algebra 1

Algebra I - Lecture ScriptProf. Ozlem Imamoglu

Mitschrift von Manuela Dubendorfer

October 19, 2008

Contents

1 Groups 11.1 Basic Definitions and Examples . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Diedral groups . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.2 Quaternion group . . . . . . . . . . . . . . . . . . . . . . . . 51.1.3 Symmetric group . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Homomorphisms and Isomorphisms . . . . . . . . . . . . . . . . . . 61.3 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3.1 Special subgroups . . . . . . . . . . . . . . . . . . . . . . . . 81.3.2 Cyclic groups an their subgroups . . . . . . . . . . . . . . . 10

1.4 Group Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.5 Normal subgroups and Quotient groups . . . . . . . . . . . . . . . . 201.6 Homomorphisms and Normal subgroups . . . . . . . . . . . . . . . 221.7 Groups acting on themselves by conjugation - class equation . . . . 271.8 Composition series and the Holder program . . . . . . . . . . . . . 301.9 Sylow’s Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321.10 Direct products and Abelian groups . . . . . . . . . . . . . . . . . . 37

1.10.1 The fundamental theorem of finitely generated abelian groups 391.10.2 Semidirect products . . . . . . . . . . . . . . . . . . . . . . . 41

1.11 Free groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2 Rings 452.1 Basic Definitions and examples . . . . . . . . . . . . . . . . . . . . 452.2 Ideals, Ring Homomorphisms and quotient rings . . . . . . . . . . . 492.3 Properties of ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . 552.4 Rings of fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592.5 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

2.5.1 Irreducible polynomials . . . . . . . . . . . . . . . . . . . . . 64

iii

iv CONTENTS

Chapter 1

Groups

1.1 Basic Definitions and Examples

Definition 1.1.1. 1) A group is an ordered pair (G, ∗) where G is a set and ∗ isa map

∗ : G×G→ G, (g, h) 7→ g ∗ h := g · h =: gh

satisfying the following axioms

(G1) For all a, b, c ∈ G, (a ∗ b) ∗ c = a ∗ (b ∗ c),(G2) There exists an element e ∈ G such that a = a∗e = e∗a ∀a ∈ G, e is calledthe identity,(G3) For all a ∈ G, there exists a−1 ∈ G such that a ∗ a−1 = a−1 ∗ a = e, a−1 iscalled the inverse of a.2) The group (G, ∗) is called abelian (or commutative) if a ∗ b = b ∗ a, for alla, b ∈ G.

Remark. (1) We’ll drop (G, ∗) and write just G.(2) We say G is a finite group if G is a finite set.(3) Note (G2) ensures that a group is always nonempty.(4) When the group G is abelian we typically write a+ b for ab and a− b for ab−1.

Example 1.1.2. Z,Q,R,C groups under +, with e = 0 and a−1 = −a for all a.

Example 1.1.3. Q\{0}, R\{0}, C\{0}, Q+, R+ groups under × with e =1, a−1 = 1

a, ∀a. Z\{0} is NOT a group under ×. (2 does not have an inverse for

example.)

Example 1.1.4. Any vector space (V,+) is an abelian group (The operation +is the vector addition).

1

2 CHAPTER 1. GROUPS

Example 1.1.5. Integers mod n : Z/nZLet n be a fixed positive integer. Define a relation on Z by a ∼ b if and only ifn|(b− a). We write a ≡ b ( mod n) if a ∼ b. a is congruent to b mod n.

∼ is an equivalence relation a ∼ a (reflexive), a ∼ b ⇒ b ∼ a (symmetric),a ∼ b, b ∼ c⇒ a ∼ c (transitive).

a = {a+ kn; k ∈ Z} = {a, a± n, a± 2n, . . . }

is the congruence class of a residue class of a. There are precisely n distinctequivalende classes mod n, namely 0, 1, 2, . . . , n− 1. The equivalence classesare called the integers modulo n, denoted by Z/nZ. We can define a sum andproduct by

a+ b := a+ b, a · b = ab.

Theorem 1.1.6. The operation of addition and multiplication on Z/nZ definedabove are both well defined, i.e. they do not depend on the choices of representativesfor the classes involved. More precisely:If a1, a2 ∈ Z, b1, b2 ∈ Z and a1 = b1, a2 = b2 then a1 + a2 = b1 + b2 and a1a2 =b1b2, i.e. if a1 ≡ b1 mod n and a2 ≡ b2 mod n then a1 + a2 ≡ b1 + b2 mod nand a1a2 ≡ b1b2 mod n.

Proof. Suppose a1 ≡ b1 mod n, i.e. a1−b1 = ns for some integer s. Similarly a2 ≡b2 mod n means a2− b2 = nt for some t ∈ Z. Then a1 + a2 = (b1 + b2) + (s+ t)n,i.e. a1 + a2 ≡ b1 + b2 mod n.

An important subset of Z/nZ consists of the collection of residue classes whichhave a multiplicative inverse in Z/nZ:

(Z/nZ)× = {a ∈ Z/nZ; ∃c ∈ Z/nZ with a− c = 1}

Proposition 1.1.7. (Z/nZ)× = {a ∈ Z/nZ; (a, n) = 1}.

Proof. Exercise.

Example 1.1.8. n = 4 : Z/4Z = {0, 1, 2, 3}, (Z/4Z)× = {1, 3}.

(Z/nZ,+) and ((Z/nZ)×,×) are groups.

Example 1.1.9. If (A, ◦), (B, ∗) are two groups, we can form a new group A×B,the direct product whose elements are

A×B = {(a, b); a ∈ A, b ∈ B}.

We define the operation componentwise

(a1, b1)(a2, b2) = (a1 ◦ a2, b1 ∗ b2).

1.1. BASIC DEFINITIONS AND EXAMPLES 3

Example 1.1.10. GLn(R) denotes the group of invertible n × n matrices undermatrix multiplication.

Example 1.1.11. Let X be a set, S(X) Permutations of X

S(X) = {f : X → X; f a bijection}

with composition as group oparation. When X = {1, . . . , n} we write Sn, theSymmetric group on n letters.

For any group G, a ∈ G, n ∈ Z+, we denote an by aa . . . a︸︷︷︸n times

, a−n by a−1 . . . a−1︸︷︷︸n times

.

a0 = 1 is the identity.

Proposition 1.1.12. Let (G, ∗) be a group then(1) The identity of G is unique,(2) For all a ∈ G, a−1 is unique,(3) (a−1)−1 = a, ∀a ∈ G,(4) (a ∗ b)−1 = (b−1) ∗ (a−1),(5) for any a1, . . . , an ∈ G, a1 ∗ · · · ∗ an is independent of how the expression isbracketed.

Proposition 1.1.13. Let G be a group, a, b ∈ G. The equations ax = b andya = b have unique solutions x, y ∈ G. In particular, the left and right cancelationlaws hold.(1) If au = av then u = v,(2) if ub = vb then u = v.

Definition 1.1.14. Let G a group, and x ∈ G. We define the order of x to be thesmallest positive integer n such that xn = 1. We denote this integer by |x|. If nopositive power of x is 1, x is said to be ∞ order.

Example 1.1.15. (1) a ∈ G has |a| = 1 if and only if a = 1.(2) (Z,+), (R; +): every non-zero element has ∞ order.(3) (R\{0},×): −1 has order 2, all other elements have ∞ order.(4) G = Z/6Z: 2 6= 0, 2 + 2 6= 0, 2 + 2 + 2 = 0, hence 2 has order 3 in Z/6Z.

Definition 1.1.16. Let G = {1 = g1, . . . , gn} be a finite group. The multiplicationtable or group table of G is the n× n matrix whose ij-entry is the group elementgigj.

Example 1.1.17. G = ((Z/3Z),+)+ 0 1 20 0 1 21 1 2 02 2 0 1

4 CHAPTER 1. GROUPS

There is another formulation of the axioms for a group, which is equivalent to theoriginal definition but somewhat simpler

Theorem 1.1.18. Let G be a set, ∗ an associative binary operation on G. Assumethat there exists an element e ∈ G such that x ∗ e = x for all x ∈ G and assumethat for any x ∈ G there exists an element y ∈ G such that x ∗ y = e. Then (G, ∗)is a group (e is called right identity, y is called right inverse)

Proof. We want to show that e is also a left identity and y is also left inverse ofx. Let x ∈ G, e ∗ e = e since x ∗ e = x, ∀x ∈ G. Let y be the right inverse of x,i.e. x ∗ y = e. Then x ∗ y = e = e ∗ e = e ∗ (x ∗ y) = (e ∗ x) ∗ y by association.x ∗ y = (e ∗ x) ∗ y. Now we use right of y to cancel y’s. x = e ∗ x, therefore e isalso a left identity. The other part is an exercise.

Note the following example

Example 1.1.19. (Z, ∗) where x ∗ y := x. Check: ∗ ist associative, 1 is a rightidentity x ∗ 1 = x, ∀x ∈ Z and 1 is a left inverse for every x ∈ Z, 1 ∗ x = 1. But(Z, ∗) is not a group since for example there is no two-sided identity element.

Now we will study some very important examples of groups

1.1.1 Diedral groups

One important family of examples of groups is the class of groups whose elementsare symmertries of geometric objects. The simplest subclass is when the geomet-ric objects are regular polygons. For each n ∈ Z+, n ≥ 3, let D2n be the setof symmetries of a regular n-gon. Fix a regular n-gon centered at the origin inthe x−y plane. Label the vertices consecutively from 1 to n in clockwise direction.

Let r denote the rotation clockwise about the origin through 2πn

and s the re-flection about the line of symmetry through vertex 1 and origin. Then one canverify(a) 1, r, r2, . . . , rn−1 all distinct and rn = 1, i.e. |r| = n,(b) |s| = 2,(c) s 6= ri for any i,(d) sri 6= srj ∀0 ≤ i, j ≤ n with i 6= j.

D2n = {1, r, r2, . . . , rn−1︸︷︷︸rotations

, s, sr, sr2, . . . , srn−1︸︷︷︸reflections

}

a group of order 2n. Every element can be written in terms of r and s. We sayr, s are generators. We’ll look at this example in more detail later. (Note alsosr = r−1s and hence D2n is not abelian).


1.1.2 Quaternion group

Let Q8 = {1,−1, i,−i, j,−j, k,−k} with the relations 1 · a = a · 1 = a, for alla ∈ Q8, (−1) · (−1) = 1, (−1) · a = a · (−1) = −a and

i · i = j · j = k · k = −1

ij = k, jk = i, ki = j, ji = −k, kj = −i, ik = −j.

Q8 is a non abelian group of order 8.

1.1.3 Symmetric group

Let X 6= ∅ be a set, and Sx be the set of all bijections from X to X. Then (Sx, ◦)is a group, where ◦ denotes the composition of functions, the permutations of X.When X = {1, 2, . . . , n}, then we write Sx = Sn, which is the symmetric group onn letters.

First note |Sn| = n! If σ is any permutation in Sn, it can send 1 to any of {1, . . . , n},hence there are n choices for σ(1), n−1 choices for σ(2), n−2 choices for σ(3), . . . .Therefore there are n! injective functions from {1, . . . , n} to itself.

An efficient way of writing elements of Sn is through their cycle decomposition.

Definition 1.1.20. A cycle is a string of integers which represents the elementsof Sn which cyclically permutes these integers and fixes all others.

r-cycle σ ∈ Sn when there exists r-element subset {x1, . . . , xr} ⊂ {1, . . . , n} suchthat(a) σ(xi) = xi+1, 1 ≤ i ≤ r,(b) σ(xr) = x1,(c) σ(y) = y ∀y ∈ {1, . . . , n}\{x1, . . . , xr}

We write (x1, x2, . . . , xr) for the r-cycle σ.

Example 1.1.21. σ = (15732) ∈ S8 is the permutation(1 2 3 4 5 6 7 85 1 2 4 7 6 3 8

)In general for every σ ∈ Sn, the numbers from 1 to n will be rearranged andgrouped into k-cycles

(a1 . . . am1)(am1+1am1+2 . . . am2) . . . (amk−1+1 . . . amk).

6 CHAPTER 1. GROUPS

Example 1.1.22. (153)(27)(68) ∈ S8(1 2 3 4 5 6 7 85 6 1 4 3 8 2 6

)We’ll study this group in more detail later.

Consider s = (123) ∈ S3, |S3| = 6 and r = (23) ∈ S3,

D6 are the symmetries of a triangle and has order 6, too. r can be represented byits effect on 3 corners as (123), s is a reflection which fixes 1 and swaps 2, 3. Infact one can show there is a bijection

φ : S3 → D6, (123) 7→ r, (23) 7→ s.

They look ”‘alike”’. Next we make this notion more precise.

1.2 Homomorphisms and Isomorphisms

Definition 1.2.1. Let (G1, ∗), (G2, ◦) be two groups. A map ϕ : G1 → G2 iscalled a homomorphism if

ϕ(x+ y) = ϕ(x) ◦ ϕ(y), ∀x, y ∈ G1.

We’ll typically not write the group operations explicity, ϕ(xy) = ϕ(x)ϕ(y),

ϕ is called an isomorphism and G1, G2 are called isomorphic (notation: G1∼= G2)

if ϕ is a bijection and ϕ is a homomorphism, i.e. G1, G2 are isomorphic if thereexists a bijection between them which preserves the group operations. (onto homo-morphism = epimorphism, 1-1 homomorphism = monomorphism).

Example 1.2.2. For any group G, G ∼= G; if G1∼= G2, G2

∼= G3 then G1∼= G3;

if G1∼= G2 then G2

∼= G1. Hence ∼= is an equivalence relation. The equivalenceclasses are called isomorphism classes.

Example 1.2.3. exp : R→ R+, x 7→ ex. ex+y = ex ·ey is a bijection with inverselog : R+ → R, hence (R,+) ∼= (R+,×).

Example 1.2.4. Let x, y be finite sets, non-empty. Then Sx ∼= Sy ⇒ |x| = |y|.

Example 1.2.5. We’ll be able to prove that any non-abelian group of order 6 isisomorphic to S3. Hence indeed D6

∼= S3.

1.3. SUBGROUPS 7

Example 1.2.6. φ : G→ G, g 7→ e ∀g ∈ G and ψ : G→ G, g 7→ g are trivialhomomomorphisms.

Example 1.2.7. φ : (Z,+) → (Z/nZ,⊕),m 7→ m = m mod n is a homo-morphism but not a monomorphism (not 1-1), because φ(0) = φ(n), but it is anepimorphism, it is onto.

Example 1.2.8. Sometimes it is easy to see that two groups are not isomorphic.For example S3 6∼= Z/6Z since Z/6Z is abelian and S3 is not.

Example 1.2.9. Fix g ∈ G, G a group. The map φg : G→ G, x 7→ gxg−1 is anisomorphism of G onto itself (automorphism). The map is called conjugation.

1.3 Subgroups

Definition 1.3.1. Let G be a group. H ⊆ G, a subset of G. H is called a subgroupof G, written H ≤ G, if H 6= ∅ and H is closed under products and inverses, i.e.a subset ∅ 6= h of a group (G, ∗) is a subgroup of G if the elements of H form agroup under ∗.

Note: e ∈ H : if x ∈ H then x−1 ∈ H and x−1x = e ∈ H since H is closedunder products and inverses. When we say H is a subgroup, we always meanthat the operation for H is the operation on G restricted to H. For example:(Q\{0},×) 6≤ (R,+) even though both are groups, Q\{0} ⊂ R as sets, but × 6= +.

Example 1.3.2. (Q,+) ≤ (R,+).

Example 1.3.3. (2Z,+) ≤ (Z,+).

Example 1.3.4. G a group, g ∈ G, H = 〈g〉 := {gn; n ∈ Z}, where g0 := 1, isa subgroup, called the cyclic subgroup generated by g. If 〈g〉 = G, G is called acyclic group. For example Z = 〈1〉.

Example 1.3.5. Let G be the group of real-valued functions on the real line,under addition of functions

G = {f : R→ R}, (f + g)(x) = f(x) + g(x).

Let H = {continuous functions} ( G. Then H is closed under addition andinverses (f, g continuous, so is f + g and −f), hence H is a subgroup.

Example 1.3.6. If G = (Z,+), n any integer, then the set nZ = 〈n〉 of multiplesof n is a subgroup of G. In particular for n = 0 we get the trivial subgroup {0},for n = 1 we get G itself.

8 CHAPTER 1. GROUPS

In fact nZ are all the subgroups of (Z,+). For if H is a subgroup of G, other than{0}, then there exists positive integers in H. Let n = min{m ∈ Z+; m ∈ H}.Then we claim H = nZ. Clearly since n ∈ H, H a subgroup nZ ⊆ H.

To see that H ⊆ nZ, note that if h ∈ H we can write h = qn + r, 0 ≤ r < nby division algorithm. Then r = h − qn ∈ H since h ∈ H, qn ∈ nZ ⊂ H. Butthis contradicts the minimality of n unless r = 0 in which case h = qn ∈ nZ. SoH ⊆ nZ.

1.3.1 Special subgroups

Definition 1.3.7. Let G be any group. The center of G, denoted by Z(G) is theset of elements that commute with everything in G, i.e.

Z(G) = {u ∈ G; zg = gz ∀g ∈ G}.

