Abstract Algebra

Paul MelvinBryn Mawr College

Fall 2011

lecture notes loosely based on Dummit and Foote’s text

Abstract Algebra (3rd ed)

Prerequisite:

Linear Algebra (203)

1

Introduction

Pure Mathematics

Algebra

Geom etry & Topology

Analysis

Foundations (set theory/logic)

What is Algebra?

• Number systems N = {1, 2, 3, . . . } “natural numbers”

Z = {. . . ,−1, 0, 1, 2, . . . } “integers”

Q = {fractions} “rational numbers”

R = {decimals} = pts on the line “real numbers”

C = {a+ bi | a, b ∈ R, i =√−1} = pts in the plane “complex nos”

‖ polar form

reiθ, where a = r cos θ, b = r sin θ

a

ba+ bi

rθ

Note N ⊂ Z ⊂ Q ⊂ R ⊂ C (all proper inclusions, e.g.√

2 6∈ Q; exercise)There are many other important number systems inside C.

2

• Structure “binary operations” + and ·

associative, commutative, and distributive properties“identity elements” 0 and 1 for + and · resp.

solve equations, e.g. i1 ax2 + bx+ c = 0 has two (complex) solutions

x =−b±

√b2 − 4ac

2a

i2 x2 + y2 = z2 has infinitely many solutions, even in N(the “Pythagorian triples”: (3,4,5), (5,12,13), . . . ).i3 xn + yn = zn has no solutions x, y, z ∈ N for any fixed n ≥ 3(Fermat’s Last Theorem, proved in 1995 by Andrew Wiles; we’ll givea proof for n = 3 at end of semester).

• Abstract systems groups, rings, fields, vector spaces, modules, . . .

A group is a set G with an associative binary operation ∗ which hasan identity element e (x ∗ e = x = e ∗x for all x ∈ G) and inverses foreach of its elements (∀x ∈ G,∃ y ∈ G such that x ∗ y = y ∗ x = e).

Examples (N,+) is not a group: no identity. (Z,+) is. (Z, ·) is not:no inverses. (Q, ·) isn’t, but (Q− {0}, ·) is.

Focus of first semester: groups and rings

Some history Theory of equations ( modern algebra)

• Quadratic equation (antiquity)

ax2 + bx+ c = 0

Divide by a and complete the square, i.e. substitute y = x+ b/2a to get

y2 + p = 0

where p = c/a − (b/2a)2. The roots are y = ±√−p which gives the usual

formula by substitution.

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Remark ∆ = b2− 4ac = −4a2p is called the discriminant of the polyno-mial ax2 + bx + c. If a, b, c ∈ R, then the equation has 2, 1 or 0 real rootsaccording to the sign of ∆:

∆ > 0 ∆ = 0 ∆ < 0

Aside (complex mult and roots) If z = reiθ and z′ = r′eiθ′, then

zz′ = rr′ei(θ+θ′),

i.e. lengths multiply and angles add (prove this using trigonometry).∗

z′

z

r′

rθ′

θ

zz′

rr′

It follows that each nonzero complex number has two square roots, threecube roots, etc. In particular, the nth roots of z = reiθ are

u, uω, . . . , uωn−1

where u = r1/neiθ/n and ω = e2πi/n. These points are equally distributedon a circle of radius r1/n about the origin (since multiplication by ω rotatesC about 0 by 2π/n radians).

• Cubic equation (16th century: del Ferro, Tartaglia Cardan, Viete)

ax3 + bx2 + cx+ d = 0

Divide by a and complete the cube, i.e. substitute y = x+ b/3a to eliminatethe quadratic term. This gives

y3 + py + q = 0∗Thus for any z, z′ in the unit circle S1, we have zz′, z−1 ∈ S1 so (S1, ·) is a group!

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which has three roots y1, y2, y3. They can be found using Viete’s magicalsubstitution

y = v(z) = z − p

3z.

Indeed, this leads to a quadratic equation in w = z3,

w2 + qw − (p/3)3 = 0

whose roots are w = −q/2±√

∆, where ∆ = (q/2)2 + (p/3)3. Thus

z =(−q/2±

√∆)1/3

(choose any one of the six possible cube roots) which yields Cardan’s formula

y1 = v(z) y2 = v(ωz) y3 = v(ω2z)

where ω = e2πi/3 = −(1/2) + (√

3/2)i. The roots of the original equationare now obtained by subtracting b/3a.

Example Find the roots of x3 + 6x2 + 9x+ 2.

Solution Substituting y = x + 2 gives y3 − 3y. Viete’s substitution isy = z + z−1, giving the quadratic w2 + 1 = 0 in w = z3, with roots w = ±i.Thus z is any cube root of ±i, say z = i. Cardan’s formula then givesy1 = i+ i−1 = 0, y2 = ωi+ (ωi)−1 = −

√3, y3 = ω2i+ (ω2i)−1 =

√3 (draw

a picture), and so x1 = −2, x2 = −√

3− 2, x3 =√

3− 2 are the roots.

Exercise (not to hand in) Verify that y1, y2, y3 are indeed roots of theeqn. Hint: r is a root ⇐⇒ (y − r) is a factor of y3 + py + q, so it sufficesto show y3 + py + q = (y − y1)(y − y2)(y − y3). But expanding out thelast expression gives y3 − (y1 + y2 + y3)y2 + (y1y2 + y1y3 + y2y3)y − y1y2y3.So you must show i1 y1 + y2 + y3 = 0, i2 y1y2 + y1y3 + y2y3 = p andi3 y1y2y3 = q. Use the fact that ω3 = 1 and 1 + ω + ω2 = 0.

Question What is the significance of the sign of ∆?

• Quartic equation 16th century: Ferrari

• Quintic equation (and higher degree) 19th century: Abel proved thatthere does not exist a formula for the roots! Galois developed general theory(second semester) birth of modern algebra.

5

Some arithmetic (study of the natural numbers)

Definition Say d divides (or is a divisor of) a , written d|a, if ∃c witha = cd. Denote the greatest common divisor of a and b by gcd(a, b).

GCD Lemma ∀ a, b ∈ N, ∃ x, y ∈ Z such that gcd(a, b) = ax+ by.

(e.g. gcd(10, 14) = 2 = 10 · 3 + 14 · (−2))

Proof Consider S = {ax + by | x, y ∈ Z}. (Remark S is closed undersubtraction, and multiplication by any integer; we say S is an ideal in Z.)Set

d = min(S ∩ N)

the smallest positive elt of S. Claim d = gcd(a, b). Must show

i1 d|a, d|b i2 e|a, e|b =⇒ e ≤ d

i1 Clearly a = qd + r for some q, r ∈ Z w/ 0 ≤ r < d, so r = a − qd ∈ S(by the remark). By the minimality of d, r = 0, and so d|a. Similarly d|b.i2 e|a, e|b =⇒ e|(ax+by) for all x, y, and in particular e|d. Thus e ≤ d.

This proof used structural prop of Z (+, ·, <: Z is an “ordered ring”),the “Well Ordering Principle” (WOP) and the “Division Algorithm” (DA):

WOP Every non-empty subset of the natural numbers has a least element(or in symbols, S ⊂ N, S 6= ∅ =⇒ ∃m ∈ S such that m ≤ s for all s ∈ S)

DA ∀a, d ∈ N, ∃q, r ∈ Z with 0 ≤ r < d such that a = qd+ r

In fact can prove DA from WOP: Let S = {a − xd | x ∈ Z, a ≥ xd} ⊂N ∪ {0}. By WOP, ∃r = minS = a− qd, i.e. a = qd+ r, for some q. Thenr < d. For if r ≥ d, then 0 ≤ r − d = a − (q + 1)d < r, contradicting theminimality of r.

Where does GCD lead? To Euclid’s Lemma and the

Fundamental Theorem of Arithmetic (FTA) Every natural no. n > 1can be written as a product of primes. This product is unique up to theorder of the factors.

(A natural number is prime if it has exactly two divisors)

6

Euclid’s Lemma If p is prime and p|ab, then p|a or p|b.

Proof Suppose p - a. Then gcd(a, p) = 1 (since p is prime) and so bythe GCD Lemma, 1 = ax+ py for some x, y. Thus b = 1 · b = (ax+ py)b =(ab)x+ pyb, which is clearly divisible by p.

Summarizing the logic so far: WOP =⇒ Euclid’s Lemma (via DA andGCD). For FTA, also need “induction”, which we use informally in thefollowing proof (see below for more formal treatment).

Proof of FTA (existence) If n is prime, there’s nothing to prove. If not,then n = ab for some a, b < n. But then by induction, each of a, b has aprime decomposition. Put these together to get one for n.

(uniqueness) Suppose p1 · · · pr = q1 · · · qs (all pi’s and qj ’s are prime).Clearly p1|p1 · · · pr, so p1|q1q2 · · · qs. By Euclid’s Lemma and induction, p1

divides at least one of the qj ’s; can assume p1|q1 by reordering. But thisimplies p1 = q1 since q1 is prime. Thus p2 · · · pr = q2 · · · qs. The resultfollows by induction.

Induction

Principle of Induction If S ⊂ N satisfiesi1 1 ∈ S and i2 n ∈ S =⇒ n+ 1 ∈ S

then S = N.

Theorem WOP ⇐⇒ Induction

=⇒ is HW #2. This shows that in fact WOP =⇒ FTA

Example Prove by induction on n that 1 + · · ·+ n = n(n+ 1)/2.

Proof Let S = {k ∈ N | 1 + · · · + k = k(k + 1)/2}. Then 1 ∈ S, since1 = 1(2)/2. Assume n ∈ S, and so

1 + · · ·+ n = n(n+ 1)/2.

We must show n+ 1 ∈ S. Adding n+ 1 to both sides gives

1 + · · ·+ n+ (n+ 1) = n(n+ 1)/2 + (n+ 1)= (n(n+ 1) + 2(n+ 1))/2= (n+ 1)(n+ 2)/2 = (n+ 1)((n+ 1) + 1)/2.

Hence n+ 1 ∈ S, and so by induction, S = N.

7

I Foundations

§0. Sets

Assume familiarity with basics of set theory:set S = {· · · | · · · }

elements x ∈ S

subsets A ⊂ S

proper subset A $ S

union S ∪ T = {x | x ∈ S or x ∈ T}

intersection S ∩ T = {x | x ∈ S and x ∈ T}

difference S − T = {x | x ∈ S and x /∈ T}

cartesian product S × T = {(s, t) | s ∈ S, t ∈ T}

cardinality |S| = # elements in S (if S is finite)

Notation: ∀,∃, !,=⇒,⇐⇒ (if and only if, or iff), =⇒⇐= (contradiction).

§1. Functions

Definition A function f : S → T (also written Sf→ T ) consists of a

pair of sets S, T (the domain and codomain of the function) and a “rule”s 7→ f(s) assigning to each element s ∈ S an element f(s) ∈ T . (To beprecise, a “rule” consists of a subset R ⊂ S × T satisfying ∀s ∈ S, ∃! t ∈ Tsuch that (s, t) ∈ R, so a function is really a triple (S, T,R ⊂ S × T )...)

Remark The domain and codomain are essential parts of the function.For example the functions R → R and R → R+ (the real nos ≥ 0), bothgiven by the rule x 7→ x2, are distinct.

Examples i1 identity functions idS) : S → S, s 7→ s.i2 inclusion of a subset A ⊂ S: A ↪→ S, a 7→ a.i3 restriction of f : S → T to a subset A ⊂ S: f |A : A→ T , a 7→ f(a).

8

i4 projections S ← S × T → T , s← (s, t) 7→ t.i5 constant functions S → T , s 7→ t0, where t0 is a fixed elt of T .i6 composition of functions Given functions g : R → S and f : S → T ,define f ◦ g : R→ T by (f ◦ g)(r) = f(g(r)). Thus f ◦ g (also written fg) isdefined by the “commutative diagram”

Rf◦g−→ T

g↘ ↗fS

Definition Given f : S → T and subsets S0 ⊂ S and T0 ⊂ T , define theimage of S0 under f to be

f(S0) := {f(s) | s ∈ S0} ⊂ T

and the preimage of T0 under f to be

f−1(T0) := {s ∈ S | f(s) ∈ T0} ⊂ S.

