Introduction to Abstract Algebra I

Nicholas Camacho

Contents

Preface 5

Chapter 1. Group Theory 71. Basic Definitions 72. Maps 83. Generators and Relations 94. Symmetric, Dihedral, and Quaternion Groups 95. Group Homomorphisms 116. Subgroups 127. Cyclic Groups 138. Subgroup Lattices for Finite Groups 159. Cosets 1510. Langrange’s Theorem 1611. Normalizer of a Subgroup 1712. Normal Subgroup 1713. Quotient Groups 1814. Building subgroups from other subgroups 2015. Counting elements of HK 2016. Isomorphism Theorems 2117. More on the Symmetric Group 2318. Composition Series 2519. Solvable Groups 2620. Group Actions 2721. Stabilizers of a Group Action 2922. Normalizers and Centralizers 2923. Orbits of a Group Action and the Orbit Lemma 3024. Groups Acting on themselves by Left Multiplication 3125. Cayley’s Theorem 3126. Groups Acting on themselves by Conjugation 3127. Class Equation 3128. Finite p-groups have non-trivial center 3129. Conjugacy of Sn 3130. A5 is simple 3131. Automorphisms 3132. An is simple for n = 3 and n > 4 3133. Sylow’s Theorems 31

Index 33

3

Preface

This document contains lecture notes from the Fall 2016 course Introduction toAlgebra I, taught by Dr. Frauke Bleher at the University of Iowa. I took the libertyin some places to change the order in which some material was actually presented,so that the sections fit more neatly in each of the chapters.

5

CHAPTER 1

Group Theory

1. Basic Definitions

Definition 1.1. A group is a nonempty set G together with a binary operation

∗ : G×G −→ G

(a, b) 7−→ a ∗ bsuch that

(a) ∗ is associative, i.e., for all a, b, c ∈ Ga ∗ (b ∗ c) = (a ∗ b) ∗ c.

(b) There exists e ∈ G (called the identity of G) such that for all a ∈ Ga ∗ e = a = e ∗ a.

(c) For all a ∈ G, there exists a−1 ∈ G (called an inverse for a) such that

a ∗ a−1 = e = a−1 ∗ a.

Notation. G or (G, ∗).

We say that G is abelian if ∗ is commutative, i.e., for all a, b ∈ Ga ∗ b = b ∗ a.

Conventions.

(1) We usually write groups multiplicatively, i.e., we write ∗ as · and a · b asab. We often write the identity as 1 or 1G.

(2) If G is abelian, we write it additively, i.e., we write ∗ as +, and the identityelement as 0 (also called the zero element). The additive inverse of a ∈ Gis denoted by −a.

Definition 1.2. The order of G, denoted by |G| or #G, is the number ofelements of G. If g ∈ G, the order of g, denoted by |g| or o(g), is the smallestpositive integer n such that

gn := g · g · · · g︸ ︷︷ ︸n factors

= 1G

if such an n exists. Otherwise, g has infinite order. (If g is additively written, thisis written as ng := g + g + · · ·+ g︸ ︷︷ ︸

n summands

= 0.)

7

8 1. GROUP THEORY

Examples.

(1) G = 1G or G = 1, the trivial group.(2) (Z,+), (Q,+), (R,+), (C,+) are abelian.(3) (Q×) = (Q−0, ·), (R×) = (R−0, ·), (C×) = (C−0, ·),Z× = U(Z) =

(±1, ·).(4) Integers Modulo n (n ∈ Z), denoted Z/nZ or Z/n, where

Z/nZ = 0, 1, . . . , n− 1

and

a = r ∈ Z | n|(a− r) = a+ n` | ` ∈ Z.Addition an multiplication, a+ b := a · b, a · b := a+ b are well-defined.

Then (Z/n,+) is an additive abelian group, and

(Z/n)× = a ∈ Z/n | a has a multiplicative inverse= a ∈ Z/n | gcd(a, n) = 1

is a multiplicative abelian group of order ϕ(n) (Euler function). For ex-ample, (Z/6)× = 1, 5 and (Z/7)× = 1, 2, 3, 4, 5, 6, 7

(5) Direct Product : Let (G, ∗) and (H, ) be groups. Define

G×H = (g, h) | g ∈ G, h ∈ H

with operation given by

(g1, h1)(g2, h2) = (g1 ∗ g2, h1 h2).

Then this is a group, called the direct product of G and H.(6) Let F be a field (eg. F = Q,R,C,Z/p when p is prime). Let n ∈ Z+.

Then

GLn(F ) := A | A is an n× n matrix with entries in F and det(A) 6= 0F

is a group under matrix multiplication, called the general linear group ofdegree n over F . Then

SLn(F ) := A ∈ GLn(F ) | det(A) = 1F

is a (sub)group under matrix multiplication, called the special linear groupof degree n over F .

2. Maps

Definition 2.1. Let A,B be nonempty sets, let ϕ : A→ B be a map

• A is called the domain of ϕ.• B is called the codomain of ϕ.• For each a ∈ A, ϕ(a) is called the image of a under ϕ.• ϕ(A) = ϕ(a) | a ∈ A is called the image of ϕ.• We say ϕ is injective (or 1− 1) if whenever ϕ(x) = ϕ(y), then x = y.• We say ϕ is surjective (or onto) if ϕ(A) = B.• We say ϕ is bijective if ϕ is both injective and surjective.• Let C ⊆ A. Define ϕ|C : C → B by ϕ|C(x) = ϕ(x) for all x ∈ C. Thenϕ|C is a map, called the restriction of ϕ to C.

• A permutation of A is a bijective map ϕ : A→ A.

4. SYMMETRIC, DIHEDRAL, AND QUATERNION GROUPS 9

Definition 2.2. Let ϕ : A→ B be a map. The there exists a map ψ : B → Asuch that ψ ϕ = idA and ϕ ψ = idB , then ψ is called the inverse map of ϕ,and is denoted by ϕ−1. (Note: idA : A → A is the identity map on A, defined byidA(x) = x.)

Recall. A map ϕ : A→ B is bijective if and only if ϕ has an inverse map.

Definition 2.3. Let ϕ : A→ B be an arbitrary map. Let Y ⊆ B and b ∈ B.

• The pre-image (or inverse image) of Y under ϕ is defined as

ϕ−1(Y ) := a ∈ A | ϕ(a) ∈ Y .• The pre-image of b under ϕ (also called the fiber of ϕ above b) is defined

asϕ−1(b) := a ∈ A | ϕ(a) = b.

Examples.

(1) ϕ : Z→ Z/n, (n ∈ Z+), a 7→ a. Then• ϕ is surjective but not injective.• ϕ−1(a) = a+ nk | k ∈ Z = a+ nZ ⊆ Z.

(2) ϕ : R→ R, x 7→ cos(x). Then• ϕ−1(2) = ∅.• ϕ−1(0) = π2 + nπ | n ∈ Z.• ϕ−1(±1 = nπ | n ∈ Z.

3. Generators and Relations

Definition 3.1. Let G be an arbitrary group. A subset S ⊆ G is called aset of generators of G if every element of G can be written as a finite product ofelements of S and their inverses. In this case, we write G = 〈S〉.

Examples.

(1) G = 〈G〉.(2) D2n = 〈r, s〉.(3) Sn 〈(1, 2), (1, 2, . . . , n)〉.