Remark. If G is abelian then G = Z(G). If G is not abelian, then Z(G) $ G.Further we can show that Z(G) is a subgroup of G:

Proof. Since if z1, z2 ∈ Z(G) then z1g = gz1 and z2g = gz2 ∀g ∈ G. It follows

(z1z2)g = z1(z2g) = z − 1(gz2) = (z1g)z2 = g(z1z2)

hence z1z2 ∈ Z(G). Further

z−11 g = (g−1z1)−1 = (z1g

−1)−1 = gz−11 ,

and so z−11 ∈ Z(G). Clearly e ∈ Z(G) and eg = ge = e by definition of e. Therefore

we have Z(G) < G.

Example 1.3.8. Let G = GL2(R) ={(a b

c d

); a, b, c, d ∈ R, ad− bc 6= 0

}. Let(

a bc d

)∈ Z(G), then since

(0 11 0

)∈ G and(

a bc d

)(0 11 0

)=

(0 11 0

)(a bc d

)we have

(b ad c

)=

(c da b

).

and hence we have a = d, b = c and therefore

(a bc d

)=

(a bb a

).

On the other hand since

(1 10 1

)∈ GL2(R) then(

a bb a

)(1 10 1

)=

(1 10 1

)(a bc d

), hence

(a a+ bb b+ a

)=

(a+ b bb+ a a

)

1.3. SUBGROUPS 9

and therefore b = 0. Hence every element in Z(G) has the form

(a 00 a

), a 6= 0.

Converse is easy clearly

(a 00 a

)= a

(1 00 1

)∈ Z(G), and therefore

Z(GL2(R)) ={(a 0

0 a

); a 6= 0

}.

Definition 1.3.9. Another imoportant subgroup is the Centralizer of an elementx in G:

CG(x) = {g ∈ G; gx = xg} = {g ∈ G; gxg−1 = x}.Remark. Check that CG(x) < G. Further if G is abelian then CG(x) = G, ∀x ∈ G.

Definition 1.3.10. Two elements x, y ∈ G are called conjugate in G if there existsg ∈ G such that gxg−1 = y.

Remark. This defines an equivalence relation on G: a ∼ b if there exists g ∈ Gsuch that a = gbg−1.

Definition 1.3.11. The equivalence class a is called the conjugacy class of a

a = {gag−1; g ∈ G}.

One more example of an important subroup

Definition 1.3.12. Let H be a subgroup of a group G. Let x ∈ G and x−1Hxdenote {x−1ax; a ∈ H}. First note x−1Hx is also a subgroup of G. Then

NG(H) = {g ∈ G; g−1Hg = H} = {g ∈ G; Hg = gH}

is called the normalizer of H, where gH = {gh; h ∈ H}.Remark. Then NG(H) < G. Note H < NG(H) trivially.

To verify that a subset H ⊂ G is a subgroup, we check that it is closed undermultiplication and inverses. These two can be combined to give

Proposition 1.3.13. Let G be a group. A non-empty subset H of G is a subgoupiff for all x, y ∈ H we have xy−1 ∈ H. If H is finite, then is suffices to check thatit is closed under muliplication.

Proof. If H ⊂ G then certainly for all x, y ∈ H we have xy−1 ∈ H. Assume ∅ 6= Hsatisfies xy−1 ∈ H ∀x, y ∈ h. We apply this with y = x then xy−1 = xx−1 = 1 ∈ H,hence the identity is in H. Further 1x−1 = x−1 ∈ H, and hence H is closed undermultiplication and therefore a subgroup of G.Suppose H is finite and closed under multiplication. Let x ∈ H. Then {x, x2, . . . }is a finite set, hence xa = xb for some a, b with b > a. If n = b− a then xn = 1, soevery element of H has finite order and x−1 = xn−1 ∈ H, hence H is also closedunder inverses.

10 CHAPTER 1. GROUPS

Definition 1.3.14. If ϕ : G → H is a group homomorphism, then the kernel ofϕ is

kerϕ = {g ∈ G; ϕ = 1H}.

Proposition 1.3.15. kerϕ is a subgroup of G.

Proof. If g1, g2 ∈ kerϕ then

ϕ(g1g2) = ϕ(g1)ϕ(g2) = 1 · 1 = 1,

hence g1g2 ∈ kerϕ.

1 = ϕ(1) = ϕ(g1g−11 ) = ϕ(g1)ϕ(g−1

1 ) = 1 · ϕ(g−11 )

and therefore ϕ(g−11 ) = 1, hence g − 1−1 ∈ kerϕ.

1.3.2 Cyclic groups an their subgroups

Definition 1.3.16. A group G is called cyclic if there exists g ∈ G such thatG = 〈g〉 = {gn; n ∈ Z}. g is called a generator.

If the group operation is additive we write this {ng; n ∈ Z}.

We’ve seen in the spacial case of the cyclic group Z = 〈1〉 that every subgroup isof the form nZ = 〈n〉. Note that in case of Z,−1 is also a generator.

Proposition 1.3.17. Let G be a cycic group, G = 〈g〉. Then |G| = |g|. Morespecifically:(1) If |G| = n < ∞ then gn = 1 and 1, g, g2, . . . , gn−1 are all distinct elements ofG.(2) If |G| =∞ then gn 6= 1 ∀n 6= 0 and ga = gb ∀a 6= b in Z.

Proof. Let |g| = n. First assume, n <∞. Then 1, g, . . . , gn−1 are all distinct sinceif ga = gb then ga−b = 1 and a − b < n contradicting that n = |g| is tha smallestpower such that gn = 1. Thus G has at least n elements 1, g, . . . , gn−1 and theseare in fact all of them.Since let gt be any element of G. Using division algorithm we can write t = nq+kwith 0 ≤ k ≤ n then gt = gnq+k = (gn)q · gk = gk ∈ {1, g, . . . , gn−1}.If |g| =∞, then no positive power of g is 1. If ga = gb for some a, b say a < b thengb−a = 1 a contradiction. Thus the distinct powers of g are all distinct elementsof G, |G| =∞.

Proposition 1.3.18. Let G be a group, g ∈ G, m,n ∈ Z. If gm = 1 and gn = 1then gd = 1 where d = gcd(m,n). In particular if gm = 1 for some m ∈ Z, then|g| | m.

1.3. SUBGROUPS 11

Proof. By Euclidean algorithm we can find r, s ∈ Z such that d = mr + ns.Therefore gd = gmr · gns = 1. The second part we leave as exercise.

Proposition 1.3.19. Let G be any group, g ∈ G, a ∈ Z\{0}(1) If |g| =∞ then |ga| =∞,(2) if |g| = n <∞ then

|ga| = n

(n, a).

Proof. (1) Assume on the contrary |g| =∞, but |ga| = m <∞.

1 = (ga)m = gam = 1, also g−am = (ga)−m = 1,

so either am,−am > 0 and this contradicts |g| =∞.(2) Let h = ga, (n, a) = d, n = dn′, a = da′ for some n′, a′ with n′ > a′. We wantto show |h| = n

d= n′.

hn′= gan

′= ga

′n′ = gda′n′ = (gdn

′)a′= 1

Applying proposition 1.3.18 |h| divides n′. Let |h| = k. Then k|n′ and gak =hk = 1. Applying proposition 1.3.18 again to 〈g〉, we have |g| = n | ak anddn′ | ak = da′k, so n′ | a′k, but (n′, a′) = 1, hence n′ | k together with k | n′ givesk = n′.

The next proposition gives a criteria to find all generators of a cyclic group.

Proposition 1.3.20. Let G = 〈g〉.(1) If |g| =∞ then G = 〈ga〉 if and only if a = ±1.(2) If |g| = n < ∞ then G =< ga > if and only if (a, n) = 1. In particular thenumber of generators of G is ϕ(n), Euler’s ϕ-function

ϕ(n) := #{k ∈ Z; 1 ≤ k ≤ n, (n, k) = 1}

Proof. Exercise.

Next we’ll see that subgroups of a cyclic subgroup as in the case of Z, have specialstucture.

Theorem 1.3.21. Let G = 〈g〉 be a cyclic group.(1) Every subgroup of G is cyclic, more precisely if H ≤ G then either H = {1}or H =

⟨gd⟩

where d is the smallest positive integer such that gd ∈ H.(2) If |G| = ∞ then for any distinct non-negative integers a and b, 〈ga〉 6=

⟨gb⟩.

For any integer m 〈gm〉 =⟨g|m|⟩

where |m| is the absolute value. The nontrivialsubgroups of G correspond bijectively with positive integers.


(3) If |G| = n < ∞. Then for each positive integer a dividing n there exists aunique subgroup H of G of order a. This subgroup H is

⟨gd⟩

where d = na

. For

every integer m, we have 〈gm〉 =⟨g(n,m)

⟩, i.e. the subgroups of G correspond

bijectively with positive divisors of n.

Proof. (1) Let H ≤ G. If H = {1} there is nothing to prove. So assume H 6= {1}.Then there exist a 6= 0 such that ga ∈ H. If a < 0 then since H is a subgroup,g−a ∈ H and −a > 0. Hence H contains some positive power of g. By wellordering principle the set

S = {a; ∈ Z+, ga ∈ H}

has a minimum, call it d. Since h is a subgroup of G = 〈g〉 every element h of His of the form gt where t = qd + r. Therefore we have gqd+r = (gd)q · gr ∈ H, but(gd)q ∈ H and gr ∈ H. This contradicts the minimality of d unless r = 0, t = qd,hence h ∈

⟨gd⟩.

(2) Is similar to the proof of (3).(3) Assume |G| = n <∞ and a|n. Let d = n

a. Then

∣∣gd∣∣ =|g|

(|g| , d)=n

d= a

by proposition 1.3.19. Hence H =⟨gd⟩

is a subgroup of order a.

To show uniqueness, suppose K is any subgroup of order a. Then by part (1)K =

⟨gb⟩

where b is the smallest power of g such that gb ∈ K.

a = |K| =∣∣gb∣∣ =

|g|(|g| , b)

=n

(n, b),

and hence a = nd

= n(n,b)

and d = (n, b). In particular d | b, so b = db′ and

gb = (gd)b′ ∈

⟨gd⟩, and therefore K =

⟨gb⟩⊂⟨gd⟩. But

∣∣< gd >∣∣ = a = |K|,

hence K =⟨gd⟩.

The final assertion of (3) follows because

〈gm〉 ≤⟨g(m,n)

⟩, |gm| = n

(n,m)and

∣∣g(m,n)∣∣ =

n

(n, (m,n))=

n

(n,m),

then 〈gm〉 =⟨g(m,n)

⟩. Since (m,n) is a divisor of n, every subgroup of G arises

from a divisor of n.

1.4. GROUP ACTIONS 13

Example 1.3.22. G = (Z/12Z,⊕). Then G =< 1 >, |G| = 12. Positive divisorsof 12 are 1,2,3,4,6,12, hence the distinct subgroups of G are

G = 〈1〉 , 〈2〉 , 〈3〉 , 〈4〉 , 〈6〉 , 〈0〉 = {e}

order1212

(2, 12)= 6

12

(3, 12)= 4

12

4= 3 2 1

Note 〈5〉 = 〈1〉 = G since (5, 12) = (1, 12). Similarly 〈7〉 = 〈11〉 = 1. We have〈8〉 = 〈4〉 since 〈8, 12〉 = 〈4, 12〉 and 〈9〉 = 〈3〉, 〈10〉 = 〈2〉. Lattice of subgroups

G = 〈1〉� �

〈2〉〈3〉� � �

〈4〉〈6〉� �〈0〉

Theorem 1.3.23. (a) Let n be a positive integer, G a cyclic group of order n.Then G ∼= (Z/nZ,⊕), i.e. any cyclic group of order n are isomorphic.(b) Let G be a ∞ cyclic group. Then G ∼= Z. Any two infinite cyclic groups areisomorphic.

Proof. Exercise Serie 2.

1.4 Group Actions

Definition 1.4.1. A group action of a group G on a set X is a map from G×X →X, written g ◦ x, for all g ∈ G, x ∈ X with the following properties(1) g1 ◦ (g2 ◦ x) = (g1g2) ◦ x ∀g1, g2 ∈ G, x ∈ X,(2) 1 ◦ x = x, ∀x ∈ X.In (1) the product (g1g2) is taken in G.

Note: For each g ∈ G, we get a map

σg : X → X, x 7→ g ◦ x.

Fact 1: σg is a permutation of X, namely σg is a bijection with inverse σg−1 . For,let x ∈ X, then

(σg−1 ◦ σg)(x) = σg−1(g ◦ x) = g−1 ◦ (g ◦ x) = (g−1g) ◦ x = 1 ◦ x = x.

Therefore σg−1 ◦ σg : X → X is the identity map. Changing the roles of g, g−1

gives that σg ◦ σg−1 is also the identity map, hence σg is a bijection.


Fact 2: The map ϕ : G → SX is a homomorphism. To see this we need toshow

ϕ(g1g2) = ϕ(g1) ◦ ϕ(g2).

The permutations ϕ(g1g2) and ϕ(g1)◦ϕ(g2) are equal iff their values agree on everyx ∈ X.

ϕ(g1g2)(x) = σg1g2(x) by definition of ϕ,

= (g1g2) ◦ x by definition of σg1g2 ,

= g1 ◦ (g2 ◦ x) by property 1. of group actions

= σg1(σg2(x)) definition of σg1 , σg2 ,

= (ϕ(g1) ◦ ϕ(g2))(x) definition of ϕ.

Definition 1.4.2. The homomorphism ϕ : G → Sx is called the permutationrepresentation associated to the given action.

This process is reversible, in the sense that if ϕ : G → Sx is any homomorphismfrom G to a symmetric group on a set X, then the map from G×X → X definedby g ◦ x = ϕ(g)(x) gives a group action of G on X. (Group action of G on Xmeans that every element g in G acts as a permutation of X in a way consistentwith the group operations in G.)

Example 1.4.3. Trivial action:

G×X → X, (g, x) 7→ x ∀x ∈ X, g ∈ G.

G is said to act trivially on X. Note distinct elements of g all induce the samepermutation on X, namely the identity permutation. The associated permutationrepresentation is

ϕ : G→ SX , g 7→ id.

Definition 1.4.4. Any action of G on X is said to be faithful if

g1 6= g2 ⇒ σg1 6= σg2 ,

i.e. an action is faithful when the associated permutation representation is injec-tive.

Definition 1.4.5. The kernel of an action of G on X is the set

{g ∈ G; g ◦ x = x ∀x ∈ X}.

These are exactly the elements of G that fix all elements of X.


Note that this kernel is the same as kerϕ, ϕ : G → SX , g 7→ σg. For trivialaction this kernel is G. Check: The kernel of an action is a subgroup of G.

Example 1.4.6. Every group acts on itself by left (right) multiplication

G×G→ G, (g, a) 7→ ga

Each fixed g ∈ G permutes elements of G by left multiplication

g : a 7→ ga

(if the group is written additively, a 7→ g + a, left translation).

Definition 1.4.7. This action is called left regular action of G on itself.

By cancelation law in groups this is a faithful action.

Example 1.4.8. Let G be a group, X = G. The map

G×G→ G, (g, a) 7→ gag−1 ∀g, a ∈ G

is a (left) group action. This action is called conjugation.

For fixed g, G→ G, a 7→ gag−1 is an isomorphism. The kernel of this action

{g ∈ G; gag−1 = a ∀a ∈ G} = {g ∈ G; ga = ag, ∀a ∈ G} = Z(G).

Let’s consider the general case, where G acts on X, x ∈ X.

Definition 1.4.9. The stabilizer of x in G is defined as

Gx = {g ∈ G; g ◦ x = x}.

Proposition 1.4.10. Gx is a subgroup of G.

Proof. Exercise.

Proposition 1.4.11. Pick x ∈ X, g ∈ G and set y = g ◦ x. Show that Gy =gGxg

−1.

Example 1.4.12. Let X = P(G) = {all subsets of G}. Then G acts on X byconjugation: for each g ∈ G, B ⊆ G

g : B 7→ gBg−1 = {gbg−1; b ∈ B}.

The stabilizer of A ⊆ G is

GA = {g ∈ G; gAg−1 = A} = {g ∈ G; gA = Ag} = NG(A),

the normalizer of A in G.


Example 1.4.13. The group G = SL(2,R) acts on the UHP (upper half plane)H = {z = x+ iy; y > 0} via

G×H→ H ,((a b

c d

), z)7→ az + b

cz + d.

Since

im(az + b

cz + d

)=

im(z)

|az + d|2> 0(

1 00 1

)◦ z = z

acts trivially. Check that((a bc d

)(A BC D

))◦ z =

(a bc d

)◦(Az +B

Cz +D

).

Let i ∈ H. The stabilizer of i is

Gi ={(a b

c d

);ai+ b

ci+ d= i}

therefore ai+ b = −c+ di and a = d, b = −c, so

Gi ={( a b−b a

); a2 + b2 = 1

},

a compact group! SO(2).

Proposition 1.4.14. Given a group G acting on a set X. The relation on Xdefined by x ∼ y iff x = g ◦ y for some y ∈ G, defines an equivalence relation.

Proof. By definition of group action 1 ◦ x = x ∀x ∈ X, hence x ∼ x.If x ∼ y, ∃g ∈ G such that x = g ◦ y, then g−1 ◦ x = g−1 ◦ g ◦ y = 1 ◦ y = y, hencey ∼ x.If x ∼ y, y ∼ z then ∃g, h ∈ G such that x = gy, y = hz, therefore x = g◦(h◦z) =gh ◦ z and x ∼ y.

