in H.


6.3 Corollary.

(i) If H is a subgroup of G of index n ([G : H] = n) and such that H doesnot 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 ifH is a subgroup of G with [G : H] = p, then H is a normal subgroup.

Proof. For (i) note that LH : G → S(G/H) ∼= Sn is a monomorphism. For(ii) consider again the action LH and its kernel K = kerLH . K is a normalsubgroup 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 onlypossibility 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|G| = pnm and p -m. The subgroups of G of order pn are called the Sylowp-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 aprime number with |G| = pnm 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 dividingthe 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.If p | |g|, then |g| = pr for some r ∈ N, and gr is an order p element.Otherwise, H = 〈g〉 is a normal subgroup of G (as G is abelian), and it canbe assumed by an inductive argument that there is an element xH ∈ G/Hof order p. If |x| = m, then we have (xH)m = eH, so p = |xH| divides m,and again xm/p has order p.

If G is not abelian, consider its class equation |G| = |Z(G)|+∑r

i=1[G :CG(xi)]. If p | |Z(G)|, then there are elements of order p in the abeliansubgroup Z(G). Otherwise there is an i such that p - [G : CG(xi)]. But thenp 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|G| = |Z(G)| +

∑ri=1[G : CG(xi)]. If p - |Z(G)|, then again there is an

i such that p - [G : CG(xi)], so that |CG(xi)| = pnm′ for some m′ (since|G| = [G : CG(xi)]|CG(xi)|). An inductive argument shows that there is asubgroup of CG(xi) of order pn, 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 = 〈z〉 is a normal subgroup of G and|G/N | = pn−1m. By an inductive argument there is a subgroup P/N ≤ G/Nof 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 Sylowp-subgroup of G, and let Q be any p-subgroup of G, then we will check thatQ is contained in a conjugate of P . Actually, consider the action

Q −→ S(G/P )

g 7→ LPg : G/P → G/P

xP 7→ gxP.

Write Ω = G/P , and let Ω1, . . . ,Ωr be the orbits of this action. Since |Ωi|divides |Q| (a power of p) for any i and m = |G/P | = |Ω| =

∑ri=1|Ωi|,

and p -m, it follows that there is an i such that |Ωi| = 1. That is, there isan element x ∈ G such that gxP = xP for any g ∈ Q. This means thatx−1gx ∈ P for any g ∈ Q, or Q ≤ xPx−1.

As for (iii), let P be a Sylow p-subgroup, and let Ω be the set of Sylowp-subgroups of G, which is the set of subgroups conjugate to P . Considerthe 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 forany g ∈ P , so that P ≤ NG(Q), but then P and Q are Sylow p-subgroupsof NG(Q), and hence they are conjugate in NG(Q). Since Q is a normalsubgroup of NG(Q), it follows that Q = P . Therefore there is just one orbitconsisting of one element, namely P. The length of any other orbit is anontrivial divisor of |P |, and hence it is a power of p. We conclude that|Ω| = 1 +

∑(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 thenumber 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 isthe 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.


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

G = SL(2,Zp) =

(a bc d

): a, b, c, d ∈ Zp, ad− bc = 1

.

We are going to compute the number of Sylow p-subgroups of G.

First, since det : GL(2,Zp)→ Z×p is a group epimorphism, and SL(2,Zp)equals ker det, it follows that

|G| = |GL(2,Zp)|p− 1

=(p2 − 1)(p2 − p)

p− 1= p(p2 − 1).

On the other hand, it is clear that

P1 =

(1 a0 1

): a ∈ Zp

and P2 =

(1 0a 1

): a ∈ Zp

are two Sylow p-subgroups. Hence P1 is not a normal subgroup, so thatNG(P1) 6= G. Also, for any 0 6= b ∈ Zp,(

b 00 b−1

)(1 a0 1

)(b−1 00 b

)=

(1 b2a0 1

),

so that H =(

b a0 b−1

): b ∈ Z×p , 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 somek ≥ 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 Sylowp-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 iscalled the direct product of the groups G1, . . . , Gn.

Note that if G = G1× · · · ×Gn is a direct product, for each i the subsetHi = (e, . . . , x, . . . , e) : x ∈ Gi (n-tuples with an element of Gi in the ithposition, and the neutral element of the group Gj in the jth position for anyj 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 notationis 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 subgroupsof G satisfying that G = 〈H1, . . . ,Hn〉 and that for any i = 1, . . . , n, Hi ∩〈H1, . . . ,Hi−1, Hi+1, . . . ,Hn〉 = 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,

hihjh−1i h−1

j =

(hihjh

−1i )h−1

j ∈ Hj since Hj E G,

hi(hjh−1i h−1

j ) ∈ Hi since Hi E G,

so that, as Hi ∩ Hj = 1, it follows that hihjh−1i h−1

j = e, or hihj = hjhi.This shows that ϕ is a group homomorphism.

Moreover, H1H2 · · ·Hn = h1h2 · · ·hn : hi ∈ Hi ∀i is then a normalsubgroup of G which contains all the Hi’s. Hence G = 〈H1, . . . ,Hn〉 =H1H2 · · ·Hn, and hence ϕ is an epimorphism.

Finally, if (h1, h2, . . . , hn) ∈ kerϕ, then h1h2 · · ·hn = e, so that for anyi, hi = h−1

1 · · · ı · · ·h−1n ∈ Hi ∩ 〈H1, . . . , ı, . . . ,Hn〉 = 1, and this shows that

ϕ is a monomorphism.

In the situation of the theorem above, G is said to be the inner directproduct of the subgroups H1, . . . ,Hn. Any element of G can be written ina 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 twonormal subgroups of a group G with HK = G and H ∩K = 1, then G isisomorphic, 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 asubgroup, and again HK = G and H∩K = 1, then G is called the semidirectproduct of the normal subgroup H by the subgroup K. In this case notethat for h1, h2 ∈ H and k1, k2 ∈ K:

(h1k1)(h2k2) =(h1(k1h2k

−11 ))(k1k2

)=(h1Ik1(h2)

)(k1k2

),

and note that we have a natural homomorphism K → AutH, given byk 7→ Ik|H .

Conversely, assume that H and K are two groups, and that ϕ : K →AutH, x 7→ ϕx, is a group homomorphism. On the cartesian product H×Kdefine a product by:

(a, x)(b, y) =(aϕx(b), xy

),


for any a, b ∈ H and x, y ∈ K. With this product H×K is a group (CHECKTHIS!), denoted by H oϕ K, and called the outer semidirect product of Hand K with respect to ϕ. This semidirect product has two distinguishedsubgroups, 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

torA = 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 freein case torA = 0.

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

9.3 Theorem. Let A be a nonzero finitely generated abelian group. Thenthere are a natural number n ∈ N, an integer r ∈ Z with 0 ≤ r ≤ n, naturalnumbers m1, . . . ,mr ≥ 2, and a system of generators a1, . . . , an of A suchthat:

• A = 〈a1〉 ⊕ · · · ⊕ 〈an〉 is the inner direct sum of the 〈ai〉’s.

• |ai| = mi for any i = 1, . . . , r, |ai| = ∞ for r + 1 ≤ i ≤ n, andm1 |m2 | · · · |mr.

Moreover, the numbers n, r,m1, . . . ,mr are uniquely determined by A, n− ris 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

group Zm1 ⊕ · · · ⊕ Zmr ⊕ Z⊕ n−r· · · ⊕ Z.

Proof. Let n be the minimal number of generators of A. First we will showthe existence of the decomposition in the Theorem by induction on n, andwe 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 mand we are done.

Assume the result is true for n − 1. If there is a generating systema1, . . . , an such that q1a1 + · · · + qnan = 0 for q1, . . . , qn ∈ Z impliesq1 = · · · = qn = 0, then it is clear that |ai| =∞ and A = 〈a1〉⊕ · · · ⊕ 〈an〉 'Z⊕ n· · ·⊕Z (because for any i 〈ai〉∩

(〈a1〉+· · ·+〈ai−1〉+〈ai+1〉+· · ·+〈an〉

)=

§ 9. FINITELY GENERATED ABELIAN GROUPS 25

0). Otherwise the set Ω of those natural numbers x ∈ N such that thereexists a generating system b1, . . . , bn of A and integers x2, . . . , xn ∈ Z suchthat xb1 +x2b2 + · · ·+xnbn = 0 is not empty. Let m1 be the minimum of Ωand let b1, . . . , bn be a generating system and s2, . . . , sn ∈ Z integers suchthat m1b1 + s2b2 + · · ·+ snbn = 0. For any i = 2, . . . , n, let qi, ri ∈ Z, with0 ≤ 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. Thereforem1a1 = 0 and, by minimality, |a1| = m1. Moreover, A is the inner directsum A = 〈a1〉⊕〈b2, . . . , bn〉, as otherwise there would exist a nonzero elementra1 ∈ 〈a1〉 ∩ 〈b2, . . . , bn〉 with 0 < r < m1. But then r ∈ Ω and r < m1, acontradiction.

By the induction argument, 〈b2, . . . , bn〉 = 〈a2〉 ⊕ · · · ⊕ 〈an〉, for someelements 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 + sand 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 ∈ Ωunless s = 0. Since s < m1, the only possibility left is s = 0, so that m1 |m2.

Let us prove the uniqueness now. Note that

torA = 〈a1〉 ⊕ · · · ⊕ 〈ar〉,

andB = A/ torA ∼= 〈ar+1〉 ⊕ · · · ⊕ 〈an〉 ∼= Z⊕ n−r· · · ⊕ Z.

HenceB/2B ∼= Z2 ⊕

n−r· · · ⊕ Z2,

which contains 2n−r elements. Therefore we obtain:

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

Now torA = 〈a1〉⊕· · ·⊕〈ar〉 and hence mr is the maximum of the ordersof 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

d tor(A) = ⊕ri=1〈dai〉


and

|d tor(A)| =r∏i=1

|dai| =

(j∏i=1

|dai|

)·

r∏i=j+1

mi

d

.

Hence |d tor(A)| = 1dr−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:

mj = mind ∈ N : d |mj+1 and |d tor(A)| = 1

dr−j(mj+1 · · · · ·mr

).

We conclude that m1, . . . ,mr are uniquely determined by tor(A), and henceby A.

9.5 Corollary. Let 0 6= A be a finite abelian group. Then there are nat-ural numbers r,m1, . . . ,mr with m1 ≥ 2 and m1 | m2 | · · · | mr, uniquelydetermined by A, such that A ∼= Zm1 ⊕ · · · ⊕ Zmr .