Definition 3.2. If G = 〈S〉 and moreover, there exists a collection of relationsRαα∈Λ in the elements of S and their inverses, for some index set Λ, such thatevery relation among the elements of S and their inverses can be obtained fromthese, we get a presentation of G:

〈S | Rα, α ∈ Λ〉 .

Example 3.3. D2n =⟨r, s | rn = 1s2, rs = sr−1

⟩.

Be careful: Usually if g ∈ G where G is a group, then gn = 1 only means thato(g) divides n. So the notation rn = 1 = s2 is sloppy, but we understand it to meanthat r has order n and s has order 2 in this context.

4. Symmetric, Dihedral, and Quaternion Groups

Definition 4.1. Let Ω 6= ∅ be a set. Then

SΩ := σ | σ : Ω→ Ω is a permutation is a group under composition, called the symmetric group on Ω.

10 1. GROUP THEORY

Special case: Ω = 1, 2, . . . , n for some n ∈ Z+. Then SΩ is denoted by Snand called the symmetric group of degree n. Note: |Sn| = #Sn = n!.

Definition 4.2. Let 1 ≤ k ≤ n. A cycle of length k (also called a k-cycle) isa permutation γ ∈ Sn of the form

i1

ik i2

i3

. . .

for certain pairwise disjoint i1, . . . , ik ∈ 1, . . . , n and γ(j) = j for all j 6∈ i1, . . . , ik.

Notation. γ = (i1, . . . , ik) = (i2, . . . , ik, i1) = · · · = (ik, i1, . . . , ik−1).

Recall.

• Every σ ∈ Sn can be written as a product of disjoint cycles. Hence, thecycles act on pairwise disjoint sets. This decomposition is unique up topermutation of the cycles.

• We denote the identity in Sn by 1 or id or (1).• Multiplication (map composition) is from right to left: For example,

(1, 2, 3, 4, 5)(3, 5, 6, 4)(7, 2, 1) = (1, 7, 4, 3)(2)(5, 6) = (1, 7, 4, 3)(5, 6).

• Disjoint cycles commute.• Non-disjoint cycles do not commute. Hence Sn for n ≥ 3 is non-abelian.• If γ = (i1, . . . , ik) is a k-cycle, then o(γ) = k and γ−1 = (ik, . . . , i1).• If σ ∈ Sn is written as σ = γ1 · · · γr where γ1, . . . , γr are disjoint cycles,

theno(σ) = lcm(o(γ1), . . . , o(γr)),

and σ−1 = γ−1k · · · γ

−11 .

Definition 4.3. The group of symmetries of a regular n-gon is called thedihedral group of order 2n and is denoted by D2n. We have

D2n = sirj | i ∈ 0, 1, j ∈ 0, 1, . . . , n− 1such that o(s) = 2 and o(r) = n. We have rs = sr−1.

When n = 4, we can write the elements of D8 as elements of S4:

id = (1) s = (1, 3)

r = (1, 2, 3, 4) sr = (1, 2)(3, 4)

r2 = (1, 3)(2, 4) sr2 = (2, 4)

r3 = (1, 4, 3, 2) sr3 = (1, 4)(2, 3)

Definition 4.4. The quaternion group, denoted by Q8, is a group of order 8with elements

Q8 = 1,−1, i,−i, j,−j, k,−kand group presentation

Q8 =⟨i, j | i2 = j2, i4 = 1, ji = i3j

⟩.

5. GROUP HOMOMORPHISMS 11

We get the above set by setting k = ij and i2 = −1, so that

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

5. Group Homomorphisms

Definition 5.1. Let (G, ∗) and (H, ) be groups. A map ϕ : G→ H is a grouphomomorphism if for all a, b ∈ G,

ϕ(a ∗ b) = ϕ(a) ϕ(b).

The image of ϕ is defined as

Im(ϕ) := ϕ(G).

This is a group under the operation of H.The kernel of ϕ is defined as

Ker(ϕ) := a ∈ G | ϕ(a) = 1H.This is a group under the operation of G.

Recall.

• If x ∈ G, define x0 = 1G, and for all m ∈ Z+,

xm = xm−1x = x · · ·x︸ ︷︷ ︸m factors

and x−m = (x−1)m.

• If ϕ : G → H is a group homomorphism, then for all a ∈ G and for alln ∈ Z, ϕ(am) = ϕ(a)m. In particular, o (ϕ(a)) divides o(a) for all a ∈ G.

Lemma 5.2. Let ϕ : G→ H be a group homomorphism.

(a) ϕ is surjective if and only if Im(ϕ) = H.(b) ϕ is injective if and only if Ker(ϕ) = 1G.

Proof.

(a) By definition of surjective.(b) (⇒) By definition of injective.

(⇐) Assume Ker(ϕ) = 1G. Let x, y ∈ G with ϕ(x) = ϕ(y). Then

1H = ϕ(y)ϕ(x)−1 = ϕ(yx−1),

so yx−1 ∈ Ker(ϕ) = 1H. Hence y = x.

Examples.

(1) ϕ : Z → Z/n where n ∈ Z+, defined by a 7→ a is a serjective grouphomomorphism with Ker(ϕ) = nZ.

(2) ϕ : GLn(F ) → F×, where n ∈ Z+, F =field, defined by A 7→ det(A) is asurjective group homomorphism with Ker(ϕ) = SLn(F ).

Definition 5.3. Let G,H be groups. A map ϕ : G→ H is called a group iso-morphism if ϕ is a bijective group homomorphism. We say G and H are isomorphic,written G ∼= H.

12 1. GROUP THEORY

Examples.

(1) G ∼= G, with isomorphism IdG.(2) If ϕ : G→ H is an injective group homomorphism, then G ∼= Im(ϕ).(3) Let Ω be a size of order n ∈ Z+. Then SΩ

∼= Sn.

(4) If G∼=→ H is an isomorphism, then |G| = |H|, o(a) = o(ϕ(a)) for all a ∈ G.

Also, G is abelian if and only if H is abelian.(5) All groups up to isomorphism of order ≤ 7:

1, Z/2, Z/3, Z/2× Z/2, Z/4, Z/5, Z/6, S3∼= D6, Z/7.

6. Subgroups

Definition 6.1. Let G be a group, let H be a subset of G. Then H is calleda subgroup of G, written H ≤ G, if H is a group using the operation in G.

Proposition 6.2 (Subgroup Criterion). Let G be a group, let H ⊆ G. ThenH ≤ G if and only if

(a) H 6= ∅, and(b) For all x, y ∈ H, xy−1 ∈ H.

Examples.

(1) If G is an arbitrary group, it always has the subgroups 1G, G.(2) If ϕ : G → H is a group homomorphism, then Im(ϕ) ≤ H and Ker(ϕ) ≤

G.(3) SLn(F ) ≤ GLn(F ).(4) Sn ≤ Sm for all n ≤ m.(5) Let G be a group, S some subset of G. Define

H = 1G ∪ x1 · · ·xn | n ∈ Z+, xi ∈ S or x−1i ∈ S∀1 ≤ i ≤ n.1

Then H is a subgroup of G, since H 6= ∅ and if x = x1 · · ·xn ∈ H andy = y1 · · · ym ∈ H (where xi ∈ S or x−1

i ∈ S, yj ∈ S or y−1j ∈ S for all

1 ≤ i ≤ n, 1 ≤ j ≤ m), then xy−1 = x1 · · ·xny−1m · · · y−1

1 ∈ H.Then H is the smallest subgroup of G containing S and is called the

subgroup of G generated by S.