(examples and pictures) Call f(S) the image of f , also denoted im(f).

Proposition 1.1 ia f−1(P ∪Q) = f−1(P ) ∪ f−1(Q)ib f−1(P ∩Q) = f−1(P ) ∩ f−1(Q)

Proof ia (Note: often prove = of sets in two steps, ⊂ and ⊃, thoughsometimes done simultaneously) x ∈ f−1(P ∪ Q) ⇐⇒ f(x) ∈ P ∪ Q ⇐⇒f(x) ∈ P or f(x) ∈ Q ⇐⇒ x ∈ f−1(P ) or x ∈ f−1(Q) ⇐⇒ x ∈ f−1(P ) ∪f−1(Q) ib Same as ia with ∪ and “or” replaced with ∩ and “and”.

Definition A function f : S → T is one-to-one (or monic or injective) ifit maps at most one element of S to any given element of T . In other words,f(x) = f(y) =⇒ x = y, or equivalently |f−1(t)| ≤ 1 for all t ∈ T .

It is onto (or epic or surjective) if it maps at least one element of S toeach element of T , i.e. im(f) = T In other words, ∀t ∈ T , ∃s ∈ S withf(s) = t, or equivalently |f−1(t)| ≥ 1 for all t ∈ T .

Exercise Classify x 7→ x2 as a map R→ R, R+ → R, R→ R+, R+ → R+

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Proposition 1.2 f : S → T isia monic ⇐⇒ ∃` : T → S with l ◦ f = idS (` is a “left inverse” of f)ib epic ⇐⇒ ∃r : T → S with f ◦ r = idT (r is a “right inverse” of f)

Proof (Sketch) ( ia =⇒) For t ∈ im(f), define `(t) to be the (unique)s ∈ f−1(t), and define `(t) arbitrarily for t /∈ im(f). ( ib =⇒) Define r(t) =any s ∈ f−1(t).

Definition If f is both one-to-one (also written 1-1) and onto, then itis called a bijection, and has an inverse function f−1 : T → S defined byf−1(t) = the unique elt s ∈ S for which f(s) = t.

Note f−1 is characterized by f ◦ f−1 = idT ; f−1 ◦ f = idS .

§2. Equivalence Relations

Definition A partition of a set S is a collection of nonempty disjointsubsets of S whose union is S.

Examples (including Banach-Tarski Paradox)

Definition A relation on S is a subset ∼ of S×S; write x ∼ y to indicate(x, y) ∈ ∼. Say ∼ is an equivalence relation if it isi1 reflexive : x ∼ x (for all x ∈ S)i2 symmetric : x ∼ y =⇒ y ∼ xi3 transitive : x ∼ y and y ∼ z =⇒ x ∼ zThe equivalence class of x ∈ S is x := {y ∈ S | x ∼ y}.

Proposition 2.1 The collection of equivalence classes of an equivalencerelation on S form a partition of S. Conversely, any partition S =

∐α Sα

gives rise to a unique equivalence relation ∼ whose equivalence classes arethe Sα’s, namely x ∼ y ⇐⇒ ∃α with x, y ∈ Sα. Proof: exercise

The set of all equivalence classes of an equivalence relation ∼ is calledthe quotient set of S by ∼

S/ ∼ := {R ⊂ S | R = x for some x ∈ S}

and the map S → S/ ∼, x 7→ x is the natural projection of S onto S/ ∼.

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Examples i1 (the integers mod n) S = Z. For each n ∈ N, define the“congruence mod n” relation ≡n by a ≡n b (also written a ≡ b (mod n))⇐⇒ n | (a − b) (i.e. a and b have the same remainder when divided by n).(Check properties i1 , i2 , i3 , e.g. i3 ... )

The equiv class

a = {a+ kn | k ∈ Z} = {. . . , a− n, a, a+ n, a+ 2n, . . . }

is sometimes called the residue class of a (mod n). The quotient set Z/ ≡n,also denoted Zn or Z/nZ, is called the integers mod n. Examples.i2 S = R2 − 0, and x ∼ y ⇐⇒ x = λy for some nonzero real number λ(picture). R2/ ∼, usually denoted RP 1, is called the real projective line.

§3. Binary Operations

Definition A binary operation on a set S is a function

S × S ∗→ S.

Write a∗b for ∗(a, b) (“infix” notation). It is associative if (a∗b)∗c = a∗(b∗c)for all a, b, c ∈ S and commutative if a ∗ b = b ∗ a for all a, b ∈ S.

Examples i1 +, · on S = N,Z,Q,R,C (associative and commutative);− on Z,Q,R,C is neitheri2 · of n× n matrices (associative but not commutative)i3 +, · on Zn (“modular arithmetic”):

a+ b := a+ b

(the LHS is defined by the RHS). Must show this is “well-defined”: Supposea ≡ a′, b ≡ b′, i.e. n|(a − a′), n|(b − b′). Then n|[(a − a′) + (b − b′)] =⇒n|[(a+ b)− (a′ + b′)]. Thus a+ b = a′ + b′. Similarly define

a · b := a · b

and show well defined. These operations are associative and commutative.Also · is distributive over +:

a(b+ c) = ab+ ac.

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Definition A morphism of sets w/ binary operations f : (S, ∗)→ (S′, ∗′)is a function f : S → S′ satisfying

f(a ∗ b) = f(a) ∗′ f(b)

for all a, b ∈ S. Say f is a monomorphism if f is 1-1, and an epimorphismif f is onto. An isomorphism is a morphism which has an inverse which isalso a morphism.†

Examples i1 The “exponential map” (R,+)exp−→ (R, ·), x 7→ ex, is a

morphism, since ex+y = exey.i2 Mn(R) det−→ R is a multiplicative morphism, since det(AB) = det(A) det(B).

†In fact, any morphism f which has an inverse f−1 (or equivalently any bijectivemorphism) is an isomorphism. You are asked to show this in the homework. This entailsshowing that f−1(x ∗′ y) = f−1(x) ∗ f−1(y) for any x, y ∈ S′. To do this, start by notingx = f(a) and y = f(b) for some a, b ∈ S. Now plug in . . .

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II Groups

§1. Basic Concepts: groups, homomorphisms, and group actions

Definition A group is a set G with a binary operation G × G∗→ G,

(a, b) 7→ a ∗ b, satisfying(G1) associativity: (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ G(G2) identity: ∃e ∈ G s.t. e ∗ a = a = a ∗ e for all a ∈ G(G3) inverses: ∀a ∈ G, ∃b ∈ G s.t. a ∗ b = e = b ∗ a.

If ∗ is commutative, say G is abelian. |G| is the order of G. If |G| <∞, sayG is finite; otherwise G is infinite. The order of an element a ∈ G is

|a| := min{n ∈ N | a ∗ · · · ∗ a︸ ︷︷ ︸n factors

= e}

unless this set is empty, in which case |a| = ∞ by convention (e.g. everynonzero integer has infinite order, and |0| = 1).

Remarks i1 Identities and inverses are uniqueProof: Suppose e1, e2 are both identities. Then

e1 = e1 ∗ e2 G2 (e2 is an identity)= e2 G2 (e1 is an identity)

Now suppose b1, b2 are both inverses of a. Then

b1 = b1 ∗ e G2= b1 ∗ (a ∗ b2) G3= (b1 ∗ a) ∗ b2 G1= e ∗ b2 G3= b2 G2

i2 Usually write

· for ∗ and a · b or ab for a ∗ b

1 (or 1G) for e

a−1 for the inverse of a, an for a ∗ · · · ∗ a (n times) when n > 0, andfor (a|n|)−1 when n < 0.

If G is abelian, sometimes write + for ∗; a+ b, 0, −a, and na for a ∗ b, e,the (additive) inverse of a, and a ∗ · · · ∗ a (n times).

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i3 Other structures ia semigroup G1 only (set with an assoc bin op)

ib monoid G1 and G2 only (set with an assoc bin op with identity)ic ring abelian group (R,+) with an add’l assoc binary op R × R ·→ Rwhich is distributive over +, i.e. a(b+c) = ab+ac and (a+b)c = ac+bc(e.g. (Z,+, ·)). Say R is commutative if · is, with identity if ∃1.id field commutative ring (F,+, ·) with 1 6= 0 such that each nonzero elthas a mult inverse (e.g. Q, R, C, and Zn for prime n; HW).

Cancellation Property If a, b, c ∈ G (a group) satisfying ab = ac, then b = c.

Proof (supply reasons) ab = ac =⇒ a−1(ab) = a−1(ac) =⇒ (aa−1)b =(aa−1)c =⇒ 1b = 1c =⇒ b = c.

Familiar examples of groups

i1 (Z,+) infinite abelian group. In fact (Z,+, ·) is a commutative ring with1, but not a field (@ inverses in gen’l)i2 Q, R, C under +, and Q− 0, R− 0, C− 0 under ·i3 (Zn,+) finite abelian group

|Zn| = n Zn = {0, 1, · · · , n− 1}

In fact (Zn,+, ·) is a comm ring with 1. (Example: Z2, mult table)

Application (Linear Diophantine Equations) For a, b, c ∈ Z, find all in-teger solutions to ax+ by = c.

Solution First reduce to the case d := gcd(a, b) = 1 (a and b relativelyprime): If d 6= 1, then either d - c in which case @ any integer solutions, ord | c in which case we simply divide through by d. So we may assume d = 1.

Then working in Zb, the equation becomes a x = c. Since d = 1, the GCDlemma =⇒ ap+bq = 1 for suitable integers p, q. Thus ap ≡ 1 (mod b), i.e. ahas an inverse, namely p, in Zb. Thus the unique solution to the equation inZb is x = p c =⇒ one solution in Z is x0 = pc, y0 = (c−ax0)/b = (c−apc)/b.

To get all the solutions, note that we can change x0 by adding anymultiple nb of b. This will change y0 by subtracting na. Thus the full set ofsolutions is {(pc+ nb, (c− apc)/b− na) |n ∈ Z}.

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Example Solve 6x+10y = 14. Equivalent equation 3x+5y = 7. Workingmod 5

3x ≡ 7 =⇒ x ≡ 3−1 · 7 ≡ 2 · 7 ≡ 14 ≡ 4

so solutions are {(x, y) = (4 + 5n,−1− 3n) | n ∈ Z}.

More examples of groups

i1 The circle group S1 = {z ∈ C | |z| = 1} infinite abelian group (in fact aLie group: a group which is also a manifold ... locally Euclidean)i2 The cyclic group of order n, Cn = {z ∈ C | zn = 1}. Note that Cn ⊂ S1.Also (Cn, ·) is “isomorphic” to (Zn,+) via e2πik/n 7→ k.i3 The Klein 4-group V4 = {1, a, b, c} with a2 = b2 = c2 = 1, and theproduct of any two of a, b, c equals the third. This is not isomorphic to C4

(why?). Every group of order 4 is iso to either C4 or V4 (study mult tables).i4 (products) G,H groups =⇒ G×H is a group under the operation (g, h) ·(g′, h′) := (gg′, hh′). (For example, S1 × S1 is the torus (picture), andC2 × C2 = {(1, 1), (1,−1), (−1, 1), (−1,−1)} ∼= V4.)

Note ia The product of abelian gps is abelian ib |G×H| = |G||H|.i5 (units) Let (R, ·) be a monoid (e.g. ignore + in any ring R), and set

R• = {a ∈ R | ∃a′ ∈ R with aa′ = a′a = 1}

i.e. the “invertible elts” of R. The elements in R• are called units in R.