9.6 Lemma. Let m = ps11 · · · psrr be the prime factorization of a naturalnumber m ≥ 2 (p1, . . . , pr are different prime numbers and s1, . . . , sr ∈ N).Then the abelian group Zm is isomorphic to Zps11 ⊕ · · · ⊕ Zpsrr .

Proof. Actually, the Chinese Remainder Theorem shows us that there is aring isomorphism Zm ∼= Zps11 ⊕· · ·⊕Zpsrr . This isomorphism is, in particular,a group isomorphism.

9.7 Example. If A is a finite abelian group with invariant factors 6 and18. 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 invariantfactors 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 directsum of cyclic groups whose orders and powers of prime numbers. Thesepowers are uniquely determined by the group (and are called the elementarydivisors of the group).

9.9 Example. How many abelian groups of order 180 do exist, up to iso-morphism?

Since 180 = 22 ·32 ·5, there are the following possibilities for the elemen-tary 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 symmetriesof 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: G′ = 〈[a, b] : a, b ∈ G〉,is called the derived subgroup of G.

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

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

sgn[x, y] = sgnx sgn y(sgnx)−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),

and hence all the 3-cycles are in S′n. 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

An ⊆ 〈cycles of length 3〉 ⊆ S′n ⊆ An,

and we conclude that the alternating group An is the derived subgroup ofthe 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 K ′. In particular G′ is anormal subgroup of G.

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

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

Proof. For (i) note that ∀g ∈ G and ∀x, y ∈ K, Ig([x, y]) = [Ig(x), Ig(y)] ∈K ′, since Ig is an automorphism. Therefore Ig(K

′) ⊆ K ′ for any g ∈ G andK ′ is a normal subgroup.


For (ii) it is enough to note the following equation in the quotient groupG/G′ valid for any x, y ∈ G:

(xG′)(yG′) = (xy)G′ = (xy)[y−1, x−1]G′ = yxG′ = (yG′)(xG′).

Hence G/G′ is abelian. Besides, if G/K is abelian, then xyK = yxK forany x, y ∈ G, and this is equivalent to [x, y] ∈ K. Thus G′ ⊆ K.

Finally, if G′ ⊆ K ⊆ G, and x ∈ K, g ∈ G, gxg−1 = [g, x]x ∈ K, because[g, x] ∈ G′ ⊆ K, so K is a normal subgroup.

We already have two subgroups related to the commutativity of a groupG:

G is abelian ⇐⇒ G′ = 1 ⇐⇒ Z(G) = G.

Define recursively the following normal subgroups of a group G:

• 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) ⊇ G′ = 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 thatG(n) = 1.

• The group G is said to be nilpotent if there is an n ∈ N such thatZn = 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, whichis cyclic of order 3. Hence its derived chain is S3 D A3 D 1 and S3 is solvabletoo.

The symmetric group S4 satisfies that its derived subgroup is A4 of order12. 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 transpositionsis a product of two disjoint transpositions. Note that V is contained inA4 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 DA4 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 DG1 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 onlyif so are both N and G/N .

Proof. (i) If G is a p-group: |G| = pn, we know (see 6.1 Corollary) thatZ1 = Z(G) is not trivial (1 & Z1). Now, if Z1 6= G, then G/Z1 is a p-grouptoo, so 1 6= Z2/Z1 = Z(G/Z1) and hence Z1 & Z2. Continuing in this waywe get 1 & Z1 & Z2 & · · · . Since G is finite, it turns out that G = Zn forsome n.

(ii) One implication is clear, because if G is solvable, its derived series is achain with abelian consecutive factors. Conversely, if G D G1 D · · · D Gm =1 is a descending chain with abelian consecutive factors, then we may proveinductively that G(i) ⊆ Gi for any i = 0, 1, 2, . . . (with G0 = G). Indeed,this is trivial for i = 0, and if G(i) ⊆ Gi, then since Gi/Gi+1 is abelian,G(i+1) =

(G(i)

)′ ⊆ G′i ⊆ Gi+1, as desired.

(iii) If G is nilpotent, the chain G = Zn D Zn−1 D · · · D Z0 = 0 is adescending chain with abelian consecutive factors, and hence G is solvable.

(iv) Note that ifN is a normal subgroup of a solvable groupG withG(n) = 1,

then1 N (n) ⊆ G(n) = 1, and also(G/N

)(n) ⊆ G(n)N/N = N/N = 1. Henceboth 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 bothN and G/N to get such a chain for G: N = N0 D N1 D · · · D Nr = 1and G/N = G0/N D G1/N D · · · D Gs/N = N/N = 1 give the descendingchain 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. Thenthe 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-groupis nilpotent, and hence solvable.

pq : We may assume p > q. If |G| = pq, the number of Sylow p-subgroupsof G is Np, and Np | q, since Np = [G : NG(P )] | [G : P ] for a fixed Sylowsubgroup 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 solvablegroup is solvable


and hence there is a unique (normal) Sylow p-subgroup P . Then |P | = p, soP is cyclic (and hence abelian and solvable) and |G/P | = q and hence G/Pis solvable too. Thus, G is solvable.

p2q : If p > q the same argument as before gives that there is a uniqueSylow 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 ofSylow q-subgroups Nq divides p2 and is ≡ 1 (mod q). Then either Nq = 1and, as before, G is solvable, or Nq = p2. In this case, there are exactlyp2 Sylow q-subgroups, and each of them is cyclic of order q, so there arep2(q − 1) elements of order q in G. But |G| = p2q = p2(q − 1) + p2, sothere is room for a unique Sylow p-subgroup, and this implies again that Gis solvable.

p2q2 : We may assume that p > q and, as before, either Np = 1 or Np = q2.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 onlypossibility is q = 2 and p = 3. Thus we have N3 = 4. Let P be a Sylow3-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 isthe largest normal subgroup contained in P is not trivial. We conclude thatker Φ has order 3, and hence it is solvable, while G/ ker Φ has order 12 andhence 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 hencethere 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 herewithout proofs:

10.8 Burnside’s Theorem (1904). Let p and q be two different primenumbers and let n,m ∈ N. Then any group of order pnqm is solvable.

10.9 Feit-Thompson’s Theorem (1963). Any finite group of odd orderis solvable.

§ 11. Simple groups

11.1 Definition. A nontrivial group G is said to be simple if it containsno proper normal subgroup.

11.2 Proposition. Let G be a solvable group. Then G is simple if and onlyif it is cyclic of prime order.

§ 11. SIMPLE GROUPS 31

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

Our purpose now is to show that the alternating groups An for n ≥ 5are 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 thewhole 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 showthat 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 checkthat N contains a cycle of length 3. Let 1 6= σ ∈ N and let r be the greatestlength 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).Then, as [σ, γ] = σ

(γσ−1γ−1

)∈ N for any γ ∈ An, we get

[σ, (a1 a2 a3)] = σ(a1 a2 a3)σ−1(a1 a2 a3)−1 =(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, thenσ = (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 andτ2 = 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 evennumber (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.


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

Proof. We know that S′n = An. Since An is simple and it is not abelian, itsderived subgroup is A′n = An. And hence the derived series of Sn is justSn D An.

Exercises

1. Let G be a semigroup, that is, a nonempty set with a binary operationwhich is associative. Assume that the following two conditions hold:

(a) There exists an element e ∈ G such that ex = x for any x ∈ G (aleft neutral element),

(b) For any x ∈ G there is an element y ∈ G such that yx = e (a leftinverse).

Prove that G is a group.

2. Let G be a semigroup containing a left neutral element e and suchthat 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 isabelian.

4. Let S be the subgroup of the cyclic group of order 154: C154 = 〈x〉,generated by x28, x88. Find a natural number n such that S = 〈xn〉.

5. Let G = 〈x〉 be an infinite cyclic group. Prove that any nontrivialsubgroup of G is an infinite cyclic group of finite index generated byxn 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 generatorsof G (that is, compute |g ∈ G : 〈g〉 = G|).

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

9. Let Cn be a cyclic group of order n and let m ∈ N. Prove that the mapαm : Cn → Cn such that x 7→ xm for any x ∈ Cn is an automorphismif and only if gcd(m,n) = 1. Prove that AutCn is an abelian group oforder φ(n). Compute IntCn.

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

EXERCISES 33

10. Consider the elements A =

(0 −11 0

)and B =

(0 −11 −1

)in GL2(Z).

Prove that A has order 4, B has order 3, while the order of AB isinfinite.

11. Give an example of a nonabelian finite group such that all its propersubgroups 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 〈x, y〉 cyclic?

(c) What is the answer to the two previous questions if we substituteQ 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 andC2 × 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 elementsof 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 ⊆ Kor 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.

18. Prove that the set of matrices(

λ −µµ 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 subgroupof a group G is normal.

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

(a) Is Φ(N) a normal subgroup of G2?


(b) Is Φ−1(K) = x ∈ G1 : Φ(x) ∈ K a subgroup of G1?

(c) If K is normal, is Φ−1(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 manyconjugates 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 planewhich fix a regular hexagon.

(a) Prove that this group is generated by the clockwise rotation x ofangle π

3 and by the reflection y through two opposite vertices.

(b) Show that x6 = 1 = y2 and yx = x−1y.

(c) Show that D6 = xnym : 0 ≤ n < 6, 0 ≤ m < 2.

28. Let G be a finite group generated by two different elements of order2. Prove that G is isomorphic to C2 × C2 or to Dn for some n ≥ 3.

29. Compute AutDn (n ≥ 3).

30. Prove that the multiplicative group Gn =(±1 λ

0 1

): λ ∈ Zn

is iso-

morphic to Dn.

31. Prove that the only groups of order 6 are, up to isomorphism, C6 andD3.

27 It can be checked that any equation satisfied by x and y is a consequence of theequations in (ii). This allows us to write

D6 = 〈x, y : x6 = y2 = 1, yx = x−1y〉.

This is a presentation by generators and relations of D6. These presentations constitutea very useful way to define groups.

The formal definition of these presentations require the definition of free groups, whichwill 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 satisfythe relations of the generators of a dihedral group.

EXERCISES 35

32. Prove that the only nonabelian groups of order 8 are, up to isomor-phism, 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 greenbeads?

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. Provethat 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. Provethat G′ equals Z(G).