Notation. H = 〈S〉

(6) Special case of (5): S = x. Then

〈x〉 = xn | n ∈ Zis called the cyclic group of G generated by x.Laws of exponents:• x0 = 1G where 0 = 0Z.• (xn)m = xnm for all n,m ∈ Z.• xnxm = xn+m for all n,m ∈ Z.

If G = (G,+) is written additively, then 〈x〉 = nx | n ∈ Z, and• 0x = 0G where 0 = 0Z.• m(nx) = (nm)x for all n,m ∈ Z.• nx+mx = (n+m)x for all n,m ∈ Z.

1We only need 1G when S 6= 0

7. CYCLIC GROUPS 13

(7) Let G be an arbitrary group.

Definition 6.3. Let Z(G) := a ∈ G | ag = ga ∀g ∈ G. This is asubgroup of G (check the subgroup criterion) called the center of G.

7. Cyclic Groups

Definition 7.1. A group H is called cyclic if there exists x ∈ H with H = 〈x〉.Note that 〈x〉 is in particular abelian.

Lemma 7.2. Let H = 〈x〉 be a cyclic group. Then |H| = o(x). More precisely,

(a) If o(x) =∞, then xi 6= xj for i 6= j ∈ Z.(b) If o(x) = n < ∞, then 1G, x, x

2, . . . , xn−1 are pairwise distinct and H =1G, x, x2, . . . , xn−1.

Proof.

(a) Suppose o(x) =∞. If xi = xj for i, j ∈ Z, then xi−j = 1+G. By definitionof order of a group element, since o(x) =∞, this means i− j = 0.

(b) Suppose o(x) = n < ∞. Let a ∈ Z. The the division algorithm, thereexists q, r ∈ Z, 0 ≤ r < n with a = qn+ r. Then

xa = xqn+r = (xn)qxr = xr.

Hence H = 1G, x, x2, . . . , xn−1. We still need to show that these ele-ments are different: Suppose xi = xj for 0 ≤ i, j < n. Say j > i. Then1G = xj−i and 0 ≤ j − i < n. By definition of order, j − i = 0, i.e. i− j.

Theorem 7.3. Every infinite cyclic group is isomorphic to (Z,+). Every finitecyclic group of order n is isomorphic to (Z/n,+).

Proposition 7.4. Let G be a group and let x ∈ G.

(a) Let m ∈ Z. Then xm = 1G ⇐⇒ o(x)|m.(b) Let a ∈ Z− 0.

• If o(x) =∞, then o(xa) =∞.• If o(x) = n <∞, then o(xa) = n

gcd(a,n) .

(c) Let a ∈ Z.• If o(x) =∞, then 〈x〉 = 〈xa〉 ⇐⇒ a ∈ ±1.• If o(x) = n <∞, then 〈x〉 = 〈xa〉 ⇐⇒ gcd(a, n) = 1.

Proof.

(a) (⇐) If o(x)|m, then there exists q ∈ Z with q · o(x) = m. Then xm =(xo(x))q = 1qG = 1G.(⇒) Suppose xm = 1G. By the division algorithm, there exists q, r ∈ Z,0 ≤ r ≤ o(x) with m = q · o(x) + r. Then

1G = xm = (xo(x))q · xr = xr.

By definition of o(x), r = 0. Hence m = q · o(x), so o(x)|m.(b) • Suppose o(x) = ∞. If o(xa) = n < ∞, then (xa)n = 1G implies

xan = 1G. So by (a), o(x)|an, a contradiction.

14 1. GROUP THEORY

• Suppose o(x) = n <∞. Let r := o(xa) and d := gcd(a, n). Then

(xa)nd = (xn)

ad = 1G.

Hence by (a), r = o(xa)∣∣nd . Conversely, we have (xa)r = xar = 1G,

so again by (a), n = o(x)|ar and so

n

d

∣∣∣ (ad

)· r,

and since gcd(ad ,

nd

)= 1, we have n

d | r.(c) • Suppose o(x) = ∞. If a ∈ ±1, then certainly 〈x〉 = 〈xa〉. Con-

versely, if a 6∈ ±1, then x 6∈ 〈xa〉 since x 6= xai for any i ∈ Z byLemma 7.2. Hence 〈x〉 6= 〈xa〉.• Suppose o(x) = n <∞. Then

〈x〉 = 〈xa〉 ⇐⇒ 〈xa〉 ≤ 〈x〉 , o(x) = o(xa),both groups have finite order

(b)⇐⇒ n =n

gcd(a, n)

⇐⇒ gcd(a, n) = 1.

Theorem 7.5.

(1) Every subgroup of a cyclic group is cyclic.(2) The subgroups of (Z,+) are 〈m〉 where m ∈ Z. Moreover, 〈m〉 = 〈m′〉 ⇐⇒

m′ ∈ ±m. Hence there exists a 1− 1 correspondence between subgroupsof Z and Z+ ∪ 0.

(3) Let n ∈ Z+. The subgroups of (Z/n,+) are

〈m〉 , m ∈ 0, 1, . . . , n− 1.

Moreover, 〈m〉 =⟨m′⟩⇐⇒ gcd(m,n) = gcd(m′, n). In particular,

the subgroups of Z/n are in 1 − 1 correspondence to the positive divisorsof n, (including 1 and n).

Proof.

(1) Let G = 〈x〉 and H ≤ G. If H = 1G, then H = 〈1G〉. SupposeH 6= 1G. Let a ∈ Z+ be the smallest such that xa ∈ H. Then 〈xa〉 ≤ H.

Claim 7.5.1. H ≤ 〈xa〉.

Proof of Claim 7.5.1. Let h ∈ H. Then h = xm for some m ∈ Z.By the division algorithm, there exists q, r ∈ Z, 0 ≤ r < a, with m =qa+ r. Then

h = xm = (xa)q · xr.Since h, (xa)q ∈ H, then xr ∈ H. By our choice of a, r = 0. Hencem = qa, and h = xm = (xa)q ∈ 〈xa〉.

So H = 〈xa〉.(2) By (1), the subgroups of (Z,+) are 〈m〉, m ∈ Z. Moreover,

〈m〉 = 〈m′〉 ⇐⇒ m|m′ and m′|m ⇐⇒ m′ ∈ ±m.

9. COSETS 15

(3) By (1), the subgroups of (Z/n,+) are of the form 〈m〉, m ∈ 0, 1, . . . , n−1. Suppose 〈m〉 =

⟨m′⟩. Then o(m) = o(m′), so by Proposition 7.4,

n

gcd(m,n)=

n

gcd(m′, n),

and hence gcd(m,n) = gcd(m′, n). Conversely, suppose gcd(m,n) =gcd(m′, n) = α. We have 〈m〉 ≤ 〈α〉 and

⟨m′⟩≤ 〈α〉. We also have

|〈m〉| = n

gcd(m,n)=n

α= |〈α〉| ,

and similarly for∣∣⟨m′⟩∣∣. Hence 〈m〉 = 〈α〉 =

⟨m′⟩.