Claim (R•, ·) is a group (Proof Must first show that · induces an opera-tion on R•, i.e. R• is closed under multiplication: if a, b ∈ R•, i.e. ∃a′, b′ ∈ Rwith aa′ = a′a = 1 = bb′ = b′b, then a′, b′ ∈ R• =⇒ (ab)(b′a′) = 1 =(b′a′)(ab) =⇒ ab ∈ R•. Associativity is inherited. 1 ∈ R, since 1 · 1 = 1, andany a ∈ R• has an inverse in R•, namely the a′ from above.)

Examples ia Z• = {+1,−1} = C2ib Z•n =def{a ∈ Zn | ∃b with ab = 1} =

claim{a ∈ Zn | gcd(a, n) = 1}

(Proof: a ∈ Z•n ⇐⇒ ∃x, y s.t. ax+ ny = 1⇐⇒GCD

gcd(a, n) = 1)

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ThusZ•n = {a | 0 < a < n, (a, n) = 1}.

For example Z•8 = {1, 3, 5, 7} is a gp of order 4 with “multiplication table”

· 1 3 5 71 1 3 5 73 3 1 7 55 5 7 1 37 7 5 3 1

We know two other gps of order 4: C4 and C2×C2. These are not isomorphicto each other: in C2 × C2, all squares = 1 (as in Z•8), but not so in C4. AreZ•8 and C2 × C2 isormorphic? Yes:

1←→ (1, 1)3←→ (1,−1)5←→ (−1, 1)7←→ (−1,−1)

The order |Z•n| is called the Euler phi function of n, denoted ϕ(n).

Fact Let n = pe11 · · · pekk where p1, . . . , pk are distinct primes. Thenia ϕ(n) = n(1− 1

p1) · · · (1− 1

pk). For example:

ϕ(76) = ϕ(22 · 19) = 76(1− 12

)(1− 119

) = 2219 · 12· 18

19= 36

ib Z•n ∼= Z•p

e11× · · · × Z•

pekk

(where ∼= means “isomorphic to”). Furthermore,

if n is a power of a prime p, then Z•n ∼= Cϕ(n) for odd p and C2×Cϕ(n)/2 forp = 2. For example, Z•76

∼= C2 × C18.Examples of groups (continued)

i6 Dihedral groups

D2n := {symmetries of a regular n-gon P ⊂ C}

(P has vertices at e2πik/n) A symmetry of P is a rigid motion (i.e. distancepreserving function f : C→ C) which carries P onto itself (i.e. f(P ) = P );

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each such is either a rotation about 0, or a reflection across a line thru 0.Clearly D2n consists of n rotations and n reflections, so |D2n| = 2n. Theoperation on D2n is composition.

Can write each elt of D2n in terms of r (rotation by 2π/n radians, i.e.mult by ζn) and s (reflection through the x-axis, i.e. conjugation):

D2n = {1, r, r2, . . . , rn−1, s, sr, . . . , srn−1}

Note the “relations” rn = s2 = 1 and rs = sr−1 (since ζnz = ζ−1n z). Since

any word in the letters r, s, r−1, s−1 can be reduced using these relations tothe unique form sarb (for some a = 0, 1 and b = 0, . . . , n− 1) we say D2n isgenerated by r, s with relations rn = s2 = 1, rs = sr−1, written

D2n = (r, s | rn = s2 = 1, rs = sr−1).

Remark D4∼= V4. Also think about D6 and D8, symmetries of the

equilateral triangle and the square, and T12, O24 and I60, symmetries ofthe (solid) tetrahedron, octahedron (or cube) and icosahedron (or dodeca-hedron).

i7 Symmetric Groups Set n = {1, . . . , n}. Then

Sn := {bijections n→ n}

is a group under composition. Its elements are called permutations (of nsymbols or letters), and (Sn, ◦) is called the symmetric group of degree n.Note that Sn is a finite, nonabelian (for n ≥ 3) group of order n!,

|Sn| = n! (why?)

(What about S1 and S2? Exercise S3∼= D6)

More generally, for any set A (possibly infinite), the set SA of bijectionsA→ A is a group under composition, the symmetric group on A.

Notation for elements σ ∈ Sn ia two-row notation – write the numbers1, . . . , n in the first row and their images σ(1), . . . , σ(n) in the second.ib cycle notation (more efficient) – break σ ∈ Sn into disjoint cycles:The k-cycle

(i1 i2 i3 · · · ik)

is the permutation which sends each ij to the next ij+1 in the list, sendsik to i1, and leaves all else fixed (draw circular picture). Note that “cyclic

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permutations” of the list, e.g. (i2 i3 · · · ik i1), represent the same cycle.A 2-cycle is also called a transposition

Obvious fact Every σ ∈ Sn can be written uniquely (up to order ofcycles and cyclic permutation in each cycle) as a product of disjoint (i.e.non overlapping) cycles, called its cycle decomposition. (Often suppress1-cycles in notation)

Examples i1 (1 2 3 4 54 5 2 1 3

)= (1 4)(2 5 3)

i2 (1 2 3 4 54 2 5 1 3

)= (1 4)(2)(3 5) = (1 4)(3 5) = (3 5)(4 1)

Exercise List all the elements of S3 and S4.

To multiply (i.e. compose) two permutations, first juxtapose their cycledecompositions. What results is a product of cycles that might not bedisjoint. To rewrite this in disjoint cycle form, work from right to left (aswith composition of functions) to see where each number 1, . . . , n maps.

For example, to compute π = στ where σ = (2 1 4 5 3) and τ = (1 5)(2 3),first see where 1 maps: we have τ(1) = 5 and σ(5) = 3, and so π(1) = 3.Similarly π(3) = 1 (giving (1 3) as one of the cycles in π), π(2) = 2 (givingthe cycle (2)), π(4) = 5 and π(5) = 4 (giving the cycle (4 5)). Thus

(2 1 4 5 3) · (1 5)(2 3) = (1 3)(2)(4 5) = (1 3)(4 5).

Note The order (as an elt of Sn) of any k-cycle is k, and in gen’l theorder of a permutation is the least common multiple (lcm) of the orders ofthe (disjoint) cycles in its cycle decomposition (why?). For example

|(2 1 4 5 3)(6 9)(7 8)| = 10 |(2 1 4 5 3)(1 5)(2 3)| = 2

(not 10 for the latter, since the cycles are not disjoint).

Two final examples

i8 Quaternion groups The set of quaternions is

H = {a+ bi+ cj + dk | a, b, c, d ∈ R}

where i2 = j2 = k2 = −1, ij = k = −ji, jk = i = −kj, ki = j = −ik. Notethat H is a group under addition, but not under multiplication. However, itis a ring, in fact almost a field; the only axiom that fails is commutativityof multiplication. Such a structure is called a division ring.

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Exercise The multiplication in H is related to the dot and cross productsin R3: Writing a quaternion a+ bi+ cj+ dk as a+ (b, c, d) = a+~v, we have

(a+ ~v)(b+ ~w) = (ab− ~v • ~w) + (a~w + b~v + ~v × ~w).

The finite quaternion group

Q8 := {±1,±i,±j,±k}

is a nonabelian (multiplicative) subgroup of H• of order 8; it lies in theinfinite subgroup

S3 := {a+ bi+ cj + dk ∈ H | a2 + b2 + c2 + d2 = 1}

called the 3-sphere which arises in topology.

i9 Matrix groups R = commutative ring with 1For each n ∈ N have the group GLn(R) of invertible n × n matrices withentries in R, called the general linear group (over R). This is just the groupof units in Mn(R) (the ring of n× n matrices/R).

Note A is invertible iff det(A) is invertible, i.e. in R• (= R− 0 for a field).

ExampleGL2(Z2) = {(

1 00 1

),

(1 10 1

),

(1 01 1

),

(0 11 0

),

(1 11 0

),

(0 11 1

)}

Exercise Show that for p prime, |GLn(Zp)| =∏n−1k=0(pn − pk).

There are many important subgps of GLn(R): SLn(R) = {A | det(A) = 1}.For R = R have O(n) (orthogonal matrices) and SO(n). For R = C haveU(n) (unitary matrices) and SU(n). Tricky exercise: Show SU(2) ∼= S3.

Homomorphisms

Definition Let G,H be groups. A function f : G → H is a grouphomomorphism (or morphism) if f(xy) = f(x)f(y) for all x, y ∈ G. Theimage and kernel of f are

im(f) := {f(x) | x ∈ G} = f(G)

ker(f) := {x ∈ G | f(x) = 1} = f−1(1)

For ex the trivial morphism, defined by f(x) = 1 for all x ∈ G, has kernel Gand image = {1}. Define mono, epi, and iso-morphisms in usual way (mono

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= 1-1, epi = onto, and iso = both). We say G and H are isomorphic if ∃and isomorphism G→ H. A morphism from a group to itself (i.e. G = H)is called an endomorphism. An automorphism is a bijective endomorphism.

Examples i1 exp : (R,+)→ (R•, ·) is a monomorphism.i2 det : GLn(R) → R• is an epimorphism (where it is understood that theoperation is multiplication in both groups).i3 For any abelian group G, the map φ− : G→ G, x 7→ x−1 is a homomor-phism since φ−(xy) = (xy)−1 = y−1x−1 = x−1y−1 = φ−(x)φ−(y). In factφ− is an automorphism which is its own inverse!

If G is nonabelian, then φ− is not a morphism: for any a, b ∈ G withab 6= ba, have φ−(ab) = (ab)−1 and φ−(a)φ−(b) = a−1b−1 = (ba)−1, but(ab)−1 6= (ba)−1 since inverses are unique.

Remarks i0 Any composition of morphisms is a morphism (verify)i1 Let f : G→ H be a homomorphism of groups. Thenia f is onto ⇐⇒ im(f) = H (true for all functions) andf is 1− 1⇐⇒ ker(f) = {1} (HW).ib f(1) = 1 (proof: f(1) = f(1 · 1) = f(1)f(1); now multiply by f(1)−1)and f(x−1) = (f(x))−1 (HW)ic |f(x)| divides |x| Proof Set n = |x| and k = |f(x)|. Then xn = 1 =⇒f(x)n = f(xn) = f(1) = 1. Now appeal to:

Order Lemma Let a be an element of order k in a group. If k is finite,then ai = aj ⇐⇒ i ≡ j (mod k). In particular an = 1 ⇐⇒ k|n. If k = ∞then ai 6= aj unless i = j.

Proof If k is finite, then for any i and j we can write i− j = qk+ r with0 ≤ k < r. Thus ai ≡ aj (mod k) ⇐⇒ ai−j = 1 ⇐⇒ ar = 1, since aqk+r =(ak)qar = ar. But ar = 1⇐⇒ r = 0, since r < k = |a|, and r = 0⇐⇒ i ≡ j(mod k). The last statement is clear since ai = aj ⇐⇒ ai−j = 1.

i2 If f : G→ H is an isomorphism of groups, thenia |G| = |H|ib G is abelian ⇐⇒ H is abelianic |f(x)| = |x| ∀x ∈ G (HW)

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Can use these to prove two groups are not isomorphic. For example:ia Cm 6∼= Cn for m 6= n since they have different orders.ib C6 6∼= S3 since C6 is abelian and S3 is not.ic Z•24 6∼= C8 since C8 has elts of order 8, but Z•24 doesn’t. Can alsocount the number of elements of a given order to distinguish groups, e.g.:D24 6∼= S4 since D24 has 13 elts of order 2 while S4 has 9 (why?).

Open Problem Classify all groups (up to isomorphism)Facts i1 There is only one of any given prime order (prove later)i2 There exist 2 of order 4 (C4, V4), 2 of order 6 (C6, S3), 5 of order 8

(C8, C4 × C2, C2 × C2 × C2, D8, Q8), 2 of order 9 (C9, C3 × C3), 2 of order10 (C10, D10), 5 of order 12, 14 of order 16 . . . (see page 168 in text)

Group Actions (a central theme in the course)

Definition An action of a group G on a set A is a map G × A → A,(g, a) 7→ g · a satisfying

(A1) g · (h · a) = (gh) · a (A2) 1 · a = a

for all g, h ∈ G, a ∈ A. The associated permutation representation is thefunction σ : G→ SA (the symmetric group on A) defined by

σ(g)(a) = g · a. (∗)

Note that σ is a group homomorphism, and conversely any homomorphismG

σ→ SA gives rise to a group action given by (∗) which has σ as its permu-tation representation (exerscise).

