Solution Dummit

Figure 0.1

ABSTRACT ALGEBRA

DAVID S. DUMMIT AND RICHARD M. FOOTE

Solutions provided by Scott Larson.

Contents

0. Preliminaries 30.1. Basics 30.2. Properties of the Integers 40.3. Z/nZ: The Integers Modulo n 6Part I – Group Theory 81. Introduction to Groups 81.1. Basic Axioms and Examples 81.2. Dihedral Groups 101.3. Symmetric Groups 111.4. Matrix Groups 111.5. Quaternion Groups 131.6. Homomorphisms and Isomorphisms 141.7. Group Actions 142. Subgroups 152.1. Definition and Examples 152.2. Centralizers and Normalizers, Stabilizers and Kernels 152.3. Cyclic Groups and Cyclic Subgroups 162.4. Subgroups Generated by Subsets of a Group 172.5. Definitions and Examples 192.6. The Lattice of Subgroups of a Group 203. Quotient Groups and Homomorphisms 203.1. Definitions and Examples 203.2. More on Cosets and Lagrange’s Theorem 213.3. The Isomorphism Theorems 233.4. Composition Series and the Holder Program 243.5. Transpositions and the Alternating Group 244. Group Actions 254.1. Group Actions and Permutation Representations 254.2. Groups Acting on Themselves by Left Multiplication – Cayley’s Theorem 254.3. Groups Acting on Themselves by Conjugation – The Class Equation 25

1

2 DAVID S. DUMMIT AND RICHARD M. FOOTE

4.4. Automorphisms 264.5. The Sylow Theorems 264.6. The Simplicity of An 275. Direct and Semidirect Products and Abelian Groups 285.1. Direct Products 285.2. The Fundamental Theorem of Finitely Generated Abelian Groups 295.3. Table of Groups of Small Order 305.4. Recognizing Direct Products 305.5. Semidirect Products 306. Further Topics in Group Theory 316.1. p-groups, Nilpotent Groups, and Solvable Groups 31Part II – Ring Theory 337. Introduction to Rings 337.1. Basic Definitions and Examples 337.2. Examples: Polynomials Rings, Matrix Rings, and Group Rings 347.3. Ring Homomorphisms and Quotient Rings 367.4. Properties of Ideals 397.5. Rings of Fractions 428. Euclidean, Principal Ideal, and Unique Factorization Domains 458.1. Euclidean Domains 459. Polynomial Rings 469.1. Definitions and Basic Properties 469.2. Polynomial Rings Over Fields I 479.3. Polynomial Rings That are Unique Factorization Domains 489.4. Irreducibility Criteria 489.5. Polynomial Rings Over Fields II 509.6. Polynomials in Several Variables Over a Field and Grobner Bases 50Part III – Modules and Vector Spaces 5310. Introduction to Module Theory 5310.1. Basic Definitions and Examples 5313. Field Theory 5413.1. Basic Theory of Field Extensions 5413.2. Basic Theory of Field Extensions 5413.3. Classical Straightedge and Compass Constructions 5613.4. Splitting Fields and Algebraic Closures 5613.5. Separable and Inseparable Extensions 5613.6. Cyclotomic Polynomials and Extensions 5714. Galois Theory 5714.1. Basic Definitions 5714.2. The Fundamental Theorem of Galois Theory 5715. Commutative Rings and Algebraic Geometry 5815.1. Noetherian Rings and Affine Algebraic Sets 5815.2. Radicals and Affine Varieties 5815.3. Integral Extensions and Hilbert’s Nullstellensatz 5815.4. Localization 58

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0. Preliminaries

0.1. Basics.

Proposition 0.1. Let f : A→ B.

(1) The map f is injective if and only if f has a left inverse.(2) The map f is surjective if and only if f has a right inverse.(3) The map f is a bijection if and only if there exists g : B → A such that f ◦ g is the identity map on B and

g ◦ f is the identity map on A.(4) If A and B are finite sets with the same number of elements (|A| = |B|), then f : A→ B is bijective if and

only if f is injective if and only if f is surjective.

Proof. (a) Suppose f is injective so that f−1(b) contains a single element for b ∈ f(A). Thus g(b) = f−1(b) iswell-defined for b ∈ f(A). Hence g(f(a)) = a for all a ∈ A. Now suppose that f has a left inverse g so thatg(f(a)) = a for all a ∈ A. If f(a1) = f(a2) = b, then g(b) = a1 and g(b) = a2. But g is a well-defined map soa1 = a2 and thus f is injective.

(b) Suppose f is surjective. Take some b ∈ B so we can choose a ∈ A such that f(a) = b. Thus we can defineg(b) = a such that f(a) = b. Hence f(g(b)) = f(a) = b. Now suppose that f has a right inverse g so that f(g(b)) = bfor all b ∈ B. So f(g(B)) = B and thus f is surjective.

(c) Suppose that f is bijective so there is a left inverse, gl, and a right inverse, gr, of f . Let f(a) = b so thata = gl(f(a)) = gl(b) and b = f(gr(b)). Since f is injective, gr(b) = a and thus gl = gr. Thus the right and leftinverses of f are the same map. Now suppose that there exists g : B → A such that f ◦ g is the identity map onB and g ◦ f is the identity map on A. Then f is surjective because f(g(B)) = B. Now let f(a1) = f(a2) so thata1 = g(f(a1)) = g(f(a2)) = a2. Thus f is injective.

(d) Suppose that f is injective. Then f takes elements of A to unique elements of B. Since |A| = |B|, f issurjective. Now suppose that f is surjective. Then since |A| = |B|, f takes elements of A to unique elements of Bso f is injective. Therefore f is either bijective or neither injective or surjective. �

Proposition 0.2. Let A be a nonepty set.

(1) If ∼ defines an equivalence relation on A then the set of equivalence classes of ∼ form a partition of A.(2) If {Ai | i ∈ I} is a partition of A then there is an equivalence relation on A whose equivalence classes are

preciselly the sets Ai, i ∈ I.

In Exercises 1 to 4 let A be the set of 2× 2 matrices with real number entries. Recall that matrix multiplicationis defined by (

a bc d

)(p qr s

)=

(ap+ br aq + bscp+ dr cq + ds

)Let

M =

(1 10 1

)and let

B = {X ∈ A |MX = XM} .0.1.1. The following elements of A lie in B:(

1 10 1

),

(0 00 0

),

(1 00 1

).

0.1.2. Prove that if P,Q ∈ B, then P +Q ∈ B (where + denotes the usual sum of two matrices).

Proof.

(P +Q)M = PM +QM = MP +MQ = M(P +Q).

�

0.1.3. Prove that if P,Q ∈ B, then P ·Q ∈ B (where · denotes the usual product of two matrices).

Proof.

(P ·Q)M = P ·M ·Q = M · P ·Q = M(P ·Q).

�


0.1.4. Find conditions on p, q, r, s which determine precisely when

(p qr s

)∈ B.

r = 0, p = s.

0.1.5. Determine whether the following functions f are well defined:

(1) f : Q→ Z defined by f(a/b) = a.(2) f : Q→ Q defined by f(a/b) = a2/b2.

Proof. (a) f is not a well defined map because 1/2 = 2/4 and 1 6= 2.(b) f is a well defined map because if a1/b1 = a2/b2, then a21/b

21 = a22/b

22. �

0.1.6. Determine whether the function f : R+ → Z defined by mapping a real number r to the first digit to theright of the decimal point in a decimal expansion of r is well defined.f is well defined because the decimal expansion of a real number is unique. Thus there is one and only one

possible first number to the right of the decimal for each real number.

0.1.7. Let f : A→ B be a surjective map of sets. Prove that the relation

a ∼ b if and only if f(a) = f(b)

is an equivalence relation whose equivalence classes are the fibers of f .

Proof. f(a) = f(a) so ∼ is reflexive. If f(a) = f(b), then f(b) = f(a) so ∼ is symmetric. If f(a) = f(b) andf(b) = f(c), then f(a) = f(c) so ∼ is transitive.

If a1, a2 ∈ f−1(b), then f(a1) = f(a2) so a1 ∼ a2. If a1 ∼ a2 then f(a1) = f(a2) so a1 and a2 are in the samefiber of f . �

0.2. Properties of the Integers.

(1) (Well Ordering of Z) If A is any nonempty subset of Z+, there is some element m ∈ A such that m ≤ a,for all a ∈ A.

(2) If a, b ∈ Z with a 6= 0, we say a divides b if there is an element c ∈ Z such that b = ac. In this case we writea | b; if a does not divide b we write a - b.

(3) If a, b ∈ Z\ {0}, there is a unique positive integer d, called the greatest common divisor of a and b, satisfying:(a) d | a and d | b,(b) if e | a and e | b, then e | d.

(4) If a, b ∈ Z\ {0}, there is a unique positive integer l, called the least common multiple of a and B, satisfying:(a) a | l and b | l,(b) if a | m and b | m, then l | m. The connection between the greatest common divisor d and the least

common multiple l of the two integers a and b is given by dl = ab.(5) The Division Algorithm: if a, b ∈ Z and b 6= 0, then there exist unique q, r ∈ Z such taht

a = qb+ r and 0 ≤ r < |b| ,where q is the quotient and r the remainder.

(6) The Euclidean Algorithm is an important procedure which produces a greatest common divisor of twointegers a and b by iterating the Division Algorithm: if a, b ∈ Z\ {0}, then we obtain a sequence of quotientsand remainders

a = q0b+ r0

b = q1r0 + r1

r0 = q2r1 + r2

r1 = q3r2 + r3

...

rn−2 = qnrn−1 + rn

rn−1 = qn+1rn

where rn is the last nonzero remainder. Such an rn exists since |b| > |r0| > |r1| > · · · > |rn| is a decreasingsequence of strictly positive integers if the remainders are nonzero and such a sequence cannot continueindefinitely. Then rn = (a, b).

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(7) One consequence of the Euclidean Algorithm which we chall use regularly is the following: if a, b ∈ Z\ {0},then there exist x, y ∈ Z such that

(a, b) = ax+ by

that is, the gcd of a and b is a Z-linear combination of a and b. This follows by recursively writing theelement rn in the Euclidean Algorithm in terms of the previous remainders.

(8) An element p of Z+ is calle a prime if p > 1 and the only posotive divisors of p are 1 and p. An integern > 1 which is not prime is called composite. An important property of primes is if p is a prime and p | ab,for some a, b ∈ Z then either p | a or p | b.

(9) The fundamental theorem of arithmetic says: if n ∈ Z, n > 1, then n can be factored uniquely into theproduct of primes, i.e., there are distinct primes p1, p2, . . . , ps and positive integers α1, α2, . . . , αs such that

n = pα11 pα2

2 · · · pαss .

This factorization is unique in the sense that if q1, q2, . . . , qt are any distinct primes and β1, β2, . . . , βt, arepositive integers such that

n = qβ1

1 qβ2

2 · · · qβtt ,then s = t and if we arrange the two sets of primes in increasing order, then qi = pi and αi = βi, 1 ≤ i ≤ s.

Suppose the positive integers a and b are expressed as products of prime powers:

a = pα11 pα2

2 · · · pαss , b = pβ1

1 pβ2

2 · · · pβsswhere p1, p2, . . . , ps are distinct and the exponents are ≥ 0. Then the gcd of a and b is

(a, b) = pmin(α1,β1)1 p

min(α2,β2)2 · · · pmin(αs,βs)

s

and the lcm is obtained by taking the maximum of the αi and βi instead of the minimum.(10) The Euler ϕ-function is defined as follows: for n ∈ Z+ let ϕ(n) be the number of positive integers a ≤ n

with a relatively prime to n, i.e., (a, n) = 1. For primes p, ϕ(p) = p− 1, and, more generally, for all a ≥ 1we have the formula

ϕ(pa) = pa − pa−1 = pa−1(p− 1).

The function ϕ is multiplicative in the sense that

ϕ(ab) = ϕ(a)ϕ(b) if (a, b) = 1.

Together with the formula above this gives a general formula for the values of ϕ: if n = pα11 pα2

2 · · · pαss , then

ϕ(n) = ϕ(pα11 )ϕ(pα2

2 ) · · ·ϕ(pαss )

= pα1−11 (p1 − 1)pα2−1

2 (p2 − 1) · · · pαs−1s (ps − 1).

0.2.1. For each of the following paris of integers a and b, determine their gcd, their lcm , and write their gcd in theform ax+ by for some integers x and y.

(a) a = 20, b = 13.(b) a = 69, b = 372.(c) a = 792, b = 275.(d) a = 11391, b = 5673.(e) a = 1761, b = 1567.(f) a = 507885, b = 60808.

Solution. �

0.2.2. Prove that if the integer k divides the integers a and b then k divides as+ bt for every pair of integers s andt.

Proof. �

0.2.3. Prove that if n is composite then there are integers a and b such that n divides ab but n does not divideeither a or b.

Proof. �


0.2.4. Let a, b and N be fixed integers with a and b nonzero and let d = (a, b) be the gcd of a and b. Suppose x0and y0 are particular solutions to ax+by = N . Prove for any integer t that the integers

x = x0 +b

dt and y = y0 −

a

dt

are also solutions to ax+ by = N (this is in fact the general solution).

Proof. �

0.2.5. Determine the value ϕ(n) for each integer n ≤ 30 where ϕ denotes the Euler ϕ-function.

Solution. �

0.2.6. Prove the Well Ordering Property of Z by induction and prove the minimal element is unique.

Proof. �

0.2.7. If p is a prime prove that there do not exist nonzero integers a and b such that a2 = pb2.

Proof. �

0.2.8. Let p be a prime, n ∈ Z+. Find a formula for the largest power of p which divides n! = n(n−1)(n−2) · · · 2 ·1.

Solution. �

0.2.9. Write a computer program to determine the greatest common divisor (a, b) of two integers a and b and toexpress (a, b) in the form ax+ by for some integers x and y.

0.2.10. Prove for any given positive integer N there exist only finitely many integers n with ϕ(n) = N where ϕdenotes Euler’s ϕ-function. Conclude in particular that ϕ(n) tends to infinity as n tends to infinity.

Proof. �

0.2.11. Prove that if d divides n then ϕ(d) divides ϕ(n) where ϕ denotes Euler’s ϕ-function.

Proof. �

0.3. Z/nZ: The Integers Modulo n.

Theorem 0.3. The operations of addition and multiplication on Z/nZ are well defined, that is, they do not dependon the choices of representatives for the classes involved. More precisely, if a1, a2 ∈ Z and b1, b2 ∈ Z with a1 = b1and a2 = b2, then a1 + a2 = b1 + b2 and a1a2 = b1b2, i.e., if

a1 ≡ b1 mod n and a2 ≡ b2 mod n

thena1 + a2 ≡ b1 + b2 mod n and a1a2 ≡ b1b2 mod n.

Proposition 0.4. Z/nZ)× = {a ∈ Z/nZ | (a, n) = 1}.0.3.1. Write down explicitly all the elements in the residue classes of Z/18Z.

Solution. �

0.3.2. Prove that the distinct equivalence classes in Z/nZ are precisely 0, 1, 2, . . . , n− 1.

Proof. �

0.3.3. Prove that if a = an10n + an−1n−1 + · · ·+ a110 + a0 is any positive integer then a ≡ an + an−1 + · · ·+ a1 + a0mod 9.

Proof. �

0.3.4. Compute the remainder when 37100 is divided by 29.

Solution. �

0.3.5. Compute the last two digits of 91500.

Solution. �

0.3.6. Prove that the squares of the elements in Z/4Z are just 0 and 1.

7

Proof. �

0.3.7. Prove for any integers a and b that a2 + b2 never leaves a remainder of 3 when divided by 4.

Proof. �

0.3.8. Prove that for any integers a and b that a2 + b2 = 3c2 has no solutions in nonzero integers a, b, and c.

Proof. �

0.3.9. Prove that the square of any odd integer always leaves a remainder of 1 when divided by 8.

Proof. �

0.3.10. Prove that the number of elements of (Z/nZ)× is ϕ(n) where ϕ denotes the Euler ϕ-function.

Proof. �

0.3.11. Prove that if a, b ∈ (Z/nZ)×, then a · b ∈ (Z/nZ)×.

Proof. �

0.3.12. Let n ∈ Z, n > 1, and let z ∈ Z with 1 ≤ a ≤ n. Prove if a and n are not relatively prime then there existsan itneger b with 1 ≤ b < n such that ab ≡ 0 mod n and deduce that there cannot be an itneger c such that ac ≡ 1mod n.

Proof. �

0.3.13. Let n ∈ Z, n > 1, and let a ∈ Z with 1 ≤ a ≤ n. Prove that if a and n are relatively prime then there isan integer c such that ac ≡ 1 mod n.

Proof. �

0.3.14. Conclude from the previous two exercises that (Z/nZ)× is the set of elements a of Z/nZ with (a, n) = 1and hence prove proposition 0.4. Verify this directly in the case n = 12.

Solution. �

0.3.15. For each of the following pairs of integers a and n, show that a is relatively prime to n and determine themultiplicative inverse of a in Z/nZ.

(a) a = 13, n = 20.(b) a = 69, n = 89.(c) a = 1891, n = 3797.(d) a = 6003722857, n = 77695236973.

Proof. �

0.3.16. Write a computer program to add and multiply mod n, for any n given as input. The output of theseoperations should be the least residues of the usms and products of two integers. Also include the feature that if(a, n) = 1, and integer c between 1 and n− 1 such taht a · c = 1 may be printed on request.


Part I – Group Theory

1. Introduction to Groups

1.1. Basic Axioms and Examples.

Proposition 1.1. If G is a group under the operation ?, then

(1) the identity of G is unique(2) for each a ∈ G, a−1 is uniquely determined(3) (a−1)−1 = a for all a ∈ G(4) (a ? b)−1 = (b−1) ? (a−1)(5) for any a1, a2, . . . , an ∈ G the value of a1 ? a2 ? · · · ? an is independent of how the expression is bracketed.

Proposition 1.2. Let G be a group and let a, b ∈ G. The equations ax = b and ya = b have unique solutions forx, y ∈ G. In particular, the left and right cancellation laws hold in G, i.e.,

(1) if au = av, then u = v,(2) if ub = vb, then u = v.

Let G be a group.

1.1.1. Determine which of the following binary operations are associative:

(a) the operation ? on Z defined by a ? b = a− b(b) the operation ? on R defined by a ? b = a+ b+ ab(d) the operation ? on Z× Z defined by (a, b) ? (c, d) = (ad+ bc, bd).

Note that (1 − 2) − 3 = −4 and 1 − (2 − 3) = 2, so (a) is not associative. We calculate that (b) and (d) areassociative below.

(a ? b) ? c = (a+ b+ ab) + c+ (a+ b+ ab)c

= a+ b+ c+ ab+ ac+ bc+ abc

= a+ b+ c+ bc+ a(b+ c+ bc)

= a ? (b ? c)

((a, b) ? (c, d)

)? (e, f) =

((ad+ bc)f + bde, bdf

)=(adf + b(cf + de), bdf

)= (a, b) ?

((c, d) ? (e, f)

)1.1.2. Decide which of the following binary operations are commutative:

(a) the operation ? on Z defined by a ? b = a− b(b) the operation ? on R defined by a ? b = a+ b+ ab(d) the operation ? on Z× Z defined by (a, b) ? (c, d) = (ad+ bc, bd).

Note that 1− 2 = −1 and 2− 1 = 1 so (a) is not commutative. We calculate that (b) and (d) are commutativebelow.

a ? b = a+ b+ ab

= b+ a+ ba

= b ? a

(a, b) ? (c, d) = (ad+ bc, bd)

= (cb+ da, db)

= (c, d) ? (a, b)

1.1.3. Prove that addition of residue classes in Z/nZ is associative.

Proof. �

1.1.4. Prove that multiplication of residue classes in Z/nZ is associative.

Proof. �

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1.1.5. Prove for all n > 1 that Z/nZ is not a group under multiplication of residue classes.

Proof. �

1.1.6. Determine which of the following sets are groups under addition:

(b) the set of rational numbers in lowest terms whose denominators are even together with 0(d) the set of rational numbers of absolute value ≥ 1 together with 0(e) the set of rational numbers with denominators equal to 1 or 2.

Note that 1/6 + 1/6 = 1/3 so addition is not a binary relation on (b) and thus (b) is not a group. Also note that−3/2 + 1 = −1/2 so addition is not a binary relation on (d) and thus (d) is note a group. We notice that additionis a well defined binary relation on (e), addition is associative on the rational numbers so this subset must also havethe associative property, the identity is 0, and the inverse of a is −a. Therefore (e) is a group.

1.1.7. Let G = {x ∈ R | 0 ≤ x < 1} and for x, y ∈ G let x?y be the fractional part of x+y (i.e., x?y = x+y−[x+y]where [a] is the greatest integer less than or equal to a). Prove that ? is a well defined binary operation on G andthat G is an abelian group under ? (called the real numbers mod 1).

Proof. Note that the sum of two real numbers has a unique fractional part, x, such that 0 ≤ x < 1. Thus ? is awell defined operation that takes x ? y into G. We show that ? is associative on G.

(x ? y) ? z = (x+ y − [x+ y]) + z − [(x+ y − [x+ y]) + z]

= x+ y + z − [x+ y] + [x+ y]− [x+ y + z]

= x+ y + z − [y + z] + [y + z]− [x+ y + z]

= x+ (y + z − [y + z])− [x+ (y + z − [y + z])]

= x ? (y ? z)

The identity element of G is clearly 0. The inverse of x ∈ G would be given by 1− x for x 6= 0 because

x ? (1− x) = x+ 1− x− [x+ 1− x] = 1− 1 = 0.

Note that 1 − x ∈ G for x ∈ G and x 6= 0. The inverse of 0 is clearly 0. We show that G is abelian by lettingx, y ∈ G and computing that ? is commutative on G.

x ? y = x+ y − [x+ y]

= y + x− [y + x]

= y ? x

�

1.1.8. Let G = {z ∈ C | zn = 1 for some n ∈ Z+}.(a) Prove that G is a gorup under multiplication.(b) Prove that G is not a group under addition.

Proof (a). �

Proof (b). �

1.1.9. Let G ={a+ b

√2 ∈ R | a, b ∈ Q

}.

(a) Prove that G is a gorup under addition.(b) Prove that the nonzero elements of G are a group under multiplication.

Proof (a). �

Proof (b). �

1.1.10. Prove that a finite gorup is abelian if and only if its group table is a symmetric matrix.

Proof. �

1.1.14. Find the orders of the following elements of the multiplicative group (Z/36Z)×: 1, −1, 5, 13, −13, 17.∣∣1∣∣ = 1,∣∣−1

∣∣ = 2,∣∣5∣∣ = 6,

∣∣13∣∣ = 3,

∣∣−13∣∣ = 6,

∣∣17∣∣ = 2.

1.1.25. Prove that if x2 = 1 for all x ∈ G then G is abelian.


Proof. Since x2 = 1 we know that x = x−1 for all x ∈ G. Thus

xy = (xy)−1 = y−1x−1 = yx.

�

1.2. Dihedral Groups. In these exercises, D2n has the usual presentation

D2n =⟨r, s | rn = s2 = 1, rs = sr−1

⟩.

1.2.1. Compute the order of each of the elements in the following groups:

(a) D6

The order of elements from (a) are: |1| = 1, |r| = 3,∣∣r2∣∣ = 3, |s| = 2, |sr| = 2,

∣∣sr2∣∣ = 2.

1.2.2. Use the generators and relations above to show that if x is any element of D2n which is not a power of r,then rx = xr−1.

Proof. If x is not a power of r, then x = srk for some 0 ≤ k < n. So

rsrk = sr−1rk = srk−1 = srkr−1.

�

1.2.3. Use the generators and relations above to show that every element of D2n which is not a power of r hasorder 2. Deduce that D2n is generated by the two elements s and sr, both of which have order 2.