8. Subgroup Lattices for Finite Groups

Definition 8.1. A subgroup lattice of G is a graph with vertices corresponding

to subgroups of G, and an edgeA

B

if B ≤ A and there does not exists a

subgroup properly contained between A and B.

Examples.

(1)

Z/12 =⟨1⟩

⟨2⟩ ⟨

3⟩

⟨4⟩ ⟨

6⟩

⟨0⟩

(2)

Q8

〈i〉 〈j〉 〈k〉

〈−1〉

〈1〉

9. Cosets

Definition 9.1. Let G be a group, let H ≤ G, let g ∈ G. Define

gH := gh | h ∈ H,called the left coset of H in G with representative g.2

Lemma 9.2. Let G be a group, let H ≤ G. The left cosets of H in G form apartition of G. Moreover, aH = bH ⇐⇒ a ∈ bH ⇐⇒ b−1a ∈ H.

Proof. (For left cosets) We have

G =⋃g∈G

gH.

Consider aH and bH which have a nontrivial intersection, i.e., there exists c ∈aH ∩ bH. We need to show aH = bH.

2the right coset of H in G with representative g is defined Hg := hg | h]inH

16 1. GROUP THEORY

There exists h, h′ ∈ H such that c = ah = bh′. Then

a = bh′h−1 =⇒ a ∈ bH =⇒ aH ⊆ bH,

and similarly, bH ⊆ aH. Hence aH = bH. This also shows aH = bH ⇐⇒ a ∈ bH.Now,

a ∈ bH ⇐⇒ there exists h ∈ H with a = bh.

⇐⇒ there exists h ∈ H with b−1a = h.

⇐⇒ ba−1 ∈ H.

Lemma 9.3. Let G be a group, let H ≤ G, let g ∈ G. Then f : H → gH,f(h) = gh is bijective.

Proof. (For left cosets) f is surjective by definition of coset. If gh = gh′, thenh = h′, so f is injective.

Definition 9.4. Let G be a group, let H ≤ G. The number of distinct leftcosets of G in G is called the index of H in G, denoted by [G : H] or |G : H|.

10. Langrange’s Theorem

Theorem 10.1 (Lagrange’s Theorem). Let G be a finite group, let H ≤ G.Then |H| divides |G|, and

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

.

Proof. Let x1, . . . , xk be a complete set of representatives of dthe distsinctleft cosets of H in G. In particular, k = [G : H], and

G =

k⊔i=1

xiH.

Since |xiH| = |H| by Lemma 9.3 for each 1 ≤ i ≤ k,

|G| =k∑i=1

xiH = k |H| .

Corollary 10.2. Let G be a finite group.

(a) For all x ∈ G, o(x)∣∣ |G|

(b) If |G| = p, where p is prime, then G is cyclic.

Proof.

(a) Let H = 〈x〉. Then o(x) = |H| divides |G| by Theorem 10.1.(b) If |G| = p is primes, then |G| > 1, so there exists x ∈ G, x 6= 1G. By

Theorem 10.1, |〈x〉|∣∣ |G| = p, and hence |〈x〉| = p. So G = 〈p〉.

12. NORMAL SUBGROUP 17

11. Normalizer of a Subgroup

Definition 11.1. Let G be a group, let x, g ∈ G, let H ⊆ G. Then gxg−1 iscalled the conjugate of x by g . The set

gHg−1 := ghg−1 | h ∈ His called the conjugate of H by g.

The setNG(H) := g ∈ g | gHg−1 = H

is called the normalizer of H in G.

Remark 11.2.

• When H ≤ G, we have NG(H) ≤ G. Indeed, NG(H) 6= ∅ since 1G ∈NG(H), and if a, b ∈ NG(H),

(ab−1)H(ab−1)−1 = a(b−1Hb)a−1 = aHa−1 = H,

where the second equality follows since bHb−1 = H ⇐⇒ H = b−1Hb.• If H ≤ G, then H ≤ NG(H).

Example 11.3. Let G = S3 and H = 〈(12)〉. Then H ≤ NH(G). By La-grange’s Theorem (10.1), either NG(H) = H or NG(H) = G.

(13)(12)(13)−1 = (23) 6∈ H,so (13) 6∈ NG(H), and hence NG(H) = H.

12. Normal Subgroup

Definition 12.1. Let G be a group, let N ≤ G. We say N is a normal subgroupof G, written N E G, if for all g ∈ G, gNg−1 = N .

Theorem 12.2. Let G be a group, N ≤ G. The following are equivalent:

(a) N E G.(b) NG(N) = G.(c) For all g ∈ G, for all n ∈ N , gng−1 ∈ N .(d) For all g ∈ G, gN = Ng.

Proof.

• (a)⇐⇒ (b) By definition of NG(N).

• (a)⇐⇒ (c) Let g ∈ G and n ∈ N . Then gng−1 ∈ gNg−1 (a)= N .

• (c)⇐⇒ (d) Let g ∈ G and n ∈ N .

gn = (gng−1)g ∈ Ng by (c) =⇒ gN ≤ Ng.ng = g(g−1ng) ∈ gN by (c) =⇒ Ng ≤ gN.

• (d)⇐⇒ (a) Let g ∈ G. Then gNg−1 (d)= (Ng)g−1 = N .

Examples.

(1) If G is abelian, then H ≤ G is normal.(2) For any group G, the subgroups 1G and G are always normal. Also

Z(G) E G.(3) If H ≤ G and [G : H] = 2, then H E G.

18 1. GROUP THEORY

Proof. We know that for all g ∈ H, gH = H. Hence (since the leftcosets of H in G partition G) the only other left coset is G−H = gH forall g ∈ G−H.

Lemma 12.3. Let ϕ : G → H be a group homomorphism. Let K = Ker(ϕ).Then K E G. In fact, we show for all h ∈ Im(ϕ), say h = ϕ(g) for some g ∈ G,we have

ϕ−1(h) = x ∈ G | ϕ(x) = h = gK = Kg.

Proof. Let x ∈ G. Then

x ∈ ϕ−1(h) ⇐⇒ ϕ(x) = h = ϕ(g)

⇐⇒ ϕ(g−1x) = 1H = ϕ(xg−1)

⇐⇒ g−1x ∈ K and xg−1 ∈ K.⇐⇒ x ∈ gK and x ∈ Kg.

13. Quotient Groups

Definition 13.1. Let G be a group, let N E G. Define G/N = gN | g ∈ G,and for all a, b ∈ G,

(aN) · (bN) = (ab)N.

Theorem 13.2. G/N with this operation is a group, called the quotient groupof G modulo N .

Proof. We first need to show that the operation is well defined. SupposeaN = a1N, bN = b1N . Then a−1

1 a, b−11 b ∈ N . We need to show abN = a1b1N , i.e.

(a1b1)−1ab ∈ N . This is true since a−11 ab ∈ Nb = bN so there exists n ∈ N with

a−11 ab = bn, and so

(a1b1)−1(ab) = b−11 a−1

1 a = b−11 bn ∈ N.

We also need to check the group axioms. The identity element in G/N is 1GN = Nand the inverse of aN in G/N is a−1N . Showing the group axioms hold is left asan exercise.

Remark 13.3. |G/N | = [G : N ]. If G is finite, by Lagrange (10.1), |G/N | =#G#H .