The kernel of the action is ker(σ) = {g ∈ G | g · a = a for all a ∈ A}.The action is said to be faithful if it has trivial kernel (=⇒ σ is one-to-one),i.e. g · a = g · b =⇒ a = b.

Examples i1 trivial action g · a = a for all a ∈ A (i.e. σ is the trivialhomomorphism). This is not faithful (unless G = 1).i2 The (faithful) action of D2n on the set of vertices of the n-gon.i3 The actions of any group G on itself by left or right multiplication: g ·a =ga or ag−1 (verify that these are both actions) or by conjugation

g · a = gag−1

(HW) Note gag−1 is called the conjugate of a by g.

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§2. Subgroups

Definition A subset H of a group G is a subgroup of G, written H < G,if it is a group under the operation induced from G, i.e.

(S1) x, y ∈ H =⇒ xy ∈ H (S2) 1 ∈ H (S3) x ∈ H =⇒ x−1 ∈ H(Note: associativity is automatic)

Examples i1 Every group has the trivial subgroups {1} and G. Anyother subgroup will be called proper.i2 For k = 0, 1, 2, . . . , the set kZ = {nk | n ∈ Z} of all multiples of k is asubgroup of Z; there are no others. (Note that + is the operation in Z, so(S2) reads 0 ∈ kZ.) 0Z = {0} and 1Z = Z are trivial, 2Z = evens, etc.i3 {1, r, . . . , rn−1} and {1, s} are subgps of D2n.i4 {1,−1} and {1, i,−1,−i} are subgroups of Q8.i5 Cn < S1 < C•.

Subgroup Criterion A subset H of a group G is a subgroup if and onlyif ia H is nonempty, and ib x, y ∈ H =⇒ xy−1 ∈ H.

Proof (=⇒) 1 ∈ H by (S2) so H 6= ∅. If x, y ∈ H, then y−1 ∈ H by (S3)so xy−1 ∈ H by (S1). Thus ib holds.

(⇐=) ∃x0 ∈ H by ia , so 1 = x0x−10 ∈ H by ib =⇒ (S2). Now

x ∈ H =⇒ x−1 = 1·x−1 ∈ H by ib =⇒ (S3). Finally x, y ∈ H =⇒ y−1 ∈ Hby (S3) =⇒ xy = x(y−1)−1 ∈ H by ib , so (S1) holds.

Remark If H is finite, ib can be replaced by (S1), since x ∈ H =⇒xn = 1 for some n (since G is finite) =⇒ 1 = xn ∈ H and x−1 = xn−1 ∈ H.

More examples of subgroups

i6 Let G be a group. Then any subgp of a subgp of G is a subgp of G, andany intersection of subgps of G is a subgp of G (why?).i7 For any homomorphism G

f→ H of groups,

ker(f) < G and im(f) < H

(check this by the def’n of subgp or using the subgp criterion)

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i8 The center Z(G) := {x ∈ G | xg = gx for all g ∈ G} of a group G isa subp of G. Proof: For x, y ∈ Z(G) and g ∈ G, we have xyg = xgy =gxy =⇒ xy ∈ Z(G) (proving S1) and xg−1 = g−1x =⇒ (taking inverses)x−1g = gx−1 =⇒ x−1 ∈ Z(G) (proving S3). For (S2), note that 1 ∈ Z(G)since 1g = g = g1 for all g. (Exercise Give an alternative proof using thesubgp criterion and the observation that xg = gx⇐⇒ x = gxg−1)

More generally, for any subset A ⊂ G, define the centralizer of A in Gto be

CG(A) := {x ∈ G | xa = ax for all a ∈ A}

(so Z(G) = CG(G)), and the normalizer of A in G to be

NG(A) := {x ∈ G | xA = Ax}

where xA = {xa | a ∈ A} and Ax = {ax | a ∈ A}. All are subgroups of G(general proof given in i9 below).

Example Z(Q8) = {1,−1} (verify). For A = {1,−1, i,−i} < Q8, haveji 6= ij, ki 6= ik, etc. and so CQ8(A) = A. However jA = {j,−j,−k, k} isthe same set as Aj = {j,−j, k,−k}, . . . and so NQ8(A) = Q8.

Remark In general, for any subsets A1, . . . , An of a group G, define

A1 · · ·An = {a1 · · · an | ai ∈ Ai for each i = 1, . . . n}.

xA and Ax are special cases of this, as is xAx−1 = {xax−1 | a ∈ A}. Notethat the conditions xa = ax and xA = Ax defining CG(a) and NG(A) canbe rewritten as xax−1 = a and xAx−1 = A.i9 If a group G acts on a set A and a ∈ A, then the stabilizer of a is

Ga = {x ∈ G | x · a = a}

This is a subgroup of G for each a. (Proof: 1 ∈ Ga by (A2), so Ga 6= ∅.Thus for any x ∈ Ga, x−1 · a = x−1 · (x · a) = (x−1x) · a = 1 · a = a sox−1 ∈ Ga. If x, y ∈ Ga then (xy) · a = x · (y · a) = x · a = a so xy ∈ Ga.)

Examples ia The stabilizer of 1 ∈ C under the usual action of thedihedral group D2n is {1, s}.ib Normalizers are examples of stabilizers for a suitable action: Let G actby conjugation on the set of all subsets of G,

x ·A = xAx−1 ( = {xax−1 | a ∈ A} ).

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Then for any A ⊂ G,

NG(A) = {x ∈ G | xA = Ax} = {x ∈ G | xAx−1 = A} = GA.

Similarly centralizers can viewed in terms of actions: For A ⊂ G, the nor-malizer NG(A) acts on A by conjugation, and CG(A) is the kernel of thisaction (do you see why?).

Thus the fact that normalizers and centralizers are subgroups can bededuced from the fact that stabilizers and kernels of homomorphisms are.

Normal subgroups

Definition A subgroup H < G is called a normal subgroup (writtenH C G) if

h ∈ H, g ∈ G =⇒ ghg−1 ∈ H

(or equivalently NG(H) = G). Thus a subgroup of G is normal iff it is“closed under conjugation” by any elt in G. For example {1,−1, i,−i} C Q8,as shown above. In generali1 Z(G) C G (why?)i2 ker(f) C G for any morphism f : G→ H (why?).i3 Any H < G with |H| = 1

2 |G| (say H is of index 2 in G) is normal in G

Proof of i3 For any x 6∈ H, the set xH is disjoint from H (if xh = h′ forsome h, h′ ∈ H then x = h−1h′ ∈ H =⇒⇐=). So xH = G −H. SimilarlyHx = G−H. Thus xH = Hx =⇒ xHx−1 = H =⇒ H C G.

Warning: K C H C G 6=⇒ K C G. ∃ examples with |G| = 8 (HW).

Cyclic groups and subgroups

Definition For any element x in a group G, denote by 〈x〉 the set of allpowers of x (or multiples of x if G is additive),

〈x〉 = {xk | k ∈ Z} (w/ repetitions deleted)

This is a subgroup of G (verify) called the cyclic subgroup generated by x.If

G = 〈x〉 for some x ∈ G

we say G is a cyclic group, and any such x is called a generator of G (theremay be many). Note: if |G| = n, then G cyclic ⇐⇒ G has an elt of order n.

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Examples i1 Z under + (or its mult analogue C∞ = {tk | k ∈ Z},where t is an indeterminant) is cyclic with generator 1 (or t). Alternatively,−1 (resp. t−1) is a generator.i2 Cn is cyclic with generator t = e2πi/n (or tk for any k rel prime to n)i3 D2n is not cyclic for n > 1: 〈rk〉 ⊂ 〈r〉 6= D2n and 〈rks〉 = {1, rks} 6= D2n

Remark The order n of any element x in a group is equal to the orderof the cyclic subgroup it generates: |x| = |〈x〉|. (Clear if n = ∞. If n < ∞then xi = xj ⇐⇒ i ≡n j (by the order lemma) so 〈x〉 consists of the ndistinct elements 1, x, . . . , xn−1.)

Classification Theorem for Cyclic Groups Every cyclic group C = 〈x〉 isisomorphic to Cn for some n = 1, 2, · · · ,∞. Thus there is up to isomorphismexactly one cyclic group of each finite order and one infinite cyclic group.

Proof If |C| = n, then the function C → Cn mapping xk to tk (for eachk ∈ Z) is a well defined isomorphism (why?).

Classification Theorem for Subgroups of Cyclic Groups Let C = (x) bea cyclic group. Thenia Every subgroup H of C is cyclic. In particular any nontrivial H =

〈xm〉, where m is the smallest positive integer for which xm ∈ H.ib If C is finite of order n, then it has a subgroup of order k ⇐⇒ k dividesn. In fact there is only one such subgroup for each divisor k.

Proof ia HWib (=⇒) If H < C with |H| = k, then H = 〈xm〉, for m as in ia . Set d =gcd(m,n). Then xd = xam+bn (for suitable a, b) = (xm)a ∈ H =⇒ m = d,which divides n =⇒ k = n/m divides n (and in fact H = 〈xn/k〉, so H isthe only subgroup of order k). (⇐=) k|n =⇒ H = 〈xn/k〉 has order k.

This theorem gives a complete picture of the “lattice” of subgroups ofany cyclic group (draw pictures, cf. §2.5 in the text).

Remarks i1 ∃ groups with non-planar lattices, e.g. D16i2 ∃ pairs of nonisomorphic groups with the same subgroup lattice, e.g.C2×C8 and the “modular group” of order 16 = (r, s | r8 = 1 = s2, rs = sr5).

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Question Does ib generalize to other finite groups G?

Answer =⇒ does (Lagranges’s theorem, below), but ⇐= does not. Forexample the group T12 of symmetries of the tetrahedron has order 12 buthas no subgroup of order 6. To see this, note that T12 consists of eight 2π/3-rotations (about lines through the vertices), three π-rotations (about linesjoining midpoints of opposite sides), and the identity element. Now supposethat H < T12 with |H| = 6. Then it must contain at least one 2π/3-rotationr. But then it must contain all of them (since they are all conjugate to r orr−1, and H C G as shown above) and this is clearly impossible.

Lagrange’s theorem The order of any subgp of a finite gp G divides |G|

Definition If H is a subgroup of a group G, then any subset of G of theform

xH = {xh | h ∈ h}

for x ∈ G, is called a left coset of H in G. Similarly define the right cosetsHx ⊂ G. The left (resp. right) cosets of H partition G into equal sizedsubsets:

Coset Lemma ia |xH| = |H| ib xH ∩ yH 6= ∅ =⇒ xH = yH, andic ∪x∈GxH = G (and similarly for right cosets)

Proof ia The map H → xH, h 7→ xh is a bijection, by cancellationib hyp =⇒ xi = yj for some i, j ∈ H, so xh = xii−1h = yji−1h ∈ Hfor any h ∈ H =⇒ xH ⊂ yH. Similarly yH ⊂ xH, so xH = yH.ic x ∈ xH.

The number of left (or right) cosets of H in G is called the index of H inG, denoted |G : H|. The coset lemma shows that |G| = |G : H||H|, whichproves Lagrange’s Theorem.

Remarks i1 The set of all left cosets is (sometimes) denoted by G/H,so |G/H| = |G : H| = |G|/|H|. Similarly for the set H\G of right cosets.i2 (coset recognition: HW) x, y ∈ G lie in the same coset of H ⇐⇒ x−1y ∈H ⇐⇒ y = xh for some h ∈ H (and similarly for right cosets)

Example of coset decomp: The subgp H = {1, (1 2)} of S3 has cosetsH, (1 3)H = {(1 3), (1 2 3)} and (2 3)H = {(2 3), (1 3 2)}.