Proof. If x ∈ D2n is not a power of r, then x = srk for some 0 ≤ k < n. Notice that sr0sr0 = 1 and suppose thatsrisri = 1 for 0 ≤ i < k. Then

srksrk = srk−1sr−1rk = srk−1srk−1 = 1,

by induction. �

1.2.7. Show that⟨a, b | a2 = b2 = (ab)n = 1

⟩gives a presentation for D2n in terms of the two generators a = s and

b = sr of order 2 computed in Exercise 1.2.3 above. [Show that the relations for r and s follow from the relationsfor a and b and, conversely, the relations for a and b follow from those for r and s.]

Proof. First we start with the relations rn = s2 = 1 and rs = sr−1. Clearly a2 = 1, Exercise 1.2.3 shows b2 = 1,and

(ab)n = (ssr)n = rn = 1.

Hence the original relations imply the new relations.Now suppose that a2 = b2 = (ab)n = 1. Then clearly s2 = 1, rn = (ssr)n = (ab)n = 1, and

rs = ssrsrr−1 = sr−1.

�

In Exercise 1.2.9 you can find the order of the group of rigid motions in R3 (also called the group of rotations)of the given Platonic solid by following the proof for the order of D2n: find the number of positions to which anadjacent pair of vertices can be sent. Alternatively, you can find the number of places to which a given face maybe sent and, once a face is fixed, the number of positions to which a vertex on that face may be sent.

1.2.9. Let G be the group of rigid motions in R3 of a tetrahedron. Show that |G| = 12.

Proof. We use the second method described above to show that |G| = 12. Since there are 4 faces of a tetrahedron,each face having 3 distinct rotations, there are 4 · 3 = 12 positions to which a vertex on each fixed face may besent. �

1.2.17. Let X2n =⟨x, y | xn = y2 = 1, xy = yx2

⟩.

(a) Show that if n = 3k, then X2k has order 6, and it has the same generators and relations as D6 when x isreplaced by r and y by s.

(b) Show that if (3, n) = 1, then x satisfies the additional relation: x = 1. In this case deduce that X2n hasorder 2. [Use the facts that xn = 1 and x3 = 1.]

11

Proof (a). Suppose we have the relations given by D6. Then clearly x3k = (x3)k = y2 = 1. We also have

xy = yx−1 = yx3−1 = yx2.

Now suppose we have the relations given by X2n. Then clearly x3 = y2 = 1. The relations xy = yx2 and x3 = 1show us

xy = yx2 = yx−1.

Therefore D6 = X6 and |X6| = 6. �

Proof (b). Since (3, n) = 1, there exist s, t ∈ Z such that 3s+ nt = 1. So

x = x3s+nt = (x3)s(xn)t = 1.

Thus X2n has order 2, namely X2n = {1, y}. �

1.3. Symmetric Groups.

1.3.4. Compute the order of each of the elments in the following groups:

(a) S3

(b) S4

For elements in S3 we calculate that |1| = 1, |(12)| = 2, |(13)| = 2, |(23)| = 2, |(123)| = 3, |(132)| = 3.For elements in S4 we calculate that |1| = 1, |(12)| = 2, |(13)| = 2, |(14)| = 2, |(23)| = 2, |(24)| = 2, |(34)| = 2,

|(123)| = 3, |(132)| = 3, |(124)| = 3, |(142)| = 3, |(134)| = 3, |(143)| = 3, |(234)| = 3, |(243)| = 3, |(1234)| = 4,|(1243)| = 4, |(1324)| = 4, |(1342)| = 4, |(1423)| = 4, |(1432)| = 4, |(12)(34)| = 2, |(13)(24)| = 2, |(14)(23)| = 2.

1.3.7. Write out the cycle decomposition of each element of order 2 in S4.(12), (13), (14), (23), (24), (34), (12)(34), (13)(24), (14)(23).

1.3.11. Let σ be the m-cycle (1 2 . . . m). Show that σi is also an m-cycle if and only if i is relatively prime to m.

Proof. First suppose that σi is an m-cycle so that

σi = (a1 a2 . . . am).

Then assume that α|i and α|m so there are s, t ∈ Z such that

αs = i, αt = m.

So σi = σαs = [(a1 aα+1 . . . αm−α+1) · · · (aα a2α . . . am)]s which can only be an m-cycle if α = 1. Thus i and mare relatively prime.

Now suppose that i is relatively prime to m so that there are s, t ∈ Z such that

si+ tm = 1.

But then σ = σsi+tm = σsiσtm = (σi)s. Since a power of σi is an m-cycle, then σi must be an m-cycle as well. �

1.3.20. Find a set of generators and relations for S3.S3 =

⟨x2, x3 | x22 = x33 = 1, x2x3 = x−13 x2

⟩1.4. Matrix Groups.

1.4.7. Let p be a prime. Prove that the order of GL2 (Fp) is p4 − p3 − p2 + p (do not just quote the order formulain this section). [Subtract the number of 2 × 2 matrices which are not invertible from the total number of 2 × 2matrices over Fp. You may use the fact that a 2× 2 matrix is not invertible if and only if one row is a multiple ofthe other.]

Proof. Notice that the total number of 2×2 matrices over Fp is p4. We count the number of non-invertible matrices

for a matrix

(a bc d

)in Figure 1.1 to be p3 + p2 − p. Thus the order of GL2 (Fp) is p4 − p3 − p2 + p.

�

1.4.8. Show that GLn (F ) is non-abelian for any n ≥ 2 and any F .


a = 0

b = 0 b 6= 0

c = 0 c 6= 0 c = 0 c 6= 0

d = 0 d 6= 0 d = 0 d 6= 0 d = 0 d 6= 0 d = 0 d 6= 0

1 p− 1 p− 1 (p− 1)2 p− 1 (p− 1)2 0 0

a 6= 0

b = 0 b 6= 0

c = 0 c 6= 0 c = 0 c 6= 0

d = 0 d 6= 0 d = 0 d 6= 0 d = 0 d 6= 0 d = 0 d 6= 0

p− 1 0 (p− 1)2 0 (p− 1)2 (p− 1)3

Figure 1.1. Number of non-invertible matrices.

Proof. First notice (1 10 1

)(0 11 0

)=

(1 11 0

)(

0 11 0

)(1 10 1

)=

(0 11 1

)so that GL2 (F ) is non-abelian for any F . Now notice that extending the matrices with zeros shows the result forn > 2. �

1.4.10. Let G =

{(a b0 c

)| a, b, c ∈ R, a 6= 0, c 6= 0

}.

(a) Compute the product of

(a1 b10 c1

)and

(a2 b20 c2

)to show that G is closed under matrix multiplication.

(b) Find the matrix inverse of

(a b0 c

)and deduce that G is closed under inverses.

(c) Deduce that G is a subgroup of GL2 (R) (cf. Exercise 26, Section 1).(d) Prove that the set of elements of G whose two diagonal entries are equal (i.e., a = c) is also a subgroup of

GL2 (R).

13

Proof (a). Computing the product of two matrices in G, we find(a1 b10 c1

)(a2 b20 c2

)=

(a1a2 a1b2 + b1c2

0 c1c2

).

Since a1, a2 6= 0 and c1, c2 6= 0, this matrix is also in G. �

(b). Notice that (a b0 c

)(1/a −b/ac0 1/c

)=

(1 −b/c+ b/c0 1

)=

(1 00 1

)(

1/a −b/ac0 1/c

)(a b0 c

)=

(1 b/a− b/a0 1

)=

(1 00 1

).

Since

(1/a −b/ac0 1/c

)is in G, then G is closed under inverses. �

(c). First notice that G 6= ∅ because

(1 00 1

)∈ G. To show that G is a subgroup of GL2 (R), we must first show

that elements of G have nonzero determinant. So let M =

(a b0 c

).

detM = ac 6= 0.

Thus G ⊆ GL2 (R). Note matrix multiplication is a binary operation on G, the identity is in G, and G is closedunder inverse. So G is a subgroup of GL2 (R). �

Proof (d). We first show that matrix multiplication is a binary operation on G′ = {G | a = c}. So(a1 b10 a1

)(a2 b20 a2

)=

(a1a2 a1b2 + b1a2

0 a1a2

).

Since a1a2 = a1a2, matrix multipliaction is indeed a binary operation G′. Notice that the identity element is in G′.The inverse is given by (

a b0 a

)−1=

(1/a −b/a20 1/a

)due to the calculation of inverse in G. Thus G′ is closed under inverse and since G′ ⊆ G ⊆ GL2 (R), we must haveG′ as a subgroup of GL2 (R). �

1.5. Quaternion Groups.

1.5.1. Compute the order of each of the elements in Q8.Let Q8 = {1,−1, i,−i, j,−j, k,−k} as given in the textbook. Then |1| = 1, |−1| = 2, |i| = 4, |−i| = 4, |j| = 4,

|−j| = 4, |k| = 4, |−k| = 4.

1.5.2. Write out the group tables for S3, D8, and Q8.

1.5.3. Find a set of generators and relations for Q8.

Q8 =⟨−1, i, j, k | (−1)2 = 1, i2 = j2 = k2 = ijk = −1

⟩


1.6. Homomorphisms and Isomorphisms.

1.6.2. If φ : G → H is an isomorphism, prove that |φ(x)| = |x| for all x ∈ G. Deduce that any two isomorphicgroups have the same number of elements of order n for each n ∈ Z+. Is the result true if φ is only assumed to bea homomorphism?

Proof. Let |φ(x)| = n so that φ(x)n = 1 and φ(x)l 6= 1 for 0 < l < n. But 1 = φ(x)n = φ(xn) and thus xn = 1. Nowsuppose that xl = 1. Then 1 = φ(xl) which is a contradiction. Thus |x| = n. Hence any two isomorphic groupshave the same number of elements of order n for each n ∈ Z+ because of the bijectivity of φ.

The result that |φ(x)| = |x| is not true for φ any homomorphism. For example the trivial map φ(x) = 1 for allx ∈ G for G non-trivial. �

1.6.7. Prove that D8 and Q8 are not isomorphic.

Proof. Note that |s| =∣∣r2∣∣ = 2. But Exercise 1.5.1 shows there are not two elements with order two in Q8. Thus

Exercise 1.5.2 shows that D8 and Q8 are not isomorphic. �

1.6.13. Let G and H be groups and let φ : G → H be a homomorphism. Prove that the image of φ, φ(G), is asubgroup of H (cf. Exercise 26 of Section 1). Prove that if φ is injective then G ∼= φ(G).

Proof. Let h1, h2 ∈ H so that there are g1, g2 ∈ G such that φ(g1) = h1 and φ(g2) = h2. Thus φ(g1g2) =φ(g1)φ(g2) = h1h2 so h1h2 ∈ H. Hence ∗H×H : H ×H → H. Notice that φ(1) = 1 so H is nonempty. Let h ∈ Hand let φ(g) = h. Then φ(g−1) = φ(g)−1 = h−1, so H is closed under inverse. Therefore H ⊆ G is a subgroup. �

1.6.14. LetG andH be groups and let φ : G→ H be a homomorphism. Define the kernel of φ to be {g ∈ G | φ(g) = 1H}(so the kernel is the set of elements in G which map to the identity of H, i.e., is the fiber over the identity of H).Prove that the kernel of φ is a subgroup (cf. Exercise 26 of Section 1) of G. Prove that φ is injective if and only ifthe kernel of φ is the identity subgroup of G.

Proof. First note that 1 ∈ kerφ so that kerφ 6= ∅. Now let g1, g2 ∈ kerφ so that φ(g1) = φ(g2) = 1. Thenφ(g1g2) = φ(g1)φ(g2) = 1 so that g1g2 ∈ kerφ. Suppose that g ∈ kerφ so that φ(g) = 1. Then φ(g−1) = φ(g)−1 = 1so g−1 ∈ kerφ and kerφ is closed under inverse. Thus kerφ ⊆ H is a subgroup.

Suppose that φ is injective. Thus there is only one element that maps to 1H . Since φ(1G) = 1H , kerφ = {1G}.Now suppose that kerφ = {1G} and let φ(g1) = φ(g2). Then φ(g1)φ(g2)−1 = 1. So φ(g1g

−12 ) = 1 and since

kerφ = {1G}, we must have g1g−12 = 1G and thus g1 = g2. Therefore φ is injective. �

1.6.18. Let G be any group. Prove that the map from G to itself defined by g 7→ g2 is a homomorphism if andonly if G is abelian.

Proof. First suppose that g 7→ g2 is a homomorphism. Let g1, g2 ∈ G so that g1g2g1g2 = (g1g2)2 = g21g22 . Thus

g2g1 = g1g2 showing that G is abelian.Now suppose that G is abelian. Let g1, g2 ∈ G so that (g1g2)2 = g1g2g1g2 = g21g

22 and thus g 7→ g2 is a

homomorphism. �

1.7. Group Actions.

1.7.4. Let G be a group acting on a set A and fix some a ∈ A. Show that the following sets are subgroups of G(cf. Exercise 26 of Section 1):

(a) the kernel of the action.(b) {g ∈ G | ga = a} –this subgroup is called the stabilizer of a in G.

Proof (a). Notice that 1 · a = a for all a ∈ A so the kernel of the action is nonempty. Let g1, g2 be two elementsin the kernel of the action. Then (g1g2)(a) = g1(g2a) = g1a = a so the group operation on G is a binary operationon the kernel of the action. Now let g be in the kernel of the action. Then g−1a = g−1(ga) = (g−1g)a = a so thekernel of the action is closed under inverse. Therefore the kernel of the action is a subgroup of G. �

Proof (b). Notice that 1 · a = a for all a ∈ A so the stabilizer of a in G is nonempty. Let g1, g2 ∈ {g ∈ G | ga = a}so that (g1g2)a = g1(g2a) = g1a = a and hence the group operation on G is a binary operation on {g ∈ G | ga = a}.Now let g ∈ {g ∈ G | ga = a} so that g−1a = g−1(ga) = (g−1g)a = a and hence {g ∈ G | ga = a} is closed underinverse. Therefore {g ∈ G | ga = a} is a subgroup of G. �

1.7.5. Prove that the kernel of an action of the group G on the set A is the same as the kernel of the correspondingpermutation representation G→ SA (cf. Exercise 14 in Section 6).

15

Proof. First suppose that ga = a for all a ∈ A. Then σg(a) = a for all a ∈ A hence σg = 1SA . Now suppose thatσg(a) = a for all a ∈ A. Then ga = σg(a) = a. Therefore these kernels are equivalent. �

1.7.10. With reference to the preceding two exercises determine:

(a) for which values of k the action of Sn on k-element subsets is faithful.

Let Sn act on subsets of order k for some 1 ≤ k ≤ n. Note if n = 1, then the group action is trivially faithful.Now let σ ∈ Sn be a non-identity element for n > 1. Since σ is not the identity, let σ(ai) = aj for ai 6= aj . If k < n,then we can choose {a1, . . . , ak} such that ai ∈ {a1, . . . , ak} and aj /∈ {a1, . . . , ak}. Thus

σ · {a1, . . . , ak} 6= {a1, . . . , ak} .Hence the homomorphism representing the group action is injective for k < n. If k = n, then σ is just a bijectionof n-element sets so the homomorphism representing the group action is trivial. Therefore the values of k givingrise to a faithful group action are all values of k < n.

1.7.13. Find the kernel of the left regular action.If g is in the kernel of the left regular action, then ga = a for all a ∈ G. Thus g = 1 so the kernel of the left

regular action is {1}.1.7.14. Let G be a group and let A = G. Show that if G is non-abelian then the maps defined by g · a = ag for allg, a ∈ G do not satisfy the axioms of a (left) group action of G on itself.

Proof. Since G is non-abelian we can find g1, g2 ∈ G such that g1g2 6= g2g1. We assume the axioms of a groupaction hold. But g1g2 = a−1ag1g2 = a−1((g1g2) · a) = a−1(g1 · (g2 · a)) = a−1(g1 · (ag2)) = a−1(ag2g1) = g2g1, acontradiction. Therefore this is not a group action. �

2. Subgroups

2.1. Definition and Examples.

Proposition 2.1 (The Subgroup Criterion). A subset H of a group G is a subgroup if and only if

(1) H 6= ∅(2) for all x, y ∈ H, xy−1 ∈ H

Furthermore, if H is finite, then it suffices to check that H is nonempty and closed under multiplication.

2.2. Centralizers and Normalizers, Stabilizers and Kernels.

2.2.2. Prove that CG(Z(G)) = G and deduce that NG(Z(G)) = G.

Proof. Notice CG(Z(G)) ={g ∈ G | gag−1 = a, ∀a ∈ Z(G)

}by definition. But a ∈ Z(G) if and only if ag = ga for

all g ∈ G, if and only if a = gag−1 for all g ∈ G. Thus CG(Z(G)) = G.Since CG(Z(G)) ≤ NG(Z(G)) and NG(Z(G)) ⊆ G, we have NG(Z(G)) = G as well. �

2.2.4. For each of S3, D8, and Q8 compute the centralizers of each element and find the center of each group. DoesLagrange’s Theorem (Exercise 19 in Section 1.7) simplify your work?

S3. CS3(1) = S3, CS3

((1 2)) = {1, (1 2)}, CS3((1 3)) = {1, (1 3)}, CS3

((2 3)) = {1, (2 3)}, CS3((1 2 3)) = {1, (1 2 3), (1 3 2)},

CS3((1 3 2)) = {1, (1 2 3), (1 3 2)}. �

D8. CD8(1) = D8, CD8(r) = 〈r〉, CD8(r2) = D8, CD8(r3) = 〈r〉, CD8(s) ={

1, r2, s, sr2}

, CD8(sr) ={

1, r2, sr, sr3}

,

CD8(sr2) =

{1, r2, s, sr2

}, CD8

(sr3) ={

1, r2, sr, sr3}

. �

Q8. CQ8(1) = Q8, CQ8

(−1) = {1,−1}, CQ8(i) = {1, i}, CQ8

(−i) = {1,−i}, CQ8(j) = {1, j}, CQ8

(−j) = {1,−j},CQ8(k) = {1, k}, CQ8(−k) = {1,−k}. �

Note that Lagrange’s theorem lets us check our answers because the centralizer is a subgroup and thus mustdivide the order of the group.

2.2.5. In each of parts (a) to (c) show that for the specified group G and subgroup A of G, CG(A) = A andNG(A) = G.

(a) G = S3 and A = {1, (1 2 3), (1 3 2)}.(b) G = D8 and A =

{1, s, r2, sr2

}.

(c) G = D10 and A ={

1, r, r2, r3, r4}

.


Proof (a). Note that CG(A) = ∩a∈ACG(a) so by Exercise 2.2.4, CS3(A) = S3∩{1, (1 2 3), (1 3 2)}∩{1, (1 2 3), (1 3 2)} =A. �

Proof (b). Similarly to part (a) by Exercise 2.2.4, CD8(A) = D8 ∩

{1, r2, s, sr2

}∩D8 ∩

{1, r2, s, sr2

}= A. �

Proof (c). Note that any power of r commutes with any power of r so A ⊆ CD10(A). Since CD10

(r) = 〈r〉, we alsomust have CD10

(A) ⊆ A. �

2.2.9. For any subgroup H of G and any nonempty subset A of G define NH(A) to be the set{h ∈ H | hAh−1 = A

}.

Show that NH(A) = NG(A) ∩H and deduce that NH(A) is a subgroup of H (note that A need not be a subset ofH).

Proof. Let x ∈ NH(A) so x ∈ H ≤ G and xAx−1 = A. Thus x ∈ NG(A) ∩ H. Now let x ∈ NG(A) ∩ H so thatx ∈ HandxAx−1 = A. Thus x ∈ GH(A) and hence NH(A) = NG(A) ∩H.

Note that NG(A) and H are both subgroups of G and the intersection of two subgroups is again a subgroup.Therefore NH(A) ≤ G. �

2.2.10. Let H be a subgroup of order 2 in G. Show that NG(H) = CG(H). Deduce that if NG(H) = G thenH ≤ Z(G).

Proof. Let H = {1, h}, x ∈ NG(H) so that x ∈ G and xHx−1 = H. Since x1x−1 = 1, we must have xhx−1 = h sothat x {1, h}x−1 = {1, h}. Thus x ∈ CG(H). Now if x ∈ CG(H), then x1x−1 = 1 and xhx−1 = h so xHx−1 = H.Thus x ∈ NG(H) and hence NG(H) = CG(H).

Note if NG(H) = G, then G = CG(H). Thus any element of G commutes with any element in H so H ⊆ Z(G).Since H and Z(G) are subgroups we get H ≤ Z(G). �

2.3. Cyclic Groups and Cyclic Subgroups.

Proposition 2.2. If H = 〈x〉, then |H| = |x| (where if one side of this equality is infinite, so is the other). Morespecifically

(1) if |H| = n <∞, then xn = 1 and 1, x, x2, . . . , xn−1 are all the distinct elements of H(2) if |H| =∞, then xn 6= 1 for all n 6= 0 and xa 6= xb for all a 6= b in Z.

Proposition 2.3. Let G be an arbitrary group, x ∈ G and let m,n ∈ Z. If xn = 1 and xm = 1, then xd = 1, whered = (m,n). In particular, if xm = 1 for some m ∈ Z, then |x| divides m.

Theorem 2.4. Any two cyclic groups of the same order are isomorphic. More specifically,

(1) if n ∈ Z+ and 〈x〉 and 〈y〉 are both cyclic groups of order n, then the map

ϕ : 〈x〉 → 〈y〉xk 7→ yk

is well defined and is an isomorphism(2) if 〈x〉 is an infinite cyclic group, the map

ϕ : Z→ 〈x〉k 7→ xk

is well defined and is an isomorphism.

Proposition 2.5. Let G be a group, let x ∈ G and let z ∈ Z\ {0}.(1) If |x| =∞, then |xa| =∞.(2) If |x| = n <∞, then |xa| = n

(n,a) .

(3) In particular, if |x| = n <∞ and a is a positive integer dividing n, then |xa| = na .

Proposition 2.6. Let H = 〈x〉.(1) Assume |x| =∞. Then H = 〈xa〉 if and only if a = ±1.(2) Assume |x| = n <∞. Then H = 〈xa〉 if and only if (a, n) = 1. In particular, then number of generators of

H is ϕ(n).

Theorem 2.7. Let H = 〈x〉 be a cyclic group.

(1) Every subgroup of H is cyclic. More precisely, if K ≤ H, then either K = {1} or K =⟨xd⟩, where d is the

smallest positive integer such that xd ∈ K.

17

(2) If |H| = ∞, then for any distinct nonnegative integers a and b, 〈xa〉 6=⟨xb⟩. Furthermore, for every

integer m, 〈xm〉 =⟨x|m|

⟩, where |m| denotes the absolute value of m, so that the nontrivial subgroups of H

correspond bijectively with the integers 1, 2, 3, . . ..(3) If |H| = n < ∞, then for each positive integer a dividing n there is a unique subroups of H of order a.

This subgroup is the cyclic group⟨xd⟩, where d = n

a . Furthermore, for every integer m, 〈xm〉 =⟨x(n,m)

⟩,

so that the subgroups of H correspond bijectively with the positive divisors of n.

2.3.1. Find all subgroups of Z45 = 〈x〉, giving a generator for each. Describe the containments between thesesubgroups.

The subgroups of Z45 are given by⟨x0⟩, 〈x〉 ,

⟨x3⟩,⟨x5⟩,⟨x9⟩,⟨x15⟩. The containments between these

subgroups are given by 〈x〉 ⊇⟨x3⟩⊇⟨x5⟩⊇⟨x9⟩⊇⟨x15⟩⊇⟨x0⟩.

2.3.5. Find the number of generators for Z/49000Z.Since Z/49000Z is cyclic of order 49000, we count the number of relatively prime integers to 49000. This is given

by Euler’s ϕ function, so

ϕ(49000) = ϕ(235372) = ϕ(23)ϕ(53)ϕ(72) = 22(1) · 52(4) · 7(6) = 16800.