40. For n ≥ 3, let Dn = 〈x, y : xn = y2 = 1, yx = x−1y〉 be the dihedralgroup of degree n.

(a) Prove that Z(Dn) = 1 for odd n, but Z(Dn) = 〈xn2 〉 for even n.

(b) Prove that D′n = 〈x2〉 (which is equal to 〈x〉 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 Gis not simple.

43. Prove that there are no simple groups of order 56, 80, 84, 132, 300 or1000.

44. Prove that, up to isomorphism, the only simple nonabelian group oforder ≤ 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 sub-group of A2m. Then argue with the possible expressions as a product of disjoint cycles ofan order 2 element x of G, taking into account that the left multiplication by x does notfix any element of G.


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

[H,K] = 〈[h, k] : h ∈ H, k ∈ K〉.

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

(c) Prove that for any n, Ln/Ln+1 ≤ Z(G/Ln+1).

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

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 subgroupN such that both N and G/N are nilpotent.

50. Give examples of nilpotent nonabelian groups and of solvable non-nilpotent groups.

51. Consider the following two groups:

G1 = 〈x, y : x5 = 1 = y4, yx = x3y〉,G2 = 〈x, y, z : x11 = 1 = z3, [x, y] = 1, zx = yz, zy = x−1y−1z〉.

(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 correspond-ing quotient groups.

(d) How many conjugacy classes do they contain?

(e) Study the solvability and nilpotency of these groups.

51 The groups G′1 and G1/G′1 are cyclic. The subgroup of G1 generated by x and y2 is

a dihedral group. These facts, together with the number of Sylow subgroups, allows youto 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.Compute [x, z] and [y, z] and deduce from here G′2 and G2/G

′2. Argue that Z(G2) cannot

contain order 3 or 11 elements. Using Sylow subgroups, check that G can only containelements of order 1, 3 or 11.

Chapter 2

Galois Theory

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

§ 1. Algebraic extensions

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

ϕ : Z −→ F

n 7→ n1 = 1+n· · · +1 (n ∈ N)

0 7→ 0

−n 7→ −n1 = (−1)+n· · · +(−1) (n ∈ N)

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

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

ψ : Q −→ Fm

n7→ ϕ(m)ϕ(n)−1

since Q is the field of fractions of Z. In this case, it is said that thecharacteristic 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 subfieldof F .

2. kerϕ = 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 primenumber, which is called the characteristic of F . In this case the fieldFp = Z/pZ is identified with its image under ϕ, which again is thesmallest subfield of F and called the prime subfield of F .Notice that p is the smallest natural number such that p1 = 0 andthat for any α ∈ F

pα = α+p· · · +α = (1+

p· · · +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 sois Fp(X) (the fraction field of Fp[X]).

1.3. Let K/F be a field extension, then 1K = 1F and the characteristics ofF 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 denotedby [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] bean 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

assume that f(X) = a0 + a1X + · · ·+ anXn with an 6= 0. Then θ is trivially

a root of f(X) in K, because

f(θ) = a0 + a1θ + · · ·+ anθn

= a0 + a1(X +(f(X)

)) + · · ·+ an(X +

(f(X)

))n

= a0 + a1X + · · ·+ anXn +

(f(X)

)= f(X) +

(f(X)

)= 0.

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

Proof. Let us show first that 1, θ, . . . , θn−1 is free. Assume that there arescalars b0, . . . , bn−1 ∈ F such that b0 + b1θ + · · ·+ bn−1θ

n−1 = 0. As before,

b0 + b1θ + · · ·+ bn−1θn−1 = (b0 + b1X + · · ·+ bn−1X

n−1) +(f(X)

),

so that we conclude that b0 + b1X + · · · + bn−1Xn−1 ∈

(f(X)

). That is,

f(X) divides b0 + b1X + · · · + bn−1Xn−1 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 el-ement 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 thatg(X) = c(X)f(X) + r(X). If r(X) = r0 + r1X + · · ·+ rn−1X

n−1, then

g(X) +(f(X)

)= r(X) +

(f(X)

)= r01 + r1θ + · · ·+ rn−1θ

n−1

is a linear span with coefficients in F of the elements in 1, θ, . . . , θn−1, asrequired.

Therefore,

K = F [X]/(f(X)

)= F1 + · · ·+ Fθn−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 theremainder of the division of r1(X)r2(X) by f(X).

Let us consider a couple of examples:

• C ∼= R[X]/(X2 + 1).

• The polynomial X3 − 2 ∈ Q[X] is irreducible (Eisenstein’s criterion).Let K = Q[X]/(X3−2) and θ = X+(X3−2). Then K = a+bθ+cθ2 :a, b, c ∈ Q, θ3 = 2, θ4 = 2θ. Therefore, the multiplication in K isgiven by:

(a+ bθ + cθ2)(a′ + b′θ + c′θ2)

= (aa′ + 2bc′ + 2cb′) + (ab′ + ba′ + 2cc′)θ + (ac′ + ca′ + bb′)θ2.

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

1 = −1

3(X3 − 2) +

1

3(X2 −X + 1)(X + 1),

so that 1 = (θ + 1)θ2 − θ + 1

3. Therefore, (1 + θ)−1 =

θ2 − θ + 1

3.


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. kerψ = 0. In this case there is no nonzero polynomial in F [X] forwhich 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 thesmallest subfield of K containing both F and S.)

2. kerψ 6= 0. Since F [X] is a principal ideal domain and imψ is anintegral domain (it is a subring of the field K), kerψ =

(m(X)

)for

a unique monic irreducible polynomial m(X) ∈ F [X]. Hence anypolynomial f(X) ∈ F [X] with f(α) = 0 is a multiple of m(X). Hencem(X) is the monic polynomial of lowest degree that annihilates α.The polynomial m(X) is called the minimal polynomial of α over Fand it is denoted by mα,F (X). In this case, α is said to be algebraicover F . Note that F (α) is isomorphic to F [X]/

(mα,F (X)

)and hence

[F (α) : F ] = degmα,F (X).The extension field K/F is said to be algebraic if for any α ∈ K, α isalgebraic over F .

Examples.

• C/R is algebraic, because for any α = a+ bi ∈ C (a, b ∈ R), α is a rootof (X − α)(X − α) = X2 − 2aX + (a2 + b2) ∈ 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 char-acteristic of F is 6= 2. Then any α ∈ K \ F is a root of a polynomialX2 + bX + c ∈ F [X]. But

X2 + bX + c =

(X − b

2

)2

− b2 − 4c

4,

so that β = 2α−b (∈ K\F ) satisfies β2 = b2−4c ∈ F . With D = b2−4c thisshows that the minimal polynomial of β is X2 −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 algebraicover 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 basisof 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 =∑ah,kαjkβih . 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 $(√

26∈Q)

Q(√

2) ⊆ Q(√

2)(√

3) = Q(√

2,√

3) (⊆ R).

Now, [Q(√

2) : Q] = 2, since m√2,Q(X) = X2 − 2 (irreducible by Eisen-

stein’s criterion). Also,√

3 is a root of X2 − 3 ∈ Q[X] ⊆ Q(√

2)[X], som√3,Q(

√2)(X) | X2−3, and either [Q(

√2,√

3) : Q(√

2)] = 2, or Q(√

2,√

3) =

Q(√

2), that is,√

3 ∈ Q(√

2). But in the latter case, there would exista, b ∈ Q such that

√3 = a+ b

√2, so 3 = (a2 + 2b2) + 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 ofQ(√

2,√

3) over Q(√

2) (because degm√3,Q(√

2)(X) = 2). By the argument

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 andonly 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 degmαi,F (X) = ni for any i = 1, . . . , r, then [K : F ] ≤n1 · · · · · nr.


Proof. If [K : F ] <∞ and α1, . . . , αr is a basis of K over F , then for anyi = 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 forany i, and let degmαi,F (X) = ni. Thus, mαr,F (X) is a polynomial inF (α1, . . . , αr−1)[X] which ‘kills’ αr, and therefore mαr,F (α1,...,αr−1)(X) di-vides 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 twoalgebraic elements over F . Then α+ β, α− β, αβ and αβ−1 (if β 6= 0) arealgebraic over F too.

In particular, the set γ ∈ K : γ is algebraic over F is a subfield of Kwhich contains F .

Proof. From the Theorem above we conclude that [F (α, β) : F ] < ∞, andtherefore the field extension F (α, β)/F is algebraic. Now everything followssince 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 .

Proof. Let α ∈ L and assume mα,K(X) = b0 + b1X + · · ·+ bn−1Xn−1 +Xn,

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 inAlgebraic Structures.

2.1 Definition. Let F be a field and 0 6= f(X) ∈ F [X]. A field extensionK of F is said to be a splitting field of f(X) over F if, as a polynomial inK[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.

• C is a splitting field of X2 + 1 ∈ R[X].

• Q(√

2) is a splitting field of X2 − 2 over Q.

• Q(√

2,√

3) is a splitting field of (X2 − 2)(X2 − 3) over Q.

• The splitting field of X3 − 2 over Q is not Q( 3√

2), because the roots

(in C) of X3−2 are 3√

2, ω 3√

2, ω2 3√

2, where ω = −1+i√

32 is a cubic root

of 1. Note that if 3√

2 and ω are in a field K containing Q, then so isi√

3 =√−3. Then K = Q( 3

√2,√−3) is a splitting field of X3−2 over

Q. Since [K : Q( 3√

2)] = 2, because√−3 is a root of X2 + 3 and K 6=

Q( 3√

2) (⊆ R), we get [K : Q] = [K : Q( 3√

2)][Q( 3√

2) : Q] = 2× 3 = 6.

2.3 Proposition. Let K be a splitting field of a polynomial f(X) of degreen 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 splittingfield 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 poly-nomials and that, up to isomorphism, they are unique.

2.4 Lemma. Let ϕ : F → F be a field isomorphism, f(X) = a0 + a1X +· · · + anX

n ∈ F [X] an irreducible polynomial, f(X) = ϕ(a0) + ϕ(a1)X +· · ·+ ϕ(an)Xn ∈ F [X] (which is irreducible too), α a root of f(X) in somefield extension of F and β a root of f(X) in some field extension of F . Thenthere exists a field isomorphism σ : F (α) → F (β) such that σ|F = ϕ andσ(α) = β.