Definition 1.4.15. The equivalence class containing x is called the orbit of x orcoset of x under G

Gx = {gx; g ∈ G}.

Example 1.4.16. G = SL(2,R), S = H, x = i. Then

Gi = {g ◦ i; g ∈ G}.


Note since let (y xi 1

)∈ GL2(R),

since y > 0 and g ◦ i = yi + x = z ∈ H for any x + iy = z ∈ H, Gi = H, so thereis just one orbit.

In General:

Definition 1.4.17. If G acts on X, and there is only one orbit, the action iscalled transitive, i.e. for all x, y ∈ X, there exists g ∈ G such that g ◦ x = y.

The orbits of a group action G on X partitions the set X into disjoint sets, i.e.

X =⋃x∈R

Gx,

where R is a set of representations for the equivalence classes. We write for theset of orbits G\X if G acts on the left and X/G if G acts on the right.

Example 1.4.18 (important example). We have seen that G acts on itself bymultiplication. One can also make a subgroup H of G act on G by multiplicationon the left or on the right

r : H ×G→ G, (h, g) 7→ hg (on the right)

l : H ×G→ G, (h, g) 7→ gh−1 (on the left).

The orbit of g ∈ G then has the form

Hg = {hg; h ∈ H}

for some action r and is called a right coset of H in G (or gH = {gh; h ∈ H} forsome action l is the left coset of H in G).

Example 1.4.19. 3Z ≤ Z act on Z by translation

3Z︸︷︷︸=G

× Z︸︷︷︸=X

→ Z, (3n,m) 7→ 3n+m.

Orbits, Cosets:

3Z+ 0 = {0,±3,±6, . . . } = 3Z+ 3 = 3Z+ 6, . . . ,

3Z+ 1 = {. . . ,−5,−2, 1, 4, 7, . . . } = 3Z+ 4 = 3Z+ 7 = 3Z− 2, . . . ,

3Z+ 2 = {. . . ,−4,−1, 2, 5, 8, . . . } = 3Z+ 5 = 3Z− 1 = 3Z+ 8, . . . .

Z = 3Z ∪ 3Z + 1 ∪ 3Z + 2 a set of representories: 3Z\Z = {0, 1, 2}. Note in thiscase 3Z+ 0 = 0 + 3Z, left and right cosets of 3Z in Z are the same.


This is not the case in general (see the next series for an example, G = S3, H =〈1, (23)〉).

Example 1.4.20. Z ≤ R under addition. Z acts on R by translation.

Z× R→ R, (n, r) 7→ n+ r.

How does the orbits look like? What is a set of representatives for this action?For any r ∈ R, the orbits or the right cosets are

Z+ r = {n+ r; n ∈ Z}.

Write r = [r]︸︷︷︸integer part

+{r}, then {r} ∈ [0, 1). Then Z+ r = Z+ {r} and if Z+ s =

Z+ {r} then assume analog {s} ≥ {r} then {s}− {r} ∈ Z and {s}− {r} ∈ [0, 1),hence {s} = {r} and R =

⋃r∈[0,1)

Z + {r}. A set of representatives is the interval

[0, 1). If one identifies 0 and 1 then we get the circle.

Example 1.4.21. Let Λ = Z + Zi = {m + ni; m,n ∈ Z} ≤ C. Λ acts on C bytranslation

Λ× C→ C, (l, z) 7→ l + z.

Orbits L + z. A set of repesentatives for C/L is {m + ni; 0 ≤ m,n < 1},C =

⋃z∈P

L+ z.

Example 1.4.22. G = (Z/2Z,⊕), H =< 4 >= {0, 4, 8}. The right cosets of Hare

H = H ⊕ 0 = {0, 4, 8} = H ⊕ 4 = H ⊕ 8

= H ⊕ 1 = {1, 5, 9} = H ⊕ 5 = H ⊕ 9

= H ⊕ 2 = {6, 10, 2} = H ⊕ 6 = H ⊕ 10

= H ⊕ 3 = {7, 11, 3} = H ⊕ 7 = H ⊕ 11

G = {0, 4, 8}︸︷︷︸0

∪{1, 5, 9}︸︷︷︸1

∪{6, 10, 2}︸︷︷︸2

∪{7, 11, 1}︸︷︷︸3

.

Note each coset has the same number of elements, i.e. 3 elements. There are 4different cosets.

Definition 1.4.23. The number of distinct right cosets of H is called the index ofH in G, denoted by [G : H].


Example 1.4.24. (1) For H = 3Z, G = Z : [G : H] = 3,(2) H = Z, G = R then [G : H] =∞,(3) And for H =< 4 >, G = Z/12Z, we have [G : H] = 4, because |G| = 12,|H| = 3.

Theorem 1.4.25 (Lagrange’s theorem). Let G be a group, H ≤ G. Then

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

In particular the order of H and the order of any element of G divides the orderof G.

Proof. G is a disjoint union of its right cosets, hence it suffices to prove that theorder of any coset is equal to the order of |H| = |He|. Let g ∈ G, define a map

φ : H → Hg, h 7→ hg.

It suffices to check that φ is a bijection. φ is clearly surjective. Suppose φ(h1) =φ(h2), then h1g = h2g. But then h1 = h2 and φ is injective and we are done.

Corollary 1.4.26. Every group G of prime order is cyclic.

Proof. By assumption |G| > 1. Pick any element g ∈ G, g 6= e. Let H = 〈g〉.Then the order of H divides |G| = p. Since p is prime, |H| = 1 or p. Sincee 6= a ∈ H, |H| 6= 1, |H| must be p, so |G| = |H| and G = H = 〈g〉 is cyclic.

Warning: Full converse of Lagrange’s theorem is not true, i.e. if G is a finitegroup, and n | |G| then G need not have a subgroup of order n.

We’ve seen for a group G, H a subgroup of G, the sets of left (or right) cosets ofH in G is

{Hg; g ∈ G} = G/H (or {Hg; g ∈ G} = H\G}).

A natural question is: Under what circumstances can we put a group structure onthe set of left cosets of H in G? Suppose we are given two left cosets g1H, g2H.Then how do we define a product? We ought to define

(g1H)(g2H) = g1g2H.

The problem with this is that to define the product we had to choose a repesen-tative of the left coset g1H, namely g1. However we could just as well chosen anyother element of this left coset as a representative and we could potentially getanother choice of left coset for the product, i.e. if aH = a′H, bH = b′H then is ittrue that abH = a′b′H? In general not!!!


Example 1.4.27. G = S3, H = {1, (23)} = {1, s}, s = (23), r = (123), thencheck that

rH = {r, rs} = (rs)H

r2H = {r2, r2s} = (r2s)H

We define(rH)(r2H) = r3H = H.

On the other hand if we choose rs as the representative for rH, then

((rs)H)(r2H) = rsr2H = (13)H = {(13), (13)(23)} 6= H.

Different representatives give rise to different answers for the procuct, which is notgood!

Solution: We can really only define a product on the set of cosets in a well-definedmanner if the group H is special! That brings us to the next section.First notethat

Lemma 1.4.28. Let H be a subgroup of G and a, b ∈ G. Then aH = bH iffa ∈ bH iff b−1a ∈ H.

Proof. Exercise.

1.5 Normal subgroups and Quotient groups

Definition 1.5.1. Let G be a group, H be a subgroup. We say that H is normalin G, and write H / G, if g−1Hg ⊂ H for all g ∈ G.

Proposition 1.5.2. Let G be a group, H ≤ G. Then it is equivalent:i) for every g ∈ G, gHg−1 ⊂ H,ii) for every g ∈ G, gHg−1 = H,iii) Ha = aH, ∀a ∈ G,iv) The set of left cosets is equal to the set of right cosets (i.e. for every a ∈G, aH = Hb for some b ∈ G).

Proof. Exercise

The significance of the concept of the normal subgroup lies in the next theorem

Theorem 1.5.3. Let G be a group, H E G. Then the set of left cosets is a groupdenoted by G/H, where the multiplication is defined by the rule

(g1H)(g2H) := (g1g2)H.

The identity element of G/H is eG/H = eH = H. The inverse of an elementgH ∈ G/H is given by g−1H.

1.5. NORMAL SUBGROUPS AND QUOTIENT GROUPS 21

Proof. It suffices to check that this rule of multiplication is well-defined. (The factthe multiplication is associative is inherited from the associativity of multiplicationinG.) Pick two left cosets g1H, g2H. Suppose giH = g′iH, i = 1, 2, i.e. g′i is anotherrepresentatives for giH.We want to show that g1g2H = g′1g

′2. If giH = g′ih then

∃hi ∈ H such that g′i = gihi. Then

g′1g′2 = (g1h1)(g2h2) = g1(h1g2)h2.

Now sinceH / G, g−1Hg ⊂ H, g−1

2 h1g2 ∈ g−1Hg ⊂ H

there exists h′1 such that g−12 h1g2 = h′1, i.e. h1g2 = g2h

′1 for some h′1 ∈ H. Therefore

we have g′1g′2 = g1(h1g2)h2 = g1g2h

′1h2 and (g′1g

′2)H = (g1g2)H as wanted.

Example 1.5.4. Check that for G = S3, H = {1, (23)} H 6C G since right cosets6= left cosets.

Example 1.5.5. G = S3, K = 〈(123)〉. Check that K / G.

Example 1.5.6. If ϕ : G → G′ a group homomorphism. Then we’ve seen thatkerϕ is a subgroup of G. In fact kerϕ / G. To see this, let H = kerϕ. Take anaribitrary element g−1hg in g−1Hg then

ϕ(g−1hg) = ϕ(g−1)ϕ(h)ϕ(g) = ϕ(g)−1eϕ(g) = e.

Therefore g−1Hg ∈ kerϕ = H. Hence g−1Hg ⊂ H and kerϕ / G.

Example 1.5.7. Let G be any group, Z(G) < G. In fact Z(G) / G and anysubgroup of Z(G) is normal in G. To see this, let H ≤ Z(G).To show that H / G we need to show that g−1hg ∈ H for all g ∈ G, h ∈ H, butsince h ∈ H ≤ Z(G) we have hg = gh, for all g ∈ G and g−1hg = g−1gh = h ∈ H.

A corollary of this is

Lemma 1.5.8. If G is abelian then every subgroup of G is normal.

Proof. If G is abelian, then Z(G) = G and every subgroup of G is a subgroup ofZ(G), hence normal by the previous result.

Example 1.5.9. It can happen thatG is not abelian, yet every subgroup is normal.For exampleG = Q8, the quaternions. The subgroups ofQ8 are {1}, {±1}, 〈i〉 , 〈j〉 , 〈k〉.{1} and {±1} are normal because they’re contained in Z(Q8). The other subgroupsare normal because they have index 2 in Q8 since

Lemma 1.5.10. Let H ≤ G. If [G : H] = 2, then H E G


Proof. Exercise.

Example 1.5.11. G = Q8, H = {±1}, H / G and G/H is a group of order

[G : H] = |G||H| = 4. Is it Z4 or V4 (klein’s 4 group)?

(Hi)2 = Hi2 = H(−1) = H = eG/H .

Similarly (Hj)2 = H, (Hk)2 = H, therefore every element inG/H = {H,Hi,Hj,Hk}has order 2 and Q8/{±1} ∼= V4.

Example 1.5.12. G = (Z,+), H = nZ. Then G/H has n elements, namelyH + 0, H + 1, . . . , H + (n− 1), which is Z/nZ.

1.6 Homomorphisms and Normal subgroups

In this section we’ll extablish a connection between the seemingly unrelated con-cepts homomorphisms and normal subgroups. We’ve seen that the kernel of ahomomorphism is a normal subgroup. More precisely

Proposition 1.6.1. Let φ : G→ G′ be a group homomorphism. Then

kerφ = {g ∈ G; φ(g) = eG′}

is a normal subgroup of G.

In fact, normal subgroups are precisely the same as the kernels of homomorphisms.Namely

Proposition 1.6.2. A subgroup N of G is normal if and only if it is the kernelof some homomorphism.

Proof. ”‘⇐”’: is Proposition 1.6.1.

”‘⇒”’: Let N / G. Then G/N is a group. Define a map

π : G→ G/N, g 7→ gN, ∀g ∈ G.Then by definition of operation in G/N

π(g1g2) = g1g2N = (g1N)(g2N) = π(g1)π(g2)

and π is a homomorphism,

kerπ = {g ∈ G; π(g) = eG/N = N} = {g ∈ G; gN = N}= {g ∈ G; g ∈ N} = N,

as wanted.

1.6. HOMOMORPHISMS AND NORMAL SUBGROUPS 23

The group G/N has the following ”‘universal property”’

Theorem 1.6.3. Let φ : G → G′ be a group homomorphism with H ⊆ kerφ.Then there exists a unique homomorphism φ : G/H → G′ such that φ = φ ◦ π,i.e. the following diagram commutes

Gφ−→ G′

π ↘ ↗φ

G/H

Moreover if H = kerφ then φ is injective.

Proof. We define a map φ : G/H → G′, gH 7→ φ(g). We first check that it iswell-defined. Suppose gH = g′H then g′ = gh for some h ∈ H and

φ(g′) = φ(gh) = φ(g)φ(h) = φ(g).

Since by assumption h ∈ kerφ we have φ(h) = e and therefore φ is well-defined.(Note clearly that φ, φ has the same image.)

From its definition it is clear that φ = φ ◦ π. The fact that φ is a homomor-phism follows easily since φ is a homomorphism. Finally if H = kerφ then φ isinjective since if φ(g1H) = φ(g2H), then φ(g1) = φ(g2) and φ(g1g

−12 ) = e, hence

g1g−12 ∈ kerφ = H and g1H = g2H.

Corollary 1.6.4 (Fundamental Theorem of Isomorphisms). If φ : G → G′ is ahomomorphism, then

G/ kerφ ∼= im(φ),

the image of φ.

Proof. The φ in Theorem 1.6.3 is injective, clearly onto, im(φ) = im(φ).

Example 1.6.5. Let φ : Z→ Cn = 〈g〉 be a cyclic group of order n with a 7→ ga.

φ(a+ b) = ga+b = ga · gb = φ(a)φ(b).

Clearly φ is surjective:

kerφ = {m ∈ Z; gm = 1} = {m ∈ Z; |g| |m}= {m ∈ Z; n|m} = nZ

hence Z/nZ ∼= Cn. The ”‘fiber”’ of φ over an element ga of Cn is

φ−1(ga) = {m ∈ Z; gm = ga} = {m ∈ Z; gm−a = 1}= {m ∈ Z; n|m− a} = {m ∈ Z; m ≡ a mod n} = a.


Fibers of φ are presicely the residue classes modulo n.

nZ nZ+ 1 . . . nZ+ a . . . nZ+ (n− 1)0 1 a n− 1±n 1± n a± n n− 1± n±2n 1± 2n . . . a± 2n . . . (n− 1)± 2n

......

......

↓ ↓ ↓ φ ↓g0 = 1 g1 ga gn−1

Example 1.6.6. Let φ : GL(1,R) → (R\{0}, ·),(a bc d

)7→ ad − bc. φ is onto

since for any r 6= 0, g =

(r 00 1

)∈ GL2(R) and φ

(r 00 1

)= r

kerφ = {g ∈ GL(2,R); detg = 1} = SL(2,R)

therefore GL(2,R)/SL2(R) ∼= (R\{0}, ·).

Example 1.6.7. Let G = (R,+), S1 = ({z ∈ C {0}; |z| = 1}, ·), i.e. S1 ismultiplicative group of complex numbers of absolute value 1. Let

φ : R→ S1, r 7→ e2πi = cos r + i sin r

thenkerφ = {r ∈ R; e2πir = 1} = Z,

so R/Z ∼= S1.

Remark. (1) N /G. The elements of the quotient group are subsets of the originalgroup G. These subsets are the fibers or cosets of the kernel N of the homomor-phism π : G → G/N . N and its cosets are projected (or collapsed) onto singleelements in the quotient group.(2) N / G. Note that the structure of G is reflected in the structure of G/N , forexample the associativity of multiplication in G/N is indeed from associativity ofG inverses in G/N are indeed from inverses in G.(3) We have seen before G acting on the set of subsets of G by conjugation.

G× P(G)→ P , (g, A) 7→ gAg−1.

Let H ≤ G, H ∈ P(G). The stabilizer of H, under this action

GH = {g ∈ G; gHg−1 ⊂ H},

1.6. HOMOMORPHISMS AND NORMAL SUBGROUPS 25

was called the normalizer of H in G and denoted by NG(H). Since stabilizers ofany action are subgroups of G we have NG(H) ≤ G.

NG(H) gives a criterion which determines precisely when a subgroup H is nor-mal, namely when NG(H) = G. NG(H) is a ”‘measure”’ of how close H to beingnormal.(4) Note being normal depends on the relation of H to G, not on the internalstructure of H. The same group H may be normal subgroup of G but not normalin a larger group containing G.

Example 1.6.8. TakeH = {e, (12)(34)}, G = {e, (12)(34), (13)(24), (14)(23)}, K =S4. Then H / G, but H 6C S4.