2.3.8. Let Z48 = 〈x〉. For which integers a does the map ϕa defined by ϕa : 1 7→ xa extend to an isomorphism fromZ/48Z onto Z48.

All integers a such that (a, 48) = 1. So a ∈ {1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47}.2.3.12. Prove that the following groups are not cyclic:

(b) Z2 × Z

Proof. Suppose Z2 × Z is cyclic and let (x, y) be a generator. Note that x, y are not identity elements in theirrespective groups. Then there exists k ∈ Z such that

(1, y) = (x, y)k = (xk, ky).

Note k 6= 1 because x 6= 1. But then y = ky shows that y = 0. This is a contradiction so this group is not cyclic. �

2.3.13. Prove that the following pairs of groups are not isomorphic:

(b) Q× Z2 and Q.

Proof. Suppose that ϕ : (Q × Z2) → Q is an isomorphism. Let (0, x) ∈ Q × Z2 be a non-identity element. Soϕ((0, x)) = y 6= 0 but

0 = ϕ((0, 1)) = ϕ((0, x)2) = 2y.

Since y 6= 0 this is a contradiction, thus no such isomorphism exists. �

2.3.18. Show that if H is any group and h is an element of H with hn = 1, then there is a unique homomorphismfrom Zn = 〈x〉 to H such that x 7→ h.

Proof. Since hn = 1, there is an m ∈ N such that |〈h〉| = m so m|n. Define a map ϕ(xk) = hk. If xk = xl sothat k ≡ l (mod n), then k − l = αn for some α ∈ Z. But then hk = hαn+l = hl so ϕ is well-defined. Now takexk, xl ∈ 〈x〉 so that

ϕ(xk)ϕ(xl) = hkhl = hk+l = ϕ(xk+l)

shows that ϕ is such a homomorphism. Now suppose that ϕ and ψ are two homomorphisms such that ϕ(x) = h =ψ(x). Take some xk ∈ 〈x〉 so that

ϕ(xk) = ϕ(x)k = hk = ψ(x)k = ψ(xk).

Therefore ϕ = ψ shows there is a unique map with these properties. �

2.4. Subgroups Generated by Subsets of a Group.

Proposition 2.8. If A is any nonempty collection of subgroups of G, then the intersection of all members of A isalso a subgroup of G.

Proposition 2.9. A = 〈A〉.2.4.8. Prove that S4 = 〈(1 2 3 4), (1 2 4 3)〉.


Proof. First note that |S4| = 4! = 24. Since 〈(1 2 3 4), (1 2 4 3)〉 ≤ S4, Lagrange’s theorem states we needto find 13 elements generated by this set so that 〈(1 2 3 4), (1 2 4 3)〉 = S4. So we have (1 2 3 4), (1 2 3 4)2 =(1 3)(2 4), (1 2 3 4)3 = (1 4 3 2), (1 2 4 3), (1 2 4 3)2 = (1 4)(2 3), (1 2 4 3)3 = (1 3 4 2), (1 2 3 4)(1 2 4 3) = (1 3 2),(1 2 3 4)(1 4)(2 3) = (2 4), (1 2 3 4)(1 3 4 2) = (1 4 3), (1 3)(2 4)(1 2 4 3) = (1 4), (1 4 3 2)(1 2 4 3) = (2 3 4), (1 4)(2 4) =(1 4 2), (1 3)(2 4)(2 4) = (1 3). �

2.4.9. Prove that SL2(F3) is the subgroup of GL2(F3) generated by

(1 10 1

)and

(1 01 1

).

Proof. First note that |SL2(F3)| = 24 so we use a method similar to exercise 2.4.8 to show this subgroup is

generated by the two matrices. So we have

(1 10 1

),

(1 10 1

)2

=

(1 20 1

),

(1 10 1

)3

=

(1 00 1

),

(1 01 1

),

(1 01 1

)2

=(1 02 1

),

(1 10 1

)(1 01 1

)=

(2 11 1

),

(1 20 1

)(1 01 1

)=

(0 21 1

),

(1 10 1

)(1 02 1

)=

(1 12 1

),

(1 01 1

)(1 10 1

)=(

1 11 2

),

(1 01 1

)(1 20 1

)=

(1 21 0

),

(1 02 1

)(1 10 1

)=

(1 12 0

),

(1 02 1

)(2 11 1

)=

(2 12 0

),

(1 11 2

)(1 21 0

)=(

2 20 2

). �

2.4.10. Prove that the subgroup of SL2(F3) generated by

(0 −11 0

)and

(1 11 −1

)is isomorphic to the quaternion

group of order 8. [Use a presentation for Q8.]

Proof. Let Q8 be given by the presentation

Q8 =⟨i, j | i4 = 1 = j4, ij = −ji

⟩.

Notice that

(0 −11 0

)4

= I =

(1 11 −1

)4

, and(0 −11 0

)(1 11 −1

)=

(−1 11 1

)= −

(1 −1−1 −1

)= −

(1 11 −1

)(0 −11 0

).

Thus the relations of the subgroup of SL2(F3) generated by the matrices are contained in the presentation of Q8.So the order of the matrix subgroup is less than or equal to 8. Since we have found 5 matrices in this subgroup and|SL2(F3)| = 24, Lagrange’s theorem shows that the order of this subgroup is indeed 8. Therefore these groups areisomorphic. �

2.4.11. Prove that SL2(F3) and S4 are two nonisomorphic groups of order 24.

Proof. First notice that there is more than one element of order two in S4, namely (1 2) and (1 3). Now let(a bc d

)∈ SL2(F3) be an element of order 2. So

(a bc d

)2

=

(a2 + bc ab+ bdac+ cd bc+ d2

)=

(1 00 1

)and ad − bc = 1. Since bc = ad − 1 we see ad − 1 + d2 = 1. So d(a + d) = 2 shows that cd(a + d) = 2c = 0 hencec = 0. So ad = a2 = d2 = 1 together with b(a + d) = 0 shows that a2b + abd = 0 hence b = 0. Therefore the only

element of order 2 is

(−1 00 −1

)and hence these groups are not isomorphic. �

2.4.12. Prove that the subgroup of upper triangular matrices in GL3(F2) is isomorphic to the dihedral group oforder 8. [First find the order of this subgroup.]

Proof. Since the diagonal must be have nonzero entries, the diagonal is a string of 1’s. Thus the order of thissubgroup is 23 = 8. We will show that the subgroup of upper triangular matrices in GL3(F2) is equal to⟨

A =

1 1 10 1 00 0 1

, B =

1 1 00 1 10 0 1

⟩

19

where AB = B−1A and B4 = I = A2. Calculating we find1 1 10 1 00 0 1

2

=

1 0 00 1 00 0 1

1 1 0

0 1 10 0 1

4

=

1 1 00 1 10 0 1

21 0 10 1 00 0 1

=

1 1 00 1 10 0 1

1 1 10 1 10 0 1

=

1 0 00 1 00 0 1

.

Thus we found five matrices generated by A and B to show that it generates the whole group by Lagrange’s theorem.Therefore these two groups are isomorphic. �

2.5. Definitions and Examples. Let G and H be groups.

2.5.1. Let ϕ : G → H be a homomorphism and let E be a subgroup of H. Prove that ϕ−1(E) ≤ G (ie., thepreimage or pullback of a subgroup under a homormorphism is a subgroup). If E E H prove that ϕ−1(E) E G.Deduce that kerϕ E G.

Proof. Let g1, g2 ∈ ϕ−1(E) so ϕ(g1) = h1 and ϕ(g2) = h2 where h1, h2 ∈ E. But ϕ(1G) = ϕ(1H) and ϕ(g1g−12 ) =

h1h−12 ∈ E so ϕ−1(E) ≤ G.

If E E H then hEh−1 ⊆ E for all h ∈ H. So take ϕ(g ϕ−1(E) g−1

)⊆ ϕ(g)E ϕ(g)−1 ⊆ E. Thus g ϕ−1(E) g−1 ⊆

ϕ−1(E) shows ϕ−1(E) E G. Since 1H is normal in H then kerϕ E G. �

2.5.5. Use the preceding exercise to prove that the order of the element gN in G/N is n, where n is the smallestpositive integer such that gn ∈ N (and gN has infinite order if no such positive integer exists). Give an example toshow that the order of gN in G/N may be strictly smaller than the order of g in G.

Proof. First note if 1 < m < n then (gN)m = gmN * N for g /∈ N so the order of gN ≥ n. But (gN)n = gnN = Nso the order of gN is n.

An example with the desired property would be D8/⟨r2⟩. Note that

∣∣r2∣∣ = 2 in D8 but∣∣r2 ⟨r2⟩∣∣ = 1 in

D8/⟨r2⟩. �

2.5.6. Define ϕ : R× → {±1} by letting ϕ(x) be x divided by the absolute value of x. Describe the fibers of ϕ andprove that ϕ is a homomorphism.

Proof. The fibers of ϕ are given by ϕ−1(1) = {α ∈ R× | α > 0} and ϕ−1(−1) = {α ∈ R× | α < 0}. To show ϕ is ahomomorphism we calculate

ϕ(αβ) =

{1 αβ > 0,−1 αβ < 0,

ϕ(α)ϕ(β) =

{1 αβ > 0,−1 αβ < 0.

�

2.5.9. Define ϕ : C× → R× by ϕ(a + bi) = a2 + b2. Prove that ϕ is a homomorphism and find the image of ϕ.Describe the kernel and the fibers of ϕ geometrically (as subsets of the plane).

Proof. We calculate,

ϕ((a+ bi)(c+ di)) = ϕ(ac+ adi+ bci− bc)= (ac− bd)2 + (ad+ bd)2

= a2c2 − 2abcd+ b2d2 + a2d2 + 2abcd+ b2d2

ϕ(a+ bi)ϕ(c+ di) = (a2 + b2)(c2 + d2)

= a2c2 + a2d2 + b2c2 + b2d2


so ϕ is a homomorphism. �

2.5.11. Let F be a field and let G =

{(a b0 c

)| a, b, c ∈ F, ac 6= 0

}≤ GL2(F ).

(a) Prove that the map ϕ :

(a b0 c

)7→ a is a surjective homomorphism from G onto F× (recall that F× is the

multiplicative group of nonzero elements in F ). Describe the fibers and kernel of ϕ.

(b) Prove that the map ψ :

(a b0 c

)7→ (a, c) is a surjective homomorphism from G onto F× × F×. Describe

the fibers and kernel of ψ.

(c) Let H =

{(1 b0 1

)| b ∈ F

}. Prove that H is isomorphic to the additive group F .

Proof (a). ϕ is clearly surjective so we show it is a homomorphism by(a1 b10 c1

)(a2 b20 c2

)=

(a1a2 a1b2 + b1c2

0 c1c2

).

The fibers of ϕ are given by ϕ−1(a) =

{(a b0 c

)| b ∈ F, c 6= 0

}. The kernel of ϕ is elements of G with a = 1. �

Proof (b). ψ is clearly surjective and the calculation from proof (a) shows it is a homomorphism. The fibers of ψ

are given by ψ−1((a, c)) =

{(a b0 c

)| b ∈ F

}. The kernel of ψ is elements of G with a = c = 1. �

Proof (c). The calculation (1 b10 1

)(1 b20 1

)=

(1 b1 + b20 1

)makes it clear that H is isomorphic to F+. �

2.5.12. Let G be the additive group of real numbers, let H be the multiplicative group of complex numbers ofabsolute value 1 (the unit circle S1 in the complex plane) and let ϕ : G→ H be the homomorphism ϕ : r 7→ e2πir.Draw the points on a real line which lie in the kernel of ϕ. Describe similarly the elements in the fibers of ϕ abovethe points −1, i, and e4πi/3 of H.

The points of the real line in the kernel of ϕ are the integers. The fibers are given by ϕ−1(−1) = (1/2)(R+/Z),ϕ−1(i) = (1/4)(R+/Z), ϕ(e4πi/3) = (2/3)(R+/Z).

2.6. The Lattice of Subgroups of a Group.

3. Quotient Groups and Homomorphisms

3.1. Definitions and Examples.

Proposition 3.1. Let G and H be groups and let ϕ : G→ H be a homomorphism.

(1) ϕ(1G) = 1H(2) ϕ(g−1) = ϕ(g)−1 for all g ∈ G.(3) ϕ(gn) = ϕ(g)n for all n ∈ Z.(4) kerϕ is a subgroup of G.(5) im (ϕ) is a subgroup of H.

Proposition 3.2. Let ϕ : G → H be a homomorphism of groups with kernel K. Let X ∈ G/K be the fiber abovea, i.e., X = ϕ−1(a). Then

(1) For any u ∈ X, X = {uk | k ∈ K}(2) For any u ∈ X, X = {ku | k ∈ K}.

Theorem 3.3. Let G be a group and let K be the kernel of some homomorphism from G to another group. Thenthe set whose eelements are the left cosets of K in G with operation defined by

uK ◦ vK = (uv)K

forms a group, G/K. In particular, this operation is well defined in the sense that if u1 is any element in uK andv1 is any element in vK, then u1v1 ∈ uvK, e.e., u1v1K = uvK so that the multiplication does not depend on thechoice of representatives for the cosets. The same statement is true with “right coset” in place of “left coset.”

21

Proposition 3.4. Let N be any subgroup of the group G. The set of left cosets of N in G form a partition of G.Furthermore, for all u, v ∈ G, uN = vN if and only if v−1u ∈ N and in particular, uN = vN if and only if u andv are representatives of the same coset.

Proposition 3.5. Let G be a group and let N be a subgroup of G.

(1) The operation on the set of left cosets of N in G described by

uN · vN = (uv)N

is well defined if and only if gng−1 ∈ N for all g ∈ G and all n ∈ N .(2) If the above operation is well defined, then it makes the set of left cosets of N in G into a group. In particular

the identity of this group is the coset 1N and the inverse of gN is the coset g−1N i.e., (gN)−1 = g−1N .

Theorem 3.6. Let N be a subgroup of the group G. The following are equivalent:

(1) N E G(2) NG(N) = G(3) gN = Ng for all g ∈ G(4) the operation on left cosets of N in G described in proposition 3.5 makes the set of left cosets into a group(5) gNg−1 ⊆ N for all g ∈ G.

Proposition 3.7. A subgroup N of the group G is normal if and only if it is the kernel of some homomorphism.

3.1.36. Prove that if G/Z(G) is cyclic then G is abelian.

Proof. If G/Z(G) is cyclic with generator xZ(G), then if gZ(G) ∈ G/Z(G) we have gZ(G) = (xZ(G))n = xnZ(G)for some n ∈ N. Thus there is some z ∈ Z(G) such that g = xnz. Now take g1, g2 ∈ G such that g1 = xmz1 andg2 = xnz2. Then

g1g2 = xmz1xnz2

= xmxnz2z1

= xnxmz2z1

= xnz2xmz1

= g2g1.

�

3.2. More on Cosets and Lagrange’s Theorem.

Theorem 3.8 (Lagrange’s Theorem). If G is a finite gorup and H is a subgroup of G, then the order of H divides

the order of G (i.e., |H| | |G|) and the number of left cosets of H in G equals |G||H| .

Corollary 3.9. If G is a finite group and x ∈ G, then the roder of x divides the order of G. In particular x|G| = 1for all x in G.

Corollary 3.10. If G is a group of prime order p, then G is cyclic, hence G ∼= Zp.

Theorem 3.11 (Cauchy’s Theorem). If G is a finite group and p is a prime dividing |G|, then G has an elementof order p.

Theorem 3.12 (Sylow). If G is a finite group of order pαm, where p is a prime and p does not divide m, then Ghas a subgroup of order pα.

Proposition 3.13. If H and K are finite subgroups of a group then

|HK| = |H| |K||H ∩K| .

Proposition 3.14. If H and K are subgroups of a group, HK is a subgroup if and only if HK = KH.

Corollary 3.15. If H and K are subgroups of G and H and H ≤ NG(K), then HK is a subgroup of G. Inparticular, if K E G then HK ≤ G for any H ≤ G.

3.2.8. Prove that if H and K are finite subgroups of G whose orders are relatively prime then H ∩K = 1.

Proof. Let |H| = m and |K| = n with (m,n) = 1. Now take x ∈ H ∩K so |x| |m and |x| |n so x = 1. �


3.2.9. This exercise outlines a proof of Cauchy’s Theorem due to James McKay. Let G be a finite group and let pbe a prime dividing |G|. Let S denote the set of p-tuples of elements of G the product of whose coordinates is 1:

S = {(x1, x2, . . . , xp) | xi ∈ G and x1x2 · · ·xp = 1} .

Define the relation ∼ on S by letting α ∼ β if β is a cyclic permutation of α.

(a) Show that S has |G|p−1 elements, hence has order divisible by p.(b) Show that a cyclic permutation of an element of S is again an element of S.(c) Prove that ∼ is an equivalence relation on S.(d) Prove that an equivalence class contains a single element if and only if it is of the form (x, x, . . . , x) with

xp = 1.(e) Prove that every equivalence class has order 1 or p (this uses the fact that p is a prime). Deduce that

|G|p−1 = k + pd, where k is the number of classes of size 1 and d is the number of classes of size p.(f) Since {(1, 1, . . . , 1)} is an equivalence class of size 1, conclude from (e) that there must be a nonidentity

element x in G with xp = 1, i.e., G contains an element of order p. [Show p|k and so k > 1.]

Proof (a). The first p − 1 coordinates have |G| choices and the pth coordinate must be the inverse of the product

of the first p− 1 coordinates. Thus |S| = |G|p−1. �

Proof (b). Let (x1, x2, . . . , xp) ∈ S and notice xpx1x2 · · ·xp−1 = 1 just by multiplication on the right by (xp)−1,

then the left by xp. Thus (xp, x1, . . . , xp−1) ∈ S and any cyclic permutation can be realized by repeating thisprocess. �

Proof (c). Reflexive is obvious. If α is a cyclic permutation of β, then β is a cyclic permutation of α so ∼ issymmetric. If β is a cyclic permutation of α and γ is a cyclic permutation of β, then γ is a cyclic permutation of αso ∼ is transitive. Thus ∼ is an equivalence relation. �

Proof (d). If some x is changed to a y 6= x, then a cyclic permutation would yield a different element of theequivalence class. If all the x are the same, then any cyclic permutation is indeed the same element of S. �

Proof (e). It is clear that every equivalence class has order less than or equal to p. Suppose the equivalence classis not of the form from part (d) and let σ ∈ Sp be given by σ(i) ≡ i + j mod p for 0 ≤ j < p and suppose that(x1, . . . , xp) ∼ (xσ(1), . . . , xσ(p)). Then xi = xσ(i) = xσ2(i) = · · · = xσp−1(i) for all i. Since not every coordinate isallowed to be equal, j = 0 because |σ| = p (for p prime, σ 6= 1).

Thus the number of equivalence classes is the sum of size 1 and size p classes. Since equivalence classes aredisjoint, |G|p−1 = k + pd. �

Proof (f). Since p divides |G|, then p divides |G|p−1 and specifically p divides k. Thus k > 1 and hence there is anelement x in G with order p. �

3.2.10. Suppose H and K are subgroups of finite index in the (possibly infinity) group G with |G : H| = m and|G : K| = n. Prove that lcm (m,n) ≤ |G : H ∩K| ≤ mn. Deduce that if m and n are relatively prime then|G : H ∩K| = |G : H| · |G : K|.

Proof by Dr. Schulze. By Problem 11, you have [G : H ∩K] = [G : K][K : H ∩K]. So n divides [G : H ∩K], andalso m divides by the same argument. Therefore lcm(m,n)⇐ [G : H ∩K].

Now note that the first isomorphism theorem works also for non-normal subgroups, but you get a statement onmaps of sets (not groups). Apply this to the map G→ G×G→ G/H ×G/K sending g 7→ (g, g) 7→ (gH, gK).

Its kernel is H ∩K, so the (set version of the) first isomorphism theorem gives an injective map G/(H ∩K)→G/H ×G/K. Thus, [G : H ∩K] = |G/(H ∩K)| ⇐ |G/H ×G/K| = |G/H||G/K| = n ·m. �

3.2.22. Use Lagrange’s Theorem in the multiplicative group (Z/nZ)× to prove Euler’s Theorem: aϕ(n) ≡ 1 mod nfor every integer a relatively prime to n, where ϕ denotes Euler’s ϕ-function.

Proof. Since a is relatively prime to n, a ∈ (Z\nZ)×. Corollary 9 states that aϕ(n) = a|(Z\nZ)×| = 1|(Z\nZ)×| ≡ 1mod n. �

23

3.3. The Isomorphism Theorems.

Theorem 3.16 (The First Isomorphism Theorem). If ϕ : G → H is a homomorphism of groups, then kerϕ E Gand G/ kerϕ ∼= ϕ(G).

Corollary 3.17. Let ϕ : G→ H be a homomorphism of groups.

(1) ϕ is injective if and only if kerϕ = 1.(2) |G : kerϕ| = |ϕ(G)|.

Theorem 3.18 (The Second Isomorphism Theorem). Let G be a group, let A and B be subgroups of G and assumeA ≤ NG(B). Then AB is a subgroup of G, B E AB, A ∩B E A and AB/B ∼= A/A ∩B.

Theorem 3.19 (The Third Isomorphism Theorem). Let G be a group and let H and K be normal subgroups of Gwith H ≤ K. Then K/H E G/H and

(G/H)/(K/H) ∼= G/K.

If we denote the quotient by H with a bar, this can be written

G/K ∼= G/K.

Theorem 3.20 (The Fourth Isomorphism Theorem). Let G be a group and let N be a normal subgroup of G.Then there is a bijection from the set of subgroups A of G which contain N onto the set of subgroups A = A/Nof G/N . In particular, every subgroup of G is of the form A/N for some subgroup A of G containing N (namely,its preimage in G under the natural projection homomorphism from G to G/N). This bijection has the followingproperties: for all A,B ≤ G with N ≤ A and N ≤ B,

(1) A ≤ B if and only if A ≤ B,(2) if A ≤ B, then |B : A| =

∣∣B : A∣∣,

(3) 〈A,B〉 =⟨A,B

⟩,

(4) A ∩B = A ∩B, and(5) A E G if and only if A E G.

3.3.3. Prove that if H is a normal subgroup of G of prime index p then for all K ≤ G either

(i) K ≤ H or(ii) G = HK and |K : K ∩H| = p.

Proof. Suppose that K � K so since K ≤ NG(H), HK = KH ≤ G. Since |K| 6= 1, |HK| > H so HK = G because|G : H| = p and Lagrange’s theorem. The fact that |K : H ∩H| = p is immediate by the diamond isomorphismtheorem. �

3.3.4. Let C be a normal subgroup of the group A and let D be a normal subgroup of the group B. Prove that(C ×D) E (A×B) and (A×B)/(C ×D) ∼= (A/C)× (B/D).

Proof. Let π1 : A → A/C and π2 : B → B/D which exist by the normality of C and D. Now let Φ: (A × B) →(A/C × B/D) be defined by π1 and π2. Thus ker Φ = C ×D and thus (C ×D) E (A× B). Since Φ is surjective,the first isomorphism theorem shows that (A×B)/(C ×D) ∼= (A/C)× (B/D). �

3.3.7. Let M and N be normal subgroups of G such that G = MN . Prove that G/(M ∩N) ∼= (G/M) × (G/N).[Draw the lattice.]

Proof. Define a homomorphism ϕ : G → (G/M) × (G/N) by ϕ(g) = (g, g). Take some (m,n) ∈ (G/M) × (G/N).Then nm′n−1 = m so take nm′ = g = mn so ϕ(g) = (g, g) = (m,n) so ϕ is surjective. The kernel of ϕ is clearlyM ∩N because g ∈M and g ∈ N if and only if (g, g) = 1G/M×G/N . Therefore the first isomorphism theorem statesthat G/(M ∩N) ∼= (G/M)× (G/N). �

3.3.9. Let p be a prime and let G be a group of order pam, where p does not divide m. Assume P is a subgroup ofG of order pa and N is a normal subgroup of G of order pbn, where p does not divide n. Prove that |P ∩N | = pb

and |PN/N | = pa−b. (The subgroup P of G is called a Sylow p-subgroup of G. This exercise shows that theintersection of any Sylow p-subgroup of G with a normal subgroup N is a Sylow p-subgroup of N .)