Proof. It is enough to concatenate the following isomorphisms:

F (α) ∼= F [X]/(f(X)

) ∼= F [X]/(f(X)

) ∼= F (β)

a ∈ F a+(f(X)

)7→ ϕ(a) +

(f(X)

) ϕ(a)

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:


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

(ii) All of them are isomorphic. More precisely, if ϕ : F → F is a fieldisomorphism, f(X) is the polynomial which results of applying ϕ tothe coefficients of f(X), E is a splitting field of f(X) over F , andE a splitting field of f(X) over F , then there is a field isomorphismσ : E → E such that σ|F = ϕ.

Proof. For (i), we use induction of deg f(X). If deg f(X) ≤ 1, then F itselfis a splitting field.

Now, if deg f(X) = n > 1, we know that there exists a field extensionof F which contains a root α1 of some irreducible factor of f(X). LetK1 = 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 ∈ Eand 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 Eis a splitting field of f(X) over F .

Induction on n = deg f(X) will be used too for (ii). Again, if n = 1, thenE = F and E = F because the roots (if any) of f(X) (respectively f(X))are in F (respectively in F ). Hence σ = ϕ. Now, if n > 1, let f1(X) be anirreducible factor of f(X) and let f1(X) be the corresponding irreduciblefactor of f(X). Let α ∈ E be a root of f1(X) and β ∈ E a root of f1(X).Thus F ≤ F (α) ≤ E and F ≤ F (β) ≤ E and, by the previous Lemma, thereexists σ1 : F (α) → F (β), a field isomorphism, extending ϕ and such thatσ1(α) = β.

In F (α)[X], f(X) = (X−α)g(X), and in F (β)[X], f(X) = (X−β)g(X),where g(X) is the quotient of f(X), which is the polynomial obtained fromf(X) by applying σ1 to all its coefficients, by (X−β), which is the polynomialobtained from (X −α) by applying σ1 to all its coefficients. Therefore g(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 E is a splittingfield of g(X) over F (β) so, by the induction hypothesis, there exists a fieldisomorphism 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 thereis a polynomial f(X) ∈ F [X] such that K is the splitting field of f(X) overF .

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

• 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 containsall the roots of mα,F (X).)

Proof. Assume first that mα,F (X) splits for any α ∈ K, and write K =F (α1, . . . , αr). ThenK is the splitting field of the polynomial

∏ri=1mαi,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 isa 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. Butσ permutes the roots of f(X) ∈ F [X], and hence σ(K) = K. It follows thatβ = 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 alge-braic closure of F if K/F is an algebraic field extension and any polynomialf(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 polynomialof degree ≥ 1 over F has a root in F . (That is to say, the irreduciblepolynomials in F [X] are the degree 1 polynomials.)

2.10 Example. C is algebraically closed by the Fundamental Theorem ofAlgebra (proved in Algebraic Structures) and hence it is an algebraic closureof 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 algebraicclosure 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 . Thenα is algebraic over F and F /F is algebraic, so that α is algebraic overF . But mα,F (X) ∈ F [X] factors over F as a product of degree 1 poly-nomials: mα,F (X) = (X − α1) · · · (X − αr), with α1, . . . , αr ∈ F . Since0 = mα,F (α) =

∏ri=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


the ring R of polynomials in the variables Xf with coefficients in F . Let I bethe 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 polyno-mials f1(X), . . . , fr(X) and polynomials g1, . . . , gr in in the variables Xf ’ssuch 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).

Let F ′ be a splitting field of f1(X) · · · fn(X) over F . In F ′ there are elementsαi such that fi(αi) = 0 for i = 1, . . . , r. Then evaluate equation (2.12) withXi 7→ αi, for i = 1, . . . , r, and Xi 7→ 0, for i = r + 1, . . . , n, to get 1 = 0, acontradiction.

Therefore the ideal I is not the whole R. Let M be a maximal ideal ofR containing I, and consider the field K1 = R/M . K1 is a field extension ofF through the embedding ι : F → R/M , a 7→ a+M , and if xf = Xf +Mfor any monic polynomial of degree ≥ 1, xf is a root of f(X) over K1. ThusK1 contains a root for each monic polynomial in F [X] of degree ≥ 1. Werepeat the process with K1 to get a field extension K2 of K1 such that anydegree ≥ 1 monic polynomial in K1[X] has a root in K2, and then we repeatthe 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 anyn ≥ 0. Consider K = ∪∞n=0Kn, which is a field extension of F . Also, givenany degree ≥ 1 monic polynomial h(X) ∈ K[X], there is an n ∈ N such thath(X) ∈ Kn[X], and hence h(X) has a root in Kn+1 ⊆ K. Therefore, K isalgebraically closed.

Now the third part of the Theorem is easy. F is a subfield ofK containingF , F /F is an algebraic field extension and if f(X) ∈ F [X] is a polynomial ofdegree ≥ 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 algebraicover 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 consistingof all the triples (L1, L2, ϕ), where L1 is a field extension of F containedin K1, L2 is a field extension of F contained in K2 and ϕ : L1 → L2 isan F -isomorphism (that is, an isomorphism which restricts to the identitymap of F ). The set S is partially ordered with (L1, L2, ϕ) ≤ (L′1, L

′2, ϕ′)

if L1 ⊆ L′1, L2 ⊆ L′2 and ϕ = ϕ′|L1 . Besides, any chain has an upperbound (obtained as the union of the first and second components, and thenatural isomorphism among these unions). By Zorn’s Lemma, there exists amaximal 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 Theorem2.5) 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 . Thenf(X) is said to be separable if it does not have multiple roots (in a splittingfield).

As proved in Algebraic Structures, under these circumstances:

f(X) is separable ⇐⇒ gcd(f(X), f ′(X)

)= 1 ⇐⇒ D

(f(X)

)6= 0,

where D(f(X)

)is the discriminant of f(X).

Recall from Algebraic Structures that if f(X) = a0 + a1X + · · ·+ anXn

and g(X) = b0 + b1X + · · · + bmXm are polynomials over a field F with

n,m ≥ 1 (although we admit that an or bm may be 0), then the resultant off(X) and g(X) is the determinant of the Sylvester matrix :

Resn,m(f, g) =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

an an−1 an−2 . . . . . . a0 0 . . . . . . 00 an an−1 . . . . . . . . . a0 0 . . . 0...

. . .. . .

......

......

. . .. . .

...0 . . . 0 an . . . . . . . . . a1 a0 00 . . . . . . 0 an . . . . . . a2 a1 a0

bm bm−1 bm−2 . . . . . . b0 0 . . . . . . 00 bm bm−1 . . . . . . . . . b0 0 . . . 0...

. . .. . .

......

......

. . .. . .

...0 . . . 0 bm . . . . . . . . . b1 b0 00 . . . . . . 0 bm . . . . . . b2 b1 b0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

m rows

n rows

︸︷︷︸n+m columns

If deg f(X) = n and deg g(X) = m (that is, if an 6= 0 6= bm), then we writesimply Res(f, g).

The discriminant of the polynomial f(X) = a0 + a1X + · · · + anXn ∈

F [X] of degree n ≥ 2 (an 6= 0) is the scalar (in F )

D(f) = (−1)(n2)a−1

n Resn,n−1(f, f ′).

In case f(X) = (X −α1) · · · (X −αn) in some field extension of F , then

D(f) =(∏

1≤i<j≤n(αi−αj))2

. Note that∏

1≤i<j≤n(αi−αj) may not belongto F , but its square is the discriminant, which does belong to F .


3.2 Examples.

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

• Over a field of characteristic 0 any irreducible polynomial is separable,as gcd

(f(X), f ′(X)

)= 1, since f(X) is irreducible, and 0 6= f ′(X) is

not a multiple of f(X).

• Assume that F is a field of characteristic p > 0. The polynomialf(X) = Xn − 1 satisfies

f ′(X) = nXn−1 =

0 if p |n,

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 irreduciblepolynomial is separable.Actually, let F be a finite field of characteristic p, and let f(X) =a0 + a1X + · · ·+ anX

n be an irreducible polynomial over F (an 6= 0).If f ′(X) 6= 0, we conclude as before that f(X) is separable. Otherwisef ′(X) = 0 and hence the powers of X which appear on f(X) with anonzero 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, forany a ∈ F there is a unique b ∈ F with bp = a. Hence

f(X) = a0 + apXp + · · ·+ arpX

rp = bp0 + bp1Xp + · · ·+ bprX

rp

= (b0 + b1X + · · ·+ brXr)p,

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

3.3 Definition.

• Given a field extension K/F , an element α ∈ K is said to be separableover 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 ofK is separable over F .

• A field F is said to be perfect if all its finite field extensions are sepa-rable.

§ 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. (Thisis 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 aboveshow that F is perfect. Conversely, if F is perfect and α ∈ F , takeK a splitting field of Xp − α over F , and take a root β in K, so thatβp = α and Xp − α = Xp − βp = (X − β)p in K[X]. Since F isperfect, mβ,F (X) is separable, but mβ,F (X) | Xp − α. We concludethat 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)

)p.

§ 4. Galois group

Let K be a field and let σ : K → K be an automorphism. The elementa ∈ K is fixed by σ if σ(a) = a. Note that we have σ(1) = 1, and henceσ(m1) = m1 for any m ∈ Z. Hence any automorphism fixes the primesubfield 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 ∈ AutK : σ(a) = a ∀a ∈ F(≤ AutK

).

(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 tobe 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 overF , 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.)

Proof. Assume mα,F (X) = a0+a1X+· · ·+an−1Xn−1+Xn, ai ∈ F for all i =

0, . . . , n−1. Then 0 = mα,F (X) = a0+a1α+· · ·+an−1αn−1+αn, and hence,

since σ(ai) = ai for any i we get 0 = a0 +a1σ(α) + · · · an−1σ(α)n−1 +σ(α)n,so that σ(α) is a root too of mα,F (X).


4.3 Examples.

(i) Take K = Q(√

2). Then any τ ∈ Gal(K/Q) satisfies τ(√

2) = ±√

2,since τ(

√2) is a root of X2 − 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.

(ii) Now let K = Q( 3√

2). Since m 3√2,Q(X) = X3 − 2 and the other two

roots of this polynomial are not in K, we conclude that τ( 3√

2) = 3√

2for 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 AutK. ThenKH = a ∈ K : σ(a) = a ∀σ ∈ H is a subfield of K, which is called thefixed 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) isa subgroup of Gal(K/F1). (Thus Gal(K/F2) ≤ Gal(K/F1) ≤ AutK.)

(ii) Let H1 and H2 be two subgroups of AutK with H1 ≤ H2. Then KH2

is a subfield of KH1.