Theorem 1.6.9 (2nd Isomorphism Theorem). Suppose H,K are normal subgroupsof G and K ⊆ H. Then K is normal in H and we have

G/K

H/K∼= G/H.

Proof. If K ≤ H / G then clearly K / H. Let h ∈ H, since h ∈ G and K / G.hKh−1 ⊂ K ⇒ K / H. We’ll apply the fundamental theorem. Define a map

φ : G/K → G/H, gK 7→ gH.

φ is well-defined, since if g1K = g2K then g−11 g2 ∈ K ⊂ H hence g1H = g2H, i.e.

φ(g1K) = φ(g2K). φ is a homomorphism since

φ((g1K)(g2K)) = φ(g1g2K) = g1g2H = (g1H)(g2H) = φ(g1K)φ(g2K)

φ is onto since if gH ∈ G/H then we have gK ∈ G/K and φ(gK) = gH. Wereally only have to check that kerφ = H/K. But

gK ∈ kerφ −→ φ(gK) = eG/H = H −→ gH = H −→ g ∈ H −→ gK ∈ H/K,

so kerφ = H/K and G/KH/K∼= G/H.

Example 1.6.10. G = (Z,+), H = 2Z, K = 6Z, then 6Z ⊂ 2Z ⊂ Z and

Z/6Z2Z/6Z

∼= Z/2Z.

Theorem 1.6.11 (3rd Isomorphism theorem). Let H,K be subgroups of G. As-sume K / G. Then

H/H ∩K ∼= HK/K,


where HK = {hk; h ∈ H, k ∈ K}.

G|

HK� �

H K� �

H ∩K|e

Proof. First note that everything in the conclusion makes sense.

Lemma 1.6.12. If H,K ≤ G, K / G then HK ≤ G.

Proof. Exercise.

K / HK follows from K / G and H ∩K / H follows from simple.

Lemma 1.6.13. H ≤ G, K / G, K / G then H ∩K / H.

Proof. Exercise.

We’ll prove the theorem by using the fundamental theorem, i.e. we’ll define ahomomorphism φ : H → HK/K and show that its kernel is H ∩K. Let

φ : H → HK/K, h 7→ Kh

(note Kh ∈ HK/K since h ∈ HK). φ is a homomorphism since

φ(h1h2) = Kh1h2(Kh1)(Kh2) = φ(h1)φ(h2).

φ is onto since any element HK/K has the form K(hk) for some h ∈ H, k ∈ Hand

K(hk) = KH ·Kk = Kh ·K = Kh.

Thus K(hK) = φ(h), hence the fundamental theorem gives H/ kerφ ∼= HK/K.Now to show kerφ = H ∩K assume x ∈ H then

x ∈ kerφ⇔ φ(x) = eHK/K = K ⇔ φ(x) = K ⇔ Kx = K ⇔ x ∈ K ⇔ x ∈ H ∩K

and we’re done.

The next theorem describes the relation between the lattice of subgroups of thequotient group G/N and the lattice of subgroups of G. There is a one-to-onecorrespondence between the subgroups K of G containing N and the subgroupsof G/N .

1.7. GROUPS ACTING ON THEMSELVES BY CONJUGATION - CLASS EQUATION27

Theorem 1.6.14. Let G be a group, N /G. Then there is a bijection from {A ⊆G; N ⊆ A} onto the set of subgroups {A = A/N ⊆ G/N}. In particular everysubgroup of G/N is of the form A/N for some subgroup A of G containing N . Thisbijection has the following properties for all A,B ⊆ G with N ≤ A and N ≤ B(1) A ⊆ B iff A ⊂ B,(2) If A ⊆ B then |B : A| =

∣∣B : A∣∣,

(3) A E G iff A / G.

Proof. Exercise.

Example 1.6.15. Let G = Q8, N =< −1 >= {±1}.

Q8

� | �〈i〉〈j〉〈k〉

� | �〈−1〉|e

, Q8/{±1} ∼= V4,

V4 = {e, a, b, c}� | �

〈a〉〈b〉〈c〉� | �

e

The lattice for G/N appears at the top of the lattice for G.

Next we go back to group actions. We have seen that if G acts on a set X. Therelation defined by a ∼ b iff a = g · b for some g ∈ G is an equivalence relation.

Lemma 1.6.16. The number of elements in the equivalence class containing a is[G : Ga], index of stabilizer if a ∈ G.

Proof. Let Ga = be the orbit of a, G/Ga the set of cosets of Ga in G,

φ : Ga→ G/Ga, b = g · a 7→ gGa.

φ is surjective since for any g ∈ G, g · a ∈ Ga. φ is injective since

g · a = h · a⇒ h−1g ∈ Ga ⇒ hGa = gGa,

hence φ is a bijection.

1.7 Groups acting on themselves by conjugation

- class equation

Consider a function G×G→ G, (g, a) 7→ gag−1. The equivalence classes underthis action are called conjugacy classes. Let a ∈ G be any element of G. Then|Ga| denotes the number of conjugators of G

|Ga| = #{gag−1 = a} = CG(a)


is called the centralizer of a in G. Then by Lemma 1.6.16 we have

|Ga| = [G : CG(a)].

Note if a ∈ Z(G) and

CG(a) = {g ∈ G; gag−1 = a} = G,

since a commutes with every g ∈ G, the conjugacy class of a has [G : CG(a)] =[G : G] = 1 element.

The converse is also true if

|Ga| = 1 ⇒ 1 = [G : CG(a)]⇒ G = CG(a)⇒ a commutes ∀g ∈ G⇒ a ∈ Z(G).

Since conjugacy classes, being equivalence classes, are disjoint, we have

G = ∪ conjugacy classes, |G| =∑|conjugacy classes| .

We have

Theorem 1.7.1 (class equation). Let G be a finite group. Let g1, g2, . . . gr berepresentatives of distinct conjugacy classes of G not contained in the center Z(G)of G. Then

|G| = |Z(G)|+r∑i=1

[G : CG(gi)].

Proof. {a} is conjugacy class of size 1 if and only if a ∈ Z(G). Let Z(G) ={1, z1, . . . , zm}. Let K1, K2, . . . , Kr be conjugacy classes of G not contained in thecenter. Let gi be a representative of Ki, i = 1, . . . , r. Then the conjugacy classesof G are

1, {z1}, {z2}, . . . , {zm}, K1, K2, . . . , Kr.

This partitions G into disjoint sets and we have

|G| =m∑i=1

1 +r∑i=1

|Ki| = |Z(G)|+r∑i=1

[G : CG(gi)].

Note that all the summands on the righthandside of class equation are divisorsof |G|. (Since they are indices of subgroups of G.) This restricts their possiblevalues.

1.7. GROUPS ACTING ON THEMSELVES BY CONJUGATION - CLASS EQUATION29

Example 1.7.2. S3 = {e, r, r2, s, rs, r2s}, r = (123), s = (23). Check thatZ(S3) = {e}, class of r : {r, r2}, class of s : {s, rs, r2s}

S3 = {e} ∪ {r, r2} ∪ {s, rs, r2s},

G = |S3| = 1 + 2 + 3, note 2 | 6, 3 | 6, 1 | 6.

Example 1.7.3. G = Q8. Note: In any group 〈g〉 ≤ CG(g). This helps tominimize the computations of conjugacy classes.In Q8 we see < i >≤ CQ8(i) ≤ Q8. Since i /∈ Z(Q8) = {±1} and [Q8 :< i >] = 2,we must have CQ8(i) =< i > (|CQ8(i)| |8, so it can be 1,2,4, since i /∈ Z(G),CQ8(i) 6= Q8). Therefore i has precisely two conjugates in Q8, i,−i = kik−1.The other conjugacy classes are {1}, {−1}, {±i}, {±j},{±k}, note 8 = |Q8| =2 + 2 + 2 + 2.

Some applications of class equationGroups of prime power order have nontrivial centers.

Theorem 1.7.4. Let p be a prime, P a group of prime power order pα for someα ≥ 1. Then P has a nontivial center, i.e. Z(G) 6= {e}.

Proof. Because of the class equation we have

|P | = |Z(G)|+r∑i=1

[P : CP (gi)],

where g1, . . . , gr are representatives of distinct non-central conjugacy classes. Bydefinition CP (gi) 6= P, i = 1, . . . , r,

[P : CP (gi)]| |P | = pα,

and it is not 1, hence p | [P : CP (gi)], i = 1, . . . , r, since p | |P | we have p | |Z(G)|and therefore |Z(G)| 6= 1, so Z(G) 6= {e}.

Corollary 1.7.5. If |P | = p2, then P is abelian. More precisely P is isomorphicto Zp2 or Zp × Zp.

Proof. Exercise.

In the next section, we’ll prove a partial converse theorem to Lagrange’s theorem.Namely if G is a finite group of order pam, p - m, then G has a subgroup of orderpa.


1.8 Composition series and the Holder program

We’ve seen Lagrage’s theorem, which says if G is a finite group, H ≤ G, then|H| | |G|. We already mentioned that the converse of thes statement is false.There are groups e.g. symmetries of regular tetrahedron which has order 12, butdoes not have a subgroup of order 6. Here are partial converses of Lagrange’stheorem

Theorem 1.8.1 (Cauchy’s Theorem). If G is a finite group, p prime dividing |G|then G has an element of order p.

Strongest converse theorem to Lagrange’s theorem is

Theorem 1.8.2 (Sylow). If G is a finite group of order pαm, where p is prime,(p,m) = 1, then G has a subgroup of order pα.

(Cauchy’s theorem follows from Sylow’s theorem.) We’ll prove Cauchy’s theoremfor finite abelian groups without using Sylow’s theorems and in fact then we’lluse Cauchy’s theorem for finte abelian groups for the proof of Sylow’s theorems.The proof of Cauchy’s theorem for finite abelian groups is also important in termsof its structure which demonstates an important technique in finite group theory,induction. The proof shows how the information on a normal subgroup N and onG/N can be used to get information about G.

Theorem 1.8.3. Let G be a finite abelian group and p prime dividing |G|. ThenG has an element (hence a subgroup) of order p.

Proof. We use induction on |G|. Assume the theorem is valid for all abelian groupsof order < |G|. Let x 6= e be an element of G and H = 〈x〉. If H = G then G iscyclic and we’re done, since we know the theorem holds for cyclic groups.

Otherwise H � G and H / G, since G is abelian. |G| = |H| · |G/H| and bothfactors on the right, |H| , |G/H| < G. Since p | |G| and p is prime then eitherp | |H| or p | |G/H|.

If p | |H| and H is a group whose order is strictly less then |G|. By inductionhypohesis it has a subgroup of order p which is of course also a subgroup of G oforder p and we’re done.

So assume p | |G/H|. Now since G is abelian all factor groups of G are alsoabelian. Hence G/H is an Abelian group of order < |G|. By induction hypothesisG/H has a subgroup of order p. Since any group of order p is cyclic, this subgroupmust be cyclic, so it is 〈Hg〉 for some g = Hg ∈ G/H of order p. Note g /∈ H

1.8. COMPOSITION SERIES AND THE HOLDER PROGRAM 31

since |Hg| 6= 1, but gp ∈ H since (Hg)p = eG/H = H, therefore 〈gp〉 6= 〈g〉 and

ord(gp) < ord(g), but ord(gp) = ord(g)(ord(g),p)

, hence (ord(g), p) > 1.

Since p is prime, p | ord(g), say (ord(g)) = kp, ord(gk) = kp(kp,k)

= p and⟨gk⟩

is the subgroup we’re looking for.

The proof worked because both N , G/N are subgroups of order smaller than G,and we could piece each information together. Clearly a basic obstruction to thismethod is the non-existence of a normal subgroup. In the Theorem 1.9.9 wecould find a normal subgroup, because G was abelian. Groups without non-trivialnormal subgroups are fundamental obstruction to this method.

Definition 1.8.4. A group G is called simple if |G| > 1 (can be ∞) and the onlynormal subgroups of G are 1 and G.

We have seen that if |G| is prime then G ∼= Zp, Zp has no non-trivial subgroups,therefore they are simple for all primes p. The smallest non-abelian simple group isA5 of order 60. Simple groups cannot be factored as N , G/N and can be thought ofbuilding blocks as the primes in Z. There is also a ”‘unique factorization theorem”’.

Definition 1.8.5. In a group G, a sequence of subgroups

1 = N0 ≤ N1 ≤ N2 ≤ · · · ≤ Nk−1 ≤ Nk = G

is called a composition series if Ni E Ni+1 and Ni+1/Ni are simple, 0 ≤ i ≤ k− 1.The quotient groups Ni+1/Ni are called composition factors.

Note we’re not assuming Ni E G, just Ni E Ni+1.

Example 1.8.6. 1 E 〈s〉 E 〈s, r2〉 / D8 or 1 E 〈r2〉 E 〈r〉 E D8. 2 differentcomposition series, 3 compsition factors each of which isomorphic to (simple) Z2,〈s〉 /1 ∼= Z2, 〈s, r2〉 / 〈s〉 ∼= Z2, D8/ 〈s, r2〉 ∼= Z2.

Theorem 1.8.7 (Jordan-Holder Theorem). Let G 6= 1 be a finite group. Then(1) G has a composition series (not unique),(2) The composition factors in a composition series are unique, i.e. if

1 = N0 E N1 E · · · E Nr = G and 1 = M0 E M1 E · · · E Ms = G,

then r = s and there exists some permutation σ of {1, . . . , r} such that

Mσ(i)/Mσ(i)−1∼= Ni/Ni−1, 1 ≤ i ≤ r.

Proof. Exercise.


Remark. (1) Every group has a ”‘factorization”’. Even though composition seriesis not unique the number and isomorphism type of composition factors are unique.(2) Note that non-isomorphic groups might have the same (up to isomorphism)list of composition factors, e.g. 1 E {±1}/ < i > /Q8.Composition factors: Q8/ 〈i〉 ∼= Z2, 〈i〉 /{±1} ∼= Z2, {±1}/{1} ∼= Z2, the same asD8, but Q8 6∼= D8.

Holder program is a two part-program for classifying all finite groups up toisomorphism(1) Classify all finite simple groups.(2) Find all ways of ”‘putting simple groups together”’ to form other groups.The classification of all finite simple groups, i.e. part (1) was completed in 1980.

Theorem 1.8.8. There is a list of 18 (infinite) families of simple groups and 26simple groups not belonging to these families (the sporadic simple groups) such thatevery finite simple group is isomorphic to one of the groups in the list.

A big theorem in this classification (whose proof 255 pages).

Theorem 1.8.9 (Feit-Thompson). If G is simple group of odd order, then G ∼= Zpfor some prime p.

Part (2) is very difficult, sometimes called extention problem. A class of groupswhich appear in the theory of polynomial equations is solvable groups.

Definition 1.8.10. G is called solvable if there exists a chain of subgroups

1 = G0 E G1 E G1 E · · · E Gs = G

such that Gi+1/Gi is abelian for i = 0, 1, . . . , s− 1.

(Finite solvable groups are precisely those groups whose composition factors areall of prime order.) Finite solvable groups satisfy the following generalization ofSylow’s Theorem.

Theorem 1.8.11 (Hall). Let G be a finite solvable group, then for every divisor

n of |G| such that (n, |G|n

) = 1, G has a subgroup of order n.

1.9 Sylow’s Theorems

Definition 1.9.1. Let G be a group, p prime. A group of order pα for some α ≥ 1is called a p-group. Subgroups of G which are p-groups are called p-subgroups. IfG is a group of order pαm where p - m, then a subgroup of order pα is called aSylow p-subgroup of G.

1.9. SYLOW’S THEOREMS 33

Theorem 1.9.2 (1st Sylow Theorem). Let G be a finite group, p prime. If pk | |G|then G has a subgroup of order pk. In particular G has a p-Sylow subgroup.

Theorem 1.9.3 (2nd Sylow Theorem). If P is a Sylow p-subgroup of G and Qis any p-subgroup of G, then there exists g ∈ G such thath Q ≤ gPg−1, i.e. Q iscontained in some conjugate of P . In particular, any two Sylow p-subgroups of Gare conjugate in G.

Theorem 1.9.4 (3rd Sylow Theorem). The number of Sylow p-subgroups of G,called np, is of the form 1 + kp, i.e. np ≡ 1 ( mod p). Further np(G) = [G :NG(P )], hence np | m.

For the proof of Sylow’s theorems recall following lemma and corollary

Lemma 1.9.5. If G acts on a set X. Let a ∈ X, a = Ga = {ga; g ∈ G} orbit ofa. Then |a| = [G : Ga] where Ga = {g ∈ G; g ◦ a = a} stabilizer of a in G.

Corollary 1.9.6. The number of conjugates of a set S in a group G is the indexof normalizer of S, [G : NG(S)]. In particular the number of conjugates of anelement s of G is [G : CG(S)].

We first prove the existence of Sylow p-subgroups

Proof of Theorem 1.9.2. We proceed by induction on |G|. If |G| = 2, the resultis trivial. Assume the statement is true for all groups of order less than |G|, andsuppose pk | |G|. If G has a proper subgroup H whose index is not divisible byp, then pk divides |H| so by induction hypothesis H has a subgroup of order pk,which is of course also a subgroup of G. Thus we may assume p | [G : H] for everyproper subgroup H of G. Recall the class equation

|G| = |Z(G)|+r∑i=1

[G : CG(gi)].