Examples. (1) IfG is an arbitrary group, 1G, G E G. ThenG/1G ∼=G and G/G ∼= 1G.

(2) If G is abelian, all H ≤ G are normal. For example, when G = Z,H = 〈n〉 = nZ, we have G/N = Z/nZ = Z/n.

(3) Let G be an arbitrary group and let Z(G) be the center of G. If H ≤ Z(G)then H E G since for all h ∈ H, ghg−1 = g ∈ H since h ∈ H ≤ Z(G).

For example, let G = D8 =⟨r, s | r4 = 1 = s2, srs = r−1

⟩. Then

Z := Z(G) =⟨r2⟩

= 1, r2. Then

G := G/Z = Z

=

1,r2

, rZ

=

r,r3

, sZ

=

s,sr2

, srZ

=

sr,sr3

,

= 1, r, s, sr.

13. QUOTIENT GROUPS 19

Now,

r2 = (rZ)2 = r2Z = Z = 1

s2 = (sZ)2 = s2Z = Z = 1

sr2 = (srZ)2 = (sr)2Z = Z = 1

Hence G ∼= Z/2× Z/2.

Proposition 13.4. Let G be a group. Then

N | N E G = Ker(ϕ) | ϕ : G→ H a group hom. for some group H..

Proof. “ ⊇ ” : We proved this in Lemma 12.3.“ ⊆ ” : Let N ≤ G. Define

π : G −→ G/N

g 7−→ gN.

Then π is a group homomorphism since for all a, b ∈ G,

π(ab) = abN = (aN)(bN) = π(a)π(b).

We have

Ker(π) = g ∈ G | gN = N = g ∈ G | g ∈ N = N.

Definition 13.5. Let G be a group, let N E G. We call π : G→ G/N , definedby π(g) = gN , the natural projection from G onto G/N . If H ≤ G/N then the fullpreimage of H in G is defined to be

π−1(H)

:= g ∈ G | gN ∈ H.

Note. H = π−1(H)/N and N ≤ π

(H).

Recall that Lagrange (10.1) says: If G is a finite group, then

H ≤ G =⇒ #H∣∣#G.

In general, the converse is not true. For example, if G = A4, then #G = 12, but Ghas no subgroup of order 6 (check). So, what is still true?

Theorem 13.6 (Cauchy’s Theorem). If G is a finite group, and p is a primewith p

∣∣#G, then there exists x ∈ G with o(x) = p.

Proof. Omitted.

Theorem 13.7 (Sylow’s Theorem). If G is a finite group of order pa ·m, wherep is a prime, a ∈ Z+, and p 6 | m, then there exists P ≤ G with #P = pa. P iscalled a Sylow p-subgroup of G.

Proof. See Section (33).

20 1. GROUP THEORY

14. Building subgroups from other subgroups

Let G be a group.

Definition 14.1. Let H,K ≤ G. Define

HK := hk | h ∈ H, k ∈ K.

When is HK a subgroup of G?

Theorem 14.2. Let G,H,K be as in the definition. Then

HK ≤ G ⇐⇒ HK = KH.

Proof. “⇒ ” : Since H ≤ HK and K ≤ HK, then KH ⊆ HK since HK isclosed under the group operation.

Let h ∈ H, k ∈ K. Since HK ≤ G and (hk)−1 ∈ HK, there exists h1 ∈ H andk1 ∈ K with (hk)−1 = h1k1. Then

hk = ((hk)−1)−1 = (h1k1)−1 = k−11 h−1

1 ∈ KH.“⇐ ” : Suppose HK = KH. To show HK ≤ G, we use the subgroup criterion:

HK 6= ∅ since 1G ∈ H and 1G ∈ K and 1G · 1G = 1oG gives 1G ∈ HK.Let h1, h2,∈ H, k1, k2 ∈ K. Then k1k

−12 h−1

2 ∈ KH = HK, so there existsh3 ∈ H, k3 ∈ K so that k1k

−12 h−1

2 = h3k3. Then

(h1k1)(h2k2)−1 = h1k1k−12 h−1

2 = h1h3k3 ∈ HK.

Corollary 14.3. Let G,H,K be as in the definition. If H ≤ NG(K) thenHK ≤ G. In particular, if K E G then HK ≤ G for all H ≤ G.

Proof. We need to show HK = KH. Let h ∈ H, k ∈ K. Then

hk = (hkh−1︸ ︷︷ ︸∈K

)h ∈ KH,

andkh = h(h−1kh︸ ︷︷ ︸

∈K

) ∈ HK,

so HK = KH.

Definition 14.4. If S ≤ NG(T ) for some T ≤ G, we say S normalizes T .

15. Counting elements of HK

Lemma 15.1. Let G be a group. Let H,K ≤ G be finite subgroups. Then

#(HK) =(#H) · (#K)

#(H ∩K).

Proof. HK =⋃h∈H hK. We need to count the distinct left cosets of the

form hK, h ∈ H.Suppose h1, h2 ∈ H. Then

h1K = h2K ⇐⇒ h−12 h1 ∈ K

⇐⇒ h−12 h1 ∈ H ∩K

⇐⇒ h1(H ∩K) = h2(H ∩K).

16. ISOMORPHISM THEOREMS 21

Hence

#distinct left cosets hK, h ∈ H = #distinct left cosets of H ∩K in H

=#H

#(H ∩K),

the last equality following from Lagrange (10.1).Since #(hK) = #K for all h ∈ H, we get

#H

#(H ∩K)·#K.

16. Isomorphism Theorems

Theorem 16.1 (First Isomorphism Theorem). Let ϕ : G → H be a grouphomomorphism. Then Ker(ϕ) E G and G/Ker(ϕ) ∼= Im(ϕ).

Proof. Let K := Ker(ϕ). We showed in Lemma 12.3 that K E G. Define

ψ : G/K −→ Im(ϕ)

gK 7−→ ϕ(g).

• ψ is well-defined: Let gK = g′K. Then there exists k ∈ K with g = g′kand

ϕ(g) = ϕ(g′k) = ϕ(g)ϕ(k) = ϕ(g′),

so ψ(gK) = ψ(gK).• ψ is a group homomorphism since ϕ is a group homomorphism.• ψ is surjective by the definition of Im(ϕ).• ψ is injective since

Ker(ψ) = gK ∈ G/K | ϕ(g) = 1H= gK ∈ G/K | g ∈ K= K= 1G/K.

Theorem 16.2 (Second (or Diamond) Isomorphism The-orem). Let G be a group, let A,B ≤ G with A ≤ NG(B).Then AB ≤ G,B E AB, A ∩B E A, and

AB/B ∼= A/(A ∩B)

AB

A B

A ∩B

‖

‖

Proof. We already know that AB ≤ G from Corollary 14.3.

• A ∩ B E A : We have A ∩ B ≤ A. Let x ∈ A ∩ B and a ∈ A. Thenaxa−1 ∈ A since A is closed, and axa−1 ∈ B since A normalizes B. Soaxa−1 ∈ A ∩B and hence A ∩B E A.

• Define

ϕ : AB −→ A/(A ∩B)

ab 7−→ a(A ∩B).