Proof. First note that |PN | = (|P | |N |)/ |P ∩N | = (pa+bn)/ |P ∩N |. But this divides pam so |P ∩N | = pbα.But then pbα divides pbn so α divides n shows α does not divide p. But now pbα divides pa so α = 1 and hence|P ∩N | = pb. The diamond isomorphism theorem shows that |PN/N | = pa−b. �


3.4. Composition Series and the Holder Program.

Proposition 3.21. If G is a finite abelian group and P is a prime dividing |G|, then G contains an element oforder p.

Theorem 3.22 (Jordan-Holder). Let G be a finite group with G 6= 1. Then

(1) G has a composition series and(2) The composition factors in a composition sereis are unique, namely, if 1 = N0 ≤ N1 ≤ · · · ≤ Nr = G and

1 = M0 ≤M1 ≤ · · · ≤Ms = G are two composition series for G, then r = s and there is some permutation,π, of {1, 2, . . . , r} such taht

Mπ(i)/Mπ(i)−1 ∼= Ni/Ni−1, 1 ≤ i ≤ r.3.5. Transpositions and the Alternating Group.

Proposition 3.23. The map ε : Sn → {±1} is a homomorphism (where {±1} is a multiplicative version of thecyclic group of order 2).

Proposition 3.24. Transpositions are all odd permutations and ε is a surjective homomorphism.

Proposition 3.25. The permutation σ is odd if and only if the number of cycles of even length in its cycledecomposition is odd.

3.5.3. Prove that Sn = 〈{(i i+ 1) | 1 ≤ i ≤ n− 1}〉.

Proof. Let σ ∈ Sn and decompose σ into a product of transpositions. But if (i j) is any transposition, then

(i j) = (j − 1 j) · · · (i i+ 1) · · · (j − 1 j).

Therefore this set generates Sn. �

3.5.5. Show that if p is prime, Sp = 〈σ, τ〉 where σ is any transposition and τ is any p-cycle.

Proof. Without loss of generality, suppose σ = (1 2) and τ = (1 · · · p). If 1 ≤ i ≤ n− 1, then

(i i+ 1) = τ i−1στ−i+1.

Exercise 3.5.3 shows this generates Sp. �

3.5.6. Show that 〈(1 3), (1 2 3 4)〉 is a proper subgroup of S4. What is the isomorphism type of this subgroup?

Proof. Notice that (1 3)(1 2 3 4)2(1 3) = (1 2 3 4)2 shows that (1 2 3 4)2 ∈ Z(〈(1 3), (1 2 3 4)〉). But

(1 2 3 4)2(1 4) = (1 2 4 3) 6= (1 3 4 2) = (1 4)(1 2 3 4)2.

Therefore 〈(1 3), (1 2 3 4)〉 < S4. �

The isomorphism type of this subgroup is D8 because we can construct 8 elements and we have

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

3.5.7. Prove that the group of rigid motions of a tetrahedron is isomorphic to A4.

Proof. First notice that both groups are order 12. Now we represent the group of rigid motions of a tetrahedron asa subgroup of Sn. The elements in this subgroup are

1, (1 2 3), (1 3 2), (1 3 4), (1 4 3), (1 2 4), (1 4 2), (2 3 4), (2 4 3), (1 3)(2 4), (1 2)(3 4), (1 4)(2 3).

Since all 12 of these elements are even, this is A4. �

3.5.10. Find a composition series for A4. Deduce that A4 is solvable.

Solution. Let N1 = {1, (1 3)(2 4)} and N2 = {1, (1 3)(2 4), (1 2)(3 4), (1 4)(2 3)}. Simple calculations show that1 = N0 E N1 E N2 E N3 = A4, while |Ni+1/Ni| = p shows that each composition factor is simple. Since the orderof each composition factor is prime, they are also abelian and hence A4 is solvable. �

3.5.17. If x and y are 3-cycles in Sn, prove that 〈x, y〉 is isomorphic to Z3, A4, A5, or Z3 × Z3.

25

Proof. If x = y, then 〈x, y〉 = 〈x〉 ∼= Z3 by order argument. Now let x = (i j k) and y = (j k l) and notice theseare both even permutations hence generate even permutations. Since 1, (i j k), (i k j), (j k l), (j l k), (i k l) and(i j)(k l) are all generated by x and y, then 〈x, y〉 ∼= A4 by Lagrange’s theorem. Now let x = (i j k), y = (k l m)and we will show that 〈x, y〉 E A5, so they are equal because A5 is simple. If we can show conjugating (i j k) by anytransposition remains in 〈x, y〉, then we are done because any permutation can be represented by a transposition, andthe argument for (k l m) will be similar. But (i j k)(i j) = (i j k)(j k) = (i j k)(i k) = (i j k)−1, (i j k)(l m) = (i j k),(i j k)(i l) = (j k l) = [(i j k)(k l m)](i j k), (i j k)(j l) = [(j i k)(k l m)](j i k), (i j k)(i m) = [(i j k)(k m l)](i j k),(i j k)(j m) = [(j i k)(k m l)](j i k), (i j k)(k l) = (i j k)(k l m), and (i j k)(k m) = (i j k)(k m l). If x and y aredisjoint 3-cycles, then it is clear that 〈x, y〉 ∼= Z3 × Z3. �

4. Group Actions

4.1. Group Actions and Permutation Representations.

4.2. Groups Acting on Themselves by Left Multiplication – Cayley’s Theorem.

4.2.2. List the elements of S3 as 1, (1 2), (2 3), (1 3), (1 2 3), (1 3 2) and label these with the integers 1,2,3,4,5,6respectively. Exhibit the image of each element of S3 under the left regular representation of S3 into S6.

Solution. �

4.2.4. Use the left regular representation of Q8 to produce two elements of S8 which generate a subgroup of S8

isomorphic to the quaternion group Q8.

Solution. �

4.2.8. Prove that if H has finite index n then there is a normal subgroup K of G with K ≤ H and |G : K| ≤ n!.

Proof. �

4.2.10. Prove that every non-abelian group of order 6 has a nonnormal subgroup of order 2. Use this to classifygroups of order 6. [Produce an injective homomorphism into S3.]

Proof. �

4.3. Groups Acting on Themselves by Conjugation – The Class Equation.

4.3.3. Find all conjugacy classes and their sizes in the following groups:

(c) A4.

Solution (c). �

4.3.6. Assume G is a non-abelian group of order 15. Prove that Z(G) = 1. Use the fact that 〈g〉 ≤ CG(g) for allg ∈ G to show that there is at most one possible class equation for G. [Use exercise 36, section 3.1.]

Proof. �

4.3.13. Find all finite groups which have exactly two conjugacy classes.

Solution. �

4.3.27. Let g1, g2, . . . , gr be representatives of the conjugacy classes of the finite group G and assume these elementspairwise commute. Prove that G is abelian.

Proof. �

4.3.29. Let p be a prime and let G be a group of order pα. Prove that G has a subgroup of order pβ , for every βwith 0 ≤ β ≤ α. [Use theorem 8 and induction on α.]

Proof. �


4.4. Automorphisms. Let G be a group.

4.4.1. If σ ∈ Aut(G) and ϕg is conjugation by g prove σϕgσ−1 = ϕσ(g). Deduce that Inn(G) E Aut(G).

Proof. Let x ∈ G, so

σϕgσ−1(x) = σ(gσ−1(x)g−1)

= σ(g)xσ(g)−1

= ϕσ(g)(x)

shows the desired equation. Since conjugating an element of Inn(G) with an element of Aut(G) is contained inInn(G), normality must hold. �

4.4.2. Prove that if G is an abelian group of order pq, where p and q are distinct primes, then G is cyclic.

Proof. We use Cauchy’s theorem to pick x, y ∈ G with |x| = p and |y| = q. But if 1 = (xy)k = xkyk for k ≥ 1, thenk | p and k | q. Since x 6= y−1, then k ≥ pq. Thus 〈xy〉 = G and therefore G is cyclic. �

4.4.3. Prove that under any automorphism of D8, r has at most 2 possible images and s has at most 4 possibleimages (r and s are the usual generators). Deduce that |Aut(D8)| ≤ 8.

Proof. Since there is only one subgroup of order 4, 〈r〉 is characteristic so if σ ∈ Aut(D8), then σ(〈r〉) = 〈r〉. Since|r| = 4, then σ(r) ∈

{r, r3

}. Since σ(〈r〉) = 〈r〉, we must have σ(s) ∈

{s, sr, sr2, sr3

}.

If x ∈ D8, then x = srk for some 0 ≤ k ≤ 3. So σ(x) = σ(srk) = σ(s)σ(r)k and thus the automorphism isuniquely determined by the value on s and r. So |Aut(D8)| ≤ 2 · 4 = 8. �

4.4.11. If p is a prime and P is a subgroup of Sp of order p, prove NSp(P )/CSp(P ) ∼= Aut(P ).

Proof. We know from a previous exercise that∣∣NSp(P )

∣∣ = p(p− 1) and from Corollary 15 that NSp(P )/CSp(P ) ∼=H ≤ Aut(P ). Since CSp(P ) = P and |Aut(P )| = p− 1, we have the desired isomorphism. �

4.4.18. This exercise shows that for n 6= 6 every automorhpism of Sn is inner. Fix an integer n ≥ 2 with n 6= 6.

(a) Prove that the automorphism group of a group G permutes the conjugacy classes of G, i.e., for eachσ ∈ Aut(G) and each conjugacy class K of G the set σ(K) is also a conjugacy class of G.

(b) Let K be the conjugacy class of transpositions in Sn and let K′ be the conjugacy class of any element oforder 2 in Sn that is not a transposition. Prove that |K| 6= |K′|. Deduce that any automorphism of Snsends transpositions to transpositions.

(c) Prove that for each σ ∈ Aut(Sn)

σ : (1 2) 7→ (a b2), σ : (1 3) 7→ (a b3), . . . , σ : (1 n) 7→ (a bn)

for some distinct integers a, b2, b3, . . . , bn ∈ {1, 2, . . . , n}.(d) Show that (1 2), (1 3), . . ., (1 n) generate Sn and deduce that any automorphism of Sn is uniquely determined

by its action on these elements. Use (c) to show that Sn has at most n! automorphisms and conclude thatAut(Sn) = Inn(Sn) for n 6= 6.

Proof (a). �

4.5. The Sylow Theorems. Let G be a finite group and let p be a prime.

4.5.1. Prove that if P ∈ Sylp(G) and H is a subgroup of G containing P then P ∈ Sylp(H). Give an example toshow that, in general, a Sylow p-subgroup of a subgroup of G need not be a Sylow p-subgroup of G.

Proof. Let |P | = pk. Then pk∣∣ |H| ∣∣ |G|, so P ∈ Sylp(H). �

Example. Let 1 = H � G and p∣∣ |G|. Then 1 ∈ Sylp(H) and 1 /∈ Sylp(G). �

4.5.3. Use Sylow’s Theorem to prove Cauchy’s Theorem. (Note that we only used Cauchy’s Theorem for abeliangroups – Proposition 3.21 – in the proof of Sylow’s Theorem so this line of reasoning is not circular.)

Proof. Suppose that p∣∣ |G| and let P ∈ Sylp(G) with |P | = pk. Now take x ∈ P\1 so |x|

∣∣pk and let |x| = pl. Then∣∣xl∣∣ = p. �

4.5.9. Exhibit all Sylow 3-subgroups of SL2 (F3).

27

Solution. Note |SL2 (F3)| = 24 so if P ∈ Syl3(SL2 (F3)) then |P | = 3. Since n3 ≡ 1 mod 3, n3 ∈ {1, 4}. But(1 10 1

),

(1 01 1

),

(0 12 2

),

(0 21 2

),

all generate different subgroups of order 3. �

4.5.13. Prove that a group, G, of order 56 has a normal Sylow p-subgroup for some prime p dividing its order.

Proof. Note 56 = 23 · 7. Suppose n2, n7 > 1. Then (⋃Syl2(G))\1 ≥ 14 and (

⋃Syl7(G))\1 ≥ 48. These are clearly

all distinct elements, so n2 or n7 is equal to 1. �

4.5.18. Prove that a group of order 200 has a normal Sylow 5-subgroup.

Proof. Note 200 = 23 · 52. But n5 ∈ {1, 6} and since 6 - 8 we must have n5 = 1. �

4.5.19. Prove that if |G| = 6545 then G is not simple.

Proof. Note 6545 = 5 · 7 · 11 · 17. But since n11 ≡ 1 mod 11 and n11 | 595, we must have n11 = 1. �

4.5.30. How many elements of order 7 must there be in a simple group of order 168?

Solution. Note 168 = 23 · 3 · 7. Since the group is simple, n7 > 1. Since n7 ≡ 1 mod 7 and n7 | 24, we must haven7 = 8. Therefore there are 6 · 8 = 48 elements of order 7. �

4.5.32. Let P be a Sylow p-subgroup of H and let H be a subgroup of K. If P E H and H E K, prove that P isnormal in K. Deduce that if P ∈ Sylp(G) and H = NG(P ), then NG(H) = H.

Proof. If P E H, then P is characteristic in H. Since P is characteristic in H and H E K, then P E K.Now if P ∈ Sylp(G) and H = NG(P ), then P ∈ Sylp(H) and P is characteristic in H. Thus if Hg = H, then

P g = P so g ∈ NG(P ) = H. Therefore NG(H) = H. �

4.5.40. Prove that the number of Sylow p-subgroups of G = GL2 (Fp) is p+ 1.

Proof. Note |GL2 (Fp)| = (p− 1)2p(p+ 1) and np ≡ 1 mod p. Since

P1 =

⟨(1 γ0 1

)⟩, P2 =

⟨(1 0γ 1

)⟩∈ Sylp(G),

for γ ∈ Fp\ {0, 1}, we must have np ≥ p+ 1.Now we show that the upper triangular matrices are a subgroup of NG(P1) with order (p − 1)2p. Simple

calculations show that for a, b, c ∈ Fp, we have(a b0 c

)−1=

(a−1 −a−1bc−10 c−1

).

Now if γ ∈ Fp\ {0, 1} we get (a b0 c

)−1(1 γ0 1

)(a b0 c

)=

(1 a−1γb− a−1bc−1b0 1

).

Since the upper diagonal matrices have nonzero diagonal elements and any upper right entry, the order must be(p− 1)2p. Thus (p− 1)2p | NG(P1).

Sylow’s theorem states that np = [G : NG(P1)], so np | p+ 1. Therefore np = p+ 1. �

4.6. The Simplicity of An.


5. Direct and Semidirect Products and Abelian Groups

5.1. Direct Products.

5.1.10. Let p be a prime. Let A and B be two cyclic groups of order p with generators x and y, respectively. SetE = A×B so that E is the elementary abelian groups of order p2 : Ep2 . Prove that the distinct subgroups of E oforder p are

〈x〉 , 〈xy〉 ,⟨xy2⟩, . . . ,

⟨xyp−1

⟩, 〈y〉

(note that there are p+ 1 of them).

Proof. Example 3 in this section shows that there are p+ 1 subgroups of order p. Note that these p+ 1 subgroupsare all distinct and nontrivial, so if the pth power of each generator is the identity element, then we are done. Butxp = 1, (xy)p = xpyp = 1, (xyk)p = xp(yk)p = 1, yp = 1. �

5.1.11. Let p be a prime and let n ∈ Z+. Find a formula for the number of subgroups of order p in the elementaryabelian group Epn .

Solution. Every nonidentity element has order p, so there are pn − 1 many elements of order p. By Lagrange’stheorem, each distinct subgroup intersects trivially. So there are (pn − 1)/(p − 1) = pn−1 + pn−2 + · · · + 1 manydistinct subgroups of order p. �

5.1.12. Let A and B be groups. Assume Z(A) contains a subgroup Z1 and Z(B) contains a subgroup Z2 withZ1∼= Z2. Let this isomorphism be given by the map xi 7→ yi for all xi ∈ Z1. A central product of A and B is a

quotient

(A×B)/Z where Z ={

(xi, y−1i ) | xi ∈ Z1

}and is denoted by A ∗B – it is not unique since it depends on Z1, Z2 and the isomorphism between them. (Thinkof A ∗ B as the direct product of A and B “collapsed” by identifying each element xi ∈ Z1 with its correspondingelement yi ∈ Z2.)

(a) Prove that the images of A and B in the quotient group A ∗ B are isomrophic to A and B, respectively,and that these images intersect in a central subgroup isomorphic to Z1. Find |A ∗B|.

(b) Let Z4 = 〈x〉. Let D8 = 〈r, s〉 and Q8 = 〈i, j〉 be given by their usual generators and realtions. LetZ4 ∗D8 be the central product of Z4 and D8 which identifies x2 and r2 (i.e., Z1 =

⟨x2⟩, Z2 =

⟨r2⟩

and the

isomorphism is x2 7→ r2) and let Z4 ∗Q8 be the central product of Z4 and Q8 which identifies x2 and −1.Prove that Z4 ∗D8

∼= Z4 ∗Q8.

Proof (a). �

Proof (b). �

5.1.18. In each of (a) to (e) give an example of a group with the specified properties:

(a) an infinite group in which every element has order 1 or 2(b) an infinite group in which every element has finite order but for each positive integer n there is an element

of order n(c) a group with an element of infinite order and an element of order 2(d) a group G such that every finite group is isomorphic to some subgroup of G(e) a nontrivial group G such that G ∼= G×G.

Solution (a). The direct sum Z2 × Z2 × · · ·. �

Solution (b). The direct sum Z2 × Z3 × · · ·. �

Solution (c). Z× Z2. �

Solution (d). S2 × S3 × · · ·. �

Solution (e). G = Z2 × Z2 × · · ·. �

29

5.2. The Fundamental Theorem of Finitely Generated Abelian Groups.

5.2.2. In each of parts (a) to (e) give the lists of invariant factors for all abelian groups of the specified order:

(a) order 270

Solution (a). Since 270 = 2 · 33 · 5, the list of invariant factors is given by (30, 3, 3), (90, 3). �

5.2.3. In each of parts (a) to (e) give the lists of elementary divisors for all abelian groups of the specified orderand then match each list with the corresponding list of invariant factors found in the preceding exercise:

(a) order 270

Solution (a). The elementary divisors of an abelian group of order 270 are either (2, 3, 3, 5) or (2, 9, 5) with listscorresponding to respective orderings. �

5.2.7. Let p be a prime and let A = 〈x1〉 × 〈x2〉 × · · · × 〈xn〉 be an abelian p-group, where |xi| = pαi > 1 for all i.Define the pth-power map

ϕ : A→ A by ϕ : x 7→ xp.

(a) Prove that ϕ is a homomorphism.(b) Describe the image and kernel of ϕ in terms of the given generators.(c) Prove both kerϕ and A/imϕ have rank n (i.e., have the same rank as A) and prove these groups are both

isomorphic to the elementary abelian group, Epn , of order pn.

Proof (a). This map is clearly well-defined, so we calculate

ϕ(xy) = (xy)p

= xpyp

= ϕ(x)ϕ(y).

�

Solution (b). The image of ϕ is given by

ϕ(A) = {(β1, . . . , βn) | βi = (xγi )p, γ ∈ Z} .

The kernel of ϕ is given by {(k1, . . . , kn)} where

ki ∈ {xγi | (xγi )p = 1, γ ∈ Z} .

�

Proof (c). The kernel of ϕ is of type (k1, . . . , kn) so it has rank n, while A/imϕ is of type (a1, . . . , an) so also has

rank n. If (xγi )p = 1, then |xγi |∣∣∣ p so

∣∣∣⟨xpαi−1

i

⟩∣∣∣ = p, with each element to the power of p clearly one. Thus

kerϕ ∼=n∏i=1

⟨xp

αi−1⟩

so it is isomorphic to Epn . Since ϕ : A→ A and∣∣kerϕ|〈xi〉

∣∣ = p, then∣∣imϕ|〈xi〉

∣∣ = αp−1i . Thus 〈xi〉 /imϕ|〈xi〉 ∼= Zp.Therefore A/imϕ ∼= Epn . �

5.2.5. Prove that An is the commutator subgroup of Sn for all n ≥ 5.

Proof. Since An E Sn and Sn/An ∼= Z2 is abelian, proposition 7 part (4) states S′n ≤ An. Also by proposition 7part (3), S′n is characteristic in Sn. Since An is simple, S′n E An, and Sn is non abelian, we must have An = S′n.

�

5.2.11. Prove that if G = HK where H and K are characteristic subgroups of G with H ∩K = 1 then Aut(G) ∼=Aut(H) × Aut(K). Deduce that if G is an abelian group of finite order then Aut(G) is isomorphic to the directproduct of the automorphism groups of its Sylow subgroups.


Proof. Let φ, ψ ∈ Aut(G) and notice that φ|H ∈ Aut(H) and φ|K ∈ Aut(K) since H,K are characteristic in G.Now let f : Aut(G)→ Aut(H)×Aut(K) be given by

f(φ) = (φ|H , φ|K).

If φ = ψ, then (φ|H , φ|K) = (ψ|H , ψ|K) so f is well-defined. Now we calculate

f(φ ◦ ψ) = ((φ ◦ ψ)|H , (φ ◦ ψ)|K)

= (φ|H ◦ ψ|H , φ|K ◦ ψ|K)

= (φ|H , φ|K)(ψ|H , ψ|K)

= f(φ)f(ψ),

so f is a homomorphism. Now if φH ∈ Aut(H) and φK ∈ Aut(K), then let φ ∈ Aut(G) be given by φ|H = φHand φ|K = φK . So f(φ) = (φ|H , φ|K) = (φH , φK) shows f is surjective. Now if (φ|H , φ|K) = (ψ|H , ψ|K), then forg = hk ∈ G,

φ(hk) = φ(h)φ(k)

= φ|H(h)φ|K(k)

= ψ|H(h)ψ|K(k)

= ψ(h)ψ(k)

= ψ(hk),

so φ = ψ. Therefore f shows Aut(G) ∼= Aut(H)×Aut(K).If G is finite abelian, then it is the direct product of its (characteristic) Sylow subgroups. Therefore Aut(G) is

isomorphic to the direct product of the automorphism groups of the Sylow subgroups. �

5.2.14. Let G = {(aij) ∈ GLn (F ) | aij = 0 if i > j, and a11 = a22 = · · · = ann}, where F is a field, be the groupof upper triangular matrices all of whose diagonal entries are equal. Prove that G ∼= D × U , where D is the groupof all nonzero multiples of the identity matrix and U is the group of upper triangular matrices with 1’s down thediagonal.

Proof. It is clear that D ∩ U = 1. Now let A,B ∈ G such that

A =

a a12 · · · a1n

a · · · a2n. . .

...a

, B =

b b12 · · · b1n

b · · · b2n. . .

...b

and note if AB = I, then ab = 1. If λI ∈ D for λ ∈ F , then

AλIB = λAIB = λAB = λI.

If U ∈ U , then note that there exist UA,UB ∈ U such that A = aUA and B = bUB and

AUB = (aUA)U(bUB) = ab (UA U UB) ∈ U.Thus D and U are normal in G and G ∼= D × U . �

5.3. Table of Groups of Small Order.

5.4. Recognizing Direct Products.

5.5. Semidirect Products.

5.5.7. This exercise describes thirteen isomorphism types of groups of order 56.

(a) Prove that there are three abelian groups of order 56.(b) Prove that every group of order 56 has either a normal Sylow 2-subgroup or a normal Sylow 7-subgroup.(c) Construct the following non-abelian groups of order 56 which have a normal Sylow 7-subgroup and whose

Sylow 2-subgroup S is as specified:one group when S ∼= Z2 × Z2 × Z2

two non-isomorphic groups when S ∼= Z4 × Z2

one group when S ∼= Z8

two non-isomorphic groups when S ∼= Q8

three non-isomorphic groups when S ∼= D8.