Since p | |G| , p | [G : CG(gi)] we know that p | Z(G). Then by Cauchy’s Theoremfor abelian groups Z(G) has a subgroup N of order p. Now since N < Z(G), N /G,G/N is a group of order |G| /p. If pk divides |G| then pk−1 | |G/N |, so by theinduction hypothesis G/N has a subgroup P of order pk−1. If we let P be thesubgroup of G containing N such that P/N = P then∣∣P ∣∣ = |P/N | ⇒ |P | = pk−1 · |N | = pk−1 · p = pk.


Before proving the 2nd and 3rd Sylow Theorems we overse some facts; By the 1stSylow Theorem, we know that there exists P ∈ Sylp(G). Let

X = {P1, P2, . . . , Pr} = {gPg−1; g ∈ G},

be the set of all conjugates of P . (Note G acts on X, G has only one orbit.) Let Qbe any p-subgroup of G. Q also acts on X by conjugation. Write X as a disjointunion of orbits under this action

X = O1 ∪ O2 ∪ · · · ∪ Os, where r = |O1|+ · · ·+ |Os| .

Note r does not depend on Q, but the number of Q-orbits s does. Renumberthe elements of X if necessary such that the Q orbits = Pi ∈ Oi, 1 ≤ i ≤ s.|Oi| = [Q : NQ(Pi)] by Lemma 1.9.5. By definition

NQ(Pi) = NG(Pi) ∩Q = Pi ∩Q,

because of the following Lemma

Lemma 1.9.7. Let P ∈ Sylp(G). If Q is any p-subgroup of G. Then

Q ∩NG(P ) = Q ∩ P.

Using these

(∗) |Oi| = [Q : NG(Pi)] = [Q : NG(Pi)∩Q] and |Oi| = [Q : Pi∩Q], 1 ≤ i ≤ s.

Since Q was arbitrary, we may take it as Q = P1, |O1| = 1. Now, for all i > 1, P1 6=Pi such that P1 ∩ Pi < P1. By (∗) we have |Oi| = [P1 : P1 ∩ Pi] > 1, 2 ≤ i ≤ s.Since P1 is a p-group, [P1 : P1∩Pi] is a power of p, hence P | Oi, 2 ≤ i ≤ s. Thusr = |O1|+ (|O2|+ · · ·+ |Os|) ≡ 1 mod p.

Now we prove Theorem 1.9.3 and Theorem 1.9.4.

Proof of Theorem 1.9.3. Let Q be any p-subgroup of G. Suppose Q is not con-tained in Pi for any i ∈ {1, . . . , r}, Q 6E gPg−1 for any g ∈ G. Then Q∩Pi < Q ∀i,hence by (*) we have |Oi| = [Q : Q ∩ Pi] > 1, 1 ≤ i ≤ s. Thus p | |Oi| , ∀i, sop divides |O1| + |O2| + · · · + |Os| = r. But this contradicts r ≡ 1 mod p (r doesnot depend on choice of Q). This proves Q ≤ gPg−1 for some g ∈ G.

To see that all Sylow p-subgroups of G are conjugate, let Q be any Sylow p-subgroup of G, by preceding argument Q < gPg−1 for some g but |Q| = |P | =|gPg−1| = pα we must have gPg−1 = Q. This proves Theorem 1.9.3.

1.9. SYLOW’S THEOREMS 35

Proof of Theorem 1.9.4. Finally since all sylow p-subgroups are conjugate, X ={all Sylow p-subgroups}. G acts on X by conjugation, P ∈ Sylp(G), orbit of Punder G is all of X. |X| = np = [G : NG(P )] for any P ∈ Sylp(G). We havealready seen that np ≡ 1 mod p.

Corollary 1.9.8. Let P be a Sylow p-subgroup of G. Then there holds equivalence(1) P is the unique Sylow p-subgroup of G, i.e. np = 1,(2) P / G,(3) p = char(G).

Examples of some applications of Sylow Theorems

Remark. No group of order 20 is simple.

Proof. If |G| = 20 = 4 · 5, G has Sylow-5 subgroups n5 ≡ 1 mod 5 and n5 | 20.Note in general np | [G : NG(P )], hence

np | [G : NG(P )] · [NG(P ) : P ] = [G : P ] =mpα

pα= m.

so in fact n5 | 4. This implies n5 = 1, hence the Sylow-p is unique and thereforenormal in G.

The following simple theorem isolates conditions under which a group G is isomor-phic to the direct product of two of its subgroups and is very useful in classificationtheorems.

Theorem 1.9.9. Let A,B ≤ G such that(1) A / G, B / G,(2) AB = G,(3) A ∩B = {e},then G ∼= A×B.

Observe (1)-(3) imply two more properties of the subgroups A and B(4) If ab = a1b1, where a, a1 ∈ A, b, b1 ∈ B, then a = a1, b = b1,(5) If a ∈ A, b ∈ B then ab = ba.

Observe that (2) says that every element of G can be written as ab for somea ∈ A, b ∈ B, (4) says this representation is unique.

Proof. To see (4) note

ab = a1b1 ⇒ a−11 a︸︷︷︸∈A

= b1b−1︸︷︷︸∈B

,


hence a−11 , b1b

−1 ∈ A ∩B = {e}, a−11 a = e = b1b

−1 and a1 = a, b1 = b. To see (5)let

bab−1a−1 = ( bab−1︸︷︷︸∈A,since A/G

)a−1 = b( ab−1a−1︸︷︷︸∈B, since B/G

),

therefore bab−1a−1 ∈ A ∩B = {e} and ab = ba.

To prove the Theorem define φ : A×B → G, (a, b) 7→ ab with

φ((a, b)(a1, b1)) = φ(aa1, bb1) = aa1bb1(5)= aba1b1 = φ(a, b)φ(a1, b1),

so φ is a homomorphism. Clearly it is onto by assumtion (2). We have

kerφ = {(a, b) ∈ A×B; φ(a, b) = ab = e},

hence (a, b) ∈ kerφ and a = b−1, a, b ∈ A ∩ B, therefore a = e, b = e and(a, b) = (e, e) = eA×B, and we have G ∼= A×B.

Theorem 1.9.10. If p, q are prime, p < q then every group G of order pq has asingle subgroup of order q, hence it is normal and G is not simple. Moreover ifq 6≡ 1 mod p then G is abelian (and in fact cyclic).

Proof. G has a Sylow-q subgroup Q. The number of such subgroups nq ≡ 1mod q and nq | p, but p < q hence nq = 1 and Q / G. Hence G is not simple.Similarly there exists P ∈ Sylp(G), np ≡ 1 mod p, np | q then np = 1 or q.Hence if q 6≡ 1 mod p then np = 1, P / G.

Every element 6= e of Q has order q (exept e) and every element 6= e of P hasorder p, hence Q ∩ P = {e}.

QP is a subgroup of G properly containing Q of order dividing pq, hence QP = G.By previous theorem G ∼= Q×P ∼= Zp×Zp. Hence G is abelian, in fact cyclic.

Example 1.9.11. Let G be a group of order 12. Then G either has a normalSylow-3 subgroup or G ∼= A4.

Proof. Suppose n3 6= 1 and P ∈ Syl3(G). n3 | 4, n3 ≡ 1 mod 3, hence n3 = 4.Since distinct Sylow-3 subgroups intersect in the identity, each Sylow-3 subgroupcontains 2 elements of order 3 and G contains 4 · 2 = 8 elements of order 3. Since[G : NG(P )] = n3 = 4 and P ≤ NG(P ) we have NG(P ) = P . Let X be the set ofSylow-3 subgroups of G, |X| = 4.

1.10. DIRECT PRODUCTS AND ABELIAN GROUPS 37

Now G acts on X by conjugation. The permutation representation of this ac-tion gives a homomorphism

ϕ : G→ SX = S4, g 7→ ϕ(g) = P → gPg−1, ∀P ∈ X.

kerϕ = {g ∈ G; ϕ(g) = identity} = {g ∈ G; gPig−1 = Pi, i = 1, 2, 3, 4}.

In particular kerϕ ≤ NG(P ) = P . Since P 6C G, kerϕ / G, kerϕ = 1, henceG ∼= ϕ(G) ≤ S4.

G contains 8 elements of order 3 and these are precisely 8 elements of order 3in S4. All these elements (of order 3) are contained in A4. Hence ϕ(G) ∩ A4

has order at least 8. But ϕ(G) ∩ A4 is a subgroup of A4, which has order|A4| = 12, |ϕ(G) ∩ A4| > 8 and divides 12, implies that |ϕ(G) ∩ A4| = 12 andhence ϕ(G) = A4 and G ∼= A4.

1.10 Direct products and Abelian groups

We’ve seen before that given two groups G1, G2 we can form another group, namelyG1 × G2, the direct product of G1 and G2 by defining the group operation com-ponentwise. In general we have

Definition 1.10.1. The direct product G1×G2× · · · ×Gn of the groups (G1, ∗1),(G2, ∗2), . . . , (Gn, ∗n) is the set of n-tuples (g1, . . . , gn), gi ∈ Gi with operationdefined as

(g1, g2, . . . , gn) ∗ (h1, . . . , hn) := (g1 ∗1 h1, g2 ∗2 h2, . . . , gn ∗n hn).

Typically even though the operation may be different in each of the factors of adirect product we drop subscrips in ”‘∗i”’s and simple write

(g1, g2, . . . , gn) ∗ (h1, . . . , hn) := (g1h1, g2h2, . . . , gnhn).

When the groups are abelian we write the operation additively and call the group

direct sum of groups G1, G2, . . . , Gn,n⊕i=1

Gi.

Direct products, aside from providing us with an easy way of building new groupsthey often also enable us to understand a given group better. This happens whenwe are able to realize that the given group is isomorphic to the direct product ofsome of its subgroups. In this way, we can break the group down into simplercomponents that are easier to deal with. We have already seen in Thm 1.9.9, theconditions under which a group G is isomorphic to the direct product of two of its


subgroups. The clue really comes from H ×K itself.

Although for given groups H,K, they are not subgroups of H ×K, there are sub-groups H∗, and K∗ of H×K such that H∗ ∼= H, K∗ ∼= K, thus H×K ∼= H∗×K∗.Namely take H∗ = H ×{eK}, K∗ = {eH}×K. What we can say about H∗, K∗?(1) H∗, K∗ are normal in H ×K

(h, k)(eH , x)(h, k)−1 = (heHh−1, kxk−1) = (eH , kxk

−1) ∈ {eH} ×K,

orπ1 : H ×K → H, (h, k) 7→ h, π2 : H ×K → K, (h, k) 7→ k

are surjective homomorphisms whose kernels are

kerπ1 = {eH} ×K = K∗, kerπ2 = H × {eK} = H∗

(2) H∗K∗ = H × K because any element (h, k) ∈ H × K can be written as(h, eK)(eH , k) with (h, eK) ∈ H∗, (eH , k) ∈ K∗.(3) Finally H∗ ∩K∗ = (eH , eK) = eH×K .And as Theorem 1.9.9 shows these 3 indicated properties captive the essense ofthe direct products.

Example 1.10.2. G = V = {e, a, b, c}, H = {e, a}, K = {e, b}. Then H,K /G, HK = {e, a, b, ab} = {e, a, b, c} = V, H ∩K = {e}, V ∼= H ×K ∼= Z2 × Z2.

Example 1.10.3. Let G = 〈x〉 cyclic group of order mn where (m,n) = 1. LetH = 〈xn〉 , K = 〈xm〉. H,K are normal subgroups of order m,n resp.

ord(xn) =ord x

(ord x, n)=

nm

(m,n)=nm

n= m.

The order of any element in H ∩K must divide |H| and |K|, hence must dividem,n but (m,n) = 1, so H ∩K = {e}.

|HK| = |H| |K||H ∩K|

=m · n

1= mn,

hence HK = G. Thus G = H ×K, i.e. Zmn ∼= Zm × Zn.

In general we have

Proposition 1.10.4. Let G1, G2, . . . Gn be groups and let G = G1 × · · · × Gn betheir direct product.(1) For fixed i, the set {(1, 1, . . . , 1, gi, 1, . . . , 1); gi ∈ Gi} is a subgroup of Gisomorphic to Gi, Gi

∼= {(1, . . . , 1, gi, 1, . . . , 1); gi ∈ G}. If we identify Gi withthis subgroup, then Gi E G and G/Gi

∼= G1 × · · · ×Gi−1 ×Gi+1 × · · · ×Gn.


(2) For each fixed i define πi : G → Gi, (g1, g2, . . . , gn) 7→ gi, then πi is asurjective homomorphism with

kerπi = {g1, . . . , gi−1, 1, gi+1, . . . , gn; gj ∈ Gj∀i 6= j}∼= G1 × · · · ×Gi−1 ×Gi+1 × · · · ×Gn.

(3) Under the identification in part (1), if x ∈ Gi, y ∈ Gj for some i 6= j, thenxy = yx.

Example 1.10.5. Let p be a prime, n ∈ Z+. Consider Epn = Zp × Zp × · · · × Zp.Then Epn is an abelian group of order pn. Every element x of Epn satisfy xp = 1.Epn is called elementary abelian group of order pn.

If n = 2, Ep2 = Zp × Zp is a group of order p2. It has exactly p + 1 subgroupsof order p (There are more than the two obvious ones). Since every non-identityelement of Ep2 has order p, each of these elements genereate a cyclic subgroup oforder p. Since distinct subgoups of order p intersect trivially, the p2−1 non-identityelements are partitioned into subsets of size p−1, i.e. each of these subsets consistsof the non-identity elements of some subgroup of order p. Hence there must bep2−1p−1

= p+ 1 subgroups of order p.

V4∼= Z2 × Z2 = {(0, 0), (0, 1), (1, 0), (1, 1)} has 3 subgroups:

{(0, 0), (0, 1)}, {(0, 0), (1, 0)}, {(0, 0), (1, 1)} of order 2.

1.10.1 The fundamental theorem of finitely generated abeliangroups

Definition 1.10.6. A group G is said to be finitely generated if there is a finiteset A of G such that G = 〈A〉 = the smallest subgroup of G containing {a; a ∈ A}.For each r ∈ Z, r ≥ 0, let Zr = Z× Z× · · · × Z︸︷︷︸

r copies

(we take Z0 = 1). The group Zr

is called the free abelian group of rank r.

Note (1) any finite group is finitely generated. Take A = G,(2) Zr is finitely generated, generated by A = {(0, 0, . . . , 0, 1, 0, . . . , 0) = ei; 1 ≤i ≤ r}.

Theorem 1.10.7 (Fundamental Theorem of Finitely Generated Abelian groups).Let G be a finitely generated abelian group. Then

G ∼= Zr × Zd1 × Zd2 × · · · × Zds


for some integers r, d1, d2, . . . , ds such that r ≥ 0, di ≥ 2 ∀i and di+1 | di for 1 ≤i ≤ s−1. Moreover the expression above is unique if G ∼= Zt×Zn1×Zn2×· · ·×Znk ,where t, nj satisfy the conditions above (t ≥ 0, nj ≥ 2 ∀j and nj+1 | nj for1 ≤ j ≤ k − 1) then t = r, s = k and di = ni ∀i.

Definition 1.10.8. The integer r is called the rank of G, d1, d2, . . . ds are calledthe invariant factors of G.

The Fundamental Theorem of Finitely Generated Abelian groups provides us witha way of listing all finite abelian groups of a given order, i.e. we must find all finitesequences of integers d1, d2, . . . , ds such that(1) dj ≥ 2 ∀j ∈ {1, . . . , s},(2) dj+1 | dh, 1 ≤ j ≤ s− 1,(3) d1d2 . . . ds = n.The Theorem says that there exists a bijection between sequences d1, d2, . . . , dssatisfying (1)-(3) and isomorphism classes of finite abelian groups of order n.

Example 1.10.9. If n = p1p2 . . . pt where pi’s are disinct then up to isomorphismthere is only one abelian group of order n,Zn.

Note that for any n, di’s satisfy d1 ≥ d2 ≥ · · · ≥ ds, i.e. d1 is the largest in-variant factor, each di | n (property (3)). If p is a prime divisor of n, then p | difor some i, by (2) p | dj ∀j ≤ i, hence every prime divisor of n divides d1.

In particular if n = p1 . . . pt, each pi | d1, hence n = p1 . . . ps | d1,but d1 | n,so n = d1. There is only one invariant factor d1 = n.

Example 1.10.10. n = 180 = 22 · 32 · 5, 2 · 3 · 5 | d1, hence the possible values ofd1 are d1 = 22 · 32 · 5, 22 · 3 · 5, 2 · 32 · 5, 2 · 3 · 5.For each of these we have to work out all possible d2. For each pair d1, d2 we workout all possible d3, etc.

Invariant factors Abelian groupsd1 = 22 · 32 · 5 Z180

d1 = 22 · 3 · 5, d2 = 3 Z60 × Z3

d1 = 2 · 32 · 5, d2 = 2 Z90 × Z2

d1 = 2 · 3 · 5, d2 = 2 · 3 Z30 × Z6

The following theorem gives a much faster way of determining all finite abeliangroups. If n = pα1

1 . . . pαkk , then all possible lists of invariant factors of abeliangroups of order n can be determined by finding all possible lists for groups oforder pαi for each i.