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Applications of Lagrange’s Theorem

i1 The order of any element x in a finite group G divides |G|.i2 Any group of prime order p is cyclic (and so there’s only one).i3 (Euler’s Theorem) If a, n ∈ N are relatively prime, then

aφ(n) ≡ 1 (mod n)

where φ(n) = |Z•n|. Special case (Fermat’s Little Theorem): If p is prime andp - a, then ap−1 ≡ 1 (mod p) (since φ(p) = p− 1). For example 34 = 81 ≡ 1(mod 5), 36 = 729 ≡ 1 (mod 7), etc.

Proofs i1 |x| = |〈x〉|, which divides |G| by Lagrange.i2 Any x 6= 1 in G has order dividing p (by i1 ) and thus equal to p, sincep is prime; so G = (x).i3 (a, n) = 1 =⇒ a ∈ Z•n (by the GCD Lemma) =⇒ aφ(n) = 1 (by i1 , sinceZ•n has order φ(n)), i.e. aφ(n) ≡ 1 (mod n).

§3. Products and Finite Abelian Groups

Recall that the product of two groups H and K is

H ×K = {(h, k) | h ∈ H, k ∈ K}

with componentwise multiplication (h, k)(h′, k′) = (hh′, kk′). It turns outthat every finite abelian group is a product of cyclic groups. More precisely,

Fundamental Theorem of Finite Abelian Groups (Primary Form) Everyfinite abelian gp is isomorphic to a product of cyclic groups of prime powerorder. Moreover, the number of factors and their orders are uniquely deter-mined by the group.

This leads to an algorithm for finding all abelian groups of order n:

• If n = pk, a pure prime power, then there is (up to isomorphism) exactlyone abelian group of order n for each partition of k (a sequence k1 ≥ · · · ≥ ksof natural numbers such that k = k1 + · · ·+ ks), namely

Cpk1 × · · · × Cpkn .

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For example, there are three abelian groups of order 125 = 53

C125 C25 × C5 C5 × C5 × C5

corresp to the three partitions 3, 2 + 1 and 1 + 1 + 1 of 3.

• For general n = pk11 pk22 · · · , there is one abelian group for each list of

partitions of k1, k2, . . . .

Exercise How many abelian groups are there of order 1500 = 22 · 3 · 53?(Answer: 6 = 2 · 1 · 3.) List them. (C4 × C3 × C125, C2 × C2 × C3 × C125,C4 × C3 × C25 × C5, C2 × C2 × C3 × C25 × C5, C4 × C3 × C5 × C5 × C5,C2 × C2 × C3 × C5 × C5 × C5).

There is another standard form for finite abelian groups: Using the fact(from homework) that

Cm × Cn ∼= Cmn

if m and n are relatively prime, can start with the primary form, then groupthe largest factors associated with each prime in the primary form, then thenext largest, etc. to get a unique form for any finite abelian group

G ∼= Cn1 × Cn2 × · · ·

where n = n1·n2 · · · and ni is divisible ni+1 for each i; the numbers n1, n2, . . .are called the invariant factors of G. For example

C4 × C2 × C3 × C5 × C5∼= C60 × C10

has invariant factors 60, 10. The group C8 × C2 × C28 × C25 × C14 has invfactors 1400, 28, 2, 2 (exercise). In summary

Fundamental Theorem of Finite Abelian Groups (Invariant Form) Anyfinite abelian gp G is isomorphic to a product Cn1 × Cn2 × · · · × Cnk

whereni is divisible by ni+1 for i = 1, . . . , k − 1. The numbers n1, . . . , nk, calledthe invariant factors of G, are unique.

We’ll prove the first statement in the fundamental theorem (primaryform). Need

Product Recognition Theorem (PRT) If H and K are normal subgroupsof a group G satisfying H ∩K = 1 and HK = G, then G ∼= H ×K.

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Proof Consider the function f : H × K → G given by f(h, k) = hk,which is onto since HK = G.

We claim that f is a homomorphism. First note that for any h ∈ H andk ∈ K, hkh−1k−1 ∈ H ∩ K (it can be written as h(kh−1k−1) ∈ H sinceH C G, or as (hkh−1)k−1 ∈ K since K C G). Since H ∩K = 1, it followsthat hkh−1k−1 = 1, i.e. hk = kh, so the elts of H commute with the elts ofK. Now f((h, k)(h′, k′)) = f(hh′, kk′) = hh′kk′ = hkh′k′ = f(h, k)f(h′, k′).

Finally observe that ker(f) = 1 (since f(h, k) = 1 =⇒ hk = 1 =⇒ h =k−1 ∈ H ∩K = 1, i.e. (h, k) = (1, 1)) so f is 1-1.

Now let G be a finite abelian group. We wish to show that G is a productof cyclic groups of prime power orders. We split off one prime at a time:Suppose |G| = pkq, where p is prime and p - q. Set

P = {x ∈ G | xpk= 1} and Q = {x ∈ G | xq = 1}.

Then P and Q are subgroups (since G is abelian) with P ∩ Q = 1 (sincep - q) and PQ = G: ∃ a, b with aq + bpk = 1, so for any x ∈ G

x = xaq+bpk

= xaqxbpk ∈ PQ

(do you see why?) Thus G ∼= P ×Q by the PRT. It remains (by inductionon n) to prove that P is a product of cyclic groups. This is the content ofthe following theorem:

Theorem If P is an abelian p-group for some prime p (meaning that eachelement in P has order a power of p) then P is a product of cyclic groups.

Proof Let H be the cyclic subgroup of P generated by an element h ofmaximal order (say ps) in P , and K be a largest possible subgroup of P forwhich H ∩K = 1. If the subgroup HK = P , then P ∼= H ×K by the PRT,and the theorem follows by induction on |P |. So assume HK 6= P . We willshow that this leads to a contradiction.

Claim ∃x ∈ P such that x /∈ HK but xp ∈ K.

Pf of claim First note that ∃ y 6∈ HK with yp ∈ HK. (Indeed, for anyz 6∈ HK, choose the smallest m such that zp

m ∈ HK, and set y = zpm−1

.)Now yp = hnk for some n ∈ Z, k ∈ K. By the maximality of |h| we have

yps

= (hnk)ps−1

= hnps−1kp

s−1= 1.

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Thus hnps−1

= 1, since H ∩ K = 1, and so p|n. Set x = h−n/py. Clearlyxp = k ∈ K, but x 6∈ HK since y 6∈ HK. This completes the proof of theclaim.

Now let K ′ be the subgroup generated by x and K,

K ′ = {xnk | n ∈ Z, k ∈ K}.

Observe that H ∩ K ′ = 1, since xnk = h ∈ H =⇒ xn = hk−1 ∈ HK =⇒p|n =⇒ xn ∈ K =⇒ h ∈ K =⇒ h = 1. But K ′ % K, which contradicts themaximality of K.

§4. Cayley’s Theorem and the Symmetric Group

Cayley’s Theorem Every group G is isomorphic to a subgroup of thesymmetric group SG.

Proof Let G act on itself by left multiplication. Then the associatedpermutation homomorphism λ : G→ SG (λ(g) = left mult by g) is 1-1 (bythe cancellation property in G) and so G ∼= im(λ) < SG.

In particular, every finite group of order n is isomorphic to a subgroupof Sn. One especially important subgroup of Sn is the alternating group Anconsisting of all even permutations (defined below).

Note that every permutation σ ∈ Sn can be written as a product oftranspositions, since any cycle can:

(i1 · · · ik) = (i1 i2)(i2 i3) · · · (ik−1 ik)

(e.g. (i j k l) = (i j)(j k)(k l)). This decomposition is not unique, e.g.1 = (1 2)(1 2) = (1 2)(3 4)(1 2)(3 4), but the parity (even or odd) of thenumber of transpositions is always the same:

Definition A permutation is even if it can be written as a product of aneven number of transpositions, and is odd if it can be written as a productof an odd number of transpositions.

In particular odd-length cycles are even and even-length cycles are odd(by the formula displayed above). Also even·even = odd·odd = even, andeven·odd = odd·even = odd.

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Remarkable Fact No permutation is both even and odd.

There are many proofs (cf. pages 109–111 in the text, or the enlighteningdance floor proof by Ty Cunningham in Math. Magazine 43 (1970) 154-5).Here is one: Consider the function

c : Sn → N

assigning to σ ∈ Sn the number of cycles in the disjoint cycle decomp of σ(counting the 1-cycles). For example c(k-cycle) = n− k + 1. Readily verify

c(στ) ≡ c(σ) + 1 (mod 2)

for any transposition τ = (i j). Indeed if i and j lie in the same cycle inσ then that cycle splits into two in στ , and if they lie in distinct cycles inσ then those cycles are joined into one in στ . The remaining cycles of σand στ coincide. It follows that if σ is a product of m transpositions thenc(σ) ≡ c(1) +m ≡ n+m, and so the parity of m is determined by σ.

Thus there is a well-defined homomorphism

sgn : Sn → C2 = {±1}

sending even permutations to +1 and odd ones to −1. The kernel is the setAn of all even permutations, known as the alternating group of degree n. Itis a normal subgroup of index 2, i.e. |An| = n!/2 (do you see why?).

Remarks i1 The parity of a permutation σ is equal to the parity ofia (algebraic) the number of even-length cycles in any cycle decompositionof σ; the number of odd-length cycles is irrelevant. Thus even permutationsare those with even number of even-length cycles: (· · ·), (· ·)(· ·), (· · · · ·),(· · ·)(· · ·), (· · · ·)(· ·), (· ·)(· ·)(· ·), ... This helps enumerate the elementsin An: A3 = {1, (1 2 3), (3 2 1)}; A4 = {1, (1 2 3), · · · , (1 2)(3 4), · · · }, etc.ib (geometric) the number of intersection points in any “picture” of σi2 Conjugation in Sn: To compute στσ−1, write τ as a product (i j · · · )of cycles; then

σ(i j · · · )σ−1 = (σ(i) σ(j) · · · )

i.e. replace each number in τ by its image under σ. This is easy to seeif τ is a single cycle, and generalizes using the trick στ1τ2 · · · τrσ−1 =(στ1σ−1)(στ2σ−1) · · · (στrσ−1). It follows that if two elements in Sn areconjugate, then they have the same cycle structure. The converse is also

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true; indeed it is an easy to determine all σ for which στσ−1 = τ ′ for any τand τ ′ which have the same cycle structure.

Conjugation in An is more complicated; one must check whether any ofthe possible conjugating permutations is even. For example τ = (2 3 4) andτ ′ = (4 3 2) are conjugate in S4 (why?) but not in A4 (since στσ−1 = τ ′ =⇒σ maps the ordered triple (2, 3, 4) to either (4, 3, 2), (3, 2, 4) or (2, 4, 3), whichforces σ = (2 4), (2 3) or (3 4), all of which are odd). This is explored furtherin the HW.

i3 A1 and A2 are trivial groups; A3∼= C3; A4

∼= T12, A5∼= I60.

i4 An is generated by 3-cycles. Indeed each element of An can be writtenas a product of an even number of transpositions, and (i j)(j k) = (i j k),(i j)(k `) = (i j k)(j k `).

i5 Recall that An C Sn. In fact if n ≥ 5, then An is the only propernormal subgroup of Sn.

Proof: Let 1 6= K C Sn. Then any σ 6= 1 in K has a disjoint cycledecomposition of the form (i j · · · ), and so does not commute with thetransposition τ = (j k), for any chosen k 6= i, j. Indeed στ(i) = j whereasτσ(i) = k. Now consider the “commutator” [σ, τ ] := στσ−1τ−1 6= 1. Clearly[σ, τ ] ∈ K (since [σ, τ ] = σ(τσ−1τ−1) and K is normal) and [σ, τ ] is a product(στσ−1)(τ−1) of two distinct transpositions (by i2 ). If these transpositionsoverlap, then [σ, τ ] is a 3-cycle =⇒ K contains all 3-cycles (since it is normal)=⇒ K ⊂ An (by i4 ) =⇒ K = An or Sn. If they don’t overlap, then Kcontains all products of pairs of disjoint transpositions (since it is normal),and in particular (1 2)(3 4) and (3 4)(2 5), whose product is the 3-cycle(1 2)(2 5) = (1 2 5) =⇒ K contains all 3-cycles and so K = An or Sn.

