Abstract Algebra Lecture Notes

0 Preliminaries

• hypothesis - what you assume, given this; the “if” part of an implication

conclusion - what you want to prove; the “then” part of an implication

• Methods of Proof - see extra handout

• The words “there exists” and “for all” are extremely important. They must be usedand used in the correct order.

• axioms - statements which we assume/hold to be true, they are not proven. Theyform the foundation of mathematics. We prove theorems (propositions, lemmas, etc)from them.

• Let A and B be sets. A is a subset of B if every element in A is an element in B.

Two sets are equal, A=B, if A is a subset of B and B is a subset of A. (Two things toprove.)

• The cardinality of a set is the number of elements in the set. It is denoted by |A|. IfA has infinitely many elements, we say it is an infinite set.

• Every finite set with |A| = n has 2n subsets. The empty set is a subset of every set.

• We read A = {:} as ‘A is the set of all (type of object) such that (property the objectmust hold).

Eg. {n ∈ Z+ : n < 6} = {1, 2, 3, 4, 5}Eg. {2k : k ∈ Z} is the set of even integers

Eg. {students at UWEC : like math 425} = everyone here plus more

• union of sets - take all elements in all of the sets: A1 ∪ · · · ∪ An = {x : x ∈Ai for some i}intersection of sets - take the elements which are in every set: A1 ∩ · · · ∩ An = {x :x ∈ Ai for all i}difference of sets - A\B = {x : x ∈ A and x /∈ B}sum of sets - A+B = {a+ b : a ∈ A, b ∈ B}

• cartesian product - A1×A2×· · ·×An = {(a1, ..., an) : ai ∈ Ai∀i}. Ordered n-tuples,related to direct product.

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• function α : A→ B, takes a 7→ α(a). A is the domain, B is the codomain. Codomainis different than the range!

Range/image is the elements in the codomain that get mapped to. It is a subset ofthe codomain.

• A map is well-defined if every element in the domain gets mapped to exactly oneelement in the codomain. Prove that a ≡ b =⇒ α(a) = α(b). A map must bewell-defined to be a function.

A function is injective (one-to-one or 1-1) if no two elements in the domain have thesame image in the range. Prove that α(a) = α(b) =⇒ a = b.

A function is surjective (onto) if every element in the codomain get mapped to; ie.the range=codomain. Prove that ∀b ∈ B, ∃a ∈ A such that α(a) = b.

A function is bijective if it is injective and surjective.

Note 1: you will need to prove these things frequently, practice a lot if you are unfa-miliar with how to prove these. Make sure you have the definitions straight in yourhead.

Note 2: notice the ‘opposite’ nature of well-defined and 1-1; their proofs will often (notalways) be the reverse of each other.

Note 3: well-defined will be especially important when we learn equivalence rela-tions/classes and cosets.

• Eg. f : R→ R×R defined by f(xy) = (x, y). NOT well-defined since f(6) = f(6∗1) =(6, 1) and f(6) = f(3 ∗ 2) = (3, 2)

• Let α : A→ B be a function where A and B are nonempty finite sets with |A| = |B|.Then α is injective if and only if it is surjective.

• read compositions of functions from right to left

• The identity map/identity function is the function which takes every element toitself: 1A : A → A defined by 1A(a) = a∀a ∈ A. Also denoted by id. This is abijection. Remember this function, it is important.

• A function α : A → B has an inverse if and only if it is bijective. αα−1 = 1B andα−1α = 1A. Inverses are unique.

• A (binary) relation on a set A compares two elements in the set.

• An equivalence relation on a set A satisfies 1) reflexive: a ≡ a∀a ∈ A 2) symmetric:if a ≡ b, then b ≡ a 3) transitive: if a ≡ b and b ≡ c, then a ≡ c. Notice the propertiesare r-s-t.

Examples: =, parallel lines, similar polygons, m-n is divisible by a fixed number d(m,n,d integers)

Non-examples: ≤, taller than

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• equivalence class of a is [a] = {x ∈ A : x ≡ a}; it is the set of all equivalent elementsin a set.

Example: m ≡ n iff m-n is even. [0] = all evens, [1] = all odds. These are the onlytwo equivalence classes of the integers under this equivalence relation.

Example: = with real numbers. Every number is in its own equivalence class.

• Theorem - Let ≡ be an equivalence relation on A. 1) a ∈ [a] 2) if a ∈ [b], then [a] = [b]3)[a] = [b] iff a ≡ b 4) if [a] 6= [b], then [a] ∩ [b] = ∅.This theorem says that ≡ partitions the set A.

• Every equivalence relation on A gives a partition of A.

Every partition of A gives an equivalence relation on A. Moreover, every equivalenceresults from a partition of the set.