Theorem 1.10.11. Let G be an abelian group of order n. Let n = pα11 . . . pαkk , pi’s

are distinct primes. Then


(1) G = A1 × A2 × · · · × Ak, where |Ai| = pαii ,(2) For each A ∈ {A1, . . . , Ak} with |A| = pα

A ∼= Zpβ1 × Zpβ2 × · · · × Zpt ,

where β1 ≥ β2 ≥ · · · ≥ βt ≥ 1, β1 + β2 + · · ·+ βt = α,(3) The decomposition in (1),(2) is unique, i.e. if G = B1× · · · ×Bm with |Bi| =pαii , ∀i, then Bi

∼= Ai and Bi, Ai have the same invariant factors.

Definition 1.10.12. The integers pβi are called the elementary divisors of G. Thedecomposition in (1) and (2) is called elementary divisor decomposition of G.

Example 1.10.13. 180 = 22 · 32 · 5. 22 : Z2 × Z2, Z4, 32 : Z3 × Z3, Z9, 5 : Z5

Z2 × Z2 × Z3 × Z3 × Z5∼= Z30 × Z6

Z2 × Z2 × Z9 × Z5∼= Z90 × Z2

Z4 × Z3 × Z3 × Z5∼= Z60 × Z3

Z4 × Z9 × Z5∼= Z180

1.10.2 Semidirect products

Semidirect products is a generalization of the notion of direct products of H andK. We relax the requirement that H and K are both normal. We’ll build a largergroup G from H and K such that G contains isomorphic copies of H and K, asin the case of direct products except that H will be normal but not necessarly K.This way we can construct a non-abelian group from abelian groups H,K.

Suppose we already have a group G containing subgroups H,K such that(a) H / G (But K not necessarily normal)(b) H ∩K = 1We still have HK ≤ G a subgroup. Every element of HK can be written uniquelyas a product hk, for some h ∈ H, k ∈ K (h1k1 = h2k2 ⇒ h−1

2 h1 = k2k−11 ∈ H ∩

K = {e} ⇒ h2 = h1 and k2 = k1), i.e. there is a bijection HK → H ×K, hk 7→(h, k) and

(h1k1)(h2k2) = h1k1h2(k−11 k1)k2 = h1(k1h2k

−11 )︸︷︷︸

h3

k1k2︸︷︷︸k3

,

since H / G. These calculations were done by the assumption that there alreadyexisted a group G containing H,K with H E G, H ∩K = 1.

The product k1k2 is obtained from the multiplication in K, hence easy to un-derstand. If we understand how the element k1h2k

−11 arises in terms of H and K


without reference to G then the group HK will have been described entirely interms of H and K.

Since H / G, K acts on H by conjugation k · h := khk−1, (h1, k1)(h2, k2) =(h1(k1 · h2), k1k2). The action of K on H gives a homomorphism ϕ : K →Aut(H) = SH , k 7→ ϕ(k) : h 7→ khk−1. Multiplication in HK depends onmultiplication in K, multiplication in H and the homomorphism ϕ.

Theorem 1.10.14. Let H,K be groups, let ϕ be a homomorphism from K intoAut(H). Let G be the set of ordered pairs (h, k) with h ∈ H, k ∈ K, define themultiplication on G by

(h1, k1)(h2, k2) := (h1ϕ(k)(h), k1k2).

Then G is a group and H ∼= {(h, 1); h ∈ H} = H∗, K ∼= {(1, k); k ∈ K} = K∗,moreover H∗ E G,H∗ ∩K∗ = {e}.

Proof. Exercise.

Definition 1.10.15. The group G is called the semidirect product of H and Kwith respect to ϕ denoted by H oϕ K. We mostly drop ϕ and write H oK whenthere is no danger of confusion.

Example 1.10.16. Let H be any abelian group, K = 〈x〉 ∼= Z2 group of order2. Define ϕ : K → Aut(H), 1 7→ identity map, x 7→ ϕ(x) : h → h−1. ThenG = H oK contains the subgroup H of index 2 and xhx−1 = h−1, ∀h ∈ H.

(1, x)(h, 1)(1, x−1) = (1ϕ(x)(h), x1)(1, x−1) = (h−1, x)(1, x−1)

= (h−1ϕ(x)(1), xx−1) = (h−11−1, 1) = (h−1, 1).

If H = Zn, then G = H oK = Zn o Z2∼= D2n. The diedral group of order n is

the group of symmetries of a regular n-gon.

1.11 Free groups

Let S be a set, {si}I , I not necessarily finite. We think S as an alphabet and siare letters of the alphabet. Any symbol of the form sni , n ∈ Z is a syllable and afinite string w of syllables written in juxtaposition is a word. The empty word 1has no syllables.

Example 1.11.1. S = {s1, s2, s3} then

s1s−13 , s3

2, s32s−11 s3s

−11 s−11

1

are all words.

1.11. FREE GROUPS 43

There are two natural way of modification of certain words(1) replacing an occurence sni s

mi by sm+n

i ,(2) Replacing the occurence of s0

i by the word 1, i.e. dropping out the word.These are the two elementary contractions. By a finite number of elementarycontractions every word can be changed to a reduced word, one for which no moreelementary contraction is possible.

Example 1.11.2. s21s

32s−12 s3s

21s−71 → s2

1s22s3s

−51

Definition 1.11.3. The set of all reduced words from an alphabet S is F (S), thefree group generated by S (S is ”‘free”’ of relations).

Theorem 1.11.4. F (S) is a group under ”‘juxtaposition”’.

F (S) has the important ”‘universal property”’ that any map from the set S to agroup G can be uniquely extented to a homomorphism from the group F (S) to G.

Theorem 1.11.5. Let G be a group, S a set and f : S → G a set map. Thenthere exists a unique group homomorphism φ : F (S)→ G such that the followingdiagram commutes

S ↪→ F (S)

f ↘ ↙ φ

G

Definition 1.11.6. The cardinality of S is called the rank of the free group.

Theorem 1.11.7 (Schreier). Subgroups of a free group are free.

Warning: The rank of a subgroup of a free group can be greater than the rankof the whole group.

Example 1.11.8. Let S = {x, y}, G = F (S), a free group on two letters. Therank of G is 2. Let yk = xkyx−k, k ≥ 0. Let H = F ({yk; k ≥ 0}) the free groupgenerated by yk. Then H ≤ G, but rank(H) =∞!

Chapter 2

Rings

2.1 Basic Definitions and examples

Definition 2.1.1. A ring R is a set together with two binary operation + and ×(called addition and multiplication) satisfying the following axioms(1) (R,+) is an abelian group(2) × is associative, i.e. (a× b)× c = a× (b× c), ∀a, b, c ∈ R,(3) The Distributive laws hold: (a + b) × c = (a × c) + (b × c) and a × (b + c) =(a× b) + (a× c), ∀a, b, c ∈ R.R is called commutative if multiplication is commutative. R is said to have identity,if there exists 1 ∈ R such that a× 1 = 1× a = a ∀a ∈ R. We write ab instead ofa× b.

Example 2.1.2. (Z,+,×), (Q,+,×), (R,+,×) are all commutative rings with1.

Example 2.1.3. (2Z,+, ·) commutative ring, but without 1.

Example 2.1.4. R = {a+ b√

2; a, b ∈ Z} with the usual +,× is a ring with 1.

Example 2.1.5. (Zn,⊕,⊗)These are interesting because they already start to show behaviour different fromthat of (Z,+, ·), e.g. in Z6, 2 ·3 = 0 even though 2 6= 0 and 3 6= 0 and in Z8 23 = 0.

Definition 2.1.6. Let R be a ring. An element a ∈ R is called a zero-divisor ifthere exists b ∈ R, b 6= 0 such that ab = 0 or ba = 0.

a is called nilpotent if ∃n such that an = 0.

At the opposite extreme from these badly behavioured elements are units.

45

46 CHAPTER 2. RINGS

Definition 2.1.7. Let R be a ring with 1. An element a ∈ R is called a unit ifthere exists b ∈ R such that ab = ba = 1. The set of units in R is denoted by R×.

It is not difficult to see that R× is a group under multiplication, and hence is calledthe group of units of R.

Definition 2.1.8. A ring R with 1, where 1 6= 0 is called a division ring (skewfield) if every non-zero element a ∈ R has a multiplicative inverse. A commutativedivision ring is called a field.

Example 2.1.9. Trivial rings: R any commutative group, define a multiplicationon R via ab = 0, for all a, b ∈ R. With this multiplication R is a commutativering. If R = {0}, trivial rings do not have 1, i.e. R = {0} is the only ring 0 = 1.That’s why we will mostly write 1 6= 0 to exclude this case. These rings give nomore information that could not be obtained from abelian group theory.

Example 2.1.10. (Real) Hamiltonian Quaternions.H = {a+ bi+ cj + dk; a, b, c, d ∈ R} with

(a+ bi+ cj + dk) + (a′+ b′i+ c′j + d′k) = (a′+ a) + (b+ b′)i+ (c+ c′)j + (d+ d′)k

multiplication is defined by expending using the distributive law and i2 = j2 =k2 = −1 and ij = k, jk = i, ki = j. H is a division ring

(a+ bi+ cj + dk)−1 =a− bi− cj − dka2 + b2 + c2 + d2

.

Example 2.1.11. R = {f : [0, 1] → R; fiscontinuous}, R× = {f ∈ R; f(x) 6=0 ∀x ∈ [0, 1]}, f−1 = 1

f. R contains many zero divisors. For example let

f(x) =

{0, 0 ≤ x ≤ 1

2

x− 12, 1

2≤ x ≤ 1.

Let g(x) = f(1− x), then fg = 0.

On the other hand f(x) = x − 12

is neither a unit nor a zero-divisor. It is nota unit since it has a zero at x = 1

2. If there exists g such that gf = 0, then g = 0

for all x 6= 12, but g is continuous. Therefore g ≡ 0. Hence f is not a zero-divisor.

Example 2.1.12. M2(R) = 2×2 matrices with real entries, non-commutative ring

with 1 =

(1 00 1

), then theres exists A,B ∈M2(R) such that AB =

(0 00 0

)6= BA,

namely

A =

(1 01 0

), B =

(0 00 1

).


This explains why in the definition of zero-divisors it has ab = 0 or ba = 0.

It also contains nilpotent elements

(0 10 0

)2

=

(0 00 0

).

Basic aritmetic holds for general rings. More precisely, we have

Proposition 2.1.13. Let R be a ring, a, b ∈ R. Then(1) a · 0 = 0 · a = 0,(2) a(−b) = (−a)b = −(ab),(3) (−a)(−b) = ab,(4) if R has an identity, then identity is unique,(5) m(ab) = (ma)b = a(mb) for any integer m.

Proof. (1) a · 0 + a · 0 = a(0 + 0) = a · 0 ⇒ a · 0 = 0.(2) a(−b) = −(ab)⇔ a(−b) + ab = 0⇔ a(−b+ b) = 0⇔ a · 0 = 0, etc.

Corollary 2.1.14. Let R be a ring with 1. Let u ∈ R be a unit. Then u is not azero-divisor in R.

Proof. If there exists r ∈ R such that ur = 0 or ru = 0 then we want to show thatr = 0. But if ur = 0 then r = (u−1u)r = u−1(ur) = u−10 = 0.

Definition 2.1.15. A commutative ring with 1 6= 0 is called an integral domainif it has no zero divisors.

Example 2.1.16. Z and every field F is an integral domain since every non-zeroelement of F is a unit.

Example of fields: Q,R,C, (Zp,⊕,⊗) p prime since any r ∈ {1, 2, . . . , p − 1}satisfies (r, p) = 1 and hence a unit in Zp. Since in general a ∈ Zn is a unit ifand only if (a, n) = 1. Every element of Zn is either a unit or a zero-divisor. If(a, n) = d > 1 let b = n

dthen 0 < b < n. Hence b 6= 0 and by construction n | ab

since ab = a · nd

=(ad

)· n. Hence ab = 0 in Zn, i.e. a is a zero divisor.

The absence of zero-divisors in integral domains gives a cancellation property.

Proposition 2.1.17. Let a, b, c ∈ R, R any ring. Assume a is not a zero-divisorIf ab = ac then either a = 0 or b = c (i.e. if a 6= 0 we can cancel a’s). In particularif R is an integral domain and ab = ac, then either a = 0 or b = c.

Proof. If ab = ac then a(b − c) = 0 if a is not a zero divisor then either a = 0 orb− c = 0.

48 CHAPTER 2. RINGS

Proposition 2.1.18. Any finite integral domain is a field.

Proof. Let a ∈ R be a non-zero element of R. By cancellation law the map x→ axis injective. Since R is finite it is also surjective. Hence a is a unit, R is a field.

Definition 2.1.19. A subring of a ring R is a subgroup of R that is closed undermultiplication.

To show that a subset S of a ring R is a subring it is enough to check that S 6= ∅and it is closed under subtraction and under multiplication.

Example 2.1.20. 2Z is a subring of Z. Note even though Z is a ring with 1, 2Zhas no unity!

Example 2.1.21. R = R⊕R, S = {(r, 0); r ∈ R}. S is a subring of R⊕R,R⊕Rhas unity (1, 1) /∈ S. S has its own unity, namely (1, 0).

Example 2.1.22. R = {f : R → R} all real-valued funtions on R under point-wise addition and multiplication,

S = {f ∈ R; f is continuous}, subring of R, T = {f ∈ R; f(0) = 0}, sub-ring of R, U = {f ∈ R; f(0) = 1}, not a subring of R: If f(0) = g(0) = 1,(f − g)(0) = 0, but f − g /∈ U .

Important examples

Example 2.1.23. Let R be a ring. We’ll take R commutative with 1, even thoughthis is not necessary. Let x be an indeterminate. Let

R[x] = {an · xn + an−1 · xn−1 + . . .+ a0 · x0; n ≥ 0, ai ∈ R}.

R[x] is the ring of polynomials in the variable x with coefficients in R. Addition,multiplication are given by the familiar operation from elementary algebra.

(an · xn + an−1 · xn−1 + . . .+ a0 · x0) + (bn · xn + bn−1 · xn−1 + . . .+ b0 · x0)

= (an + bn)xn + · · ·+ (a0 + b0)

(an · xn + an−1 · xn−1 + . . .+ a0 · x0) · (bn · xn + bn−1 · xn−1 + . . .+ b0 · x0)

= a0b0 + (a0b1 + b0a1)x+ · · ·+( l∑k=0

akbl−k

)xl + . . . .

Polynomials g(x) = an · xn + an−1 · xn−1 + . . .+ a0 · x0 is said to be of degree n ifan 6= 0, anx

n is called the leading term. H is called monic if an = 1.

R is a subring of R[x] (R= constant poynomials). The ring in which the co-efficients are taken makes a big difference in the behaviour of polynomials, e.g.x2 + 1 is not a perfect square in Z[x], but x2 + 1 = (x+ 1)2 in Z/2Z[x].

2.2. IDEALS, RING HOMOMORPHISMS AND QUOTIENT RINGS 49

Proposition 2.1.24. Let R be an integral domain and p(x), q(x) ∈ R[x]. Then(1) degree(p(x)q(x)) = degree(p(x)) + degree(q(x)),(2) units of R[x] are just the units of R,(3) R[x] is an integral domain.

Proof. If R has no zero-divisors, if p(x), q(x) has leading coefficient anxn, bmx

m

resp. then the leading term of p(x)q(x) = anbmxn+m and anbm 6= 0. Hence R[x] is

an integral domain and (1) holds.

Finally to see (2): If p(x) is a unit then there exists q(x) such that p(x)q(x) =1 hence degree(p(x)) + degree(q(x)) = 0 so deg(p(x)) = deg(q(x)) = 0 andp(x), q(x) ∈ R, hence p(x), q(x) are units in R.

Example 2.1.25. Group Rings. Let R be a commuative ring with 1 6= 0, G afinite group. The group ring of G, RG is defined as the set of all formal sumsa1g1 + a2g2 + · · · + angn, ai ∈ R, 1 ≤ i ≤ n, where G = {1 = g1, . . . , gn}, ai ∈R, agi = ai ∈ RG. Similarly 1g ∈ RG, ∀g ∈ G. Addition is defined componen-twise. To define multiplication we first define (agi)(bgj) = (ab)gk where gk = gigjthe product in G. Then extend this multiplication to all formal sums by distribu-tive laws. Coefficients of gk in (

∑aigi)(

∑bjgj) is

∑gigj

= gkaibj.

If RG is commutative then G is also commutative. If |G| > 1, RG has alwayszero-devisors. If g ∈ G, ord(g) = m then (1− g)(1 + g+ · · ·+ gm−1) = 1− gm = 0,hence 1− g is a zero divisor. If S is a subgroup of R, SG is a subring of RG.

2.2 Ideals, Ring Homomorphisms and quotient

rings

In dealing with groups, we have seen that some subgroups are ”‘better”’ thanothers. For example if H < G, then G/H has a group structure iff H / G. Weencounter a similar situation with rings.

Let (R,+,×) be a ring, S a subring. Since (R,+) is abelian (S,+) is a nor-mal subgroup, and (R/S,+) is a group whose elements are true cosets of S in R,{a+ S; a ∈ R} with addition defined as (a+ S) + (b+ S) = (a+ b) + S.