22 1. GROUP THEORY

∗ ϕ is well-defined: Let ab = a′b′ for a, a′ ∈ A, b, b′ ∈ B. Then

(a′)−1a = b′b−1 =⇒ (a′)−1a ∈ A ∩B,

so a(A ∩B) = a′(A ∩B).∗ ϕ is a group homomorphism:

ϕ(a1b1a2b2) = ϕ(a1a2 (a−12 b1a2)b2︸ ︷︷ ︸∈B

)

= a1a2(A ∩B)

= a1(A ∩B) · a2(A ∩B)

ϕ(a1b1)ϕ(a2b2).

∗ ϕ is surjective by definition of A/(A ∩B).∗ Ker(ϕ) = ab ∈ AB|a(A∩B) = A∩B = ab ∈ AB | a ∈ A∩B = B.

Hence B E AB, and by the First Isomorphism Theorem (16.1),

AB/B ∼= A/(A ∩B).

Theorem 16.3 (Third (or Cancellation) Isomorphism Theorem). Let G be agroup, let H,K E G with H ≤ K. Then K/H E G/H and

(G/H)/

(K/H) ∼= G/K.

Proof. Define

ϕ : G/H −→ G/K

gH 7−→ gK.

• ϕ is well defined: Suppose gH = g′H. Then (g′)−1g ∈ H ⊆ K, so

ϕ(gH) = gK = g′K = ϕ(g′K).

• ϕ is a group homomorphism: Let a, b ∈ G Then

ϕ(aH · bH) = ϕ(abH) = abK = aK · bK = ϕ(aH)ϕ(bH).

• ϕ is surjective by definition of G/K.• ϕ is injective:

Ker(ϕ) = gH ∈ G/H | gK = K= gH ∈ G/H | g ∈ K= K/H.

Hence K/H E G/H and by the First Isomorphism Theorem (16.1)

(G/H)/

(K/H) ∼= G/K.

Theorem 16.4 (Fourth (or Lattice) Isomorphism Theorem). Let G be a group,let N E G. Define

G = H | N ≤ H ≤ G and G = H | H ≤ G/N.

17. MORE ON THE SYMMETRIC GROUP 23

Then the map

f : G −→ GH 7−→ H/N

is a bijection. Moreover, define G := G/N . If A,B ∈ G define A := A/N,B :=B/N . We have:

(1) A ≤ B ⇐⇒ A ≤ B.(2) If A ≤ B then [B : A] = [B : A].

(3) 〈A,B〉 =⟨A,B

⟩.

(4) A ∩B = A ∩B.(5) A E G ⇐⇒ A E G.

Proof. Let π : G→ G/N be the natural projection homomorphism.

• f is surjective: Let H ≤ G/N . Define H := π−1(H) = x ∈ G | xN ∈ H.Then H ≤ G: 1GN = 1G/N ∈ H since H is a group, so 1G ∈ H, and so

H 6= ∅. If x, y ∈ H, then xy−1N = xN ·y−1N ∈ H since H is a group, soxy−1 ∈ H. Notice that N ≤ H since for all n ∈ N , nN = N = 1G/N ∈ H.Then

f(H) = H/N = hN | h ∈ H = hN | hN ∈ H = H.

• f is injective: Let H ∈ G. Then

π−1(H/N) = x ∈ G | π(x) ∈ H/N= x ∈ G | xN ∈ H/N= HN

= H.(since N ≤ H)

So if f(H1) = f(H2) for H1, H2 ∈ G, then π−1(H1/N) = π−1(H2/N),and hence H1 = H2. Parts (a) – (e) are left as an exercise.

Important: If G is a group, N E G, π : G→ G/N the natural projection, then

• If H ≤ G/N then H = π(H)/N .• If H ≤ G then π(H) = hN | h ∈ H = HN/N ∼= H/(H ∩N).

17. More on the Symmetric Group

Let n ≥ 2.

Claim 17.1. Every σ ∈ Sn can be written as a product of 2-cycles (not neces-sarily disjoint).

Proof. We only need to show this for cycles:

(i1, i2, . . . , ik) = (i1, ik)(i1, ik1) . . . (i1, i3)(i1, i2)

Definition 17.2. A 2-cycle is called a transposition. An element σ ∈ Sn iscalled an even permutation if σ can be written as a product of an even number oftranspositions. . Otherwise σ is called an odd permutation.

24 1. GROUP THEORY

Examples.

• 1 = (12)(12) is an even permutation.• A k-cycle is even iff k is odd.

Lemma 17.3. Let σ ∈ Sn (n ≥ 2). If

σ = τ1 · · · τk = ρ1 · · · ρ`where τi, ρj are transpositions, then k ≡ ` mod 2.

Proof. It suffices to prove: If τ1 · · · τk = 1Snwhere τ1, . . . , τk are transposi-

tions, then k ≡ 0 mod 2. Define

f(x1, . . . , xn) =∏

1≤i<j≤n

(xi − xj).

So f(x1, . . . , xk) is a polynomial in x1, . . . , xn with integer coefficients. For allσ ∈ Sn, define

σ(f) =∏

1≤i<j≤n

(xσ(i) − xσ(j)).

Consider τ = (i1, i2).

Claim 17.3.1. τ(f) = f .

Proof of Claim 17.3.1.

τ(f) =

∏1≤i<j≤n

i,j∩i1,i2=∅

(xi − xj)

∏

i=i1≤i1<j≤i2

(xi2 − xj)

∏i=i1j>i2

(xi2 − xj)

(xi2 − xi1)

∏i=i2j>i2

(xi1 − xj)

∏j=i1i<i1

(xi − xi2)

∏

j=i2i1<i<i2

(xi − xi1)

∏j=i2i<i2

(xi − xi1)

= (−1)i2−i1−1 · (−1)(−1)i2−i1−1f

= −f.

Hence if τ1 · · · τk = 1Snand τi are transpositions, then (τ1 · · · τk)(f) = f implies

(−1)kf = f by the claim, and hence k ≡ 0 mod 2.

Remark 17.4. As a consequence of Lemma 17.3, the definition of even andodd permutations is well-defined. Also,

(even permutation) · (even permutation) = even

(odd permutation) · (even permutation) = odd

(even permutation) · (odd permutation) = odd

(odd permutation) · (odd permutation) = even

18. COMPOSITION SERIES 25

Definition 17.5. Let n ≥ 2. Define

ε : Sn −→ ±1

σ 7−→

1 if σ is even

−1 if σ is odd.

By Remark 17.4, ε is a group homomorphism, called the sign character . ε(σ) iscalled the sign of σ for σ ∈ Sn.

18. Composition Series

Definition 18.1. A group G (finite or infinite) is called simple if G 6= 1 andthe only normal subgroups of G are 1G and G.

Example 18.2. For all prime numbers p, Z/[ is a simple group.

Notation. • Z/p = (Z/pZ,+).• Zp = 〈x〉, the cyclic group of order p written multiplicatively.

Definition 18.3. Let G be a group. A composition series of G is a finite chainof subgroups

1 = G0 ≤ G1 ≤ · · · ≤ Gk = G

such that for all 0 ≤ i ≤ k − 1:

• Gi E Gi+1 and• Gi+1/Gi is simple.

The groups Gi+1/Gi, 0 ≤ i ≤ k − 1, are called the composition factors of G.

Careful! Gi E Gi+1 but usually Gi 6E G.

Examples.

(1) G = D8 =⟨r, s | r4 = 1 = s2, srs = r−1

⟩.