Therefore the two maps:

subfields of K → subgroups of AutKF 7→ Gal(K/F ),

and

subgroups of AutK → subfields of KH 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 “linearlyindependent”. That is, if a1, . . . , an ∈ K satisfy

∑ni=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 thatthere are different group homomorphisms σ1, . . . , σn : G→ K× and scalarsa1, . . . , an ∈ K, not all of them different from 0, such that

a1σ1(g) + · · ·+ anσn(g) = 0,


for any g ∈ G. If one of these scalars is 0, then we get a contradiction withthe n−1 case. Hence ai 6= 0 for any i. Now, as σ1 6= σn, there is an elementh ∈ G such that σ1(h) 6= σn(h). Then we have:

n∑i=1

aiσi(hg) = a1σ1(h)σ1(g) + · · ·+ anσn(h)σn(g) = 0

σn(h)( n∑i=1

aiσi(g))

= a1σn(h)σ1(g) + · · ·+ anσn(h)σn(g) = 0.

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 induc-

tion hypothesis.

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

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

Proof. Let F = KG and G = σ1 = 1, . . . , σn. Assume first [K : F ] < nand let ω1, . . . , ωm be a basis of K over F (m < n). Consider the followinghomogeneous 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. Notethat σi(λj) = λj for any i, j, because F = KG, so if we multiply in thesystem above, with the xi’s substituted by the αi’s, the first row by λ1, thesecond 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 haveto consider σ1, . . . , σn as group homomorphisms K× → K×.)

Note that we have not used here that G is a group, only that σ1, . . . , σnare different automorphisms of K.

Assume now [K : F ] > n, and let ω1, . . . , ωn+1 ∈ K be a family oflinearly independent elements (over F ). Again there is a nontrivial solution


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 arein F (look at the first row of the system). Among all the possible nontrivialsolutions we take one (α1, . . . , αn+1) with the lowest possible number r ofnonzero elements. After reordering the ωi’s, we may assume that this solu-tion 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 tothe 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 thetuple (σk(β1), . . . , σk(βr), 0, . . . , 0) is another solution of the homogeneouslinear 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 withthe first component different from 0. This contradicts our assumption onthe 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 ofAutK. 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 istrivial.

4.10 Corollary. Let K be a field and let G be a finite subgroup of AutK.Then Gal(K/KG) = G.


Proof. The containment G ≤ Gal(K/KG) is clear, and both subgroups havethe 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 isseparable 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 ∈ Kbe a root of f(X). Consider the different elements α1, . . . , αr in the setσ(α) : σ ∈ G (which is contained in the set of roots of f(X)) and take thepolynomial g(X) = (X − α1) · · · (X − αr) ∈ K[X]. Each σ ∈ G permutesthe αi’s, and hence the coefficients of g(X) are fixed by σ. By (i) we getg(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) impliesthat mωi,F (X) is separable for any i and all its roots are in K. HenceK is the splitting field of the least common multiple of these polynomialsmωi,F (X), which is separable.

(iv) ⇒ (ii): We know that if ϕ : F → F ] is a field isomorphism, K is asplitting field of a polynomial f(X) ∈ F [X] over F , and K] is a splittingfield of the polynomial f ](X) = ϕ

(f(X)

)over F ] (this is the polynomial

obtained from f(X) by applying ϕ to all its coefficients), then there is afield isomorphism ψ : K → K] which extends ϕ. We will prove by inductionon [K : F ] that the number of ψ’s extending ϕ is at most [K : F ], and thatit 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 degreeextensions. 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 aroot of h](X) and τ = ψ|F (α) : F (α)→ F ](α]) is an isomorphism extending


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 (respectivelyK]) is a splitting field of f(X) (respectively f ](X)) over F (respectivelyF ]), there is an isomorphism ψ : K → K] extending τ . Therefore thenumber of extensions ψ : K → K] of ϕ equals the product of the number ofextensions τ : F (α)→ F ](α]) of ϕ by the number of extensions of each suchτ to an isomorphism ψ : K → K]. This last number is, by the inductionhypothesis, at most [K : F (α)], with equality if f(X) is separable, whilethe first number (of extensions τ) is the number of roots of h(X), whichis at most deg h(X) = [F (α) : F ], again with equality if f(X) (and henceh(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 equivalentconditions 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 Galoisextension 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 sep-arable 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 split-ting field of a monic polynomial f(X) ∈ F [X]. Consider its factorizationinto irreducible factors: f(X) = f1(X)m1 · · · fs(X)ms (the fi(X)′s are ir-reducible and coprime). Then K is also the splitting field over F of thepolynomial f1(X) · · · fs(X). Now, if K/F is separable, each fi(X) is sepa-rable, and hence so is its product. Hence item (iv) of the previous theoremis 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 charac-teristic of F is p > 0, then for any α ∈ K there is an n ∈ N∪0 suchthat αp

n ∈ S.

(iii) If L/K and K/F are two finite field extensions, then L/F is separableif 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, ofthe least common multiple of themαi,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. Ifthe characteristic is 0 any field extension is separable, so the last assertionfollows. Assume now that the characteristic is p, and take α ∈ K andf(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 g′(X) 6= 0. Since f(X) is irreducible,so is g(X), and hence g(X) is separable. It follows that αp

n ∈ S.

For (iii) note that if L/F is separable, then so is K/F . Also for anyα ∈ 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/Kand K/F are separable, then if the characteristic is 0 obviously L/F isseparable, 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 thatαp

n ∈ S. Then mα,S(X) |Xpn − αpn = (X − α)pn. But L/K is separable,

and hence so is L/S. Therefore mα,S(X) = X − α and α ∈ S. Thus L = Sis 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 fieldof (X2 − 2)(X2 − 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) (whichis a root of X2 − 2) and ξ(

√3) (a root of X2 − 3). Hence there are four


possibilities:√2 7→

√2,

√3 7→

√3,

√2 7→ −

√2,

√3 7→

√3,

√2 7→

√2,

√3 7→ −

√3,

√2 7→ −

√2,

√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 =〈σ〉 × 〈τ〉 ' C2 × C2.

Let us have a look at the subgroups of Gal(K/F ) and the correspondingfixed subfields:

Subgroups of Gal(K/F ) Fixed subfields

1 K = Q(√

2,√

3)

〈σ〉 Q(√

3)

〈τ〉 Q(√

2)

〈στ〉 Q(√

6)〈σ, τ〉 = Gal(K/F ) Q

Here is the diagram of subgroups:

1

〈σ〉〈στ〉〈τ〉

〈σ, τ〉

22

2 2

2

2

..................

..................

..................

..................

..................

..

............................................................................................

............................................................................................

............................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

...

..................................................................

and here the diagram of fixed subfields:

K

Q(√

3) Q(√

6) Q(√

2)

Q

22

2 2

2

2

..................

..................

..................

..................

..................

..

............................................................................................

............................................................................................

............................................................................................

.......

.......

.......

.......

.......

.......

.......

.......

.......

...

..................................................................

The Fundamental Theorem of Galois Theory establishes a bijection be-tween the set of intermediate subfields in a Galois extension K/F and theset 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 infi-nite objects), while the second set is a finite discrete set, more suitable fordirect 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 Galoisextension with Galois group G = Gal(K/F ). Then the maps

Subfields of K containing F ϕ−−→ Subgroups of GE 7→ Gal(K/E),

and

Subgroups of G ψ−−→ Subfields of K containing FH 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/Fis a Galois extension if and only if H is a normal subgroup of G. In thiscase, Gal(E/F ) is isomorphic to G/H.

Proof. We already know that if H1 and H2 are subgroups of G and KH1 =KH2 , 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], andhence K is a splitting field too of f(X) over E. Therefore, K/E is a Galoisextension. On the other hand, the extension K/E is a Galois extension if andonly 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 abijection and ϕ = ψ−1.

Moreover, since K/E is a Galois extension, we get [K : E] = |H|, withH = Gal(K/E). But

|G| =

[G : H]|H|,[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 theequivalence 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 overF 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 ∈ Gand 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, τ

(σ(γ)

)=

σ(σ−1τσ)(γ) = σ(γ), because σ−1τσ is in H. Hence σ(γ) ∈ KH = E andσ(E) = E.


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

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 anyautomorphism τ : E → E can be extended to K because K is a splittingfield over F . By the First Isomorphism Theorem G/H is isomorphic toGal(E/F ) and |Gal(E/F )| = [G : H] = [E : F ]. We conclude that E/Fis a Galois extension. (Alternatively, for any α ∈ EGal(E/F ) and σ ∈ G,σ(α) = σ|E(α) = α, so α ∈ KG = F . Hence F = EGal(E/F ), and E/F is aGalois extension.)

5.3 Corollary. Let K/F be a Galois extension. Then there is a finitenumber 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 multi-plicative group F×. Then G is cyclic.

Proof. G is a finite abelian group, and hence G is isomorphic to a directproduct of cyclic groups G ' Cm1 × · · · × Cmr , with m1 | m2 | · · · | mr.Therefore any x ∈ G is a root of the polynomial Xmr − 1. But F containsat 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 ndenotes 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 thereexists 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 groupof order n. (Notation: F = GF (pn).)

Proof. If F is a field with pn elements, then F× is a cyclic group of orderpn−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 polynomialf(X) = Xpn − X over Fp (note that the derivative of this polynomial is

§ 7. PRIMITIVE ELEMENTS 59

−1, and hence gcd(f(X), f ′(X)) = 1 and it contains no multiple roots). LetF = x ∈ E : f(x) = 0. Then since f(X) has no multiple roots, |F | = pn.But F is a field, because (a+ b)p

n= ap

n+ bp

n, (ab)p

n= ap

nbp

nand, if b 6= 0,

(bpn)−1 = (b−1)p

nfor 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 anautomorphism (called the Frobenius automorphism). Since F× is cyclic,F× = 〈α〉 for some 0 6= α ∈ F , with αp

n−1 = 1 and αi 6= 1 for anyi < pn − 1. Then σn(α) = αp

n= 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) = 〈σ〉 is cyclic.

6.3 Remark.

(i) By the Fundamental Theorem of Galois Theory, the subfields ofGF (pn)are in one-to-one correspondence with the subgroups of Gal(F/Fp) =〈σ〉, and hence with the set of divisors of n. For any divisor d of n,〈σd〉 is a subgroup of 〈σ〉 = Gal(F/Fp) and the fixed subfield by σd is

β ∈ F : βpd

= β = GF (pd). Thus the set of subfields of GF (pn) isGF (pd) : d |n.