31

(d) Let G be a group of order 56 with a non-normal Sylow 7-subgroup. Prove that if S is the Sylow 2-subgroupof G then S ∼= Z2 × Z2 × Z2.

(e) Prove that there is a unique group of order 56 with a non-normal Sylow 7-subgroup.

Proof (a). Suppose that G is an abelian group of order 56 = 23 · 7. The invariant factors are given by (23 · 7), (22 ·7, 2), (2 · 7, 2, 2). Thus there are three abelian groups of order 56 given by Z56, Z28 × Z2, Z14 × Z2 × Z2. �

Proof (b). Suppose that n2, n7 > 1. Since np ≡ 1 mod p, we have n2 ≥ 3 and n7 ≥ 8. Counting |⋃Syl2(G)\1| +|⋃Syl7(G)\1| = 3 · 7 + 8 · 6 > 56 reaches a contradiction. Therefore every group of order 56 has either a normalSylow 2-subgroup or a normal Sylow 7-subgroup. �

Solution (c). Note for all of the Sylow 2-subgroup isomorphism types, if P ∈ Syl7(G), we have P ∩ S = 1 andP E G so G is a semidirect product.

If S ∼= Z2 × Z2 × Z2, then we consider the non-trivial homomorphisms ϕ : S → Aut(P ), where Aut(P ) ∼= Z6.Thus |kerϕ| = 4 and |ϕ(S)| = 2 for any such ϕ. Therefore P oϕ S is the unique isomorphism type.

If S ∼= Z4 × Z2, then kerϕ ∈{Z4, Z

22

}. Thus there are two isomorphism types.

If S ∼= Z8, then kerϕ = Z4 giving rise to one isomorphism type.If S ∼= Q8, then the trivial map gives rise to the non-abelian Z7×Q8. The non-trivial maps ϕ : S → Aut(P ) give

rise to a second isomorphism type.If S ∼= D8, then the trivial map gives rise to the non-abelian Z7 ×D8. Two possible kernels of non-trivial maps

are 〈r〉 ∼= Z4 and⟨r2, s

⟩ ∼= Z2 × Z2. Since these kernels are not isomorphic, the comment in part (c) states thereare three isomorphism types. �

Proof (d). If n7 > 1, then the map ϕ : P → Aut(S) is non-trivial so that P is not normal. Thus kerϕ = 1 and

ϕ(P ) ∼= Z7. Let ϕ(P ) = 〈x〉 so if s ∈ S\1, then sxk

shows all elements in S\1 have the same order. Since everynon-identity element in S must have order 2, it is given by the elementary abelian group, S ∼= Z3

2 . �

Proof (e). If S ∼= Z32 , then Aut(S) ∼= GL3 (F2) with |Aut(S)| = 7 · 6 · 4 = 168. Let P1, P2 ∈ Syl7(G) so that for any

non-trivial ϕ : Pi → Aut(S), kerϕ = 1. Thus ϕ(Pi) ∈ Syl7(Aut(S)), so P1 = Pσ2 for some σ ∈ Aut(S). Thereforeby exercise 6, S oϕ P1

∼= S oϕ P2. �

5.5.20. Let p be an odd prime. Prove that if P is a non-cyclic p-group then P contains a normal subgroup U withU ∼= Zp × Zp. Deduce that for odd primes p a p-group that contains a unique subgroup of order p is cyclic.

Proof. Let |P | = pn with P non-abelian. If n = 2, then P ∼= Zp×Zp since P is not cyclic. Now suppose there existsU C P with U ∼= Zp × Zp for all k < n. Since p | |Z(P )|, let Z ≤ Z(P ) with |Z| = p. Thus the non-cyclic quotient

group, P/Z, has a subgroup U E P/Z with U ∼= Zp × Zp. Now we have the non-cyclic U E P with |U | = p3, sothere is a U ′ E P with U ′ ∼= Zp × Zp. �

6. Further Topics in Group Theory

6.1. p-groups, Nilpotent Groups, and Solvable Groups.

6.1.3. If G is finite prove that G is nilpotent if and only if it has a normal subgroup of each order dividing |G|, andis cyclic if and only if it has a unique subgroup of each order dividing |G|.

Proof. Suppose G is nilpotent so theorem 3 states G ∼= P1×· · ·×Ps for Pi ∈ Sylpi . But if n | |G|, then n = pk11 · · · pkss .

Theorem 1 states each Pi has a normal subgroup, P ′i , with order pkii . Thus |P ′1 × · · · × P ′s| = n with the desirednormality property.

Now suppose G has a normal subgroup of each order dividing |G|. Then each Sylow subgroup is normal in G,so theorem 3 states G is nilpotent.

If G is cyclic, then theorem 2.7 states it has a unique subgroup of each order dividing |G|.If G has a unique subgroup of each order dividing |G|, then proposition 5 shows G is cyclic. �

6.1.4. Prove that a maximal subgroup of a finite nilpotent group has prime index.

Proof. Let M < G be a maximal subgroup of G. Since G is nilpotent, M /G. Since G/M is nilpotent, exercise 6.1.3shows it has a normal subgroup of each order dividing |G/M |. If P E G/M with

∣∣P ∣∣ = p for p prime, thenM ≤ P ≤ G so P = G by maximality of M . Therefore [G : M ] = p. �

6.1.8. Prove that if p is a prime and P is a non-abelian group of order p3 then |Z(P )| = p and P/Z(P ) ∼= Zp×Zp.


Proof. Since |P | = p3 we know P is nilpotent. Since P is a p-group Z(P ) 6= 1, since P is non-abelian, Z(P ) 6= P .Thus |Z(P )| ∈

{p, p2

}. If |Z(P )| = p2, then P/Z(P ) is cyclic contradicting P being non-abelian. Thus |Z(P )| = p.

Since P/Z(P ) is not cyclic, we must have P/Z(P ) ∼= Zp × Zp. �

6.1.12. Find the upper and lower central series for A4 and S4.

Solution. The upper central series of A4 is Zi(A4) = 1 and of S4 is Zi(S4) = 1 for all i since the center of S4 istrivial.

Since s−1a−1sa is an even permutation, S14 ≤ A4. But if α ∈ A4, we can choose s ∈ S4 such that s−1a−1s = αa

so that α ∈ S14 . The same argument shows S2

4 = A4. Since S14 = A4 and S2

4 = A4, we have found the lower centralseries for S4.

Now N = {1, (12)(34), (13)(24), (14)(23)} is the only proper normal subgroup of A4. Thus Ai4 = N for all i sothis is the lower central series for A4. �

33

Part II – Ring Theory

7. Introduction to Rings

7.1. Basic Definitions and Examples.

Proposition 7.1. Let R be a ring. Then

(1) 0a = a0 = 0 for all a ∈ R.(2) (−a)b = a(−b) = −(ab) for all a, b ∈ R.(3) (−a)(−b) = ab for all a, b ∈ R.(4) if R has an identity 1, then the identity is unique and −a = (−1)a.

Proposition 7.2. Assume a, b and c are element of any ring with A not a zero divisor. If ab = ac, then eithera = 0 or b = c. In particular, if a, b, c are any elements in an integral domain and ab = ac, then either a = 0 orb = c.

Corollary 7.3. Any finite integral domain is a field.

7.1.3. Let R be a ring with identity and let S be subring of R containing the identity. Prove that if u is a unit inS then u is a unit in R. Show by example that the converse if false.

Proof. If u is a unit in S, then there is a v ∈ S such that uv = 1 = vu. But then v ∈ R and hence u is a unit in R.Now consider 2 ∈ Q and notice that it is a unit because 1/2 ∈ Q. But Z ⊆ Q is a subring of Q and 2 is not a

unit. Thus the converse is false. �

7.1.5. Decide which of the following (a)-(f) are subrings of Q:

(a) the set of all rational numbers with odd denominators (when written in lowest terms)(b) the set of all rational numbers with even denominators (when written in lowest terms)(c) the set of nonnegative rational numbers(d) the set of squares of rational numbers(e) the set of all rational numbers with odd numerators (when written in lowest terms)(f) the set of all rational numbers with even numerators (when written in lowest terms).

Solution. (a), (f). �

7.1.6. Decide which of the following are subrings of the ring of all functions from the closed interval [0, 1] to R.

(a) the set of all functions f(x) such that f(q) = 0 for all q ∈ Q ∩ [0, 1](b) the set of all polynomial functions(c) the set of all functions which have only a finite number of zeros, together with the zero function(d) the set of all functions which have an infinite number of zeros(e) the set of all functions f such that limx→1− f(x) = 0(f) the set of all rational linear combinations of the functions sinnx and cosmx, where m,n ∈ Z+.

Solution. (a), (b), (c), (e). �

7.1.7. The center of a ring R is {z ∈ R | zr = rz, ∀r ∈ R} (i.e., is the set of all elements which commute with everyelement of R). Prove that the center of a division ring is a field.

Proof. �

7.1.12. Prove that any subring of a field which contains the identity is an integral domain.

Proof. Let F be a field and let R ⊆ F be a subring of F with 1 ∈ R. Then since R ⊆ F , every element of R is aunit. Therefore there are no zero divisors in R. �

7.1.13. An element x in R is called nilpotent if xm = 0 for some m ∈ Z+.

(a) Show that if n = akb for some integers a and b then ab is a nilpotent element of Z/nZ.(b) If a ∈ Z is an integer, show that the element a ∈ Z/nZ is nilpotent if and only if every prime divisor of n

is also a divisor of a. In particular, determine the nilpotent elements of Z/72Z explicitly.(c) Let R be the ring of functions from a nonempty set X to a field F . Prove that R contains no nonzero

nilpotent elements.

7.1.14. Let x be a nilpotent element of the commutative ring R (cf. Exercise 7.1.13).

(a) Prove that x is either zero or a zero divisor.


(b) Prove that rx is nilpotent for all r ∈ R.(c) Prove that 1 + x is a unit in R.(d) Deduce that the sum of a nilpotent element and a unit is a unit.

Proof (a). Suppose x 6= 0 and since xm = 0, let n be the smallest positive integer such that xn = 0. Then xxn−1 = 0where xn−1 6= 0 so x is a zero divisor. �

Proof (b). Since R is commutative, we have (rx)m = rmxm = 0. Thus rx is nilpotent for all r ∈ R. �

Proof (c). Observing the calculation

(1 + x)(1− x+ x2 − x3 + · · ·xm−1) = 1

shows that 1 + x is a unit. �

Solution (d). Let u be a unit so vu = 1 and let x be our nilpotent element. Then part (b) shows vx is nilpotentand part (c) shows 1 + vx is a unit. Thus v(u+ x) is a unit and hence u+ x is a unit. �

7.1.23. Let D be a squarefree integer, and let O be the ring of integers in the quadratic field Q(√D). For any

positive integer f prove that the set Of = Z[fω] = {a+ bfω | a, b ∈ Z} is a subring of O containing the identity.Prove that [O : Of ] = f (index as additive abelian groups). Prove conversely that a subring of O containing theidentity and having finite index f in O (as additive abelian group) is equal to Of . The ring Of is called the order

of conductor f in the field Q(√D). The ring of integers O is called the maximal order in Q(

√D).

Proof. Let a + bfω ∈ Of . Since bf ∈ Z, then a + bfω ∈ O so Of ⊆ O. Now (a + bfω)(c + dfω) = ac + (ad +bd)fω + bdf2ω2 ∈ Of and (a+ bfω)− (c+ dfω) = (a− c) + (b− d)fω ∈ Of . Thus Of is a subring of O.

Let a+ bω+Of = a′+ b′ω+Of . Then (a+ bω)− (a′+ b′ω) ∈ Of and (a− a′) + (b− b′)ω ∈ Of . So b− b′ = b′′fthus b ≡ b′ mod f . Now let b ≡ b′ + α for 1 ≤ α < f and suppose ∃a, a′ such that a + bω +Of = a′ + b′ω +Of .But this is a contradiction because f would have to divide b− b′. Thus a+ bω+Of and a′+ b′ω+Of are the samecoset if and only if b ≡ b′ mod f hence [O : Of ] = f .

Suppose that a subring, S, of O contains the identity and has finite index f . Since 1 ∈ S then a ∈ S for alla ∈ Z because S is closed under addition. Since |O/S| = f , |ω + S| divides f . But then S = f(ω + S) = fω + S sofω ∈ S. Since S is closed under addition then bfω ∈ S for all b ∈ Z. Now we have a+ bfω ∈ S for all a, b ∈ Z soOf ⊆ S. Since |Of | = |S|, we must have Of = S.

�

7.1.26. Let K be a field. A discrete valuation on K is a function ν : K× → Z satisfying

(i) ν(ab) = ν(a) + ν(b)(ii) ν is surjective, and(iii) ν(x+ y) ≥ min {ν(x), ν(y)} for all x, y ∈ K× with x+ y 6= 0.

The set R = {x ∈ K× | ν(x) ≥ 0} ∪ {0} is called the valuation ring of ν.

(a) Prove that R is a subring of K which contains the identity. In general, a ring R is called a discrete valuationring if there is some field K and some discrete valuation ν on K such that R is the valuation ring of ν.

(b) Prove that for each nonzero element x ∈ K either x or x−1 is in R.(c) Prove that an element x is a unit of R if and only if ν(x) = 0.

Proof (a). �

Proof (b). �

Proof (c). �

7.2. Examples: Polynomials Rings, Matrix Rings, and Group Rings.

Proposition 7.4. Let R be an integral domain and let p(x), q(x) be nonzero elements of R[x]. Then

(1) degree p(x)q(x) = degree p(x) + degree q(x),(2) the units of R[x] are just the units of R,(3) R[x] is an integral domain.

Let R be a commutative ring with 1.

7.2.1. Let p(x) = 2x3 − 3x2 + 4x − 5 and let q(x) = 7x3 + 33x − 4. In each of parts (a), (b), and (c) computep(x) + q(x) and p(x)q(x) under the assumption that the coefficients of the two given polynomials are taken fromthe specified ring (where the integer coefficients are taken mod n in parts (b) and (c)):

35

(a) R = Z,(b) R = Z/2Z,(c) R = Z/3Z.

Solution (a).

p(x) + q(x) = 9x3 − 3x2 + 37x− 9

p(x)q(x) = 14x6 − 21x5 + 94x4 − 142x3 + 144x2 − 181x+ 20

�

Solution (b).

p(x) + q(x) = x3 − x2 + x− 1

p(x)q(x) = x5 − x

�

Solution (c).

p(x) + q(x) = x

p(x)q(x) = 2x6 + x4 + x3 + 2x

�

7.2.6. Let S be a ring with identity 1 6= 0. Let n ∈ Z+ and let A be an n × n matrix with entries from S whosei, j entry is aij . Let Eij be the element of Mn(S) whose i, j entry is 1 and whose other entries are all 0.

(a) Prove that EijA is the matrix whose ith row equals the jth row of A and all other rows are zero.(b) Prove that AEij is the matrix whose jth column equals the ith column of A and all other columns are zero.(c) Deduce that EpqAErs is the matrix whose p, s entry is aqr and all other entries are zero.

Proof (a). �

Proof (b). �

Solution (c). �

7.2.7. Prove that the center of the ring Mn(R) is the set of scalar matrices (cf. Exercise 7.1.7). Use Exercise 7.2.6.

Proof. Let [zij ] = Z ∈ Z(Mn(R)) so ZEij = EijZ for all Eij . If i 6= j, then the (j, j)th element of EjiZ is zij andthe (j, j)th element of ZEji is 0, so zij = 0. Now we observe that the (j, i)th element of EjiZ is zii and the (j, i)th

element of ZEji is zjj , so zii = zjj . Thus Z must be a scalar matrix.If Z = rI, then

AZ = ArI = rAI = rIA = ZA,

for all A ∈Mn(R). �

7.2.10. Consider the following elements of the integral group ring ZS3:

α = 3(1 2)− 5(2 3) + 14(1 2 3) and β = 6(1) + 2(2 3)− 7(1 3 2)

where (1) is the identity of S3. Compute the following elements:

(a) α+ β,(b) 2α− 3β,(c) αβ,(d) βα,(e) α2.

Solution (a).

α+ β = 6(1) + 3(1 2)− 3(2 3) + 14(1 2 3)− 7(1 3 2)

�


Solution (b).

2α− 3β = −18(1) + 6(1 2)− 16(2 3) + 28(1 2 3) + 21(1 3 2)

�

Solution (c).

αβ = −108(1) + 81(1 2)− 21(1 3)− 30(2 3) + 90(1 2 3)

�

Solution (d).

βα = −108(1) + 18(1 2) + 63(1 3)− 51(2 3) + 84(1 2 3) + 6(1 3 2)

�

Solution (e).

α2 = 34(1)− 70(1 2)− 28(1 3) + 42(2 3)− 15(1 2 3) + 196(1 3 2)

�

7.2.12. Let G = {g1, . . . , gn} be a finite group. Prove that the element N = g1 + g2 + · · · + gn is in the center ofthe group ring RG (cf. Exercise 7.1.7).

Proof. Let X = x1g1 + · · ·+ xngn for xi ∈ R. Then

NX =

n∑k=1

∑gigj=gk

xj

gk

=

n∑k=1

n∑j=1

xj

gk

=

n∑k=1

∑gigj=gk

xi

gk

= XN.

�

7.3. Ring Homomorphisms and Quotient Rings.

Proposition 7.5. Let R and S be rings and let ϕ : R→ S be a homomorphism.

(1) The image of ϕ is a subring of S.(2) The kernel of ϕ is a subring of R. Furthermore, if α ∈ kerϕ then rα and αr ∈ kerϕ for every r ∈ R, i.e.,

kerϕ is closed under multiplication by elements from R.

Proposition 7.6. Let R be a ring and let I be an ideal of R. Then the (additive) quotient group R/I is a ringunder the binary operations:

(r + I) + (s+ I) = (r + s) + I and (r + I)× (s+ I) = (rs) + I

for all r, s ∈ R. Conversely, if I is any subgroup such that the above operations are well defined, then I is an idealof R.

Theorem 7.7.

(1) (The First Isomorphism Theorem for Rings) If ϕ : R → S is a homomorphism of rings, then the kernel ofϕ is an ideal of R, the image of ϕ is a subring of S and R/ kerϕ is isomorphic as a ring to ϕ(R).

(2) If I is any ideal of R, then the map

R→ R/I defined by r 7→ r + I

is a surjective ring homomorphism with kernel I (this homomorphism is called the natural projection of Ronto R/I). Thus every ideal is the kernel of a ring homomorphism and vice versa.

Theorem 7.8. Let R be a ring.

(1) (The Second Isomorphism Theorem for Rings) Let A b ea subring and let B be an ideal of R. ThenA+B = {a+ b | a ∈ A, b ∈ B} is a subring of R, A ∩B is an ideal of A and A+B/B ∼= A/(A ∩B).

37

(2) (The Third Isomorphism Theorem for Rings) Let I and J be ideal of R with I ⊆ J . Then J/I is an idealof R/I and (R/I)/(J/I) ∼= R/J .

(3) (The Fourth Isomorphism Theorem for Rings) Let I be an ideal of R. The correspondence A↔ A/I is aninclusion preserving bijection between the wset of subrings A of R that contain I and the set of subrings ofR/I. Furthermore, A (a subring containing I) is an ideal of R if and only if A/I is an ideal of R/I.

7.3.12. Let D be an integer that is not a perfect square in Z and let S =

{(a bDb a

)| a, b ∈ Z

}.

(a) Prove that S is a subring of M2(Z).

(b) If D is not a perfect square in Z prove that the map ϕ : Z[√D]→ S defined by ϕ(a+ b

√D) =

(a bDb a

)is

a ring isomorphism.

(c) If D ≡ 1 mod 4 is squarefree, prove that the set

{(a b

(D − 1)b/4 a+ b

)| a, b ∈ Z

}is a subring of M2(Z)

and is isomorphic to the quadratic integer ring O.

Proof (a). S is clearly a group under addition, so we show that S is closed under multiplication. But(a1 b1Db1 a1

)(a2 b2Db2 a2

)=

(a1a2 +Db1b2 a1b2 + a2b1D(a2b1 + a1b2) Db1b2 + a1a2

)so S is a subring of M2(Z). �

Proof (b). We calculate

ϕ((a1 + b1√D) + (a2 + b2

√D)) = ϕ((a1 + a2) + (b1 + b2)

√D)

=

(a1 + a2 b1 + b2

D(b1 + b2) a1 + a2

)=

(a1 b1Db1 a1

)+

(a2 b2Db2 a2

)ϕ((a1 + b1

√D)(a2 + b2

√D)) = ϕ((a1a2 + b1b2D) + (a1b2 + a2b1)

√D)

=

(a1a2 + b1b2D a1b2 + a2b1D(a1b2 + a2b1) a1a2 + b1b2D

)=

(a1 b1Db1 a1

)(a2 b2Db2 a2

)so ϕ is a homomorphism of rings. Now let

(a1 b1Db1 a1

)=

(a2 b2Db2 a2

), so a1 = a1, b1 = b2 and thus

a1 + b1√D = a2 + b2

√D. Now if

(a bDb a

)∈ S, then ϕ maps a+ b

√D to this element. Therefore ϕ is an

isomorphism. �

Proof (c). This is clearly a group under addition, so we show it is closed under multiplication. But(a1 b1

(D − 1)b1/4 a1 + b1

)(a2 b2

(D − 1)b2/4 a2 + b2

)=

(a1a2 + (D − 1)a1b2/4 a1b2 + a2b1 + b1b2

(D − 1)(a1b2 + b1b2)/4 (D − 1)b1b2/4 + a1a2 + a1b2 + a2b1 + b1b2

)so this is a subring of M2(Z). Similarly to the proof in part (b), this is isomorphic to the quadratic integerring O. �

7.3.15. Let X be a nonempty set and let P(X) be the Boolean ring of all subsets of X defined in exercise 21 ofsection 1. Let R be the ring of all functions from X into Z/2Z. For each A ∈ P(X) define the function

χA : X → Z/2Z by χA(x) =

{1, x ∈ A,0, x /∈ A,

χA is called the characteristic function of A with values in Z/2Z. Prove that the map P(X) → R defined byA 7→ χA is a ring isomorphism.


Proof. Let A = A′ so that ϕ(A) = χA = χA′ = ϕ(A′), so ϕ is well defined. Now let A,B ⊆ X so that

ϕ(A+B) = ϕ((A\B) ∪ (B\A)) = χ(A\B)∪(B\A).

But

χ(A\B)∪(B\A) =

1, x ∈ A\B,1, x ∈ B\A,0 x ∈ A ∩B,0, x ∈ X\(A ∩B)

But simple calculations show this is ϕ(A) + ϕ(B). Similarly,

ϕ(A×B) = ϕ(A ∩B)

= χA∩B

=

0, x ∈ A\B,0, x ∈ B\A,1, x ∈ A ∩B,0, x ∈ X\(A ∩B),

= ϕ(A)ϕ(B).

Now let f ∈ R so let A ⊆ X be the set such that f(A) = 1 and f(X\A) = 0. So ϕ(A) = χA = f and ϕ issurjective. Now let f = g ∈ R so there exist A,B such that ϕ(A) = f = g = ϕ(B). So χA = χB , thus A = B andϕ is injective. Therefore ϕ is a ring isomorphism. �

7.3.17. Let R and S be nonzero rings with identity and denote their respective identities by 1R and 1S . Letϕ : R→ S be a nonzero homomorphism of rings.