We would like to define a multiplication× on the quotientR/S such that (R/S,+,×)becomes a ring. There is a natural choice, namely (a + S)(b + S) := ab + S. Weneed to check that this multiplication is well-defined, i.e. if a + S = a′ + S andb+S = b′+S, then ab+S = a′b′+S, i.e. we want to show if a−a′ ∈ S, b− b′ ∈ S

50 CHAPTER 2. RINGS

then ab− a′b′ ∈ S for any choice a, b, a′, b′ ∈ R.

For example if we take a ∈ S, a′ = 0, b arbitrary and b = b′ then we havea− a′ ∈ S, b− b′ = 0 ∈ S. We want ab− 0 · b ∈ S, i.e. we want if a ∈ S, b ∈ R,then ab ∈ S. Similary we want if a ∈ R, b ∈ S then ab ∈ S. In fact these twoconditions are enough. To see this assume a− a′ ∈ S, b− b′ ∈ S. Then

ab− a′b′ = (a− a′)b︸︷︷︸∈S

+ a′(b− b′)︸︷︷︸∈S

by the two conditions on S, hence ab − a′b′ ∈ S and our multiplication is well-defined. Subrings that have the special properties required to make multiplicationof the additive cosets well-defined are called ideals.

Definition 2.2.1. A subring S of a ring R is called an ideal of R if for all s ∈S, r ∈ R, we have rs ∈ S, sr ∈ S.

Definition 2.2.2. Let R be a ring, I a subset of R. Let r ∈ R.(1) rI = {ra; a ∈ I}, Ir = {ar; a ∈ I}(2) A subset I of R is a left ideal of R if

(i) I is a subring of R(ii) rK ⊆ I, ∀r ∈ R (closed under left multiplication). Similary for left ideals,

if Ir ⊆ I, ∀r ∈ R, i.e. I is an ideal (two-sided ideal) if it is both, a left andright ideal.

Theorem 2.2.3. Let (R,+,×) be a ring, (I,+,×) be an ideal of R. Then the setof cosets R/I is a ring under the operation (I + a) + (I + b) := I + (a + b), (I +a)(I + b) := I + ab.

Proof. Exercise.

Definition 2.2.4. R/I is called the quotient ring or factor ring of R by I.

The next theorem gives an efficient characteritic of ideals.

Theorem 2.2.5. Let R be a ring, S be a subset. Then S is an ideal of R iff thefollowing holds1) S is an additive subgroup of R (≡ S 6= ∅ and S is closed under subtraction),2) For all r ∈ R, s ∈ S, we have rs ∈ S, sr ∈ S.

Proof. Exercise.

Example 2.2.6. (Z,+,×), (nZ,+,×) is an ideal since if we multiply an elementof mZ by an element of Z, we get an element of mZ.


Example 2.2.7. (Z,+,×) is a subring of (Q,+,×) but it is not an ideal. Forexample 1 ∈ Z, 1

3∈ Q, but 1 · 1

3/∈ Z.

In fact if R is a ring with 1 then the only ideal of R that contains 1 is R itself.

Definition 2.2.8. R is always an ideal of R, R is called improper ideal of R, allother ideals are called proper ideals, {0} is called the trivial ideal.

Example 2.2.9. Let I = {(r, 0); r ∈ R} then I is an ideal of R ⊕ R. Since(r, 0)(a, b) = (ra, 0) ∈ I for any r, a, b ∈ R, then (a, b)(r, 0) ∈ I. Note I hasa unity, yet I is a proper ideal of R. Why does this not contradict the abovestatement? (Note, unity in R is not unity in I!)

With groups we have seen that if φ : G → H is a group homomorphism, thenN = kerφ is normal in G, and infact, every normal subgroup is the kernel ofa homomorphism, namely the canonical projection π : G → G/N . We have asimilar situation for rings. But first we define ring homomorphisms.

Definition 2.2.10. Let R, S be rings.(1) A ring homomorphism is a map ϕ : R→ S such that

(i) ϕ(a + b) = ϕ(a) + ϕ(b), i.e. ϕ is a group homomorphism of additive groups(R,+), (S,+),

(ii) ϕ(ab) = ϕ(a)ϕ(b) ∀a, b ∈ R.(2) The kernel of the ring homomorphism is kerφ = {r ∈ R; φ(r) = 0S}, (i.e. itis the kernel of φ viewed as homomorphism of additive groups).(3) A bijective ring homomorphism is called an isomorphism.

Example 2.2.11.

ϕ : Z→ Z/2Z, n 7→

{0, if n is even

1, if n is odd

ϕ is multiplicative since the product of even integers is even, the product of an evenand an odd integer is even and the product of two odd integers is odd, kerϕ = 2Z,which is an ideal of Z.

Example 2.2.12. Fix n ∈ Z. The maps ϕn : Z → Z, x 7→ nx are not ring ho-momorphisms in general (except for n = 0, 1). Since ϕn(xy) = nxy, ϕn(x)ϕn(y) =n2xy, hence ϕn is a ring homomorphism if n2 = n. Hence only when n = 0, 1. Butnote

ϕn(x+ y) = n(x+ y) = nx+ ny = ϕn(x) + ϕn(y),

hence ϕn is a homomorphism of groups.

52 CHAPTER 2. RINGS

Example 2.2.13. ϕ : (M2(R),+,×)→ (R,+,×),

(a bc d

)7→ det

(a bc d

).

ϕ(AB) = ϕ(A)ϕ(B), but ϕ(A+B) 6= ϕ(A)ϕ(B), hence ϕ is not a ring homomor-phism.

Example 2.2.14. Let R be the ring of real valued functions on R under pointwiseaddition and multiplication. Let α ∈ R, ϕα : R→ R, f 7→ f(α) evaluation at α,then ϕα is a ring homomrophism, since

ϕα(f + g) = (f + g)(α) = f(α) + g(α) = ϕα(f) + ϕα(g).

Similarly for ϕα(fg) = ϕα(f)ϕα(g).

Proposition 2.2.15. Let R, S be rings. ϕ : R → S be a ring homomorphism.Then

(1) The image of ϕ is a subring of S,

(2) kerϕ is an ideal of R.

Proof. (1) If s1, s2 ∈ im(ϕ), then s1 = ϕ(r1), s2 = ϕ(r2) for some r1, r2 ∈ R.Then ϕ(r1− r2) = s1− s2, ϕ(r1r2) = s1s2, hence s1− s2, s1s2 ∈ im(ϕ), so im(ϕ)is a subring of S.

(2) If a, b ∈ kerϕ then ϕ(a) = ϕ(b) = 0. Hence ϕ(a − b) = ϕ(a) − ϕ(b) =0, ϕ(ab) = ϕ(a)ϕ(b) = 0, so kerϕ is a subring of R. To see that it is an ideal,let r ∈ R, α ∈ kerϕ then ϕ(rα) = ϕ(r)ϕ(α) = ϕ(r) · 0 = 0, hence rα ∈ kerϕ.Similarly αr ∈ kerϕ, so kerϕ is an ideal.

The following theorem is the analog of the first isomorphism theorem of groups.

Theorem 2.2.16 (1st isomorphism theorem). (1) If ϕ : R → T is an onto ho-momorphism then

R/ kerϕ ∼= T.

(2) If π is the canonical homomorphism from R to R/ kerϕ then there exists anisomorphism ϕ : R/ kerϕ→ T , such that ϕ ◦ π = ϕ.

Proof. Exercise.

If ϕ : R → T is an onto homomorphism, then we have a one-to-one correspon-dence between subrings of T and subrings of R which contain kerϕ, with idealscorresponding to ideals.


Example 2.2.17. LetR = Z[x], I = {p(x) ∈ Z[x]; each term of p(x) has degree ≥ 2}∪{0}, i.e. p(x) ∈ I has no constant term and has no ”‘x”’ term. Then I is an idealsince the product of p(x) ∈ I and any q ∈ Z[x],

(a2x2 + · · ·+ anx

n)(b0 + bx+ · · ·+ bxm) = b0a2x2 + · · ·+ anbmx

n+m

has no constant term and has no x-term, hence p(x)q(x) ∈ I. Two polynomialsp, q are in the same subset ⇔ p(x)− q(x) ∈ I, i.e. they differ by a polynomial ofdegree at least 2, e.g. 1 + 3x− x4, 1 + 3x− x5 are in the same coset.

A complete set of representatives of R/I are given by polynomials of degree at most1, i.e. {a+ bx; a, b ∈ Z} = R/I, for example 1 + 3x− x4 = 1 + 3x = 1 + 3x− x5.Addition and multiplication in the quotient are performed by representatives. e.g.(1 + 3x) + (2 + 5x) = 3 + 8x, (1 + 3x)(2 + 5x) = (2 + 11x+ 15x2) = 2 + 11x.Since 15x2 = 0, i.e. 15x2 ∈ I = eR/I .

Note in R/I we have zero-divisors for example xx = x2 = 0 even though x = 0,even though R has no zero-divisors. This example shows that the structure of thequotient ring may seem worse than the original ring.

This is also the case for

Example 2.2.18. Z, 8Z is an ideal of Z. Z/8Z ∼= Z8, Z8 has zero-divisors, Z doesnot, but

Example 2.2.19. I = {0, 3} ⊆ Z6 is an ideal of Z6, Z6/I has 3 elements

0 + I = {0, 3}, 1 + I = {1, 4}, 2 + I = {2, 5}, Z6/I ∼= Z3

under the correspondence 0 + I ↔ 0, 1 + I ↔ 1, 2 + I ↔ 2. R = Z6 haszero-divisors, Z6/I = Z3 is a field.

Example 2.2.20. Reduction Homomorphism. The canonical projection map

π : Z→ Z/nZ, m 7→ m = m mod n.

is a ring homomorphism and has important applications to number theory,

e.g. x2 + y2 = 3z2 to be solved in integers (Diophantic equation). Suppose suchsolutions exists, i.e. there exist x, y, z ∈ Z such that x2 + y2 = 3z2. First weassume x, y, z have no common factors. Since if they did, we could find solutionsx′, y′, z′ that are relatively prime. If this equation holds in the ring Z, it must holdin any quotient ring Z/nZ. Choose n = 4. This is a conveniant choice since the

54 CHAPTER 2. RINGS

only squares mod 4 are 0 and 1 (02 = 0, 12 = 1, 22 = 0, 32 = 1). Therefore thepossible solutions in Z/4Z are{

01

}+

{01

}≡ 3

{01

}=

{03

}mod 4.

Checking all possibilities shows that we must take 0 each time, hence x, y, z areeven integers ((2n + 1)2 ≡ 1 (4)). But we assumed x, y, z to have no commonfactors, contradiction follows. Hence the equation x2 + y2 = 3z2 has no integersolution. (There are equations which have a solution modulo every integer but donot have integer solutions, e.g. 3x2 + 4y3 + 5z3 = 0.)

Example 2.2.21. R = {f : R→ R} under pointwise addition and multiplication.Fix c ∈ C and define

ϕc : R→ R, f 7→ f(c)

is an onto homomorphism since for every r ∈ R the constant polynomial f(x) = rhas ϕc(f) = r. Since

kerϕc = Ic = {f ∈ R; f(c) = 0}, then R/ kerϕc ∼= R.

Example 2.2.22. Let R be a commutative ring with 1, G = {g1, . . . , gn} a finitegroup and RG the group ring. Define

ϕ : RG→ R,n∑i=1

aigi 7→n∑i=1

ai,

the augmentation map. In = kerϕ = augmentation ideal = elements in RG whosecoefficients add to 0, for example gi . . . gj ∈ IA. ϕ is surjective (ae 7→ a), henceRG/IA ∼= R.

The remaining isomorphism theorems can be proved as in the case of groups.

Theorem 2.2.23 (2nd Isomorphism Theorem). Let R be a ring, S a subring, Ian ideal of R. Then

S + I = {a+ b; a ∈ S, b ∈ I}is a subring of R, S ∩ I is an ideal of S and

(S + I)/I ∼= S/(S ∩ I).

(

Theorem 2.2.24 (3rd Isomorphism Theorem). Let R be a ring, I, J ideals of R.Suppose I ⊆ J . Then J/I is an ideal of R/I and

R/I

J/I∼= R/J.

2.3. PROPERTIES OF IDEALS 55

2.3 Properties of ideals

Let R be a ring with 1 6= 0.

Definition 2.3.1. Let A be a subset of R.

(1) (A) denotes the smallest ideal of R containing A, called the ideal generated byA

(A) =⋂

I an ideal, A⊆I

I.

(2) RA denotes the set of finite terms of elements of the form ra, with r ∈ R, a ∈A

RA = {r1a1 + · · ·+ rnan; ri ∈ R, ai ∈ A, n ∈ Z+}.

Similarly for AR.

RAR = {r1a1s1 + · · ·+ rnansn; ri, si ∈ R, ai ∈ A, n ∈ Z+}.

(3) (a) denotes the ideal generated by a single element and is called the principalideal (generated by a).

(4) An ideal generated by a finite set is called finitely generated.

If R is commutative then RA = AR = RAR = (A) and (a) = {ra; r ∈ R}. Ina commutative ring, principal ideals is a particularly easy way of forming ideals.Similar to generating cyclic subgroups of a group. Note for b ∈ R, b ∈ (a) iffb = ra for some r ∈ R, i.e. b ∈ (a) iff b is a ”‘multiple”’ of a or iff a ”‘devides”’ b(i.e. divide is to contain). Note also b ∈ (a) iff (b) ⊆ (a).

Commutative rings in which all ideals are principal are among the easiest to studyand have arithmetic properties similar to Z.

Definition 2.3.2. An integral domain with the property that every ideal is prin-cipal is called a principal ideal domain (PID).

Example 2.3.3. Z is a principal ideal domain, since every ideal in Z is of theform nZ for some n.

We have observed that for a ring R and an ideal I the structure of R/I can be”‘better”’ or ”‘worse”’ than that of R.

Example 2.3.4. (a) Z has no zero-divisors, Z/6Z has zero-divisors.

(b) I = {0, 3} ∈ Z6, Z6/I ∼= Z3 has no zero-divisors.

56 CHAPTER 2. RINGS

Given R a ring, I an ideal we can sensibly ask which properties of R translate intoR/I?

For example if a ∈ R, when is I + a a zero-divisor in R/I? I + a is a zero-divisor iff there exists I + b ∈ R/I such that I + b 6= 0R/I = I + 0 and either(I + a)(I + b) = I or (I + b)(I + a) = I. Hence I + a is a zero-divisor iff thereexists b /∈ I such that either ab ∈ I or ba ∈ I. I + a is a (non-zero) zero-divisor inR/I iff a /∈ I and there exists b /∈ I such that ab ∈ I or ba ∈ I.

Example 2.3.5. 2 /∈ 6Z, 3 /∈ 6Z but 6 ∈ 6Z, hence 2 + 6Z is a zero-divisor inZ/6Z.

To rule out zero-divisors in R/I the condition we need on I is

Definition 2.3.6. Let R be a ring, I an ideal. Then I is called prime if whenevera, b ∈ R and ab ∈ I, then at least one of a or b is in I.

Example 2.3.7. p is prime in Z. Then pZ = (p) is a prime ideal because ab ∈ (p)iff p | ab iff p | a or p | b.

Theorem 2.3.8. R/I has no zero-divisors iff I is a prime ideal.

Proof. Clear from the discussion above.

Corollary 2.3.9. If R is a ring with 1, I an ideal in R. Then R/I is an integraldomain iff I is a (proper) prime ideal.

Natural question: When is R/I a field? Let R be a ring with 1. R/I is a fieldiff it is non trivial and each of its non-zero elements is a unit. R/I is nontrivial iffI is a proper ideal.Every non-zero element of R/I is a unit iff for all a ∈ R\I,there exists b ∈ R suchthat

(I + a)(I + b) = I + 1

iff(∗) ∀a /∈ I ∃b ∈ R such that ab− 1 ∈ I.

Claim 2.3.10. (∗) holds iff I is maximal.

Definition 2.3.11. An ideal I of R is called maximal if I is proper and there isno proper ideal J % I.

Proof of the claim. ”‘⇐”’ Suppose I is maximal. Take a /∈ I. Then the set

J = {ar + x; r ∈ R, x ∈ I}

2.3. PROPERTIES OF IDEALS 57

is an ideal that properly includes I, hence J = R. In particular there existsr0 ∈ R, x0 ∈ I such that ar0 + x0 = 1. Hence ar0 − 1 = −x0 ∈ I, i.e. (∗) holds.

”‘⇒”’ Suppose (∗) holds. Let J be an ideal such that I $ J . Take a ∈ J\I andb ∈ R such that y := ab− 1 ∈ I. Then a ∈ J , y ∈ I ⊂ J . Hence 1 = ab− y ∈ J .But this implies r · 1 ∈ J for all r ∈ R, therefore J = R and I is maximal.

We have shown

Theorem 2.3.12. Let R be a commutative ring with 1, I an ideal in R. ThenR/I is a field iff I is maximal.

Corollary 2.3.13. Let R be a commutative ring with 1, I an ideal. Then

I maximal⇒ I prime.

Proof. If I is maximal then R/I is a field, therefore R/I is an integral domain andhence I is prime.

Warning! I prime 6⇒ I maximal.

Example 2.3.14. R = Z ⊕ Z, I = {(a, 0); a ∈ Z} non-trivial prime ideal.((a, b), (A,B) ∈ R with (aA, bB) ∈ I then bB = 0 and hence either b = 0 orB = 0.) Let J = {(a, 2b); a, b ∈ Z} be a proper ideal, I $ J . Hence I is notmaximal.