(ii) If α is a generator of the cyclic group GF (pn)×, then GF (pn) = Fp(α)and degmα,Fp(X) = n. In particular, for any n ∈ N there is anirreducible 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 anelement α ∈ K such that K = F (α). In this case, the element α is said tobe 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 ifand 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 thepolynomial mα,E(X) = a0 +a1X+ · · ·+arXr+Xr+1 divides f(X) (in E[X],and hence in K[X]). Consider the subfield E′ = F (a0, a1, . . . , ar). Then F ⊆E′ ⊆ E ⊆ K and mα,E(X) |mα,E′(X) in E[X], but mα,E(X) ∈ E′[X], andit is irreducible over E, and hence over E′, so equality follows: mα,E(X) =mα,E′(X). But then we have [K : E] = [E(α) : E] = degmα,E(X) = [K :E′] and E = E′.


Therefore the intermediate subfields are generated over F by the coeffi-cients of monic factors of f(X), and there are only a finite number of suchfactors.

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 anyα, β ∈ K, there is an element c ∈ F such that F (α, β) = F (α+cβ). Since Fis infinite and there is only a finite number of intermediate subfields, thereare elements c1 6= c2 in F such that F (α + c1β) = F (α + c2β), but thenα+ c1β, α+ c2β ∈ F (α+ c2β), and it follows that F (α+ c2β) contains bothα and β, so F (α+ c2β) = F (α, β).

7.3 Primitive Element Theorem. Let K/F be a finite separable fieldextension. 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 henceit 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 betweenF and E is finite. Hence so is the number of intermediate fields between Fand K and Steinitz’s Theorem applies.

§ 8. Ruler and compass constructions

Recall from Algebraic Structures that an element x ∈ R≥0 is said to beconstructible 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 isconstructible if so is −x.

It was shown that the set C = x ∈ R : x is constructible is a subfieldof 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 ∪ 0and 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 extensionwhose 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 someα ∈ E with mα,K(X) = X2 − µ for some µ ∈ K. Write G = Gal(K/F ).The Primitive Element Theorem shows that there is an element β ∈ Ksuch that K = F (β). Write mβ,F (X) = f(X). Let µ1, . . . , µm be the dif-ferent elements in the set σ(µ) : σ ∈ G, and consider the polynomialg(X) =

∏mj=1

(X2 − µj

). This is a polynomial in F [X], because its coeffi-

cients are fixed by the elements in G (which permute the µj ’s). Let L bea splitting field over F of the separable polynomial lcm

(f(X), g(X)

)con-

taining E. Hence L/F is a Galois extension. Let α1 = α, α2, . . . , αr bethe 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 extensionK(α1, . . . , αi)/K(α1, . . . , αi−1) is ≤ 2. Hence [L : K] is a power of 2, and sois [L : F ] = [L : K][K : F ].

8.3 Theorem. Let α be a real number algebraic over Q, and let K be asplitting 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 forany i = 1, . . . , n, and α ∈ Fn. An inductive argument using the previouslemma 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 asplitting field of mα,Q(X) contained in L. The uniqueness of splitting fieldsgive the result.

Conversely, assume that α belongs to K, which may be assumed to becontained in C, and K/Q is a Galois extension of degree a power of 2. ThenG = 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:

Q = K0 = KGr ≤ K1 = KGr−1 ≤ · · · ≤ Kr = K = KG0 ,

such that [Kj : Kj−1] = 2 for any j = 1, . . . , r. The problem is that for anyj, Kj is contained in C, and not necessarily in R. Write Kj = Kj−1(aj+bji),with (aj + bji)

2 ∈ Kj−1. For each j consider the field Fj generated over Qby the real and imaginary parts of the elements of Kj . It is easy to checkthat F0 = Q and Fj = Fj−1(aj , bj) for any j ≥ 1. Note that α ∈ R ∩K, sothat α ∈ Fr.

But (aj + bji)2 ∈ Kj−1, which implies a2

j − b2j , ajbj ∈ Fj−1, and hence wehave two extensions

Fj−1 ≤ Fj−1(a2j + b2j ) ≤ Fj−1(aj , bj),


whose degrees are either 1 or 2, because (a2j+b2j )

2 = (a2j−b2j )2+(2ajbj)

2 is inFj−1, and if, for instance, aj 6= 0, then bj ∈ Fj−1(aj) (as ajbj ∈ Fj−1), anda2j = 1

2(a2j + b2j ) + 1

2(a2j − b2j ) ∈ Fj−1(a2

j + b2j ). (The same argument applies ifbj 6= 0.) Therefore the tower of subfields Q = F0 ≤ · · · ≤ Fr can be refinedto 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 overF , 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/Fis a Galois extension.

9.1 Proposition. Let F be a field and let f(X) ∈ F [X] be a polynomialof degree ≥ 1. Let E be a splitting field of f(X) over F . Consider thecomplete factorization f(X) = af1(X)n1 · · · fr(X)nr , where 0 6= a is theleading coefficient and f1(X), . . . , fr(X) are the different monic irreduciblefactors of f(X). Denote by Ωi the set of roots of fi(X) in E, for anyi = 1, . . . , r, and by Ω = Ω1 ∪ · · · ∪ Ωr (disjoint union) the set of roots off(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 anyσ ∈ Gal(E/F ). Hence σ(α) ∈ Ωi too. On the other hand, if an elementσ ∈ 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 asplitting field of f(X) over F , and hence this τ extends to an automorphismσ of E. Hence there is an element σ ∈ Gal(E/F ) such that σ(α) = β.

9.2 Example. For the polynomial f(X) = (X2 − 2)(X2 − 3) ∈ Q[X], thesplitting field is E = Q(

√2,√

3), and the orbits in the proposition aboveare 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 groupsof 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 differentunknowns. The elements s1, . . . , sn ∈ F (x1, . . . , xn) defined by:

s1 = x1 + · · ·+ xn,

s2 =∑

1≤i<j≤nxixj ,

s3 =∑

1≤i<j<k≤nxixjxk,

...

sn = x1x2 · · ·xn,

are called the elementary symmetric polynomials.The polynomial pn(X) = (X−x1)(X−x2) · · · (X−xn) ∈ F (x1, . . . , xn)[X]

is said to be the general polynomial of degree n.

9.4 Remark. Note that

pn(X) = (X − x1)(X − x2) · · · (X − xn)

= Xn − s1Xn−1 + s2X

n−2 − · · ·+ (−1)nsn ∈ 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 iso-

morphic 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, . . . , ni 7→ j if σ(xi) = xj ,

is a one-to-one group homomorphism. But for any τ ∈ Sn we may definethe automorphism:

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

f(x1, . . . , xn) 7→ f(xτ(1), . . . , xτ(n)).

Then τ(si) = si for any i, so we have τ ∈ 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 irreducibleover F (s1, . . . , sn).

Proof. Just notice that Sn acts transitively on 1, . . . , n.


9.7 Definition. A rational function f(x1, . . . , xn) ∈ F (x1, . . . , xn) is saidto be symmetric if for any τ ∈ Sn, f(x1, . . . , xn) = f(xτ(1), . . . , xτ(n)).

9.8 Example.((x1−x2)(x1−x3)(x2−x3)

)2is symmetric in F (x1, x2, x3).

9.9 Proposition. A rational function f(x1, . . . , xn) ∈ F (x1, . . . , xn) is sym-metric 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 thefixed 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 willbe assumed to be 6= 2.

Note that the discriminant of the general polynomial is

D =( ∏

1≤i<j≤n(xi − xj)

)2.

We will identify any σ ∈ Sn with the corresponding element in the Galoisgroup Gal

(F (x1, . . . , xn)/F (s1, . . . , sn)

). Write ∆ =

∏1≤i<j≤n(xi − xj).

Then for any σ ∈ Sn, σ(∆) = ±∆, and we have.

σ(∆) = ∆ ⇐⇒ σ ∈ An.

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

∏1≤i<j≤n(αi − αj), and ∆2 = D

(f(X)

)). Let G = Gal(E/F ) be

the associated Galois group, which will be identified with a subgroup of Snthrough its action on the roots of f(X), and let H be the intersection of Gwith the alternating group An: H = G ∩An.

Then we have EH = F (∆), so ∆ is in F if and only if G is containedin An.

Proof. For any σ ∈ G, σ(∆) = ∆ if and only if σ ∈ H. Thus we getGal(E/F (∆)

)= H, and EH = F (∆), because E/F is a Galois extension.

Galois groups of degree 2 polynomials.

If f(X) = X2 + bX + c is a monic degree 2 polynomial over a field F ,then D = D

(f(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. ThenG . S2 ' C2, so either G = 1 and ∆ ∈ F , or G is the cyclic group of order2, ∆ 6∈ F and E = F (∆).


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

Let f(X) = X3 + aX2 + bX + c be a monic degree 3 polynomial overa field F of characteristic 6= 2, 3. Let E be a splitting field of f(X) over Fand let G be the associated Galois group.

If f(X) is reducible, then either it is a product of polynomials of degree1 over F , and hence E = F and G = 1, or f(X) is a product of a polynomialof degree 1 and an irreducible polynomial of degree 2. The considerationsabove 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 charac-teristic being 6= 3 assures that f(X) is separable, so there are three differentroots α1, α2, α3 ∈ E with f(X) = (X − α1)(X − α2)(X − α3). The Galoisgroup G is, up to isomorphism, a subgroup of S3 which acts transitively onthese roots. Therefore either G = A3 or G = S3 and by the previous propo-

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:

D(f(X)

)= −

∣∣∣∣∣∣∣∣∣∣1 a b c 00 1 a b c3 2a b 0 00 3 2a b 00 0 3 2a b

∣∣∣∣∣∣∣∣∣∣= −

∣∣∣∣∣∣∣∣∣∣1 a b c 00 1 a b c0 −a −2b −3c 00 0 −a −2b −3c0 0 3 2a b

∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣1 a b c0 a2 − 2b ab− 3c ac0 a 2b 3c0 3 2a b

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣a2 − 2b ab− 3c aca 2b 3c3 2a b

∣∣∣∣∣∣= a2b2 − 4b3 − 4a3c− 27c2 + 18abc.