(a) Prove that if ϕ(1R) 6= 1S then ϕ(1R) is a zero divisor in S. Deduce that if S is an integral domain thenevery ring homomorphism from R to S sends the identity of R to the identity of S.

(b) Prove that if ϕ(1R) = 1S then ϕ(u) is a unit in S and that ϕ(u−1) = ϕ(u)−1 for reach unit u of R.

Proof (a). If ϕ(1R) 6= 1S , then

(ϕ(1R)− 1S)ϕ(1R) = ϕ(1R)ϕ(1R)− 1Sϕ(1R)

= ϕ(1R)− ϕ(1R)

= 0.

So ϕ(1R) is a zero divisor in S. If S is an integral domain, then there are no zero divisors. Since an integral domainhas a 1 6= 0 and ϕ(1R) 6= 0 if ϕ(R) 6= 0, then ϕ(1R)− 1S = 0. Therefore ϕ(1R) = 1S . �

Proof (b). If u ∈ R is a unit, there exists v ∈ R such that uv = 1R. So ϕ(u)ϕ(v) = ϕ(uv) = ϕ(1R) = 1S . Thusϕ(u) is a unit, with ϕ(u)−1 = ϕ(v) = ϕ(u−1). �

7.3.27. Prove that a nonzero Boolean ring has characteristic 2 (cf. exercise 15, section 1).

Proof. Let a = −1 so that0 = (−1)2 + 1 = 1 + 1.

Since 1 6= 0, then R has characteristic 2. �

7.3.29. Let R be a commutative ring. Recall (cf. exercise 7.1.13) that an element x ∈ R is nilpotent if xn = 0 forsome n ∈ Z+. Prove that the set of nilpotent element form an ideal – called the nilradical of R and denoted byN (R).

Proof. �

7.3.30. Prove that if R is a commutative ring and N (R) is its nilradical (cf. exercise 7.3.29) then zero is the onlynilpotent element of R/N (R) i.e., prove that N (R/N (R)) = 0.

Proof. Let x ∈ R/N (R) with xn = 0 for some n ∈ N. Then there exists r ∈ R such that x = r +N (R). But then0 = xn = (r +N (R))n = rn +N (R). This implies rn ∈ N (R) so r ∈ N (R) and thus x = 0. �

7.3.33. Assume R is commutative. Let p(x) = anxn + an−1x

n−1 + · · ·+ a1x+ a0 be an element of the polynomialring R[x].

(a) Prove tha tp(x) is a unit in R[x] if and only if a0 is a unit and a1, a2, . . . , an are nilpotent in R. [Seeexercise 7.1.14.]

39

(b) Prove that p(x) is nilpotent in R[x] if and only if a0, a1, . . . , an are nilpotenet elements of R.

Proof (a). �

Proof (b). �

7.4. Properties of Ideals.

Proposition 7.9. Let I be an ideal of R.

(1) I = R if and only if I contains a unit.(2) Assume R is commutative. Then R is a field if and only if its only ideals are 0 and R.

Corollary 7.10. If R is a field then any nonzero ring homomorphism from R into another ring is an injection.

Proposition 7.11. In a ring with identity every proper ideal is contained in a maximal ideal.

Proposition 7.12. Assume R is commutative. The ideal M is a maximal ideal if and only if the quotient ringR/M is a field.

Proposition 7.13. Assume R is commutative. Then the ideal P is a prime ideal in R if and only if the quotientring R/P is an integral domain.

Corollary 7.14. Assume R is commutative. Every maximal ideal of R is a prime ideal.

Let R be a ring with identity 1 6= 0.

7.4.2. Assume R is commutative. Prove that the augmentation ideal in the group ring RG is generated by{g − 1 | g ∈ G}. Prove that if G = 〈σ〉 is cyclic then the augmentation ideal is generated by σ − 1.

7.4.3.

(a) Let p be a prime and let G be an abelian group of order pn. Prove that the nilradical of the group ringFpG is the augmentation ideal (cf. exercise 7.3.29).

(b) Let G = {g1, . . . , gn} be a finite group and assume R is commutative. Prove that if r is any element of theaugmentation ideal of RG then r(g1 + · · ·+ gn) = 0.

Proof (a). Suppose there exists m such that (∑pn

i=1 aigi)m = 0. If ϕ is the augmentation map, then

0 = ϕ(

pn∑i=1

aigi)m = (

pn∑i=1

ai)m

. Since Fp\ {0} is a multiplicative group, we must have∑pn

i=1 ai = 0. Thus∑pn

i=1 aigi ∈ kerϕ.

Now let g−1 ∈ kerϕ. Then (g−1)pn

= gpn

+p(. . .)−1 = 0 in FpG. Since exercise 7.4.2 states kerϕ is generatedby {g − 1 | g ∈ G}, we must have kerϕ ⊆ N (FpG). Therefore kerϕ = N (FpG). �

Proof (b). By exercise 7.4.2 we may assume r = gi − 1. But then

(gi − 1)(g1 + · · ·+ gn) = gig1 + · · · gign − g1 − · · · − gn= g1 + · · ·+ gn − g1 − · · · − gn= 0.

�

7.4.7. Let R be a commutative ring with 1. Prove that the principal ideal generated by x in the polynomial ringR[x] is a prime ideal if and only if R is an integral domain. Prove that (x) is a maximal ideal if and only if R is afield.

Proof. Let ϕ : R[x] → R be given by ϕ(p(x)) = p(0) and notice kerϕ = (x). Thus R[x]/(x) ∼= R so by proposi-tion 7.13, (x) is prime if and only if R is an integral domain.

Similarly, proposition 7.12 shows (x) is maximal if and only if R is a field. �

7.4.10. Assume R is commutative. Prove that if P is a prime ideal of R and P contains no zero divisors then R isan integral domain.

Proof. Since P is prime, we know that R = R/P is an integral domain. So if r, a ∈ R\P , then ra = r a 6= 0. Thusra 6= 0 and R is an integral domain. �

7.4.13. Let ϕ : R→ S be a homomorphism of commutative rings.


(a) Prove that if P is a prime ideal of S then either ϕ−1(P ) = R or ϕ−1(P ) is a prime ideal of R. Apply thisto the special case when R is a subring of S and ϕ is the inclusion homomorphism to deduce that if P is aprime ideal of S then P ∩R is either R or a prime ideal of R.

(b) Prove that if M is a maximal ideal of S and ϕ is surjective then ϕ−1(M) is a maximal ideal of R. Give anexample to show that this need not be the case if ϕ is surjective.

Proof (a). Let r ∈ ϕ−1(P ), so ϕ(rs) = ϕ(r)ϕ(s) ∈ P because ϕ(r) ∈ P and P is an ideal. Since R is commutative,ϕ−1(P ) is an ideal. Now let rs ∈ ϕ−1(P ), so ϕ(r)ϕ(s) = ϕ(rs) ∈ P so ϕ(r) or ϕ(s) is in P because P is prime. Sor or s is in ϕ−1(P ) so ϕ−1(P ) is prime or ϕ−1(P ) = R.

Specifically if R ⊂ S and ϕ is the inclusion homomorphism, then ϕ−1(P ) is either R or a prime ideal of R. Butϕ−1(P ) = P ∩R. �

Proof (b). Let π1 : S → S/M and π2 : R/ ker(π1 ◦ ϕ) be the canonical projection maps. Since π1 ◦ ϕ is surjective,R/ ker(π1 ◦ ϕ) ∼= S/M . But ker(π1 ◦ ϕ) = ϕ−1(M). So R/ϕ−1(M) is a field and ϕ−1(M) is maximal. �

7.4.15. Let x2 + x+ 1 be an element of the polynomial ring E = F2[x] and use the bar notation to denote passageto the quotient ring F2[x]/(x2 + x+ 1).

(a) Prove that E has 4 elements: 0, 1, x and x+ 1.(b) Write out the 4× 4 addition table for E and deduce that the additive group E is isomorphic to the Klein

4-group.

(c) Write out the 4× 4 multiplication table for E and prove that E×

is isomorphic to the cyclic group of order3. Deduce that E is a field.

Proof (a). Let∑ni=1 aix

i be an element of F2[x]. If n = 1 then∑1i=0 = a0 + a1x. Suppose there exists c0, c1 such

that∑n−1i=0 aix

i = c0 + c1x. Then if n is even,

n∑i=0

aixi =

n−1∑i=0

aixi + anxn

=

n−1∑i=0

aixi + an(x+ 1)n/2

=

n−1∑i=0

cixi

= c0 + c1x.

Now if n > 2 is odd,

n∑i=0

aixi =

n−1∑i=0

aixi + anxn

=

n−1∑i=0

aixi + anx(x+ 1)(n−1)/2

=

n−1∑i=0

bixi

= c0 + c1x.

So there exist c0, c1 such that∑ni=0 aix

i = c0 + c1x for all n. Since c0, c1 ∈ {0, 1}, F2[x]/(x2+x+1) ⊆{

0, 1, x, x+ 1}

.�

Solution (b). The addition table in table 7.1 shows F2[x]/(x2 + x+ 1) ∼= V4. �

Solution (c). The multiplication table in table 7.2 shows E× ∼= Z3. Therefore E is a field. �

7.4.30. Let I be an ideal of the commutative ring R and define

rad I ={r ∈ R | rn ∈ I for some n ∈ Z+

}called the radical of I. Prove that rad I is an ideal containing I and that (rad I)/I = N (R/I) (cf. exercise 7.3.29).

41

Addition Table

0 1 x 1 + x

0 0 1 x 1 + x1 1 0 1 + x xx x 1 + x 0 11 + x 1 + x x 1 0

Table 7.1. F2[x]/(x2 + x+ 1).

Multiplication Table

0 1 x 1 + x

0 0 0 0 01 0 1 x 1 + xx 0 x 1 + x 11 + x 0 1 + x 1 x

Table 7.2. F2[x]/(x2 + x+ 1).

7.4.31. An ideal I of the commutative ring R is called a radical ideal if rad I = I.

(a) Prove that every prime ideal of R is a radical ideal.(b) Let n > 1 be an integer. Prove that 0 is a radical ideal in Z/nZ if and only if n is a product of distinct

primes to the first power (i.e., n is square free). Deduce that (n) is a radical of Z if and only if n is aproduct of distinct primes in Z.

Proof (a). Note P ⊆ radP . Let r ∈ R so if rn ∈ P , either r or rn−1 is in P . Induction shows r ∈ P , soradP = P . �

Proof (b). Suppose rad 0 = 0 and n = mp2. Then mp 6= 0 mod n and (mp)2 = 0 mod n which is a contradiction.Thus n is a product of distinct primes to the first power.

Now suppose n is a product of distinct primes to the first power. If αm = 0 mod n, then αm contains all of theprime factors of n. Thus α must contain all prime factors of n so α = 0 mod n. Therefore rad 0 = 0. �

7.4.39. Following the notation of exercise 7.1.26, let K be a field, let ν be a discrete valuation on K and let R bethe valuation ring of ν. For each integer k ≥ 0 define Ak = {r ∈ R | ν(r) ≥ k} ∪ {0}.

(a) Prove that Ak is a principal ideal and that A0 ⊇ A1 ⊇ A2 ⊇ · · ·.(b) Prove that if I is any nonzero ideal of R, then I = Ak for some k ≥ 0. Deduce that R is a local ring with

unique maximal ideal A1.

Proof (a). Since ν is surjective, Ak is nonempty. Since −1 ∈ R×, ν(−1) = 0. So ν(a − b) ≥ min {ν(a), ν(−b)} =min {ν(a), ν(b)} ≥ k. Also ν(ab) = ν(a) + ν(b) ≥ k so Ak is a subring. If r ∈ R and a ∈ Ak then ν(ra) =ν(r) + ν(a) ≥ k and similarly ν(ar) ≥ k. So Ak is an ideal.

Now let ν(a) = k by surjectivity of ν. Notice that 0 = ν(1) = ν(aa−1) = ν(a) + ν(a−1) so ν(a) = −ν(a−1). Ifb ∈ Ak then ν(ba−1) = ν(b)−ν(a) ≥ 0 so ba−1 ∈ R. Thus b = ba−1a ∈ (a), so Ak ⊆ (a). Therefore Ak is a principalideal and it is clear that A0 ⊇ A1 ⊇ A2 ⊃ · · ·. �

Proof (b). Let k = min {ν(I)} with ν(r) = k. Then Ak = (r) = I. From part (a) any proper ideal is contained inA1 so A1 is the unique maximal ideal of R. �

7.4.41. A proper ideal Q of the commutative ring R is called primary if whenever ab ∈ Q and a /∈ Q then bn ∈ Qfor some positive integer n. Establish the following facts about primary ideals.

(a) The primary ideal of Z are 0 and (pn), where p is a prime and n is a positive integer.(b) Every prime ideal of R is a primary ideal.(c) An ideal Q of R is primary if and only if every zero divisor in R/Q is a nilpotent element of R/Q.(d) If Q is a primary ideal then rad (Q) is a prime ideal (cf. exercise 7.4.30).


Proof (a). Let I ⊂ Z be a primary ideal so I = (n) for some n ∈ Z since every ideal of Z is principal. Let n = pafor p some prime, so p /∈ (n), a /∈ (n) with (n) a primary ideal implies that pi ∈ (n). So n|pi shows that n = pk forsome k ≥ 1. Thus (n) = (pk).

Now consider (pk) ⊆ Z. Let ab ∈ (pk) with a /∈ (pk). Then pk|ab and pk - a, so p|b. Thus pk|bk, so bk ∈ (pk) and(pk) is a primary ideal. �

Proof (b). Let P be a prime ideal of R with ab ∈ P . If a /∈ P then b ∈ P so P is a primary ideal. �

Proof (c). Let a b = 0 in R = R/Q. Then ab ∈ Q with ak ∈ Q and bl ∈ Q for k, l > 1. So ak = 0 thus every zerodivisor is nilpotent in R.

Now suppose every zero divisor in R is a nilpotent element. Let ab ∈ Q with a /∈ Q. Then b is a zero divisor in

R so bk

= 0. Thus bk ∈ Q and Q is a primary ideal of R. �

Proof (d). Let ab ∈ rad (Q) and a /∈ radQ, so (ab)n ∈ Q and am /∈ Q for all m ∈ Z+. Since Q is primary thenbnk ∈ Q so b ∈ rad (Q). Therefore rad (Q) is a prime ideal. �

7.5. Rings of Fractions.

Theorem 7.15 (cf. theorem 15.36). Let R be a commutative ring. Let D be any nonempty subset of R that doesnot contain 0, does not contain any zero divisors and is closed under multiplication (i.e., ab ∈ D for all a, b ∈ D).Then there is a commutative ring Q with 1 such that Q contains R as a subring and every element of D is a unitin Q. The ring Q has the following additional properties.

(1) every element of Q is of the form rd−1 for some r ∈ R and d ∈ D. In particular, if D = R\ {0} then Q isa field.

(2) (uniqueness of Q) The ring Q is the “smallest” ring containing R in which all elements of D become units,in the following sense. Let S be any commutative ring with idenitty and let ϕ : R→ S be any injective ringhomomorhpism such that ϕ(d) is a unit in S for every d ∈ D. Then there is an injective ring homomorphismΦ: Q → S such that Φ|R = ϕ. In other words, any ring containing an isomorphic copy of R in which allthe elements of D become units must also contain an isomorphic copy of Q.

Corollary 7.16. let R be an integral domain and let Q be the field of fractions of R. If a field F contains a subringR′ isomorphic to R then the subfield of F generated by R′ is isomorphic to Q.

Let R be a commutative ring with identity 1 6= 0.

7.5.1. Prove theorem 7.15.

Proof. Let F = {(r, d) | r ∈ R, d ∈ D} and define the relation ∼ on F by

(r, d) ∼ (s, e) if and only if re = sd.

It is immediate that this relation is reflexive and symmetric. Suppose (r, d) ∼ (s, e) and (s, e) ∼ (t, f). Thenre − sd = 0 and sf − te = 0. Multiplying the first of these equations by f and the second by d and adding themgives rf − td)e = 0. Since e ∈ D is neither zero nor a zero divisor we must have rf − td = 0, i.e., (r, d) ∼ (t, f).This proves ∼ is transitive, hence an equivalence relation. Done the equivalence class of (r, d) by r

d :

r

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

Let Q be the set of equivalence classes under ∼. Note that rd = re

de in Q for all e ∈ D, since D is closed undermultiplication.

We now define an additive and multiplicative structure on Q:

a

b+c

d=ad+ bc

bdand

a

b× c

d=ac

bd.

In order to prove that Q is a commutative ring with identity there are a number of things to check:

(1) these operations are well defined(2) Q is an abelian group under addition, where the additive identity is 0

d for any d ∈ D and the additive

inverse of ad is −ad ,

(3) multiplication is associative, distributive and commutative, and(4) Q has an identity.

43

To check that addition is well defined assume ab = a′

b′ (i.e., ab′ = a′b) and cd = c′

d′ . We must show thatad+bcbd = a′d′+b′c′

b′d′ , i.e.,

(ad+ bc)(b′d′) = (a′d′ + b′c′)(bd).

The left hand side of this equation is ab′dd′ + cd′bb′ substituting a′b for ab′ and c′d for cd′ gives a′bdd′ + c′dbb′,which is the right hand side. Hence addition of fractions is well defined.

To check that multiplication is well defined assume ab = a′

b′ and cd = c′

d′ . We must show

(ac)(b′d′) = (a′c′)(bd).

The left hand side of this equation is (ab′)(cd′) substituting a′b for ab′ and c′d for cd′ gives (a′b)(c′d) which is theright hand side. Hence multiplication of fractions is well defined.

To show Q is an abelian group under addition note that 0d ∈ Q with 0

d1+ r

d2= d1r

d1d2= r

d2so there exists an

additive identity in Q. Now if r1d1, r2d2 ,

r3d3∈ Q, then

(r1d1

+r2d2

) +r3d3

=r1d2 + r2d1

d1d2+r3d3

=r1d2d3 + r2d1d3 + r3d1d2

d1d2d3

=r1d1

+r2d3 + d1d2

d2d3

=r1d1

+ (r2d2

+r3d3

).

Thus addition is associative. Now if rd ∈ Q, then −rd ∈ Q with

r

d+−rd

=rd− rdd2

=0

d.

So Q is closed under inverse. Since addition is clearly commutative in Q, we see Q is an abelian group underaddition.

To show multiplication is associative, let r1d1, r2d2 ,

r3d3∈ Q so

(r1d1× r2d2

)× r3d3

=r1r2d1d2

× r3d3

=r1r2r3d1d2d3

=r1d1× r2r3d2d3

=r1d1× (

r2d2× r3d3

).

To show multiplication is distributive, let r1d1, r2d2 ,

r3d3∈ Q so

(r1d1

+r2d2

)× r3d3

=r1d2 + r2d1

d1d2× r3d3

=r1r3d2 + r2r3d1

d1d2d3

=r1d1× r3d2d2d3

+r2d2× r3d1d1d3

=r1d1× r3d3

+r2d2× r3d3

r1d1× (

r2d2

+r3d3

) =r1d1× r2d3 + r3d2

d2d3

=r1r2d3 + r1r3d2

d1d2d3

=r1d1× r2d3d2d3

+r1d1× r3d2d2d3

=r1d1× r2d2

+r1d1× r3d3.


To show multiplication is commutative, let r1d1, r2d2 ∈ Q so

r1d1× r2d2

=r1r2d1d2

=r2r1d2d1

=r2d2× r1d1.

To show Q has an identity, notice that dd ∈ Q for any d ∈ D. Now let r

d1∈ Q so

r

d1× d

d=

rd

d1d=

r

d1.

Next we embed R into Q by defining

ι : R→ Q by ι : r 7→ rd

dwhere d is any element of D.

Since rdd = re

e for all d, e ∈ D, ι(r) does not depend on the choice of d ∈ D. Now let r1, r2 ∈ R so

ι(r1 + r2) =(r1 + r2)d

d

=r1d

d+r2d

d= ι(r1) + ι(r2)

ι(r1 × r2) =(r1r2)d

d

=r1d

d× r2d

d= ι(r1)ι(r2),

so ι is a ring homomorphism. Furthermore, ι is injective because

ι(r) = 0⇔ rd

d=

0

d⇔ rd2 = 0⇔ r = 0

because d is neither zero nor a zero divisor. The subring ι(R) of Q is therefore isomorphic to R. We henceforthidentify each r ∈ R with ι(r) and so consider R as a subring of Q.

Next note that each d ∈ D has a multiplicative inverse in Q: namely, if d is represented by the fraction dee then

its multiplicative inverse is ede . One then sees that every element of Q may be written as r · d−1 for some r ∈ R and

some d ∈ D. In particular, if D = R\ {0}, every nonzero element of Q has a multiplicative inverse and Q is a field.It remains to establish the uniqueness property of Q. Assume ϕ : R→ S is an injective ring homomorphism such

that ϕ(d) is a unit in S for all d ∈ D. Extend ϕ to a map Φ: Q → S by defining Φ(rd−1) = ϕ(r)ϕ(d)−1 for allr ∈ R, d ∈ D. This map is well defined, since rd−1 = se−1 implies re = sd, so ϕ(r)ϕ(e) = ϕ(s)ϕ(d), and then

Φ(rd−1) = ϕ(r)ϕ(d)−1 = ϕ(s)ϕ(e)−1 = Φ(se−1).

Now let r1d−11 , r2d

−12 ∈ Q. Then

Φ(r1d−11 + r2d

−12 ) = Φ((r1d2 + r2d1)(d1d2)−1)

= ϕ(r1d2 + r2d1)ϕ(d1d2)−1

= ϕ(r1)ϕ(d2)ϕ(d1)−1ϕ(d2)−1 + ϕ(r2)ϕ(d1)ϕ(d1)−1ϕ(d2)−1

= ϕ(r1)ϕ(d1)−1 + ϕ(r2)ϕ(d2)−1

= Φ(r1d−11 ) + Φ(r2d

−12 )

Φ(r1d−11 r2d

−12 ) = Φ((r1r2)(d1d2)−1)

= ϕ(r1r2)ϕ(d1d2)−1

= ϕ(r1)ϕ(d1)−1ϕ(r2)ϕ(d2)−1

= Φ(r1d−11 )Φ(r2d

−12 ).

So Φ is a ring homomorphism. Finally, Φ is injective because rd−1 ∈ ker Φ implies r ∈ ker Φ∩R = kerϕ; since ϕ isinjective this forces r and hence also rd−1 to be zero. �

45

7.5.2. Let R be an integral domain and let D be a nonempty subset of R that is closed under multiplication.Prove that the ring of fractions D−1R is isomorphic to a subring of the quotient field of R (hence is also an integraldomain).

Proof. Let Q be the quotient field of R. Since R is a domain, D\ {0} is also a multiplicative set of R. By theorem 7.15there exists an injective ring homomorhpism ι : R→ Q with ι(R\ {0}) ⊆ Q×. Since D\ {0} ⊆ R\ {0}, theorem 7.15states there exists an injective ring homomorhpism ψ : D−1R→ Q with ψ|R = ι. �

8. Euclidean, Principal Ideal, and Unique Factorization Domains

8.1. Euclidean Domains.

Proposition 8.1. Every ideal in a Euclidean Domain is principal. More precisely, if I is any nonzero ideal in theEuclidean Domain R then I = (d), where d is any nonzero element of I of minimum norm.

Proposition 8.2. If a and B are nonzero element sin the commutative ring R such that the ideal generated by aand b is a principal ideal (d), then d is a greatest common divisor of a and b.

Proposition 8.3. Let R be an integral domain. If two elements d and d′ of R generate the same principal ideal,i.e., (d) = (d′), then d′ = ud for some u ∈ R×. In particular, if d and d′ are both greatest common divisors of aand b, then d′ = ud for some unit u.