Note We have repeatedly used the observation that a normal subgroupthat contains an element h must contain all conjugates of h. In fact H isnormal in G if and only if it is a union of conjugacy classes in G. (Theconjugacy class of x ∈ G is the set of all elts in G that are conjugate to g.)

In general, one can understand a lot about the structure of a group froma knowledge of its normal subgroups. A group which has no proper normalsubgroups is called a simple group; these are the building blocks of all finitegroups. (More on this later)

Examples i1 Any group of prime order is simple (indeed such a grouphas no proper subgroups whatsoever by Lagrange’s Theorem).

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i2 Abel’s Theorem An is simple for all n ≥ 5.

Proof Suppose ∃K C An, but K 6= An or 1. Then An is the normalizerof K in Sn, by remark i5 above, and so K has exactly two “conjugates”in Sn; i.e. setting H = (1 2)K(1 2), we have τKτ−1 = K or H accordingto whether τ is even or odd (check this). But then H ∩ K C Sn and soH ∩ K = 1, by i5 again. It follows that the elements of H commutewith the elements of K (note H C An). But if σ 6= 1 ∈ K, then ∃ odd τsuch that σ and τστ−1 do not commute (for example, if σ = (i j k · · · ) or(i j)(k `) · · · , take τ = (k m) for any m 6= i, j, k, ` and consider the imageof j or `, respectively), which is a contradiction.

§5. Quotient Groups

Recall that normal subgroups can be characterized in a variety of ways.

Normality Lemma A subgroup H of G is normal in G if and only if anyof the following conditions holds:ia H is a union of conj classes in G ib xhx−1 ∈ H for all h ∈ H, x ∈ Gic xHx−1 ⊂ H for all x ∈ G id xHx−1 = H for all x ∈ Gie xH = Hx for all x ∈ G if xHyH = xyH for all x, y ∈ G

Proof The equivalence of ia – ie was noted previously, so it suffices toshow ie =⇒ if =⇒ ic : Assuming ie , have xHyH = xyHH = xyH(HH ⊂ H since H < G, and HH ⊃ H since 1 ∈ H). Assuming if , havexHx−1H = H =⇒ xHx−1 ⊂ H since 1 ∈ H.

Using if , can make the set G/H of left cosets of H in G into a group,called the quotient group of G by H, provided H is normal in G:

Theorem If H C G, then G/H is a group under the operation defined byxH · yH = xyH. Furthermore the map

p : G→ G/H

defined by p(x) = xH, called the natural projection of G onto G/H, is anepimorphism.

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Proof The operation is well defined (by the lemma), associative, withidentity element 1H = H and (xH)−1 = x−1H (verify). Clearly p is ahomomorphism, since p(xy) = xyH = xHyH = p(x)p(y), and is onto sinceevery coset is the image of any element in it.

Remarks i1 The product of two cosets H1, H2 in G/H can be describedin words as follows: Choose one element from each. Then H1H2 is the cosetcontaining the product of these elements. The lemma shows that this is“well-defined”, i.e. independent of which elements you choose.i2 Be very careful when G is an “additive” group, i.e. the operation is +.Then cosets of H < G are of the form x + H = {x + h | h ∈ H}, and theoperation in G/H is addition: (x+H)+(y+H) = (x+y)+H. For example,if H = nZ = {multiples of n} < Z, then Z/nZ = {k+nZ | k = 0, . . . , n−1}.

Note: If G′ is another additive group, then the product G × G′ will beoften be called the direct sum, written

G⊕G′ = {(g, g′) | g ∈ G, g′ ∈ G′}

with the operation (g, g′) + (h, h′) = (g + g′, h+ h′). In other words G⊕G′is the same group as G×G′, but written additively (cf. example i2 below).i3 Any quotient group G/H of an abelian group G is abelian, since xHyH =xyH = yxH = yHxH, for any x, y ∈ G.

Examples i1 Z/nZ ∼= Zn, via the isomorphism k + nZ 7→ k.i2 Let H := 〈4〉 = {0, 4, 8} C Z12. Then H has order 3 =⇒ Z12/H ={H, 1 + H, 2 + H, 3 + H} has order 4 =⇒ Z12/H ∼= C4 or V4. Since 1 + Hhas order 4 (why?) we have in fact Z12/H ∼= C4.i3 Let S := 〈(0, 1)〉 C G := Z2⊕Z4. Note that |S| = 4 (since (0, 1) has order4) so G/S has order |G/S| = |G|/|S| = 4/2 = 2, and therefore G/S ∼= Z2.G also has three subgroups of order two, H = 〈(1, 0)〉, J = 〈(1, 2)〉 andK = 〈(0, 2)〉, with G/H ∼= H/J ∼= Z4 and G/K ∼= Z2 ⊕ Z2; verify this bycomputing the orders of elements in the quotient, e.g. (0, 1) + H has order4 in G/H.i4 Let H be the cyclic subgroup of S3 generated by the 3-cycle (1 2 3),H = 〈(1 2 3)〉 = {1, (1 2 3), (1 3 2)} C S3. Then H has two cosets, H andH ′ = (1 2)H = {(1 2), (2 3), (1 3)}, so S3/H = {H,H ′} ∼= C2.i5 Z(D8) = {1, r2} C D8, and D8/Z(D8) ∼= C2×C2 (compute orders again).

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Most of the important properties of quotient groups follow from the

Universal Property of Quotient Groups Let K C G and p : G → G/Kbe the natural projection. Then for any homomorphism f : G → H whosekernel contains K, there exists a unique homomorphism g : G/K → H suchthat f = g ◦ p, i.e. the following diagram commutes

Gp−→ G/ ker(f)

f ↘ ↓ g

H

In particular g(xK) = f(x), i.e. g(coset) = f(any element in the coset).Also ker(g) = ker(f)/K and im(g) = im(f).

Proof First check that g is well defined (and thus unique by the com-mutativity of the diagram): if x, y lie in the same coset, then x−1y ∈ K.But K ⊂ ker(f), by hypothesis, so f(x−1y) = 1 = f(x)−1f(y) =⇒ f(x) =f(y). It is now straightforward to show g is a homomorphism (g(xKyK) =g(xyK) = xy = g(xK)g(yK)) with ker(g) = {xK | f(x) = 1} = ker(f)/Kand im(g) = {f(x) | x ∈ G} = im(f).

The UPQG can be used to construct morphisms from quotient groups.For example, you can show that Z2 ⊕ Z4/〈(0, 2)〉 is isomorphic to Z2 ⊕ Z2

(see example i2 above) by constructing an explicit iso. One such arisesfrom the UPQG via the map Z2⊕Z4 → Z2⊕Z2, (a, b) 7→ (a, b) (check this).

It has the following important consequences:

The Isomorphism Theoremsi1 If f : G→ H is a homomorphism, then G/ ker(f) ∼= im(f).i2 If H < G and K C G, then H ∩K C H and HK/K ∼= H/H ∩K.

HK� ��

H K�� �

H ∩K

i3 If H,K C G with H ⊂ K, then K/H C G/H and G/K ∼= (G/H)/(K/H).

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Proof Apply the Universal Property toi1 G −→ G/ ker(f)

f ↘ ↓

im(f) ⊂ H

i2 (note that HK is a subgroup of G)

H −→ H/H ∩K

incl ↓ ↓

HK −→ HK/K

i3 G −→ G/H

↘ ↓ ← now apply i1 to this

G/K

The Correspondence Theorem Let f : G → H be a homomorphism ofgroups, G be the set of all subgroups of G containing K = ker(f) and H bethe set of all subgroups of H contained in I = im(f). Then the map

f : G → H , S 7→ f(S)

is a bijection which respects containment, indices, normality and quotients.In other words,

(a) S < T ⇐⇒ f(S) < f(T ), in which case |T : S| = |f(T ) : f(S)|

(b) S C T ⇐⇒ f(S) C f(T ), in which case T/S ∼= f(T )/f(S)

Proof The inverse of f is the “preimage” map f−1 : H → G sendingeach subgroup A ∈ H to its full preimage f−1(A) ∈ G.

Indeed f−1f(S) = S: The inclusion ⊃ holds in general (s ∈ S =⇒f(s) ∈ f(S), i.e. s ∈ f−1f(S)) and ⊂ holds since S ⊃ K (if x ∈ f−1f(S),i.e. f(x) ∈ f(S) so f(x) = f(s) for some s ∈ S, then f(xs−1) = 1, i.e.xs−1 ∈ K =⇒ x ∈ Ks ⊂ S). Similarly ff−1(A) = A: ⊂ holds in gen’l, and⊃ since A ⊂ I. Thus f is a bijection.

The rest is straightforward, and is left as an exercise for the reader; theUPQG is used to construct the isomorphism of quotient groups.

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Examples i1 applied to GLn(R) det→ R• gives GLn(R)/SLn(R) ∼= R•.i2 applied to M,N C G for which M ∩N = 1 and MN = G gives isomor-phisms

G/M ∼= N and G/N ∼= M.

Note: if M and N are simple groups which are “maximal” proper normalsubgroups of G (i.e. there are no larger such subgroups) then the conditionsM ∩N = 1 and MN = G are automatic.

Application : Classification of Finite Groups

Definition A composition series for a finite group G is a sequence ofsubgroups

G = G0 B G1 B G2 B · · · B Gn−1 B Gn = 1

each a maximal proper normal subgroup of the preceeding.

The quotients Gi/Gi+1 are simple, by the correspondence theorem, andG can be viewed as being built up from these composition factors, whichare unique up to order by the “Jordan Holder Theorem” (proved using theisomorphism theorems, cf. the special case in i2 above, where n = 1).

Unfortunately ∃ distinct groups with identical composition factors, sothe classification of finite groups is a two-step program: the Holder Program(late 19th century):i1 Classify all simple finite groupsi2 Find all ways to “put simple groups together” to form other groups

The first step was recently achieved (∼1980): 18 infinite families

1) Cp (p prime)

2) An (n ≥ 5)

3) PSLn(F ) (F a finite field, n ≥ 2) . . . etc.

and 26 “sporadic” (exceptional) simple groups. The smallest sporadic group,of order 7920, was discovered by Mattieu in 1861. The largest, of order

808, 017, 424, 794, 512, 875, 886, 459, 904, 961, 710, 757, 005, 754, 268, 000, 000, 000

was constructed by Fischer-Griess in 1981, the “monster” group.

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§6. Sylow Theory

Throughout this section

G will denote a finite group of order n.

Recall Lagrange’s Theorem: H < G, |H| = k =⇒ k|n. The converse fails:k|n 6=⇒ ∃H < G, |H| = k. For example

A4 (∼= T12) has no subgroup of order 6

as proved above. (Another pf using quotient groups: Sps H < A4, |H| = 6.Then H C A4, so A4/H ∼= C2 =⇒ H contains all eight 3-cycles τ in A4,since τH = (τH)3 = τ3H = H =⇒ τ ∈ H. This contradicts |H| = 6.)

Partial converses hold:

Cauchy’s Theorem If p is prime and p|n, then ∃H < G with |H| = p.

More generally

Sylow Theorem I If p is prime and pk|n, then ∃H < G with |H| = pk.

Definition Let p be a prime. Any group P of order pα is called a p-group,and if P < G then P is called a p-subgroup of G. If in addition pα is thelargest power of p that divides |G|, that is

n = pαm and p - m

then P is called a Sylow p-subgroup of G. They always exist by Sylow I.

For example If n = 500 = 2253, then G has subgroups of orders2, 4, 5, 25 and 125. The ones of order 4 are the Sylow-2 subgroups, andthose of order 125 are the Sylow-5 subgroups.