Recall from Algebraic Structures that in order to get the roots, we mayproceed as follows. By means of the change of variables given by Y = X+ 1

3a,we are left with a degree 3 equation of the form

Y 3 + pY + q = 0,

with p = 13(3b− a2) and q = 1

27(2a3 − 9ab+ 27c). Write Y = u+ v to get

u3 + v3 + (3uv + p)(u+ v) + q = 0,

which, with uv = −p3 , gives u3 + v3 = −q,

uv = −p3.

And this allows us to compute u3 and v3, and hence u and v, by solving adegree 2 equation. Let ω be a primitive cubic root of 1 in an extension field


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. Wethus 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 polynomialover a field F of characteristic 6= 2, 3. Let E be a splitting field of f(X) overF and let G be the associated Galois group.

To simplify later computations, we can always perform the change ofvariables Y = X + 1

4a, and hence assume that our polynomial is f(X) =X4 + pX2 + qX + r.

If f(X) is reducible, the computation ofG reduces to the computation forlower 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 beidentified, through its action on the roots, with a subgroup of S4. ConsiderKlein’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. LetH = 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 symmetricpolynomials on the θi’s are in EG = F . Our hypotheses imply that α1 +α2 + α3 + α4 = 0 and a few easy computations yield:

θ1 + θ2 + θ3 = 2∑i<j

αiαj = 2p,


θ1θ2+θ1θ3 + θ2θ3

=(

(α1 + α2)(α1 + α3))2

+(

(α1 + α3)(α1 + α4))2

+(

(α1 + α4)(α1 + α2))2

=(α2

1 + α1(α2 + α3) + α2α3

)2+(α2

1 + α1(α3 + α4) + α3α4

)2+(α2

1 + α1(α2 + α4) + α2α4

)2=(α2α3 − α1α4

)2+(α3α4 − α1α2

)2+(α2α4 − α1α3

)2=

∑1≤i<j≤4

(αiαj)2 − 6α1α2α3α4 =

∑1≤i<j≤4

(αiαj)2 − 6r

=( ∑

1≤i<j≤4

αiαj)2 − 2

4∑i=1

α2i

( ∑1≤j≤k<4j,k 6=i

αjαk)− 12r

= p2 − 2

4∑i=1

(αi

∑1≤j≤k<4j,k 6=i

αiαjαk

)− 12r

= p2 − 2

4∑i=1

(αiq − r)− 12r

= p2 − 2( 4∑i=1

αi)q − 4r = p2 − 4r,

θ1θ2θ3 = −(

(α1 + α2)(α1 + α3)(α1 + α4))2

= −(α3

1 + α21(α2 + α3 + α4) + α1(α2α3 + α2α4 + α3α4) + α2α3α4

)2

= −q2.

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

h(X) = (X − θ1)(X − θ2)(X − θ3) = X3 − 2pX2 + (p2 − 4r)X + q2,

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:

D(f(X)

)=( ∏

1≤i<j≤4

(αi − αj))2

=( ∏

1≤i<j≤3

(θi − θj))2

= D(h(X)

),

and ∆ =∏

1≤i<j≤4(αi − αj) = −∏

1≤i<j≤3(θi − θj).We conclude that θ1, θ2, θ3 are different elements and now it is easy to

check that for any σ ∈ G, σ(θi) = θi for any i = 1, 2, 3 if and only ifσ ∈ H = G ∩ V . (Note that σ(θi) = θi for any i implies σ(∆) = ∆, so thatσ ∈ A4 = V ∪ (1 2 3)V ∪ (1 3 2)V , but (1 2 3)θ1 = θ3, ....)


Therefore we have F (∆) = EG∩A4 and F (θ1, θ2, θ3) = EG∩V :

E⋃| [E : E1] divides 4 = |V |

E1 = F (θ1, θ2, θ3)⋃| [E1 : E2] divides 3 = [A4 : V ]

E2 = F (∆)⋃| [E2 : F ] divides 2 = [S4 : A4]

F

We are left with several possibilities:

• If h(X) is irreducible, then |G| = [E : F ] is a multiple of both 3 sinceE contains the subfield F (θ1) and [F (θ1) : F ] = deg h(X) = 3, and4 since [F (α1) : F ] = 4. Hence |G| is a multiple of 12, and G is asubgroup of S4, whose order is 24. The only subgroup of S4 of order12 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, andG = 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 (asD4 ∩ A4 = V ), so we get ∆ 6∈ F . Therefore, G is a group whoseorder is a multiple of 4, contained in D4, different from V and whichacts transitively on 1, 2, 3, 4. There are only two possibilities: eitherG = D4 or G = 〈(1 3 2 4)〉 = C4. In the first case G∩A4 = G∩V = Vand the action of Gal

(E/F (∆)

)= V on the roots is transitive, so

that f(X) is irreducible over F (∆). In the second case the action ofGal(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

To obtain the roots of the equation X4 + pX2 + 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 =√−θ1 and α3+α4 = −

√−θ1. In a similar vein, α1+α3 =√

−θ2 and α2 + α4 = −√−θ2, and α1 + α4 =

√−θ3 and α2 + α3 = −

√−θ3.

Besides,√−θ1

√−θ2

√−θ3 = (α1+α2)(α1+α3)(α1+α4) = α3

1+α21(α2+α3+

α4) +∑

1≤i<j<k≤4 αiαjαk = −q (recall α1 + α2 + α3 + α4 = 0), and hence

once we have chosen the square roots√−θ1 and

√−θ2, the value (among

the two possible ones) of√−θ3 is completely determined. Using again that

α1 + α2 + α3 + α4 = 0, we get

2α1 = (α1 + α2) + (α1 + α3) + (α1 + α4) =√−θ1 +

√−θ2 +

√−θ3,

and, in a similar vein,

2α1 =√−θ1 +

√−θ2 +

√−θ3,

2α2 =√−θ1 −

√−θ2 −

√−θ3,

2α3 = −√−θ1 +

√−θ2 −

√−θ3,

2α4 = −√−θ1 −

√−θ2 +

√−θ3.

9.11 Remark. The group S5 is not solvable, so we do not have a chain ofnormal 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 ob-tained 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 elementα ∈ K and a natural number n ∈ N such that K = F (α) and αn ∈ 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 . Thenf(X) is said to be solvable by radicals if there exists a ‘tower’ of fieldextensions:

F = K0 ⊆ K1 ⊆ K2 ⊆ · · · ⊆ Kr = K

such that each Ki/Ki−1 is purely radical (i = 1, . . . , r) and there areα1, . . . , αs ∈ K with f(X) = a(X − α1) · · · (X − αs). The extensionK/F is said to be a radical extension.


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

i ∈F (γ1, . . . , γi−1).

10.3 Lemma. Let F be a field and let µn be the set of nth roots of unity inF : µn = α ∈ F : αn = 1. Then µn is a cyclic group whose order dividesn.

Proof. It is clear that µn is a finite subgroup of F×, hence it is cyclic (Lemma6.1). If γ is a generator of µn, then γn = 1, so that the order of µn, whichequals the order of γ, divides n.

In the event that |µn| = n, the generators of µn are called the primitiventh roots of 1. Note that if µn = 〈ζ〉 with |ζ| = n, then ζi is a primitive nth

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 polynomialXn − 1 ∈ F [X] has n different roots (in a splitting field over F ) if and onlyif the characteristic of F does not divide n.

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

10.5 Proposition. Let F be a field, n ∈ N, and let ζ be a primitive nth rootof 1 in a field extension of F (so that the characteristic does not divide n).Then F (ζ)/F is a Galois extension and Gal

(F (ζ)/F

)is an abelian group.

Proof. Since ζ ∈ F (ζ) and the roots of Xn−1 are the powers of ζ, it followsthat 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 ∈ Gal(F (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 character-istic of F does not divide n and such that F contains all the nth roots ofunity (that is, |µn| = n). Let K/F be a finite field extension. Then K/F isa Galois extension with Gal(K/F ) a cyclic group whose order is a divisor ofn if and only if there is an element α ∈ K such that K = F (α) and αn ∈ F .


Proof. If K = F (α) and αn = a ∈ F , then either a = 0 and hence K = Fand Gal(K/F ) is trivial, or K/F is a splitting field over F of the separablepolynomial 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/nZσ 7→ ı such that σ(α) = ζiα,

is a one-to-one group homomorphism. Therefore Gal(K/F ) is isomorphicto a subgroup of the cyclic group of order n, and it is thus cyclic of order adivisor of n.

Conversely, assume that K/F is a Galois field extension whose Galoisgroup 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), thereis an element α ∈ K such that (α, ξ) 6= 0. Write β = (α, ξ). Then forany i, σi(β) = ξ−iβ, so that β is not fixed under any nontrivial elementin Gal(K/F ). Therefore β does not belong to any proper subfield of Kcontaining 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 interme-diate 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 ⊆ Kand 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):

To prove this note that K = F (u1, . . . , ur) with umii ∈ 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 ). Weget the containments F ⊆ E ⊆ K ⊆ K.


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 anisomorphism σ : K → K. Then F ⊆ σ(K) ⊆ K and v ∈ σ(K). Moreover,σ being an isomorphism, we have vmj = σ(umj ) ∈ F

(σ(u1), . . . , σ(uj−1)

).

Then with Gal(K/F ) = σ1 = 1, σ2, . . . , σm, the field K is generated overF 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 re-striction 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 asplitting field, any τ ∈ Gal(E/F ) extends to a σ ∈ Gal(K/F ). Hence thismap is onto. If Gal(K/F ) is solvable, so is Gal(E/F ), so it is enough toprove that Gal(K/F ) is solvable. Applying again the reduction 1), we mayassume that K/F is a radical Galois extension.

4) Assume then that K/F is a radical Galois extension. Without loss ofgenerality we may assume too that there are elements r ∈ N, u1, . . . , ur ∈ Kand prime numbers m1, . . . ,mr such that K = F (u1, . . . , ur) and umi

i ∈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).)

Also note that if some mi equals the characteristic of F , then umii ∈

Ki−1 = F (u1, . . . , ui−1), so that mui,Ki−1(X) divides Xmi − umii = (X −

ui)mi . 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 andwe can get rid of ui in the expression K = F (u1, . . . , ur). Hence we mayassume that none of the prime numbers mi equals the characteristic of F .

Take now m = m1 · · ·mr and ζ a primitive mth root of 1 in a fieldextension of K. Take K = K(ζ) = F (ζ, u1, . . . , ur). The extension K/F isagain radical. Moreover, K is the splitting field of a separable polynomialf(X) ∈ F [X] over F , and hence K is a splitting field of the separablepolynomial lcm

(Xm − 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 maychange K to K and assume ζ ∈ K.