Note we have a simple lemma

Lemma 2.3.15. Let I be an ideal of R with 1.(1) I = R iff I contains a unit.(2) Assume R is commutative. Then R is a field iff its only ideals are 0 and R.

Proof. (1) We’ve already seen.(2) Let R be a field, then every non-zero ideal contains a unit, therefore I = R by(a). Conversly if 0, R are the only ideals of R. Let u ∈ R, u 6= 0. By hypothesis(u) = R, so 1 ∈ (u). This means there exists r ∈ R such that 1 = ru, and henceu is a unit. Since u was arbitrary, R is a field.

Hence using lemma 2.3.15 we could also prove thoerem 2.3.12 using IsomorphismTheorems

R/I a field ⇔ R/I has no proper ideals.

By isomorphism theorems the ideals of J containing I correspond bijectively withthe ideals of R/I, therefore there exists no proper ideal of R/I ⇔ 6 ∃ a proper idealJ such that I $ J ⊆ R ⇔ I is maximal. Theorem 2.3.12 tells us how to constructfields. Take a ring R with 1 (commutative) and take its quotient by a maximalideal.

58 CHAPTER 2. RINGS

Proposition 2.3.16. In a ring with 1, every proper ideal is contained in a maximalideal.

Proof. We skip the proof. Uses Zorn’s Lemma. (If S is a partially ordered set suchthat every chain in S has an upper bound in S then S has at least one maximalelement.)

Definition 2.3.17. A partial ordering of a set S is given by a relation ≤ definedfor certain ordered pairs of elements of S such that the following is true(1) a ≤ a, ∀a ∈ S (reflexive),(2) if a ≤ b, b ≤ a then a = b (antisymmetric),(3) if a ≤ b, b ≤ c then a ≤ c (transitive).

Remark. In a partially ordered set, not every two elements need to be comparable,i.e. for a, b ∈ S we need not have either a ≤ b or b ≤ a.

Definition 2.3.18. A subset T of a partially ordered set S is a chain if every twoelements a, b ∈ T are comparable, i.e. either a ≤ b or b ≤ a.

Example 2.3.19. The ideal (x) in Z[x] is not maximal because x ⊂ (2, x) ⊂ Z[x],where (2, x) is the ideal generated by 2 and x in Z[x]. Observe

(2, x) = {2p(x) + xq(x); p(x), q(x) ∈ Z[x]},

the polynomials with integer coefficients whose constant term is even.

(2, x) = ker(Z[x]→ Z/2Z, p(x) 7→ p(0) mod 2

)= {p(x) ∈ Z[x]; p(0) ∈ 2Z}.

Note (2, x) is not principal since if it were, then (2, x) = (a(x)) for some a(x) ∈Z[x]. Since 2 ∈ (a(x)), there exists p(x) such that 2 = p(x)a(x).

0 = deg(p(x)a(x)) = deg(p) + deg(a),

hence both p(x), a(x) have degree 0. Since 2 is prime a(x), p(x) ∈ {±1,±2} ifa(x) = ±1} then (a(x)) = Z[x], but this is a contradiction since (2, p(x)) is proper.

If a(x) = ±2, then since x ∈ (2, x) = (a(x)) = (2) = (−2) we have x = 2q(x) forsome q(x) ∈ Z[x] and this is clearly impossible since q is an integer polynomial.

Since (x) is not maximal Z[x]/(x) is not a field. In fact Z[x]/(x) ∼= Z since

(x) = ker(ϕ : Z[x]→ Z, p(x) 7→ p(0)

).

The ideal (2, x) is maximal in Z[x] because Z[x]/(2, x) ∼= Z/2Z is a field.

2.4. RINGS OF FRACTIONS 59

Let F be a field. Using the homomorphism

ϕ : Z→ F, n 7→ 1 + 1 + · · ·+ 1︸︷︷︸n times

= n · 1

one can show that

Theorem 2.3.20. Let F be a field. Then F has a subfield that is isomorphic toeither Q or some Zp and F has only one such subfield.

Proof. Exercise.

Definition 2.3.21. The subfield of F that is isomorphic to Q or some Zp is calledthe prime subfield of F . If F is either Q or Zp, then F is its own prime subfield.Q and Zp are called prime fields.If the prime subfield of F is isomorphic to Q equivalently 1F has infinite order in(F,+), we say F has characteristic 0.If F ’s prime subfield is isomorphic to Zp, F is said to have characteristic p.

Every finite field must have characteristic p for some prime p.

The nature of the solutions of equations involving elements of F can be heav-ily influenced by whether F has characteristic 0 or p. For example 0 = x2 + 1 hasno solutions in R, but has solutions in Z2 : 12 + 1 = 0 in Z2.

2.4 Rings of fractions

Let R be a commutative ring. In Proposition 2.1.17 we’ve seen that if a is nota zero-divisor and a 6= 0 then we have: if ab=ac then b = c. Thus a non zeroelement which is not a zero-divisor enjoys some of the properties of a unit with-out necessarily possesing an inverse. (Recall that we have also seen that if a is azero-divisor then it can not be a unit.)

It turns out that a commutative ring R can always be made a subring of a largerring Q, in which every non zero element of R which is not a zero-divisor is a unitin Q. In the case that R is an integral domain Q will be a field, called its field offractions or quotient field. The construction of Q from R takes inspiration fromthe construction of rationals numbers Q from Z. What are the essential featuresof construction of Q from Z?

Any rational number q ∈ Q is a quotient of two integers, but may be repre-sented in many different ways as a quotient of two integers 1

3= 2

6= 3

9= . . . .

60 CHAPTER 2. RINGS

These reprsentations are related by

a

b=c

d⇔ ad = bc,

more precisely each fraction ab

is the equivalence class of ordered pairs of integers(a, b), b 6= 0 under the equivalence relation

(a, b) ∼ (c, d)⇔ ad = bc.

Addition and multiplication on fractions are given by

a

b+c

d=ad+ bc

bd,

a

b× c

d=ac

bd.

Z is identified with the subring {a1; a ∈ Z} of Q. Every non-zero integer n has an

inverse 1n

in Q.

Now turning to general ring R. If we allow arbitrary ”‘denominators”’ in con-structing Q we run into problems. If b is a zero-divisor in R say bd = 0 and if weallow b as a denominator then we will have (assume R has 1)

d =d

1=bd

b=

0

b= 0 in Q,

i.e. if we allow zero-divisors or 0 in the denominators, there will be some collapsingand we cannot expect R to appear naturally as a subring of ”‘ring of fractions”’.(1) Hence the set of denominators should not have zero-divisors.(2) A second restriction is more obviously imposed by laws of addition and multi-plication: if b, d are allowed as denominators, then bd should also be allowed, i.e.the set of denominators must be closed under multiplication.The next theorem shows these two conditions are sufficient to construct a ring offractions of R.

Theorem 2.4.1. Let R be a commutative ring. Let D be any non-empty subset ofR that does not contain 0 or any zero-divisors, and is closed under multiplication,i.e. for all a, b ∈ D we have ab ∈ D. Then there exists a commutative ring Q with1 such that R is a subring of Q and every element of D is a unit in Q. The ringQ has the properties(1) Every element of Q is of the form rd−1 for some r ∈ R, d ∈ D. If D = R\{0},i.e. R has no zero-divisors, then Q is a field.(2) (Uniqueness of Q) The ring Q is the smallest ring containing R in which allelements of D become units. Q is the smallest in the following sense:Let S be any commutative ring with 1. Let ϕ : R → S be any injective ring

2.5. POLYNOMIALS 61

homomorphism such that ϕ(d) is a unit in S for all d ∈ D. Then there existisan injective homomorphism Φ : Q → S such that Φ

∣∣R

= ϕ. In other words, anyring containing an isomorphism copy of R in which all elements of D become unitsmust also contain an isomorphism copy of Q.

Proof. Let Q = {(r, d); r ∈ R, d ∈ D} and ∼: (r, d) ∼ (s, e)⇔ re = sd. Then

(r, d) = {(a, b); a ∈ R, b ∈ D and rb = ad}.

a

b+c

d=ad+ bc

bd,a

b× c

d=ac

bd.

Show1) These operations are well-defined.2) Q is an abelian group under addition. Additive identity is 0

dfor all d ∈ D,

inverse of ad

is −ad.

3) Multiplication is associative, distributive and commutative.4) Q has an identity (= d

dfor any d ∈ D).

Definition 2.4.2. The ring Q is called the ring of fractions of D with respect toR and is denoted by D−1R. If R is an integral domain and D = R\{0} then Q iscalled field of fractions or quotient field of R.

Example 2.4.3. (1) If R is a field, its ring of fractions is R.(2) The ring of fractions of Z is Q.(3) 2Z also has no zero-divisors (no identity), its field of fractions is also Q.(4) Let R be an integral domain. R[x] is also an integral domain and its field offractions is the field of rational functions in x, R[x].

2.5 Polynomials

Let R be a ring, then R[x] denotes the ring of polynomials in x with coefficientsin R. We’ve seen that

Proposition 2.5.1. Let R be an integral domain, p(x), q(x) ∈ R[x], then1) deg

(p(x)q(x)

)= deg

(p(x)

)+ deg

(q(x)

),

2) units of R[x] are the units of R,3) R[x] is an integral domain.

The property being an integral domain is clearly passed from R to R[x]. Oneproperty that will obviously not be passed on is that of being a field. If F is afield, then F [x] is not a field. Clearly x is not a unit in F [x]. It is still reasonableto expect that assuming F to be a field will have some beneficial impact on F [x].

62 CHAPTER 2. RINGS

Theorem 2.5.2. Let F be a field and f(x), g(x) ∈ F [x]. If g(x) 6= 0 then thereexist q(x), r(x) ∈ F [x] such that

f(x) = q(x)g(x) + r(x)

and either f(x) = 0 or deg(r) < deg(g). (Division algorithm for F [x].)

Proof. If f(x) = 0 or if f(x) 6= 0 and deg(f) < deg(g) we write f(x) = 0·g(x)+f(x)and we are done.

We proceed induction on deg(f). If deg(f) = 0 then we are done by above unlessdeg(g) = 0. In this case both f, g are constant polynomials, f(x) = a0, g(x) = b0

and thenf(x) = (a0b

−10 )g(x) + 0.

Note b−10 exists because F is a field and b0 6= 0. Assume the result has been proved

for deg(f) < n and suppose

f(x) = a0 + a1x+ · · ·+ anxn, g(x) = b0 + b1x+ · · ·+ bmx

m, an, bm 6= 0.

If n < m we are done by the beginning of the proof. If n ≥ m we write

f(x) = anb−1m xn−mg(x) + h(x)

where h(x) = 0 or deg(h) < deg(f) = n. If h(x) = 0 we are done. If h(x) 6= 0then by inductive hypothesis we can write

h(x) = q(x)g(x) + r(x)

with r(x) = 0 or deg(r) < deg(g). Then

f(x) = anb−1m xn−mg(x) + q(x)g(x) + r(x)

=(anb−1m xn−m + q(x)

)g(x) + r(x)

and we are done.

This proof really boils down to ”‘long division”’ of polynomials. (In fact thequotient and remainder (q, r) are uniquely determined by f and g.) There aresome important corollaries of the division algorithm in F [x].

Definition 2.5.3. Let F be a field, f(x) = a0+a1x+· · ·+anxn ∈ F [x]. An elementa ∈ F is called a root (or zero) of f(x) if f(a) = 0, i.e. if a0 +a1a+ · · ·+ana

n = 0in F .

2.5. POLYNOMIALS 63

Theorem 2.5.4 (Factor theorem). Let F be a field, a ∈ F , f(x) ∈ F [x]. Thenf(a) = 0 iff (x − a) divides f(x) in F [x] (i.e. ∃h(x) ∈ F [x] such that f(x) =(x− a)h(x)).

Proof. If f(x) = (x− a)h(x) then f(a) = 0 · h(a) = 0. Let

ϕa(f) = ϕa((x− a)(h(x))

)= ϕa(x− a)ϕa(h(x)) = 0 · h(a)

be the evaluation homomorphism. Conversly suppose f(a) = 0. Apply divisionalgorithm to f(x), (x− a) to get

f(x) = (x− a)q(x) + r(x)

where either r(x) = 0 or deg(r) < deg(x− a) = 1. Thus deg(r(x)) = 0 or r = 0 ineither case there exists c ∈ F such that r(x) = c and f(x) = (x− a)q(x) + c.

Applying evaluation homomorphism ϕa again gives

1 = f(a) = ϕa(f(x)) = ϕa((x− a)q(x) + c) = 0 · q(a) + c

and therefore c = 0, so (x− a) divides f(x).

Corollary 2.5.5. A non-zero polynomial f(x) ∈ F [x] of degree n can have at mostn zeroes in a field F .

Proof. Use induction on n. If n = 0 then f(x) is a non-zero constant polynomial,hence has no roots in F . Suppose n = m + 1, the result holds for polynomials ofdegree m. Suppose f has m+ 2 roots in F say a1, a2, . . . am+2.

If we write f(x) = (x − a1)h(x) then since F is a field, in particular an inte-gral domain, i.e. has no zero-divisors. Each of a1, . . . , am+2 must be a root ofeither x − a1 or h(x) (f(ai) = 0 ⇒ either ai − a1 = 0 or h(ai) = 0). In partic-ular a2 . . . am+1 are m + 1 distinct roots of h(x). Since h(x) has degree m thiscontradicts the inductive hypothesis and completes the proof.

Corollary 2.5.6. Let F be an infinite field. S an infinite subset of F . If f(x) ∈F [x] and f(s) = 0, ∀s ∈ S. Then f ≡ 0.

Corollary 2.5.7. Let F be an infinite field, f(x), g(x) ∈ F [x], S an infinite subset.If f(s) = g(s), for all s ∈ S then f(x) = g(x) for all x ∈ F .

64 CHAPTER 2. RINGS

2.5.1 Irreducible polynomials

Definition 2.5.8. Let F be a field. A non-constant polynomial f(x) ∈ F [x] iscalled irreducible in F [x] (or irreducible over F ) if f cannot be written as theproduct of two non-constant polynomials in F [x].

In general F [x] may contain irreducible polynomials of degree higher than 2. Itis still true that every non-constant polynomial can be written as a product ofirreducible factors. Note that the concept of irreducibility over F is a conceptthat depends on the field. A polynomial f(x) may be irreducible over F but notirreducible over a larger field containing F .

Example 2.5.9. x2 − 2 is a polynomial in Q[x] but has no zeroes in Q. Hencex2 − 2 is irreducible over Q. However x2 − 2 is also a polynomial in R[x] whereit is reducible, namely (x −

√2)(x +

√2). x2 = 2 has no solutions in Q, since if

mn∈ Q, (m,n) = 1 and

(mn

)2

= 2 then m2 = 2n2, hence we have

2 | 2n2 ⇒ 2 | m2 ⇒ 2 | m⇒ 4 | m2 ⇒ 4 | 2n2

⇒ 2 | n2 ⇒ 2 | n, but (m,n) = 1,

and hence we have the contradiction.

Theorem 2.5.10. Let F be a field, f(x) a non-constant polynomial in F [x]. Thenthere exist irreducible polynomials f1(x), . . . , fk(x) ∈ F [x] such that

f(x) = f1(x) . . . fk(x).

Proof. Use induction on deg(f). If deg(f) = 1 then f(x) itself is irreducible andwe’re done. Suppose deg(f) = n, and the theorem is proved for all polynomialsof degree less than n. If f(x) is irreducible we are done. Otherwise we can writef(x) = g(x)h(x), where g, h have degree at least 1; hence each have degree strictlyless than n. So by inductive hypothesis we can factor g(x) and h(x) into irreduciblefactors. This yields the desired factorization.

If f has degree 2 or 3, f ∈ F [x], F a field, then f is reducible in F [x]⇔ f(x) has aroot in F . Note this is not true for deg(f) > 3. For example (x2 +1)(x2 +1) ∈ R[x]is clearly reducible but has no root in R. There are special techniques that can beused in case of checking irreducibility in Q[x]. In fact we can confine our attentionto polynomials with integer coefficients. Since if

f(x) =a0

b0

+a1

b1

x+ · · ·+ anbnxn ∈ Q[x],

then the irreducibility of f(x) is equivalent to the irreducibility of g(x) whereg(x) ∈ Z[x] obtained from f(x) by multiplying f(x) by b0b1 . . . bn. We have

2.5. POLYNOMIALS 65

Lemma 2.5.11. Let f(x) ∈ Z[x]. If f(x) can be written as the product of twonon-constant polynomials in Q[x] then f(x) can be written as the product of twonon-constant polynomials in Z[x].

To prove this lemma we first define

Definition 2.5.12. The content of a non-zero polynomial in Z[x] is the greatestcommon divisor of its coefficients. A polynomial is called primitive if its contentis 1.

The essential fact about those concepts is

Lemma 2.5.13 (Gauss’ lemma). The product of two primitive polynomials isprimitive.

Proof. Exercise.

Proof of lemma 2.5.11. Exercise too!

algebra 1

Documents

b mod n

group g

a1 b1 mod n

a2 b2 mod n

bya b

integers mod n

znzlet n

element e g