(a) 1 E 〈s〉 E⟨r2, s

⟩E D8 is a composition series with each composition

factor isomorphic to Z/2.(b) 1 E

⟨r2⟩E 〈r〉 E D8 is a composition series with each composition

factor isomorphic to Z/2.(2) G = S4.

1 E 〈(12)(34)〉 E 〈(12)(34), (13)(24)〉 E A4 E S4

is a composition series with A4/ 〈(12)(34), (13)(24)〉 ∼= Z/3 and all othercomposition factors isomorphic to Z/2.

Theorem 18.4 (Jordan-Holder Theorem). Let G be a finite group. Then

(a) G has a composition series.(b) If 1 = G0 E G1 E · · · E Gk = G and 1 = H0 E H1 E · · · E H` = G are

two composition series, then k = ` and there exists σ ∈ Sn such that

Gσ(j)

/Gσ(j)−1

∼= Hj/Hj−1

for all 1 ≤ j ≤ k.

26 1. GROUP THEORY

19. Solvable Groups

Definition 19.1. Let G be a group (finite or infinite). G is called solvable ifthere exists a finite chain of subgroups

1 = G0 E G1 E · · · E Gk = G

such that Gi+1/Gi is abelian for all 0 ≤ i ≤ k− 1. (Gi E Gi+1 but usually Gi 6E Gfor all 0 ≤ i ≤ k − 1.)

Examples. (1) If G is abelian, then G is solvable: 1 E G.(2) For n ≥ 2, D2n is solvable since 1 E 〈r〉 E D2n, each successive quotient

isomorphic to Z/2.(3) S4 is solvable since

1 E 〈(12)(34), (13)(24)〉 E A4 E S4,

and

〈(12)(34), (13)(24)〉 /1 ∼= Z/2× Z/2A4/ 〈(12)(34), (13)(24)〉 ∼= Z/3

S4/A4∼= Z/2.

We will see later (Section 32) than Sn is not solvable for n ≥ 5.

Proposition 19.2. Let G be a group, N E G. If N and G/N are solvable thenG is solvable.

Proof. Since N is solvable, we have

1 = N0 E N1 E · · · E Nr = N,

with Ni+1/Ni abelian for all 0 ≤ i ≤ r − 1. Since G/N is solvable, we have

N= G0 E G1 E · · · E Gs = G/N,

with Gi+1/Gi abelian for all 0 ≤ i ≤ s− 1.Define Hi := Ni for all 0 ≤ i ≤ r, and Hi := π−1(Gi−r) for all r ≤ i ≤ r + s

where π : G → G/N is the natural projection homomorphism. By the FourthIsomorphism Theorem (16.4), Hi/N = Gi−r for all r ≤ i ≤ r + s.

Since Gi−r E Gi−r+1 for all r ≤ i ≤ r + s − 1, then also by the FourthIsomorphism Theorem, Hi E Hi+1 for all r ≤ i ≤ r + s − 1. By the ThirdIsomorphism Theorem (16.3),

Hi+1/Hi∼= (Hi+1/N)

/(Hi/N) = Gi−r+1/Gi−r

is abelian. Hence we get a chain

1 = H0 E H1 E · · · E Hr = N E · · · E Hr+s = G

with Hi+1/Hi abelian for all 0 ≤ i ≤ r + s− 1. Hence G is solvable.

20. GROUP ACTIONS 27

20. Group Actions

Definition 20.1. Let G be a group, let M 6= ∅ be a set. A (left) group actionof G on M is a map

· : G×M −→M

(g,m) 7−→ g·msuch that for all g, h ∈ G and for all m ∈M :

(1) g·(h·m = (gh)·m.(2) 1G·m = m.

A (right) group action of G on M is a map

· : G×M −→M

(g,m) 7−→ m·gsuch that for all g, h ∈ G and for all m ∈M :

(1) (m·g)·h = m·(gh).(2) m·1G = m.

A permutation representation of G on M is a group homomorphism ϕ : G →SM , where SM is the symmetric group on M .

Proposition 20.2. Let G be a group, let M 6= ∅ be a set. There is a bijectionbetween the left group actions of G on M and the permutation representations ofG on M .

Proof. Let · : G×M →M be a left group action. Define a map

ϕ : G −→ SM

g 7−→ σg, where σm : M →M

m 7→ g·m• ϕ is well-defined: i.e.,σg ∈ SM for all g ∈ G. σg is injective since ifg·m1 = g·m2, then

m1 = 1G·m1 = g−1·(g·m1) = g−1·(g·m2) = 1G·m2 = m2,

σg is surjective since is m ∈M , then

σg(g−1·m) = g·(g−1·m) = 1G·m = m.

• ϕ is a group homomorphism: Let m ∈M, g, h ∈ G. Then

σgh(m) = (gh)·m = g·(h·m) = σg(σh(m)),

hence σgh = σg σh, so

ϕ(gh) = σgh = σg σh = ϕ(g) ϕ(h).

Therefore, ϕ is a permutation representation on M .Let ϕ : G→ SM be a permutation representation. Define

· : G×M −→M

(g,m) 7−→ g·m := ϕ(g)(m).

28 1. GROUP THEORY

Let g, h ∈ G, let m ∈M . Then

g·(h·m) = ϕ(g)(ϕ(h)(m))

= (ϕ(g) ϕ(h))(m)

= ϕ(gh)(m)(ϕ is a group hom.)

= (gh)·m.Also 1G·m = ϕ(1G)(m) = idM (m) = m. Hence · is a left group action of G on M .

We leave it as an exercise to show that the two above maps between groupactions and permutation representations are inverses of each other.

Notation. Let · : G×M →M be a left group action of G on M . Then thisaction induces a permutation representation ϕ : G → SM , g 7→ σg. ϕ is called thepermutation representation associated to the group action.

Definition 20.3. The kernel of a group action is defined as

g ∈ G | g·m = m ∀ m ∈M = g ∈ G | σG = idM = Ker(ϕ).

We say the action if faithful is the kernel is 1G.

Examples. Let G be a group, let M 6= ∅ a set.

(1) The trivial group action: For all g ∈ G for all m ∈ M , g·m = m. So thekernel of the trivial action in G.

(2) Let M = G.(a) G acts on itself by left multiplication:

G×M −→M

(g,m) 7−→ gm

This is called the left regular action of G on itself. The kernel of thisaction is 1G.

(b) G acts on itself by conjugation:

G×M −→M

(g,m) 7−→ gmg−1.

The kernel is

g ∈ G | gmg−1 = m∀ m ∈ G = Z(G).

(3) GLn(F ) acts on

Fn =

a1

...an

∣∣∣∣∣ ai ∈ F

by left multiplication, with kernel In.(4) For all n ∈ Z+, D2n and Sn act on 1, 2, . . . , n with kernel 1.

22. NORMALIZERS AND CENTRALIZERS 29

21. Stabilizers of a Group Action

Definition 21.1. Given a group action of G on M , and m ∈M , define

Gm := g ∈ G | g·m = mcalled the stabilizer of m in G

Note 21.2. The kernel of a group action is

g ∈ G | g·m = m ∀ m ∈M =⋂m∈M

Gm,

the intersection of all the stabilizers.

Claim 21.3. For all m ∈M , Gm ≤ G.