Thus, we assume K = F (ζ, u1, . . . , ur), with umii ∈ F (ζ, u1, . . . , ui−1),

the mi’s being prime numbers different from the characteristic. Take K0 =


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

Kr = K⋃|

...⋃|

K1 = F (ζ, u1)⋃|

K0 = F (ζ)⋃|

F

Galois−→

correspondence

1⋂|

...⋂|

G1 = Gal(K/K1)⋂|

G0 = Gal(K/K0)⋂|

G = Gal(K/F )

But K0/F is a Galois extension, and hence G0 is a normal subgroup ofG with G/G0 ' Gal(K0/F ), which is abelian by Proposition 10.5. Sinceζ ∈ Ki−1 for any i ≥ 1, the extension Ki/Ki−1 is a cyclic Galois exten-sion (Proposition 10.6), and hence Gi is a normal subgroup of Gi−1 andGi−1/Gi ' Gal(Ki/Ki−1) is a cyclic group.

We then have the descending series of subgroupsG 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 fieldF 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 solvableby 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 generalpolynomial of degree n ≥ 5 is not solvable by radicals.

For a more concrete example, take f(X) = X5 − 4X + 2 ∈ Q[X]. Thenf(X) is irreducible (Eisenstein’s Criterion). Let E be the splitting field off(X) with Q ⊆ E ⊆ C, and let G = Gal(E/Q). Let us check that G is notsolvable.

The derivative f ′(X) = 5X4 − 4, considered as a function of a realvariable, grows for x ≥ 0 and decreases for x ≤ 0. It follows then easilythat f(X) has at most three real roots. But f(−1) > 0 and f(1) < 0, whilelimx→−∞ f(x) = −∞, limx→∞ =∞. It follows that f(X) has exactly threedifferent real roots: α1, α2, α3, and two non real complex roots: α4, α5 = α4.

By irreducibility of f(X), 5 = [Q(α1) : Q] divides |G|, and hence Gcontains an element of order 5. Identify G with a subgroup of S5 throughits 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 givesa transposition in G. But an arbitrary cycle of length 5 and an arbitrarytransposition generate S5 (you have essentially proved this as an exercise inChapter 1). Hence G is the whole S5, which is not solvable.


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 isa 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 amultiple of the characteristic) in some extension of E. Now E is a splittingfield of some separable polynomial f(X) ∈ F [X] over F , so that E(ζ) is asplitting field of the separable polynomial lcm

(f(X), X [E:F ]−1

)over F , and

hence the extension E(ζ)/F is a Galois extension, and so is the extensionE(ζ)/F (ζ). Note also that the map

Gal(E(ζ)/F (ζ)

)−→ Gal(E/F )

σ 7→ σ|E ,

is a one-to-one group homomorphism, which shows that Gal(E(ζ)/F (ζ)

)is

solvable and [E(ζ) : F (ζ)] divides [E : F ], so it is not a multiple of thecharacteristic. Finally, if K is an extension of E(ζ) such that K/F (ζ) isradical, then K/F is radical too. (Note that ζ [E:F ]/[E(ζ):F (ζ)] is a primitiveroot of unity of order [E(ζ) : F (ζ)].)

2) Assuming that F contains the roots of unity of order [E : F ], takeG = Gal(E/F ). Since G is solvable, there is a normal subgroup H ofG such that [G : H] is a prime number p (just take for H any maximalsubgroup containing G′). Using an inductive argument, since E/EH is aGalois 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, whichimplies 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 afield F , and let E be a splitting field of f(X) over F . Then if p is a primefactor 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) inF (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 fieldF whose characteristic does not divide n! (that is, either the characteristicis 0 or greater than n). Then f(X) is solvable by radicals if and only if itsGalois group is solvable.

Proof. If f(X) is solvable by radicals, then we already know that its Galoisgroup is solvable (Corollary 10.8).

Conversely, assume that the Galois group of f(X) is solvable, and letE be a splitting field of f(X) over F . Take a complete factorization off(X): f(X) = ag1(X)r1 · · · gm(X)rm , with 0 6= a ∈ F and the gi(X)’sdifferent monic irreducible polynomials. Because of our restrictions on thecharacteristic g′i(X) 6= 0 for any i, so that the gi(X)’s are separable, andE 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 givesthe 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 thepolynomial f(X) = Xp−X−t has no multiple roots, and that if α is aroot 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 followingassertions 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 fieldF : X2 + 2X−1, X3 + 1, and X6 +X5 +X4 +X3 +X2 +X+ 1, whereF 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 ofthe 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 ofeach of these polynomials:

• X2 − 2,

• X3 − 2,


• 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 bea splitting field of f(X) over R.

(c) Prove that K(i)/R is a Galois extension, where i denotes a rootof X2 + 1 in some extension of K.

(d) Write [K(i) : R] = 2nm for n ≥ 0 and odd m. Prove that m = 1by considering the fixed field of a Sylow 2-subgroup of the Galoisgroup.

(e) Prove that n = 1 by arguing with index 2 subgroups of the Galoisgroup.

(f) Conclude that f(X) splits over C.

8. Prove that an arbitrary finite field has irreducible polynomials of ar-bitrary degree.

9. Prove that the polynomial Xpn−X is the product of all the irreduciblemonic 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 irreduciblepolynomial. Then prove the following assertions:

(a) D(f(X)) > 0 if and only if the three roots of f(X) are real.

(b) D(f(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 degree3 polynomial whose discriminant is a square in F . Prove that eitherf(X) is irreducible or f(X) splits in F .

EXERCISES 77

12. Prove that over any field F either the cubic polynomial X3 − 3X + 1is 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. Provethe 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 , thenG ' C4.

(c) If neither c nor c(b2 − 4c) are squares in F , then G is isomorphicto 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)where ζ1, . . . , ζr are the primitive nth roots of unity in C. Φn(X) iscalled the nth cyclotomic polynomial. Prove the following assertions:

(a) Xn − 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

primitive nth root of unity, and Gal(Q(ζ)/Q

)'(Z/nZ

)×.

16. Prove that if n ≥ 3, the regular polygon of n vertices is constructiblewith ruler and compass if and only if n = 2mp1 · · · pr, where m, r ≥ 0and p1, . . . , pr are Fermat primes (that is, of the form 22m + 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 thatthere is a primitive root ζ and a prime number p which does not divide n such thatf(ζ) = 0 6= f(ζp). Conclude that f(X) divides g(Xp). Now reduce modulo p (and usebars to denote the reductions) and show that Φn(X) has multiple roots. Get from here acontradiction.

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 normalsubgroup 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 andn ∈ N. Prove that there exists a chain of subgroups 1 = G0 ≤G1 ≤ · · · ≤ Gn = G, such that for each i = 1, . . . , n, Gi is anormal subgroup of G and |Gi| = pi. (7 points)

3. Consider the polynomial f(X) = X4 − 2X2 − 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 unityζ = e2πi/n inside C, and the element θ = ζ + ζ−1.

(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 =e2πi/n in C, and the subgroup D of the general linear group of degree2 over C generated by the elements

x =

(ζ 00 ζ−1

), y =

(0 1−1 0

).

(a) Show that 〈x〉 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 groupDn.

2. (a) Prove that any group of order 400 is solvable.

(b) Give an example of a solvable but not nilpotent group of order400.

3. Consider the polynomial f(X) = X4 +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 com-

pute 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 notsufficient 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.

3. Consider the polynomial f(X) = X5 − 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) overF2.

(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 anexample to show that this fails for |N | = 3.

(b) Assume that |G| is odd, and that |N | = 3. Prove that N is centraltoo in this case.

2. (a) Prove that if p and q are different prime numbers, any group oforder pq or p2q is solvable.

(b) How many groups of order 62 are there, up to isomorphism?

3. Consider the polynomial f(X) = X4 − 3 ∈ Q[X].

(a) Prove that f(X) is irreducible. (3 points)

(b) Compute its Galois group. (7 points)

4. (a) Prove that the roots of the polynomial X3 − 3X + 1 ∈ Q[X] areall 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 = 〈x, y, z | x2 = y2 = z2 = 1, xy = yx, xz = zx, yz = zxy〉.

(a) What is the order of G?

(b) Give an explicit isomorphism from G into a well-known group.

3. How many subfields does Q( 4√

3) contain? Why?

4. Let K be a splitting field of X15 − 1 over F2. Compute [K : F2] andGal(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 〈P1, P2〉 = G and P1 ∩ P2 ≤ Z(G).

(b) Prove that G is solvable.

2. Let O(n) be the orthogonal group:

O(n) =A ∈ Matn(R) : AtA = I

,

and consider its natural action on Rn (identified with the set of n× 1matrices).

(a) What are the orbits of this action?

(b) What is the stabilizer of

(10...0

)?

3. Let K be the splitting field of X8 − 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 rootof f(X) in a field extension K of F2, then the cyclic group generatedby 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 givea precise description of them.

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

(c) Prove that any subgroup of order 6 of S4 is the normalizer of asubgroup of order 3.

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

2. Let G be a group of order 8×pn, where p is an odd prime number andn ∈ 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 homo-morphism G→ S4 with solvable kernel.

(c) If p = 7, prove that either G contains a unique Sylow 7-subgroupor 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 polynomialX4−4X3+6X2−4X−1 ∈Q[X].

4. Consider the field F7 of 7 elements.

(a) Compute the cubes of the elements in F7.

(b) Find a monic polynomial f(X) ∈ F7[X] such that GF (73) =F7[X]/

(f(x)

).

(c) Describe explicitly the elements in Gal(GF (73)/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 subgroupof G, and let x ∈ G be an element of order p. Prove the followingassertions:

(a) N ∩ Z(G) 6= 1.

(b) 〈x〉 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 order80.

3. Let K be a splitting field of f(X) = X4 +X2 − 1 over Q.

(a) Compute the degree [K : Q] and give a basis of K as a vectorspace over Q.

(b) What is the Galois group of f(X) over Q? Describe its elements.

4. (a) Show that the polynomial X4 + 1 is irreducible over Q.

(b) Show that, on the other hand, X4 + 1 is reducible over F2, F3,and F5.

(c) Show that, in general, X4 + 1 splits in GF (p2) for any primenumber p, and deduce that X4 + 1 is reducible over Fp.

groups and galois theorypersonal.unizar.es/elduque/files/groupsgalois2019.pdfvi syllabus some more...

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