Theorem 8.4. Let R be a Euclidean Domain and let a and b be nonzero elements of R. Let d = rn be the lastnonzero remainder in the Euclidean Algorithm for a and b described at the beginning of this chapter. Then

(1) d is a greatest common divisor of a and b, and(2) the principal ideal (d) is the ideal generated by a and b. In particular, d can be written as an R-linear

combination of a and b, i.e., there are elements x and y in R such that

d = ax+ by.

Proposition 8.5. Let R be an integral domain that is not a field. If R is a Euclidean Domain then there areuniversal side divisors in R.

8.1.3. Let R be a Euclidean Domain. Let m be the minimum integer in the set of norms of nonzero elements of R.Prove that every nonzero element of R of norm m is a unit. Deduce that a nonzero element of norm zero (if suchan element exists) is a unit.

Proof. Let a ∈ R with N(a) = m. Since R is a Euclidean Domain, there exists q, r ∈ R such that 1 = qa+ r withN(r) < N(a) or r = 0. Since a is chosen to be minimum norm, r = 0. Thus a is a unit. �

8.1.5. Determine all integer solutions of the following equations:

(a)(b) 17x+ 29y = 31(c)

Proof (b). Notice that 372 · 17− 217 · 29 = 31. So any solution to this equation has form

x = 372 +m · 899, y = −217−m · 527,

for any m ∈ Z. �

8.1.8. Let F = Q(√D) be a quadratic field with associated quadratic integer ring O and field norm N as in section

7.1.

(a) Suppose D is −1,−2,−3,−7 or −11. Prove that O is a Euclidean Domain with respect to N .(b) Suppose that D = −43,−67 or −163. Prove that O is not a Euclidean Domain with respect to any norm.

8.1.10. Prove that the quotient ring Z[i]/I is finite for any nonzero ideal I of Z[i].

Proof. By exercise 8.1.8(a), Z[i] is a Euclidean Domain. So I = (α) for some α ∈ Z[i]. Let x + I be a coset of I.Then there exists q, r such that

x+ α = qα+ r.

Since x /∈ (α), then r 6= 0 so N(r) < N(α). But now x+ I = r+ I, so any coset of I has a representative with normless than N(α).

Let F = {f ∈ R | N(f) < N(α)}. Then if f = f1 + f2i and α = a1 + a2i we have f21 + f22 < a21 + a22. Since thereare only a finite number of (f1, f2) ∈ Z× Z with this property, Z[i]/I is finite for any nonzero ideal I of Z[i]. �


8.1.11. Let R be a commutative ring with 1 and let a and b be nonzero elements of R. A least common multipleof a and b is an element e of R such that

(i) a | e and b | e, and(ii) if a | e′ and b | e′ then e | e′.

(a) Prove that a least common multiple of a and b (if such exists) is a generator for the unique largest principalideal contained in (a) ∩ (b).

(b) Deduce that any two nonzero elements in a Euclidean Domain have a least common multiple which is uniqueup to multiplication by a unit.

(c) Prove that in a Euclidean Domain the least common multiple of a and b is ab(a,b) , where (a, b) is the greatest

common divisor of a and b.

Proof (a). First suppose there exists a least common multiple e of a and b. Since a | e and b | e we must have(e) ⊆ (a) and (e) ⊆ (b). If (e′) is a principal ideal such that (e′) ⊆ (a) ∩ (b) then e′ | a and e′ | b. Thus e | e′ and(e′) ⊆ (e) so (e) is the largest principal ideal contained in (a) ∩ (b).

Now suppose there is a largest principal ideal (e) contained in (a) ∩ (b). Then e | a and e | b. If a | e′ and b | e′,then (e′) ⊆ (a) ∩ (b). Since (e) is the largest such principal ideal, then (e′) ⊆ (e). But then e | e′. �

Proof (b). If a and b are nonzero elements of R then (a) ∩ (b) is the largest ideal contained in (a) ∩ (b). Since R isa Euclidean Domain, there exists e ∈ R with (e) = (a) ∩ (b). So part (a) shows e is a least common multiple of aand b.

Now let e, e′ be two least common multiples of a and b. Then e | e′ and e′ | e. So e = se′ and et = e′. Since aand b are nonzero then e is also nonzero so e = ste and e(1− st) = 0 implies 1 = st because R is a domain. Thusleast common multiples are unique up to multiplication by a unit. �

Proof (c). First note that ( ab(a,b) ) ⊆ (a) ∩ (b) so ( ab

(a,b) ) ⊆ (e). Now let ax = e = by. Then abx = eb and aby = ea

so ab | eb and ab | ea. Thus ab | (ea, eb) and ab(a,b) | e. So we have (e) ⊆ ( ab

(a,b) ). Therefore ab(a,b) is a least common

multiple of a and b. �

8.1.12 (A Public Key Code). Let N be a positive integer. Let M be an integer relatively prime to N and let dbe an integer relatively prime to ϕ(N), where ϕ denotes Euler’s ϕ-function. Prove that if M1 ≡Md mod N then

M ≡Md′

1 mod N where d′ is the inverse of d mod ϕ(N) : dd′ ≡ 1 mod ϕ(N).

Proof. Since (M,N) = 1 we know that Mϕ(N) ≡ 1 mod N . Thus

Md′

1 ≡Mdd′ mod N

≡Mαϕ(N)+1 mod N

≡M mod N.

�

9. Polynomial Rings

9.1. Definitions and Basic Properties.

Proposition 9.1. Let R be an integral domain. Then

(1) degree p(x)q(x) = degree p(x) + degree q(x) if p(x), q(x) are nonzero(2) the units of R[x] are just the units of R(3) R[x] is an integral domain.

Proposition 9.2. Let I be an ideal of the ring R and let (I) = I[x] denote the ideal of R[x] generated by I (theset of polynomials with coefficients in I). Then

R[x]/(I) ∼= (R/I)[x].

In particular, if I is a prime ideal of R then (I) is a prime ideal of R[x].

9.1.13. Prove that the rings F [x, y]/(y2 − x) and F [x, y]/(y2 − x2) are not isomorphic for any field F .

Proof. Since y2−x2 = (y−x)(y+x) then F [x, y]/(y2−x2) is not a domain. Not let ϕ : R[x, y]→ R[y] be given by

ϕ(p(x, y)) = p(y2, y).

Let p(x, y)(y2 − x) ∈ (y2 − x) so that

ϕ(p(x, y)(y2 − x)) = p(y2, y)(y2 − y2) = 0.

47

Thus (y2 − x) ⊆ kerϕ. Now let p(x, y) ∈ kerϕ so that 0 = ϕ(p(x, y)) = p(y2, y). Note that we can write p(x, y) =r(x)+ys(x)+q(x, y)(y2−x) by considering cosets of F [x, y]/(y2−x). But then 0 = ϕ(r(x)+ys(x)) = r(y2)+ys(y2).So r = 0 and s = 0. Therefore p(x, y) = q(x, y)(y2 − x) so kerϕ ⊆ (y2 − x). �

9.1.14. Let R be an integral domain and let i, j be relatively prime integers. Prove that the ideal (xi − yj) is aprime ideal in R[x, y].

Proof. �

9.1.15. Let p(x1, x2, . . . , xn) be a homogeneous polynomial of degree k in R[x1, . . . , xn]. Prove that for all λ ∈ Rwe have

p(λx1, λx2, . . . , λxn) = λkp(x1, x2, . . . , xn).

Proof. First we write

p(x1, x2, . . . , xn) =

m∑i=0

ai

n∏j=1

xkjj ,

where∑nj=1 kj = k. Then

p(λx1, λx2, . . . , λxn) =m∑i=0

ai

n∏j=1

(λxj)kj

= λkm∑i=0

ai

n∏j=1

xkjj

= λkp(x1, x2, . . . , xn).

�

9.1.17. An ideal I in R[x1, . . . , xn] is called a homogeneous ideal if whenever p ∈ I then each homogeneouscomponent of p is also in I. Prove that an ideal is a homogeneous ideal if and only if it may be generated byhomogeneous polynomials.

Proof. �

9.2. Polynomial Rings Over Fields I.

Theorem 9.3. Let F be a field. The polynomial ring F [x] is a Euclidean Domain. Specifically, if a(x) and b(x)are two polynomials in F [x] with b(x) nonzero, then there are unique q(x) and r(x) in F [x] such that

a(x) = q(x)b(x) + r(x) with r(x) = 0 or deg r(x) < deg b(x).

Corollary 9.4. If F is a field, then F [x] is a Principal Ideal Domain and a Unique Factorization Domain.

9.2.4. Let F be a finite field. Prove that F [x] contains infinitely many primes.

Proof. Suppose on the other hand that p1(x), . . . , pn(x) are all of the primes in F [x]. Let p(x) =∏ni=1 pi(x) and

notice 1 + p(x) ∈ F [x]. Since F [x] is a unique factorization domain we can write

1 + p(x) =

m∏i=1

pαi(x)

where each pαi(x) is irreducible. But now

1 =

m∏i=1

pαi(x)− p(x)

= pα1(x)

(m∏i=2

pαi(x)− p(x)

pα1(x)

).

This is a contradiction because a prime is not a unit. �

9.2.5. Exhibit all the ideals in the ring F [x]/(p(x)), where F is a field and p(x) is a polynomial in F [x].


Proof. Since F is a field, F [x] is a Euclidean domain and in particular a unique factorization domain. So let

p(x) =

n∏i=1

pi(x),

where each pi(x) is irreducible. Let I/(p(x)) ⊆ F [x]/(p(x)) be an ideal. By the lattice isomorphism theorem,(p(x)) ⊆ I, and I is an ideal of F [x]. Since F [x] is a Euclidean domain, it is also a principal ideal domain so let I =

(q(x)). Since (p(x)) ⊆ (q(x)), p(x) = r(x)q(x) for some r(x) ∈ F [x]. Let r(x) =∏si=1 ri(x) and q(x) =

∏ti=1 qi(x)

for ri(x), qi(x) irreducible. Then α∏si=1 ri(x)

∏ti=1 qi(x) =

∏ni=1 pi(x), for some α ∈ F . So there exists σ ∈ Sn

such that αiqi(x) = pσ(i)(x) for all 1 ≤ i ≤ t, αi ∈ F . Therefore (q(x)) = (∏ti=1 qi(x)) = (

∏ti=1 pσ(i)(x)). �

9.2.6. Describe the ring structure of the following rings:

(d) Z[x, y]/(x2, y2, 2).

Show that α2 = 0 or 1 for every α in the last ring and determine those elements with α2 = 0. Determine thecharacteristics of each of these rings.

Proof (d). The highest power of x and y is 1 along with the largest coefficient. Let α = α(x, y) + (x2, y2, 2) soα2 = (α(x, y))2 + (x2, y2, 2) ∈

{0, 1}

because if there is no constant term then every term has even degree and if

there is a constant term, let α(x, y) = β(x, y)+1 where β(x, y) has no constant term. So α(x, y)2 = (β(x, y)+1)2 =β(x, y)2 + 2β(x, y) + 1 so this is in 1. The elements with α = 0 are the elements with no constant term. �

9.2.10. Determine the greatest common divisor of a(x) = x3 + 4x2 + x − 6 and b(x) = x5 − 6x + 5 in Q[x] andwrite it as a linear combination in Q[x] of a(x) and b(x).

Solution. Using division of polynomials we calculate

x5 − 6x+ 5 = (x3 + 4x2 − x− 6)(x2 − 4x+ 15)− (50x2 + 45x− 95)

x3 + 4x2 + x− 6 = (50x2 + 45x− 95)(x/50 + 31/500) + (31/10x2 + 1395/500x− 2945/500)

50x2 + 45x− 95 = (31/10x2 + 1395/500x− 2945/500)(500/31).

So we write the greatest common divisor as

31/10x2 + 1395/500x− 2945/500 =

= (x3 + 4x+ x− 6)[(x2 − 4x+ 15)(x/50 + 31/500)− 1]− (x5 − 6x+ 5)[x/50 + 31/500].

�

9.3. Polynomial Rings That are Unique Factorization Domains.

Proposition 9.5 (Gauss’ Lemma). Let R be a Unique Factorization Domain with field of fractions F and letp(x) ∈ R[x]. If p(x) is reducible in F [x] then p(x) is reducible in R[x]. More precisely, if p(x) = A(x)B(x) for somenonconstant polynomials A(x),B(x) ∈ F [x], then there are nonzero elements r, s ∈ F such that rA(x) = a(x) andsB(x) = b(x) both lie in R[x] and p(x) = a(x)b(x) is a factorization in R[x].

Corollary 9.6. Let R be a Unique Factorization Domain, let F be its field of fractions and let p(x) ∈ R[x]. Supposethe greatest common divisor of the coefficients of p(x) is 1. Then p(x) is irreducible in R[x] if and only if it isirreducible in F [x]. In particular, if p(x) is a moni polynomial that is irreducible in R[x], then p(x) is irreduciblein F [x].

Theorem 9.7. R is a Unique Factorization Domain if and only if R[x] is a Unique Factorization Domain.

Corollary 9.8. If R is a Unique Factorization Domain, then a polynomial ring in an arbitrary number of variablewith coefficients in R is also a Unique Factorization Domain.

9.4. Irreducibility Criteria.

Proposition 9.9. Let F be a field and let p(x) ∈ F [x]. Then p(x) has a facto of degree one if and only if p(x) hasa roote in F , i.e., there is an α ∈ F with p(α) = 0.

Proposition 9.10. A polynomial of degree two or three over a field F is reducible if and only if it has a root in F .

Proposition 9.11. Let p(x) = anxn + an−1x

n−1 + · · ·+ a0 be a polynomial of degree n with integer coefficients. Ifr/x ∈ Q is in lowest terms (i.e., r and s are relatively prime integers) and r/s is a root of p(x), then r divides theconstant term and s divides the leading coefficient of p(x). In particular, if p(x) is a monic polynomial with integercoefficients and p(d) 6= 0 for all itnegers d dividing the constant tterm of p(x), then p(x) has no roots in Q.

49

Proposition 9.12. Let I be a proper ideal in the integral domain R and let p(x) be a nonconstant monic polynomialin R[x]. If the image of p(x) in (R/I)[x] cannot be factored in (R/I)[x] into two polynomials of smaller degree,then p(x) is irreducible in R[x].

Proposition 9.13 (Eisenstein’s Criterion). Let P be a prime ideal of the integral domain R and let f(x) =xn + an−1x

n−1 + · · ·+ a1x+ a0 be a polynomial in R[x] (here n ≥ 1). Suppose an−1, . . . , a1, a0 are all elements ofP and suppose a0 is not an element of P 2. Then f(x) is irreducible in R[x].

Corollary 9.14 (Eisenstein’s Criterion for Polynomial Ring over Z). Let p be a prime in Z and let f(x) =xn + an−1x

n−1 + · · ·+ a1x+ a0 ∈ Z[x], n ≥ 1. Suppose p divides ai for all i ∈ {0, 1, . . . , n− 1} but that p2 does notdivide a0. Then f(x) is irreducible in both Z[x] and Q[x].

9.4.1. Determine whether the following polynomials are irreducible in the rings indicated. For those that arereducible, determine their factorization into irreducibles. The notaion Fp denotes the finite field Z/pZ, p a prime.

(b) x3 + x+ 1 in F3[x].(c) x4 + 1 in F5[x].

Solution (b). �

Solution (c). �

9.4.2. Prove that the following polynomials are irreducible in Z[x]:

(c) x4 + 4x3 + 6x2 + 2x+ 1

(d) (x+2)p−2px , where p is an odd prime.

Proof (c). �

Proof (d). �

9.4.3. Show that the polynomial (x− 1)(x− 2) · · · (x− n)− 1 is irreducible over Z for all n ≥ 1.

Proof. �

9.4.6. Construct fields of each of the following orders: (a) 9, (b) 49, (c) 8, (d) 81.

Solution (a). �

Solution (b). �

Solution (c). �

Solution (d). �

9.4.8. Prove that K1 = F11[x]/(x2 + 1) and K2 = F11[y]/(y2 + 2y + 2) are both fields with 121 elements. Provethat the map which sends the element p(x) of K1 to the element p(y + 1) of K2 (where p is any polynomial withcoefficients in F11) is well defined and gives a ring isomorphism from K1 to K2.

Proof. �

9.4.16. Let F be a field and let f(x) be a polynomial of degree n in F [x]. The polynomial g(x) = xnf(1/x) iscalled the reverse of f(x).

(a) Describe the coefficients of g in terms of the coefficients of f .(b) If f(0) 6= 0 prove that f is irreducible if and only if g is irreducible.

Solution (a). �

Proof (b). �

9.4.19. Let F be a field and let f(x) = anx + an−1xn−1 + · · · + a0 ∈ F [x]. The derivative, Dx(f(x)), of f(x) is

defined byDx(f(x)) = nanx

n−1 + (n− 1)an−1xn−2 + · · ·+ a1

where, as usual, na = a+ · · ·+ a (n times). Note that Dx(f(x)) is again a polynomial with coefficients in F .The polynomial f(x) is said to have a multiple root if there is some field E containing F and some α ∈ E such

that x−α)2 divides f(x) in E[x]. For example, the polynomial f(x) = (x−1)2(x−2) ∈ Q[x] has α = 1 as a multipleroot and the polynomial f(x) = x4 + 2x2 + 1 = (x2 + 1)2 ∈ R[x] has α = ±i ∈ C as multiple roots. We shall provein section 13.5 that a nonconstant polynomial f(x) has a multiple root if and only if f(x) is not relatively prime to


its derivative (which can be detected by the Euclidean Algorithm in F [x]). Use this criterion to determine whetherthe following polynomials have multiple roots:

(a) x3 − 3x− 2 ∈ Q[x](b) x3 + 3x+ 2 ∈ Q[x](c) x6 − 4x4 + 6x3 + 4x2 − 12x+ 9 ∈ Q[x](d) Show for any prime p and any a ∈ Fp that the polynomial xp − a has a multiple root.

9.5. Polynomial Rings Over Fields II.

Proposition 9.15. The maximal ideals in F [x] are the ideal (f(x)) generated by irreducible polynomials f(x). Inparticular, F [x]/(f(x)) is a field if and only if f(x) is irreducible.

Proposition 9.16. Let g(x) be a nonconstant monic element of F [x] and let

g(x) = f1(x)n1f2(x)n2 · · · fk(x)nk

be its factorization into irreducibles, where the fi(x) are distinct. Then we have the following isomorphism of rings:

F [x]/(g(x)) ∼= F [x]/(f1(x)n1)× F [x]/(f2(x)n2)× · · · × F [x]/(fk(x)nk).

Proposition 9.17. If the polynomial f(x) has roots α1, α2, . . . , αk in F (not necessarily distinct), then f(x) has(x− α1) · · · (x− αk) as a factor. In particular, a polynomial of degree n in one variable over a field F has at mostn roots in F , even counted with multiplicity.

Proposition 9.18. A finite subgroup of the multiplicative group of a field is cyclic. In particular, if F is a finitefield, then the multiplicative group F× of nonzero elements of F is a cyclic group.

Corollary 9.19. Let p be a prime. The multiplicative group (Z/pZ)× of nozero residue classes mod p is cyclic.

Corollary 9.20. Let n ≥ 2 be an integer with factorization n = pα11 · · · pαrr in Z, where p1, . . . , pr are distinct

primes. We have the following isomorphisms of multiplicative groups:

(1) (Z/nZ)× ∼= (Z/pαii Z)× × · · · × (Z/pαrr Z)×

(2) (Z/2αZ)× is the direct product of a cyclic group of order 2 and a cyclic group of order 2α−2, for all α ≥ 2(3) (Z/pαZ)× is a cyclic group of order pα−1(p− 1), for all odd primes p.

9.6. Polynomials in Several Variables Over a Field and Grobner Bases.

Theorem 9.21 (Hilbert’s Basis Theorem). If R is a Noetherian ring then so is the polynomial ring R[x].

Corollary 9.22. Every ideal in the polynomial ring F [x1, x2, . . . , xn] with coefficients from a field F is finitelygenerated.

Theorem 9.23. Fix a monomial ordering on R = F [x1, . . . , xn] and suppose {g1, . . . , gm} is a Grobner basis forthe nozero ideal I in R. Then

(1) Every polynomial f ∈ R can be written uniquely in the form

f = fI + r

where fI ∈ I and no nonzero monomial term of the remainder r is divisible by any of the leading termsLT (g1), . . . , LT (gm).

(2) Both fI and r can be computed by general polynomial division by g1, . . . , gm and are independent of theorder in which these polynomials are used in the division.

(3) The remainder r provides a unique representative for the coset of f in the quotient ring F [x1, . . . , xn]/I. Inparticular, f ∈ I if and only if r = 0.

Proposition 9.24. Fix a monomial ordering on R = F [x1, . . . , xn] and let I be a nonzero ideal in R.

(1) If g1, . . . , gm are any elements of I such that LT (I) = (LT (g1), . . . , LT (gm)), then {g1, . . . , gm} is a Grobnerbasis for I.

(2) The ideal I has a Grobner basis.

Lemma 9.25. Suppose f1, . . . fm ∈ F [x1, . . . , xn] are polynomials with the same multidegree α and that the linearcombination h = a1f1 + · · ·+ amfm with constants ai ∈ F has strictly smaller multidegree. Then

h =

m∑i=2

biS(fi−1, fi), for some constants bi ∈ F.

51

Proposition 9.26 (Buchberger’s Criterion). Let R = F [x1, . . . , xn] and fix a monomial ordering on R. If I =(g1, . . . , gm) is a nonzero ideal in R, then G = {g1, . . . , gm} is a Grobner basis for I if and only if S(gi, gj) ≡ 0mod G for 1 ≤ i < j ≤ m.

Theorem 9.27. Fix a monomial ordering on R = F [x1, . . . , xn]. Then there is a unique reduced Grobner basis forevery nonzero ideal I in R.

Corollary 9.28. Let I and J be two ideal in F [x1, . . . , xn]. Then I = J if and only if I and J have the samereduced Grobner basis with respect to any fixed monomial ordering on F [x1, . . . , xn].

Proposition 9.29 (Elimination). Suppose G = {g1, . . . , gm} is a Grobner basis for the nonzero ideal I in F [x1, . . . , xn]with respect to the lexicographic monomial ordering x1 > · · · > xn. Then G ∩ F [xi+1, . . . , xn] is a Grobner basisof the ith elimination ideal Ii = I ∩ F [xi+1, . . . , xn] of I. In particular, I ∩ F [xi+1, . . . xn] = 0 if and only ifG ∩ F [xi+1, . . . , xn] = ∅.Proposition 9.30. If I and J are any two ideals in F [x1, . . . , xn] then tI + (1− t)J is an ideal in F [t, x1, . . . , xn]and I ∩ J = (tI + (1− t)J) ∩ F [x1, . . . , xn]. In particular, I ∩ J is the first elimination ideal of tI + (1− t)J withrespect to the ordering t > x1 > · · · > xn.


9.6.24. Use reduced Grobner bases to show that the ideal I = (x3−yz, yz+y) and the ideal J = (x3z+x3, x3 +y)in F [x, y] are equal.

Proof. By corollary 9.28 two ideals in F [x, y, z] are equal if and only if I, J have the same reduced Grobner basisfor any monomial ordering. So fix the lexicographical ordering x > y > z for F [x, y, z]. To find a Grobner basis forI,

S(f1, f2) = yzf1 − x3f2 = −y2z2 − x3y ≡ 0 mod {f1, f2} .Since x3 − yz = −1(yz + y) + (x3 + y), then our reduced Grobner basis for I is G =

{x3 + y, yz + y

}.

Now to find the reduced Grobner basis for J ,

S(g1, g2) = g1 − zg2 = x3 − yz ≡ −yz − y mod {g1, g2}S(g1, g3) = −yg1 − x3g3 = −x3y + x3y ≡ 0 mod {g1, g2, g3}S(g2, g3) = x3g2 + yzg3 = x3 − y2z ≡ 0 mod {g1, g2, g3} .