Proof. Use the subgroup criterion: Gm 6= ∅ since 1G ∈ Gm. Let g, h ∈ Gm.Then g·m = m = h·m, and so

(gh−1)·m = (gh−1)·(h·m) = (gh−1h)·m = g·m = m.

Hence gh−1 ∈ Gm.

22. Normalizers and Centralizers

Definition 22.1. Let G be a group.

(1) Let M = P(G) = A | A ⊆ G. G acts on P(G) by conjugation: For allA ∈ P(G), for all g ∈ G, g·A := gAg−1. The stabilizer of A under thisaction

GA := g ∈ G | gAg−1 = Ais called the normalizer of A in G, denoted NG(A). Since NG(A) = GA,NG(A) ≤ G.

(2) Let A ∈ P(G), let NG(A) act on A by conjugation: For all a ∈ A, for allg ∈ NG(A), g·a := gag−1. The kernel of this action is

g ∈ NG(A) | gag−1 = a ∀ a ∈ Ais call the centralizer of A in G .

Note 22.2. CG(A) ≤ NG(A) ≤ G and Z(G) = CG(A). If A ≤ G, thenZ(A) = CG(A) ∩A.

Examples.

(1) If G is abelian, then for all A ⊆ G, NG(A) = G = CG(A).(2) G = D8 =

⟨r, s | r4 = s2 = 1, srs = r−1

⟩, A = 〈r〉. Since A is abelian,

A ≤ CG(A). We have srs−1 = srs = r−1 6= r and so s 6∈ CG(A). SinceA ≤ CG(A) and [G : A] = 2, CG(A) = A or G. Since s 6∈ CG(A),CA(A) = A. Since A E G (as a subgroup of index 2), NG(A) = G.

Definition 22.3. Let G be a group acting on a set M 6= ∅. We define a

relation ∼ on M by: If m1,m2 ∈ M then m1 ∼ m2def⇐⇒ there exists g ∈ G with

m1 = g·m2.

Lemma 22.4. The above relation is an equivalence relation.

Proof.

• reflexive: For all m ∈M , m ∼ m since m = 1G·m.

30 1. GROUP THEORY

• symmetric: If m1,m2 ∈ M with m1 ∼ m2 then there exists g ∈ G withm1 = g·m2. So m2 = g−1·m1 gives m2 ∼ m1.

• transitive: If m1,m2,m3 ∈M with m1 ∼ m2,m2 ∼ m2, then there existsg, h ∈ G with m1 = g·m2 and m2 = h·m3. So m1 = g·(h·m3) = (gh)·m3,and hence m1 ∼ m3.

23. Orbits of a Group Action and the Orbit Lemma

Definition 23.1. If G acts on M and ∼ is the corresponding equivalencerelation , then the equivalence class containing m ∈M is called the G-orbit of M ,denoted by G.m = g.m | g ∈ G. (Other notation: Om).

If there is only one orbit, we call the action transitive, i.e. for all m1,m2 ∈M ,there exists g ∈ G with m1 = g.m2, i.e. Om = G.m = M for all m ∈M .

Careful! We have Gm ≤ G (stabilizer subgroup), but G.m ⊆M (G-orbit).

Lemma 23.2 (Orbit Lemma). Given a (left) group action of G on M , we havefor all m ∈M

|G.m| = [G : Gm].

Proof. Let C = gGm | g ∈ G be the set of left cosets of Gm in G. Define

f : G.m −→ Cg.m 7−→ gGm.

f is well defined and injective: Let g, h ∈ G. Then

g.m = h.m ⇐⇒ (h−1g).m = m

⇐⇒ h−1g ∈ Gm⇐⇒ gGm = hGm.

f is surjective by the definition of C. So |G.m| = |C| = [G : Gm].

Examples.

(1) G = Sn acts on M = 1, . . . , n by: For all σ ∈ Sn, for all i ∈ 1, . . . , n,σ.i := σ(i). This is a left group action (exercise). The kernel is

σ ∈ Sn | σ(i) = i ∀ 1 ≤ i ≤ n = id,

so this is a faithful action.Stabilizers: Let i ∈ 1, . . . , n. Then

Gi = σ ∈ Sn | σ(i) = i = S1,2,...,i−1,i+1,...,n ∼= Sn−1.

Orbits: This action is transitive since for all i 6= j in 1, . . . , n the trans-position τ = (i, j) ∈ Sn gives τ.i = τ(i) = j.

Note 23.3. By the Orbit Lemma (23.2), for all i ∈ 1, . . . , n,

|G.i| = [G : Gi] =#G

#Gi=

n!

(n− 1)!= n.

Hence we see each orbit has length n, so there can only be one orbit.

33. SYLOW’S THEOREMS 31

(2) For n ≥ 2, let σ ∈ Sn. Then G := 〈σ〉 acts on 1, . . . , n by

σa.i = σa(i) ∀ i ∈ 1, . . . , n, ∀ a ∈ Z.The G-orbits in 1, . . . , n give the cycles in the cycle decomposition of σ:

G.i = i, σ(i), σ2(i), σ2(i), . . . , σr−1(i)where r is the smallest positive integer with σr ∈ Gi (i.e σr(i) = i). Sincethe G-orbits partition 1, . . . , n, the cycle decomposition is unique up topermutation of the cycles.

24. Groups Acting on themselves by Left Multiplication

Definition 24.1. Let G be a group, let M = G. Then G acts on itself by leftmultiplication, the left regular action: For all g ∈ G, for all m ∈M , g.m := gm. Weemphasize here that g.m denotes the group action and gm is the group operation.

25. Cayley’s Theorem

26. Groups Acting on themselves by Conjugation

27. Class Equation

28. Finite p-groups have non-trivial center

29. Conjugacy of Sn

30. A5 is simple

31. Automorphisms

32. An is simple for n = 3 and n > 4

33. Sylow’s Theorems

Index

centerof a group, 13

centralizer, 30composition factors, 26composition series, 26conjugate

of a subgroup, 17of an element, 17

coset, 16cycle

in a symmetric group, 10

dihedral group, 10direct product, 8

even permutation, 24

general linear group, 8generators

of a group, 9group, 7

trivial, 8abelian, 7cyclic, 13identity element, 7inverse element, 7

group action, 27left regular group action, 29faithful, 28kernel, 28orbit, 30stabilizer, 29transitive, 30trivial, 28

homomorphismof groups, 11

image of a group hom., 11

index, 16integers modulo n, Z/n, 8inverse image , see also pre-imageisomorphism

of groups, 11isomorphism theorems

for groups, 21

Jordan-Holder Theorem, 26

kernelof a group hom., 11

Lagrange’s Theorem, 16laws of exponents, 12

mapidentity, 9inverse, 9bijective, 8codomain, 8domain, 8image, 8injective, 8permutation, 8restriction of, 8surjective, 8

natural projectionof groups, 20

normal subgroup, 18normalizer, 17, 29

odd permutation, 24order

of a group, 7of a group element, 7

permutation representation, 27pre-image, 9

33

34 INDEX

presentationof a group, 9

quaternion group, 10quotient group, 19

relations, 9

signcharacter (of Sn), 25

of a permutation, 25solvable group, 26special linear group, 8subgroup, 12subgroup lattice, 15subgroup lattices, 15Sylow’s Theorem, 20, 31symmetric group, 9

transposition, 24