Sylow Theorem II (Conjugacy) Any two Sylow p-sugroups P,Q of G areconjugate, i.e. ∃x ∈ G such that Q = xPx−1.

Sylow Theorem III (Counting) The total number np = np(G) of Sylowp-subgroups of G satisfies ia np|m and ib np ≡ 1 (mod p).

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Example If G has order 12 = 223, then n3|4 and n3 ≡ 1 (mod 3), son3 = 1 or 4. Either case may arise: n3(C12) = 1 and n3(A4) = 4 (the fourSylow 3-subgroups of A4 are 〈(123)〉, 〈(124)〉, 〈(134)〉 and 〈(234)〉, whichare all conjugate). Similarly n2 = 1 or 3, e.g. n2(C12) = n2(A4) = 1 whilen2(D12) = 3 (can you find the three Sylow 2-subgps of D12?).

Application (non-simplicity results)

Observe that if a finite group G has only one Sylow p-subgroup P , forsome prime divisor p of |G|, then P C G (since any xPx−1 is also a Sylowp-subgroup of G). Thus

np(G) = 1 =⇒ G is not simple

if G is not a p-group. (It can also be shown that the only p-group that issimple is Cp, see below). It’s often not hard to show some np = 1:i1 |G| = pq for distinct primes p and q =⇒ G is not simple.

Proof: We may assume p > q, and then np|q and np ≡ 1 (mod p), bySylow III, which forces np = 1.†

For example no groups of order 6, 10, 14, 15, 21, 22, 26, 33, . . . are simple.

i2 |G| = 30 = 2 · 3 · 5 =⇒ G is not simple.

Proof: The possibilities (using Sylow III) are n2 = 1, 3, 5, 15, n3 = 1, 10and n5 = 1, 6. Thus if G is simple, then n2 ≥ 3, n3 = 10 and n5 = 6.

Now observe that if P and Q are distinct Sylow subgroups of G, thenP ∩ Q = 1. This is true in general if P and Q are associated with distinctprimes, since P ∩Q is a subgroup of both P and Q, and so |P ∩Q| dividesboth |P | and |Q| =⇒ |P ∩ Q| = 1 (since |P | and |Q| are relatively prime).If P and Q are associated with the same prime p and are of prime order(this is the case for n = 30 since 30 is square free), then P ∩ Q must betrivial since it is a proper subgroup of P and |P | = p. (Note that two Sylowp-subgroups that are not of prime order may overlap nontrivially!)

Finally, count the nonidentity elements in G, using the fact that theSylow subgroups don’t overlap: |G| > 6 · 4 + 2 · 10 + 1 · 3 = 47 =⇒⇐=. ThusG is not simple.†In fact, Sylow theory gives more information. For example, if q - (p − 1), then G

is in fact cyclic! Indeed Sylow III then gives nq = 1 as well, so G has a unique Sylowp-subgroup P ∼= Cp and a unique Sylow q-subroup Q ∼= Cq. Thus P,Q C G and P ∩Q = 1,and Lagrange’s Theorem shows that PQ = G, and so G ∼= P ×Q ∼= Cpq, by the productrecognition theorem. This shows that all groups of order 15, 33, ... are cyclic.

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More is known: No group of order pqr is simple, where p, q and r aredistinct primes (HW; counting argument as in i2 ). Similarly groups oforder p2q and pα (for any α > 1) are never simple. Much deeper results:

Burnside Theorem |G| = pαqβ =⇒ G is not simple.

Feit-Thompson Theorem |G| odd (but not prime) =⇒ G is not simple

Proofs of Sylow Theorems (using group actions)

Let G ×X → X, (g, x) 7→ g · x be an action of a group G on a set X.For any x ∈ X, define the orbit of x to be

Gx = {g · x | g ∈ G} ⊂ X

and the stabilizer (or isotropy subgroup) of x to be

Gx = {G ∈ G | g · x = x} < G.

(note the subscript). Call x a fixed point of the action if g · x = x for allg ∈ G, or equivalently Gx = {x}, or Gx = G. Let XG = {all fixed points}.

The most important theorem in the subject is:

Orbit Stabilizer Theorem (OST) If G and X are finite, then the sizeof any orbit Gx is equal to the index of the corresponding stabilizer Gx. Insymbols |Gx| = |G : Gx|, or equivalently |G| = |Gx||Gx|.

Proof The function G/Gx → Gx, gGx 7→ g ·x is a well defined bijection:gGx = hGx ⇐⇒ g−1h ∈ Gx ⇐⇒ g−1 · h · x = x⇐⇒ h · x = g · x.

Examples i1 Let G act on itself by conjugation, g · x = gxg−1. ThenGx = C(x) (the conjugacy class of x), Gx = Z(x) (the centralizer of x) andGG = Z(G) (the center of G). The OST says

|C(x)| = |G : Z(x)|.

i2 Let G act on the set X of all its subgroups by conjugation, g ·S = gSg−1.Then GS = C(S) = {conjugates of S}, GS = N(S) (the normalizer of S)and XG = {normal subgroups of G}. The OST says

|C(S)| = |G : N(S)|.

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Now observe that the orbits of an action of G on X partition X (sinceGx ∩Gy 6= ∅ =⇒ g · x = h · y for some g, h ∈ G =⇒ k · y = (kh−1g) · x =⇒Gy ⊂ Gx, and similarly Gx ⊂ Gy, so Gx = Gy). Thus

|X| =∑|Gxi|

where the sum is over a set of representatives xi, one chosen from each orbit.Since |Gx| = |G : Gx| by the OST, the right hand side can be rewritten as∑|G : Gxi | = |XG| +

∑|G : Gxi | where the last sum is only over those xi

that are not fixed points. This gives the class equation (for finite X and G)

|X| = |XG|+∑|G : Gxi |

summed over representatives xi of the non-trivial orbits. In the special casewhen G acts on itself by conjugation, as in Example i1 above, this reads

|G| = |Z(G)|+∑|G : Z(xi)|

summed over representatives of the non-trivial conjugacy classes.

One very useful consequence of the general class equation, which is thekey to our proof of Sylow’s Theorems, is

Lemma If a p-group P acts on a finite set X, then |X| ≡ |XP | (mod p).

Proof The class equation says |X| = |XP | +∑|P : Hi|, where the Hi

are proper subgps of P , and each |P : Hi| ≡ 0 (mod p) since P is a p-group,so |X| ≡ |XP | (mod p).

Corollary The center Z of any nontrivial p-group P is nontrivial.†

Proof Let P act on itself by conjugation. Then the fixed point set is Z,which by the lemma has order divisible by p, so cannot be trivial.

Now we are ready to prove Sylow’s Theorems.

†It follows that if |P | = p2, then P is abelian. Indeed |G/P | = 1 or p by the Corollary,and so P/Z is cyclic, generated say by some xZ. But then any element in P is of the formxku for some k ∈ Z, u ∈ Z and so P is abelian: ∀u, v ∈ Z, xkux`v = xk+`uv = x`vxku.

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Proof of Sylow I The result is obvious if G is the trivial group, so wesuppose G nontrivial and induct on its order |G| = pαm (with p - m)assuming the result for groups of smaller order. Set Z = center of G.

Case 1: p divides |Z|. Then it follows easily from the structure theoremfor finite abelian groups that Z has a subgroup P of order p =⇒ |G/Z| =pα−1m =⇒ (by induction) G/P has subgroups of order 1, p . . . , pα−1 =⇒(by the correspondence theorem) G has subgroups of order 1, p, . . . , pα.

Case 2: p does not divide |Z|. Then by the class equation, p does notdivide the index of the centralizer Z(x) of some x 6∈ Z =⇒ pα||Z(x)| =⇒ (byinduction) Z(x) has subgroups of order 1, p, . . . , pα =⇒ G does as well.

Proof of Sylow II For any two Sylow p-subgroups P and Q of G, let Pact by left multiplication on G/Q. Since |G/Q| = m 6≡ 0 (mod p), it followsfrom the lemma that ∃ a fixed point xQ, i.e. for all g ∈ P , have

gxQ = xQ =⇒ gx ∈ xQ =⇒ g ∈ xQx−1.

Thus P ⊂ xQx−1, and so they are equal since they have the same order.

Note : This proof shows that any p-subgroup of G is a subset of someSylow p-subgroup.

Proof of Sylow III Let X = {Sylow p-subgroups of G}, so

np = |X|.

Choose P ∈ X (by Sylow I) so X = C(P ), the set of all subgroups conjugateto P (by Sylow II). Thus |X| = |G : N(P )| (see example i2 on page 40),which is a divisor of m = |G : P | = |G : N(P )||N(P ) : P |, i.e. np|m.

To see np ≡ 1 (mod p), consider the action of P on X by conjugation.Since np|m, we have np 6≡ 0 (mod p) =⇒ ∃ a fixed point Q ∈ X, i.e.P ⊂ N(Q). But this forces P = Q, since Q C N(Q) (by definition) and soQ is the only Sylow p-subgroup of N(Q). Thus P is the unique fixed pointof this action, so np ≡ 1 (mod p) by the lemma.

Exercises i1 Find all orders n < 100 for which Sylow III applies directly(without extra counting arguments as above) to produce a normal Sylowsubgroup. (Answer: all except n = 12, 24, 30, 36, 48, 56, 60, 72, 80, 90, 96)i2 Show that S4 (of order 24) has no normal Sylow subgroups, but thatevery group of order less than 24 has at least one such subgroup.

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III Rings

§1. Basic Concepts

Definition A ring is a set R with two binary operations + and · satisfyingia (R,+) is an abelian group (with identity 0),ib (R, ·) is a semigroup, andic · is distributive over + (on both sides)

Make sure you know what this entails (e.g. distributivity says that r(s+t) =rs+ rt and (s+ t)r = sr+ tr, ∀r, s, t ∈ R). Say R is a ring with 1 if ∃1 ∈ Rsuch that 1r = r1 = r (∀r ∈ R), and R is commutative if rs = sr (∀r, s ∈ R).

A subset S ⊂ R is a subring of R, denoted S < R, if 0 ∈ S ands, t ∈ S =⇒ −s, s + t and st ∈ S. Can show (as in group theory) thatS < R⇐⇒ S is nonempty and closed under subtraction and multiplication.

Examples i1 The trivial ring R = {0} is commutative with 1 = 0.i2 Z < Q < R < C < H are all rings with 1, and all but the last are comm.i3 Zn is a commutative ring with 1, for any n ∈ N.i4 The set Z[i] = {a + bi | a, b ∈ Z} of Gaussian integers is a subring of C.Similarly

Z[ζ] := {a+ bζ | a, b ∈ Z} < C

where ζ = e2πi/3 ∈ C; this ring will play a special role in the proof belowof Fermat’s last theorem for n = 3. (Draw “pictures” of these rings assquare/triangular lattice points in the plane)i5 (new rings from old) For any rings R and S have

(a) the product R× S with component-wise operations

(b) matrix rings Mn(R) of n×n matrices with entries in R with the usualaddition and multiplication of matrices

(c) polynomial rings R[x] = {∑n

i=0 rixi | ri ∈ R} with the usual addition

and multiplication of polynomials, and in more variablesR[x, y], R[x, y, z],etc.

(d) power series rings R[[x]] = {∑∞

i=0 rixi | ri ∈ R}, R[[x, y]], etc.

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(e) group rings RG = {∑n

=0 rigi | ri ∈ R, gi ∈ G} of G (any group) overR, with the operations

∑ni=0 rigi +

∑ni=0 sigi =

∑ni=0(ri + si)gi and

(∑n

i=0 rigi)(∑n

j=0 sjgj) =∑n

i,j=0(risj)gigj

(f) function rings The set [X,R] of all functions from a set X to R underthe operations (f + g)(x) = f(x) + g(x) and (fg)(x) = f(x)g(x).