Thus {g1, g2, g3} is a Grobner basis for J . Since LT (g2) mod LT (g1), G ={x3 + y, yz + y

}is the reduced Grobner

basis for J . �

9.6.32. Use Grobner bases to show that (x, z) ∩ (y2, x− yz) = (xy, x− yz) in F [x, y, z].

Proof. Let I = (x, z) and J = (y2 − yz). Then proposition 9.30 states

I ∩ J = (tI + (1− t)J) ∩ F [x, y, z].

By calculating the S(fi, fj) we find

tI + (1− t)J = (tx, tz, ty2 − y2, tx− tyz − x+ yz, xy2, tyz + x− yz, x2 − xyz, y2z, xz − yz2, x− yz)is given by a Grobner basis. Thus

tI + (1− t)J = (tz, ty2 − y2, tyz + x− yz, y2z, x− yz)and finally I ∩ J = (y2z, x− yz).

Now to find a Grobner basis for (xy, x− yz), we calculate

S(f1, f2) = f1 − yf2 = y2z

S(f1, f3) = yzf1 − xf3 = 0

S(f2, f3) = y2zf2 − xf3 = −y3z2 ≡ 0 mod {f1, f2, f3} .Thus

{xy, x− yz, y2z

}is a Grobner basis so

{x− yz, y2z

}is the reduced Grobner basis because x | xy. Since the

reduced Grobner basis is the same for both ideals, corollary 9.28 states the ideals are equal. �

53

Part III – Modules and Vector Spaces

10. Introduction to Module Theory

10.1. Basic Definitions and Examples.

Proposition 10.1 (The Submodule Criterion). Let R be a ring and let M be an R-module. A subset N of M is asubmodule of M if and only if

(1) N 6= ∅, and(2) x+ ry ∈ N for all r ∈ R and for all x, y ∈ N .

10.1.8. An element m of the R-module M is called a torsion element if rm = 0 for some nonzero element r ∈ R.The set of torsion elements is denoted

Tor(M) = {m ∈M | rm = 0 for some nonzero r ∈ R} .(a) Prove that if R is an integral domain then Tor(M) is a submodule of M called the torsion submodule of

M .(b) Give an example of a ring R and an R-module M such that Tor(M) is not a submodule.(c) If R has zero divisors show that every nonzero R-module has nonzero torsion elements.

Proof (a). Tor(M) 6= ∅ because r0 = r(m−m) = rm− rm = 0. If r1x = r2y = 0, r ∈ R, then

r1r2(x+ ry) = r2r1x+ rr1r2y = 0

so Tor(M) is a submodule of M because r1r2 6= 0. �

Example (b). If R = 0, then this is such an example. �

Proof (c). Let x, y ∈ R\ {0} with xy = 0. If ym 6= 0, then xym = 0 so either way there are nonzero torsionelements. �

10.1.14. Let z be an element of the center of R, i.e., zr = rz for all r ∈ R. Prove that zM is a submodule of M ,where zM = {zm | m ∈M}. Show that if R is the ring of 2× 2 matrices over a field and e is the matrix with a 1 inposition 1, 1 and zeros elsewhere then eR is not a left R-submodule (where M = R is considered as a left R-moduleas in example 1) – in this case the matrix e is not in the center of R.

Proof. zM is clearly nonempty. Let x = zm1, y = zm2, r ∈ R so

x+ ry = zm1 + rzm2 = zm1 + zrm2 = z(m1 + rm2)

so zM is a submodule of M .

If a ∈ eR then a =

(f1 f20 0

)but (

1 11 1

)(f1 f20 0

)=

(f1 f2f1 f2

)so eR is not a left R-submodule. �

10.1.16. Prove that the submodules Uk described in the example of F [x]-modules are all of the F [x]-submodulesfor the shift operator.

Proof. Note since F is a field, then if a component contains a nonzero element then that component of the submodulemust contain all of F . But if a component is zero to the left of some nonzero component, then it would not beT -stable. Thus it is clear that the Uks are all of the F [x]-submodules by the bjiection of F [x]-submodules andsubspaces invariant under T . �

10.1.22. Suppose A is a ring with identity 1A that is a (unital) left R-module satisfying r · (ab) = (r · a)b = a(r · b)for all r ∈ R and a, b ∈ A. Prove that the map f : R→ A defined by f(r) = r ·1A is a ring homomorphism mapping1R to 1A and that f(R) is contained in the center of A. Conclude that A is an R-algebra and that the R-modulestructure on A induced by its algebra structure is precisely the original R-module structure.

Proof. If f(x) = f(y) then x = x · 1A = y · 1A = y so f is well defined. Now f(x+ y) = (x+ y) · 1A = f(x) + f(y),f(xy) = (xy) · 1A = f(x)f(y) and f(1R) = 1R · 1A = 1A. Since (r · 1A) · a = (r · a) · 1A = a · (r · 1A). Thusby definition A is an R-algebra and the R-module structure on A induced by its algebra structure is precisely theoriginal R-module structure. �


13. Field Theory

13.1. Basic Theory of Field Extensions.

Proposition 13.1. The characteristic of a field F , ch (F ), is either 0 or a prime p. If ch (F ) = p then for anyα ∈ F ,

p · α = α+ α+ · · ·+ α = 0.

Proposition 13.2. Let ϕ : F → F ′ be a homomorphism of fields. Then ϕ is either identically 0 or is injective, sothat the image of ϕ is either 0 or isomorphic to F .

Theorem 13.3. Let F be a field and let p(x) ∈ F [x] be an irreducible polynomial. Then there exists a field Kcontaining an isomorphic copy of F in which p(x) has a root. Identifying F with this isomorphic copy shows thatthere exists an extension of F which p(x) has a root.

Theorem 13.4. Let p(x) ∈ F [x] be an irreducible polynomial of degree n over the field F and let K be the fieldF [x]/(p(x)). Let θ = x mod (p(x)) ∈ K. Then the elements

1, θ, θ2, . . . , θn−1

are a basis for K as a vector space over F , so the degree of the extension is n, i.e., [K : F ] = n. Hence

K ={a0 + a1θ + a2θ

2 + · · ·+ an−1θn−1 | a0, a1, . . . , an−1 ∈ F

}consists of all polynomials of degree < n in θ.

Corollary 13.5. Let K be as in theorem 13.4, and let a(θ), b(θ) ∈ K be two polynomials of degree < n in θ. Thenaddition in K is defined simply by usual polynomial addition and multiplication in K defined by

a(θ)b(θ) = r(θ)

where r(x) is the remainder (of degree < n) obtained after dividing the polynomial a(x)b(x) by p(x) in F [x].

Theorem 13.6. Let F be a field and let p(x) ∈ F [x] be an irreducible polynomial. Suppose K is an extension fieldof F containing a root α of p(x): p(α) = 0. Let F (α) denote the subfield of K generated over F by α. Then

F (α) ∼= F [x]/(p(x)).

Corollary 13.7. Suppose in theorem 13.6 that p(x) is of degree n. Then

F (α) ={a0 + a1α+ a2α

2 + · · ·+ an−1αn−1 | a0, a1, . . . , an−1 ∈ F

}⊆ K.

Theorem 13.8. Let ϕ : F→F ′ be an isomorphism of fields. Let p(x) ∈ F [x] be an irreducible polynomial and letp′(x) ∈ F ′(x) be the irreducible polynomial obtained by applying the map ϕ to the coefficients of p(x). Let α bea root of p(x) (in some extension of F ) and let β be a root of p′(x) (in some extension of F ′). Then there is anisomorphism

σ : F (α)→F ′(β)

mapping α to β and extending ϕ, i.e., such that σ restricted to F is the isomorphism ϕ.

13.2. Basic Theory of Field Extensions.

Proposition 13.9. Let α be algebraic over F . Then there is a unique monic irreducible polynomial mα,F (x) ∈ F [x]which has α as a root. A polynomial f(x) ∈ F [x] has α as a root if and only if mα,F (x) divides f(x) in F [x].

Corollary 13.10. If L ⊂ F is an extension of fields and α is algebraic over both F and L, then mα,L(x) dividesmα,F (x) in L[x].

Proposition 13.11. Let α be algebraic over the field F and let F (α) be the field generated by α over F . Then

F (α) ∼= F [x]/(mα(x))

so that in particular[F (α) : F ] = degmα(x) = degα,

i.e., the degree of α over F is the degree of the extension it generates over F .

Proposition 13.12. The element α is algebraic over F if and only if the simple extension F ⊂ F (α) is finite.More precisely, if α is an element of an extension of degree n over F then α satisfies a polynomial of degree at mostn over F and if α satisfies a polynomial of degree n over F then the degree of F (α) over F is at most n.

Corollary 13.13. If the extension F ⊂ K is finite, then it is algebraic.

55

Theorem 13.14. Let F ⊆ K ⊆ L be fields. Then

[L : F ] = [L : K][K : F ],

i.e., extension degrees are multiplicative, where if one side of the equation is infinite, the other side is also infinite.

Corollary 13.15. Suppose F ⊂ L is a finite extension and let K be any subfield of L containing F , F ⊆ K ⊆ L.Then [K : F ] divides [L : F ].

Lemma 13.16. F (α, β) = (F (α))(β), i.e., the field generated over F by α and β is the field generated by β overthe field F (α) generated by α.

Theorem 13.17. The extension F ⊂ K is finite if and only if K is generated by a finite number of algebraicelements over F . More precisely, a field generated over F by a finite4 number of algebraic elements of degreesn1, n2, . . . , nk is algebraic of degree ≤ n1n2 · · ·nk.

Corollary 13.18. Suppose α and β are algebraic over F . Then α ± β, αβ, α/β (for β 6= 0), (in particular α−1

for α 6= 0) are all algebraic.

Corollary 13.19. Let F ⊂ L be an arbitrary extension. Then the collection of elements of L that are algebraicover F form a subfield K of L.

Theorem 13.20. If K is algebraic over F and L is algebraic over K, then L is algebraic over F .

Proposition 13.21. Let K1 and K2 be two finite extensions of a field F contained in K. Then

[K1K2 : F ] ≤ [K1 : F ][K2 : F ]

with equality if and only if an F -basis for one of the fields remains linearly independent over the other field.If α1, α2, . . . , αn and β1, β2, . . . , βm are bases for K1 and K2 over F , respectively, then the elements αiβj fori = 1, . . . , n and j = 1, . . . ,m span K1K2 over F .

Corollary 13.22. Suppose that [K1 : F ] = n, [K2 : F ] = m in proposition 13.21, where n and m are relativelyprime: (n,m) = 1. Then [K1K2 : F ] = [K1 : F ][K2 : F ] = nm.

13.2.1. Let F be a finite field of characteristic p. Prove that |F| = pn for some positive integer n.

Proof. Let ϕ : Z→ F be the ring map given by ϕ(n) = n · 1F. Since pZ ⊆ kerϕ then ϕ : Fp → F is well defined. Butϕ is a ring map of fields so it is trivial or injective. Notice ϕ(1Fp) = 1F so ϕ is injective. Thus we can consider F asa vector space over Fp. Since F is finite then

|F| = |Fp|[F : Fp] = pn

for some n ∈ Z+. �

13.2.7. Prove that Q(√

2+√

3) = Q(√

2,√

3). Conclude that [Q(√

2+√

3) : Q] = 4. Find an irreducible polynomial

satisfied by√

2 +√

3.

Proof. Since Q,√

2,√

3 are contained in Q(√

2,√

3), then Q,√

2 +√

3 are contained in Q(√

2,√

3). Hence by

definition Q(√

2 +√

3) ⊆ Q(√

2,√

3).

To prove the other inclusion notice that((√

2 +√

3)3 − 9(√

2 +√

3))/2 =

√2. Thus

√2 ∈ Q(

√2 +√

3) thus√3 ∈ Q(

√2 +√

3). By definition we have Q(√

2,√

3) ⊆ Q(√

2 +√

3) and thus equality.

Since Q(√

2,√

3) = Q(√

2 +√

3) then [Q(√

2 +√

3) : Q] = [Q(√

2,√

3) : Q] = 4. Notice that p(√

2 +√

3) = 0where

p(x) = x4 − 10x2 + 1.

If p were to be reducible, then there would be an irreducible polynomial with degree less than 4. This contradictsthe minimality of the degree of m√2+

√3 = 4. Therefore p is irreducible in Q. �

13.2.10. Determine the degree of the extension Q(√

3 + 2√

2) over Q.

Solution. First note that [Q(√

3 + 2√

2) : Q] = [Q(√

3 + 2√

2: Q(√

2)][Q(√

2) : Q]. But x2 − 3 + 2√

2 = (x − 1 +

2√

2)(x+ 1 + 2√

2) so [Q(√

3 + 2√

2) : Q(√

2)] = 1. Thus [Q(√

3 + 2√

2) : Q] = 2. �

13.2.13. Suppose F = Q(α1, . . . , αn) where α2i ∈ Q for i = 1, . . . , n. Prove that 3

√2 /∈ F .

Proof. Since [Q(α1, . . . , αk) : Q(α1, . . . , αk−1)] ∈ {1, 2}, then [Q(α1, . . . , αn) : Q] = 2l for some l ∈ Z+. Now if3√

2 ∈ F then Q ⊂ Q( 3√

2) ⊆ F then [Q( 3√

2: Q)] | [F : Q]. Since 3 does not divide 2l for l ∈ Z+ then 3√

2 /∈ F . �


13.2.16. Let F ⊂ K be an algebraic extension and let R be a ring contained in K and containing F . Show thatR is a subfield of K containing F .

Proof. Let α ∈ R\ {0} so since α ∈ F then there is some irreducible p ∈ R[x] such that p(α) = 0. Let p(x) be givenby

p(x) = anxn + · · ·+ a0,

and notice since p is irreducible, a0 6= 0. Since p(α) = 0 we have

α−1 = −a−10 (anαn−1 + · · ·+ a1).

Since ai ∈ F ⊆ R and α ∈ R then α−1 ∈ R and R is a field. �

13.3. Classical Straightedge and Compass Constructions.

Proposition 13.23. If the element α ∈ R is obtained from a field F ⊆ R by a series of compass and straightedgeconstructions then [F (α) : F ] = 2k for some integer k ≥ 0.

Theorem 13.24. None of the classical Greek problems: (I) Doubling the Cube, (II) Trisecting an Angle, and (III)Squaring the Circle, is possible.

13.4. Splitting Fields and Algebraic Closures.

Theorem 13.25. For any field F , if f(x) ∈ F [x] then there exists an extension K of F which is a splitting fieldfor f(x).

Proposition 13.26. A splitting field of a polynomial of degree n over F is of degree at most n! over F .

Theorem 13.27. Let ϕ : F→F ′ be an isomorphism of fields. Let f(x) ∈ F [x] be a polynomial and let f ′(x) ∈ F ′[x]be the polynomial obtained by applying ϕ to the coefficients of f(x). Let E be a splitting field for f(x) over F andlet E′ be a splitting field for f ′(x) over F ′. Then the isomorphism ϕ extends to an isomorphism σ : E→E′, i.e., σrestricted to F is the isomorphism ϕ.

Corollary 13.28 (Uniqueness of Splitting Fields). Any two splitting fields for a polynomial f(x) ∈ F [x] over afield F are isomorphic.

Proposition 13.29. Let F be an algebraic closure of F . Then F is algebraically closed.

Proposition 13.30. For any field F there exists an algebraically closed field K containing F .

Proposition 13.31. Let K be an algebraically closed field and let F be a subfield of K. Then the collection ofelements F of K that are algebraic over F is an algebraic closure of F . An algebraic closure of F is unique up toisomorphism.

Theorem (Fundamental Theorem of Algebra). The field C is algebraically closed.

Corollary 13.32. The field C contains an algebraic closure for any of its subfields. In particular, Q, the collectionof complex numbers algebraic over Q, is an algebraic closure of Q.

13.5. Separable and Inseparable Extensions.

Proposition 13.33. A polynomial f(x) has a multpile root α if and only if α is also a root of Dx f(x), i.e., f(x)and Dx f(x) are both divisible by the minimal polynomial for α. In particular, f(x) is separable if and only if it isrelatively prime to its derivative: (f(x), Dx f(x)) = 1.

Corollary 13.34. Every irreducible polynomial over a field of characteristic 0 is separable. A polynomial over sucha field is separable if and only if it is the product of distinct irreducible polynomials.

Proposition 13.35. Let F be a field of characteristic p. Then for any a, b ∈ F ,

(a+ p)p = ap + bp, and (ab)p = apbp.

Put another way, the pth-power map defined by ϕ(a) = ap is an injective field homomorphism from F to F .

Corollary 13.36. Suppose that F is a finite field of characteristic p. Then every element of F is a pth power in F,i.e., F = Fp.

Proposition 13.37. Every irreducible polynomial over a finite field F is separable. A polynomial in F[x] is separableif and only if it is the product of distinct irreducible polynomials in F[x].

57

Proposition 13.38. Let p(x) be an irreducible polynomial over a field F of characteristic p. Then there is a uniqueinteger k ≥ 0 and a unique irreducible separable polynomial psep(x) ∈ F [x] such that

p(x) = psep(xpk).

Corollary 13.39. Every finite extension of a perfect field is separable. In particular, every finite extension of eitherQ or a finite field is separable.

13.6. Cyclotomic Polynomials and Extensions.

Lemma 13.40. The cyclotomic polynomial Φn(x) is a monic polynomial in Z[x] of degree ϕ(n).

Theorem 13.41. The cyclotomic polynomial Φn(x) is an irreducible monic polynomial in Z[x] of degree ϕ(n).

Corollary 13.42. The degree over Q of the cyclotomic field of nth roots of unity is ϕ(n):

[Q(ζn) : Q] = ϕ(n).

14. Galois Theory

14.1. Basic Definitions.

Proposition 14.1. Aut(K) is a group under composition and Aut(K/F ) is a subgroup.

Proposition 14.2. Let K/F be a field extension and let α ∈ K be algebraic over F . Then for any σ ∈ Aut(K/F ),σα is a root of the minimal polynomial for α over F i.e., Aut(K/F ) permutes the roots of irreducible polynomials.Equivalently, any polynomial with coefficients in F having α as a root also has σα as a root.

Proposition 14.3. Let H ≤ Aut(K) be a subgroup of the group of automorphisms of K. Then the collection F ofelements of K fixed by all the elements of H is a subfield of K.

Proposition 14.4. The association of groups to fields and fields to groups defined above is inclusion reversing,namely

(1) if F1 ⊆ F2 ⊆ K are two subfields of K then Aut(K/F2) ≤ Aut(K/F1), and(2) if H1 ≤ H2 ≤ Aut(K) are two subgroups of automorphisms with associated fixed fields F1 and F2, respec-

tively, then F2 ⊆ F1.

Proposition 14.5. Let E be the splitting field over F of the polynomial f(x) ∈ F [x]. Then

|Aut(E/F )| ≤ [E : F ]

with equality if f(x) is separable over F .

Corollary 14.6. If K is the splitting field over F of a separable polynomial f(x) then K/F is Galois.

14.2. The Fundamental Theorem of Galois Theory.

Theorem 14.7 (Linear Independence of Characters). If χ1, . . . , χn are distinct characters of G with values in Lthen they are linearly independent over L.

Corollary 14.8. If σ1, . . . , σn are distinct embeddings of a field K into a field L, then they are linearly independentas functions on K. In particular distinct automorphisms of a field K are linearly independent as functions on K.

Theorem 14.9. Let G = {σ1 = 1, σ2, . . . , σn} be a subgroup of automorphisms of a field K and let F be the fixedfield. Then

[K : F ] = n = |G| .Corollary 14.10. Let K/F be any finite extension. Then

|Aut(K/F )| ≤ [K : F ]

with equality if and only if F is the fixed field of Aut(K/F ). Put another way, K/F is Galois if and only if F isthe fixed field of Aut(K/F ).

Corollary 14.11. Let G be a finite subgroup of automorphisms of a field K and let F be the fixed field. Then everyautomorphism of K fixing F is contained in G, i.e., Aut(K/F ) = G, so that K/F is Galois, with Galois group G.

Corollary 14.12. If G1 6= G2 are distinct finite subgroups of automorphisms of a field K then their fixed fields arealso distinct.

Theorem 14.13. The extension K/F is Galois if and only if K is the splitting field of some separable polynomialover F . Furthermore, if this is the case then every irreducible polynomial with coeffiecients in F which has a rootin K is separable and has all its roots in K (so in particular K/F is a separable extension).

Theorem 14.14 (Fundamental Theorem of Galois Theory).


15. Commutative Rings and Algebraic Geometry

15.1. Noetherian Rings and Affine Algebraic Sets.

15.2. Radicals and Affine Varieties.

15.3. Integral Extensions and Hilbert’s Nullstellensatz.

15.4. Localization.

Theorem 15.36 (cf. theorem 7.15). let R be a commutative ring with 1 6= 0 and let D be a multiplicatively closedsubset of R containing 1. Then there is a commutative ring D−1R and a ring homomorphism π : R → D−1Rsatisfying the following universal property: for any homomorphism ψ : R → S of commutative rings that sends 1to 1 such that ψ(d) is a unit in S for every d ∈ D, there is a unique homomorphism Ψ: D−1R → S such thatΨ ◦ π = ψ.

Corollary 15.37. In the notation of theorem 15.36,

(1) kerπ = {r ∈ R | xr = 0 for some x ∈ D}; in particular, π : R → D−1R is an injetion if and only if Dcontains neither zero nor any zero divisors of R, and

(2) D−1R = 0 if and only if 0 ∈ D, hence if and only if D contains nilpotent elements.

15.4.18. Let R be any commutative ring with 1 and let f be any element of R. Let D be the multiplicative set{fn | n ≥ 0} of nonnegative powers of f in R. Define Rf = D−1R. Note that Rf = 0 if and only if f is nilpotent.If f is not nilpotent, then f becomes a unit in Rf . Prove that Rf ∼= R[x]/(fx − 1) where R[x] is the polynomialring in the variable x, if f is not nilpotent in R.

Proof. �

15.4.21. Suppose ϕ : R → S is a ring homomorphism with ϕ(1R) = 1S and D′ is a multiplicatively closed subsetof S. Let D = ϕ−1(D′). Prove D is a multiplicatively closed subset of R and the map ϕ′ : D−1R → D′−1S givenby ϕ′(r/d) = ϕ(r)/ϕ(d) is a ring homomorphism.

Proof. Let d1, d2 ∈ D with ϕ(d1d2) = ϕ(d1)ϕ(d2) so d1d2 ∈ ϕ−1(D′) because D′ is a multiplicative set.Let r1

d1= r2

d2. Then r1d2 = r2d1 so

ϕ′(r1d1

) = ϕ′(r1r2d2r2d1d2

)

=ϕ(r1r2d2)

ϕ(r2d1d2)

=ϕ(r2d1)ϕ(r2)

ϕ(r2d1)ϕ(d2)

= ϕ′(r2d2

)

59

So ϕ′ is well defined. Now let r1d1, r2d2 ∈ D

−1R. Then

ϕ′(r1d1

+r2d2

) = ϕ′(r1d2 + r2d1

d1d2)

=ϕ(r1d2 + r2d1)

ϕ(d1d2)

=ϕ(r1)

ϕ(d1)+ϕ(r2)

ϕ(d2)

= ϕ′(r1d1

) + ϕ′(r2d2

)

ϕ′(r1d1

r2d2

) = ϕ′(r1r2d1d2

)

=ϕ(r1r2)

ϕ(d1d2)

=ϕ(r1)

ϕ(d1)

ϕ(r2)

ϕ(d2)

= ϕ′(r1d1

)ϕ′(r2d2

).

�

15.4.22. Suppose P ⊆ Q are prime ideals in R and let RQ be the localization of R at Q. Prove that the localizationRP is isomorphic to the localization of RQ at the prime ideal P RQ (cf. exercise 15.4.21).

Proof. �

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