Remark Many familiar properties of Z generalize to all rings, for exampleia 0r = 0, ib (−r)s = r(−s) = −(rs), ic −1 · r = −r (in rings with 1);see the text for proofs. But not all, e.g. commutativity. Also, it is not truein general (even in commutative rings) that

rs = 0 =⇒ r or s = 0

e.g. 2 · 2 = 0 in Z4. Commutative rings with 1 6= 0 where this holds arecalled integral domains (more on this below).

Definition A function f : R → S, where R and S are rings, is called aring homomorphism if

f(r + s) = f(r) + f(s) and f(rs) = f(r)f(s)

for all r, s ∈ R. If R and S are rings with identity, often also require thatf(1) = 1. The kernel of f is defined by ker(f) = f−1(0) (not f−1(1)!) andthe image im(f) is defined in the usual way. Both are subrings (of R and Srespectively). Have the usual criterion f is 1-1 ⇐⇒ ker(f) = {0}.

Example i1 f : Z→ Zn, f(k) = k is an epimorphism with kernel nZ.i2 g : Z→ Zm × Zn, g(k) = (k, k) (e.g. if m = 5, n = 9 then f(13) = (3, 4))is a homomorphism with kernel `Z, where ` is the lcm of m and n.

§2. Ideals and Quotient Rings

Definition A subring A of a ring R is an ideal in R, denoted A C R, ifa ∈ A, r ∈ R =⇒ ra and ar ∈ A, i.e. A is closed under multiplication on theleft or right by arbitrary elements of R.

ideals are the ring theory analogue of normal subgroups in group theory

Examples i0 Every ring R has the trivial ideals {0} and R; all otherideals are called proper.

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i1 nZ C Z for any n (these are the only ideals in Z).i2 Let R be a commutative ring with 1 and a ∈ R. Then

aR = {ar | r ∈ R} (also denoted 〈a〉)

is an ideal in R containing a (verify this) called the principal ideal generatedby a. An ideal J C R is called a principal ideal if J = 〈a〉 for some a ∈ R.i3 Let f : R→ S be a ring morphism. Then ker(f) C R (verify this).

Definition Given an ideal J in a ring R, define the quotient ring R/J tobe the set of cosets {r + J | r ∈ R} with operations + and · defined in theobvious way

(r + J) + (s+ J) = (r + s) + J and (r + J)(s+ J) = (rs) + J.

These operations are well-defined (e.g. for ·: if r+J = r′+J , i.e. r− r′ ∈ J ,then rs − r′s = (r − r′)s ∈ J , so rs + J = r′s + J ; similarly indep of thechoice of rep for s + J). They make R/J into a ring; the additive identityis 0 + J = J and the negative of r + J is (−r) + J .

There’s a universal property, as for gps, and isomorphism theorems, e.g.:

First Isomorphism Theorem If f : R→ S is a ring homomorphism, thenR/ ker(f) ∼= im(f).

Examples i1 Using f : Z→ Zn of example i1 in §1, have Z/nZ ∼= Zn.i2 Using g : Z → Zm×Zn of example i2 in §1, have im(g) ∼= Z/`Z, where` =lcm(m,n). In particular, if m and n are relatively prime, so ` = mn,then (counting elements) we see that g induces an isomorphism

Zmn ∼= Zm × Zn if (m,n) = 1

mapping k to (k, k). This =⇒ the “Chinese Remainder Theorem” that statesthat for any a, b ∈ Z, there exists x ∈ Z satisfying the system of congruences

x ≡ a (mod m)x ≡ b (mod n)

and all other solutions are of the form x+kmn for some k ∈ Z. For example,if m = 5, n = 9 and a = 2, b = 8, then have the soln set {17 + 45k | k ∈ Z}.

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§3. Integral Domains and Fields

Throughout this section, assume R is a commutative ring with 1 6= 0.

Definition A nonzero element a ∈ R is ia a zero divisor if ∃ b ∈ R − 0such that ab = 0 ib a unit if ∃ b ∈ R such that ab = 1 (Note that bis unique, if it exists, and is denoted a−1.) Two elements a, b ∈ R areassociates, written a ∼ b, if a = bu for some unit u.

Set R◦ = {zero divisors in R} and R• = {units in R}. Then

R◦ ∩R• = ∅

since a ∈ R◦ ∩ R• =⇒ 0 = ab for some b 6= 0 =⇒ b = a−1ab = a−10 = 0which is a contradiction.

Definition‡ R is called an integral domain (or just a domain) if it hasno zero divisors (i.e. R◦ = ∅), and a field if all nonzero elements are units(i.e. R• = R− {0}).

Clearly every field is a domain (since R◦ ∩ R• = ∅) but not conversely(e.g. Z is a domain but not a field). There is a useful characterization offields in terms of ideals:

Lemma† R is a field ⇐⇒ R has no proper ideals.

Proof (=⇒) For any nonzero ideal J , choose a 6= 0 in J . Then for anyr ∈ R have r = ra−1a ∈ J , so J = R. (⇐=) For any a 6= 0 in R, theprincipal ideal aR 6= 0, so by hypothesis aR = R. Thus ∃r ∈ R such thatar = 1 so a is a unit. Thus R is a field.

There are two special kinds of ideals that relate to these notions:

Definition† Let J C R, J 6= R. Then J is prime if ab ∈ J =⇒ a ∈ J orb ∈ J , and J is maximal if J ⊂ K C R =⇒ K = J or K = R.

Exercise R is a domain ⇐⇒ {0} is a prime ideal.

Theorem Let R be a commutative ring with 1 6= 0 and J C R. Thenia J is prime ⇐⇒ R/J is a domain ib J is maximal ⇐⇒ R/J is a field.

Proof ia HW ib By the correspondence theorem J is maximal ⇐⇒R/J has no proper ideals, and this is equivalent to R/J being a field by theprevious lemma.‡Recall that we are assuming that R is a commutative ring with 1 6= 0.

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Examples i1 nZ C Z is prime ⇐⇒ maximal ⇐⇒ n is a prime number.i2 In general, maximal ideals are always prime (HW) but not conversely.For example 〈x〉 C Z[x] (consisting of all polynomials with 0 constant term)is prime but not maximal: Z[x]/〈x〉 ∼= Z (and now apply the theorem).

Coda

Fermat’s Last Theorem (1637, proved by A. Wiles in 1995) For n ≥ 3,the equation

xn + yn = zn

has no non-zero integral solutions x, y, z. (Call this FLTn for any given n.)

Remarks i1 FLTn =⇒ FLTkn, since any solution x, y, z for kn wouldgive the solution xk, yk, zk for n. Fermat knew a proof for n = 4, and so thisreduces the proof to the case of odd primes, i.e. it suffices to prove that

xp + yp = zp

has no non-zero integral solutions x, y, z for any odd prime p.i2 A solution x, y, z is primitive if x, y and z are nonzero and have no commonfactor. Enough to prove 6 ∃ primitive solutions (since any soln cx, cy, cz givesanother x, y, z). Note x, y, z primitive solution =⇒ pairwise relatively prime.i3 Proof for p = 3 (which we give below) was known to Euler, for p = 5 toDirichlet & Legendre, for p = 7 to Gabriel Lame. Indeed Lame and Cauchy(independently) thought they had it all. Their approach was to show 6 ∃ anysolutions in the bigger ring

Λp := Z[ζp] = {polys in ζp with Z coeffs}

where ζp = e2πi/n; their mistake, uncovered by Kummer, was to assumethat Λp has unique factorization into primes (as Z does) which in fact itdoes only for p ≤ 19. Kummer’s work led him to introduce “ideals” whichthen led to modern ring theory.

Proof (for n = 3) Set ζ = ζ3 and Λ = Λ3 = Z[ζ]. We will show that forany unit u ∈ Λ•,

x3 + y3 = uz3

has no primitive solutions in Λ (FLT is the case u = 1). Note that

Λ• = {±1,±ζ,±ζ2}

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as seen by inspection (since these are the elements of complex norm 1 in Λ,and all other elements have norm > 1).

It is a fact (which we won’t prove here) that any nonzero, nonunit inΛ can be factored uniquely (up to order and replacement by associates) asa product of primes (nonunits whose only divisors are their associates andunits). One such prime is p := ζ − 1. Why is p prime? Well, |p| =

√3

which is the smallest norm achieved by a nonunit in Λ, so if p = ab then|p| = |a||b| =⇒ |a| or |b| = 1, i.e. a or b ∈ Λ•.

Remarks i1 3 is not prime in Λ, but factors as 3 = p2u, where u = −ζ2.Thus p2|3 (also written 3 ≡p2 0) but p3 - 3 (3 6≡p3 0).i2 Each a ∈ Λ can be written uniquely in the form

a = a+ rp

for some r ∈ Λ where a = 0, 1 or −1. This defines an isomorphism

Λ/pΛ→ Z3 , a+ pΛ 7→ a

To show this, write a as a poly w/ integral coeff in ζ, and thus in p bysubstituting p+ 1 for ζ. Now reduce the constant term to 0 or ±1 using i1For example u = ±1 for any u ∈ Λ•; in particular ζ = ζ2 = 1.i3 For any a ∈ Λ, have a3 ≡p3 a. Proof: a = a + rp, by i2 , so a3 =a3 + 3a2rp+ 3ar2p2 + r3p3 ≡p3 a3 (by i1 ) = a.

Now suppose xo, yo, zo is a primitive solution to

x3 + y3 = uz3

for some unit u. This is Fermat’s equation when u = 1, but as we shall see,it is easier to prove simultaneously that none of these six equations (for thevarious units u) have solutions.

Case 1 p - xoyozo. Then by the remarks above, xo, yo, zo = ±1 =⇒0 = x3

o + y3o − uz3

o ≡p3 xo + yo ± zo = ±1 or ±3 6≡p3 0 =⇒⇐=.

Case 2 p | xoyozo. Then p divides exactly one of xo, yo, zo, since theyare pairwise relatively prime.

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Case 2a Suppose p | zo. Then xo = −yo = ±1, so without loss of gener-ality, xo = rp+ 1 and yo = sp− 1 for suitable r, s ∈ Λ.

Let k be the largest natural number for which pk | zo, called the p-orderof the solution. A simple calculation (reducing mod p4) shows that in factk ≥ 2.† We assume that the solution was chosen so that k is minimal.

Now (and this is the magic!) define

a =xo + yo

pb =

ζxo + ζ2yop

c =ζ2xo + ζyo

p.

The numerators are all divisible by p (e.g. for b, ζxo + ζ2yo = xo + yo = 0)and so a, b, c ∈ Λ. A straightforward argument, using the fact that ζ3 = 1and 1 + ζ + ζ2 = 0, showsi1 a+ b+ c = 0i2 abc = (x3

o + y3o)/p

3 = u(zo/p)3i3 a, b, c are pairwise relatively prime, i.e. have no common prime factors(to see this, note that xo, yo are linearly related to any pair of a, b, c, and soa common factor for such a pair would yield one for xo, yo).i2 and i3 show that each of a, b, c is a unit times a cube, and that thesecubes are relatively prime. It follows from i1 that ∃x1, y1, z1 with x3

1, y31, z

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associates of a, b, c, in some order, such that p - x1, p - y1, p | z1, pk - z1,and

x31 + vy3

1 + wz31 = 0

for suitable units v, w. It is easy to check that v = ±1 (since the left handside is congruent mod p3 to x1 + vy1 = ±1 ± v =⇒ v ≡p3 ±1 =⇒ v = ±1)and so this gives a new solution x1,±y1, z1 of smaller p-order to one of theoriginal equations, contradicting the minimality of k.

Case 2b Suppose p |xo (or yo). Then uz ≡p3 x + y =⇒ u ≡p3 ±1 =⇒u = ±1 =⇒ (−yo)3 + (±zo)3 = x3

o, which is handled in case 2a.

Thus Fermat’s Last Theorem for n = 3 is proved.

†x30 + y3

0 = (rp+ 1)3 + (sp− 1)3 ≡p4 r3 + s3 − ζ2(r + s))p3 ≡p4 (r + s− (r + s))p3 =0 =⇒ z ≡p2 0.

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