• The set of all equivalence classes is called the quotient set, A≡.

The function A→ A≡ defined by a 7→ [a] is the quotient map (natural mapping).It is surjective (NOT injective in general).

• Maps involving equivalence classes must be checked for well-defined - no matter whatrepresentative I use, I get the same answer.

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1 Integers and Permutations

1.1 Induction

Theorem 1.1 (Principle of Mathematical Induction). Let pn be a statement for each integern ≥ j (typically j = 1). Suppose the following are satisfied: 1) pj is true 2) k ≥ j andpk =⇒ pk+1. Then pn is true for all n ≥ j.

We use induction when we want to prove a statement true for the natural numbers. Touse induction, we prove the base case, assume the statement is true for n = k (write out thewhole statement), and then prove it is true for n = k+ 1. The proof will use the phrase “byinduction hypothesis”.

Example 1.1. Prove 1 + 5 + 9 + · · ·+ (4n− 3) = n(2n− 1) for all n ≥ 1.

Theorem 1.2 (Strong Induction). - Let pn be a statement for each integer n ≥ j. Supposethe following are satisfied: 1) pj is true 2) if k ≥ j and pj, pj+1, ..., pk =⇒ pk+1. Then pn istrue for all n ≥ j.

The difference between strong induction and induction is that strong induction assumesthat the statement is true for n = 1 through n = k, whereas induction only assumes it is truefor n = k. It has stronger assumptions, so is a weaker statement (or equal); i.e. inductionimplies strong induction.

Example 1.2. Prove that for all n ≥ 2, n is a product of primes.

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Well-ordering axiom - every nonempty subset of the natural numbers has a leastelement.

Note: the well-ordering axiom (or principle) is equivalent to the Axiom of Choice as wellas many other statements in mathematics. Not all mathematicians accept this axiom, butwe do in this class.

Theorem 1.3. The well-ordering axiom, induction, and strong induction are equivalentstatements (all true or all false; iff between each one).

1.2 Divisors and Prime Factorization

Goal: Prove every natural number has a unique factorization as a product of primes.Note: this is an existence and uniqueness theorem. We already proved existence.The theorems and techniques in this section will be used again in a more abstract setting.

So, make sure you understand them for the integers before we try to generalize them toabstract sets.

Theorem 1.4 (Division Algorithm). Let n ∈ Z and d ∈ N. There exist uniquely determinedintegers q and r such that n = qd+ r, 0 ≤ r < d.

Proof. Existence:

Uniqueness:

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Definition (divides). Let n, d be integers. d divides n if there exists q ∈ Z such that n = qd.In other words, the remainder is zero. Write d|n. If d does not divide n, write d - n.

Theorem 1.5. Let m,n, d ∈ Z.1) n|n ∀n2) If d|m and m|n, then3) If d|n and n|d, then4) If d|n and d|m, then d|(xn+ ym) ∀x, y

Proof. Prove 1,2 on your own.

Definition (greatest common divisor). Let m,n ∈ Z where not both m,n are zero, andd ∈ N. Then d is the greatest common divisor of m and n (d=gcd(m,n)) if:

1) d|m and d|n - common divisor2) If k|m and k|n, then k|d. - greatest

Proposition 1.6. If m = qn+ r, then gcd(m,n)=gcd(n,r).

Proof. .

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Theorem 1.7 (Euclidean Algorithm). used to find the greatest common divisor. Idea - keepdividing using the previous remainder as next divisor until you get a remainder of 0. Lastnonzero remainder is the answer.

Proof. .

Proposition 1.8. For any nonzero integers m,n, there exist integers x,y such that gcd(m,n) =xm+ yn. The gcd is the smallest positive integer of this form.

Proof. .

Example 1.3. Find the gcd(2520,154). Write the gcd as a linear combination of 2520 and154.

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Definition (relatively prime). Two integers are relatively prime if their gcd is 1. gcd(m,n)=1.

Theorem 1.9. m and n are relatively prime if and only if 1 = xm + yn for some integersx, y.

Proof. .

Theorem 1.10. Let m, n be relatively prime. Then 1) if m|k and n|k, mn|k. Also, 2) ifm|kn for some k, then m|k.

Proof. See page 34 in your book. Know the proof of (2).

Definition (prime number). An integer p ≥ 2 such that its only divisors are 1 and p.

Proposition 1.11. Let p be prime and n be a nonzero integer. Then either p|n or gcd(p, n) =1.

Proof. Homework

Theorem 1.12 (Euclid’s Lemma). Let p be prime and m,n,mi ∈ Z.1) If p|mn, then p|m or p|n.2) If p|(m1...mr), then p|mi for some i.

Proof. 1)

2) use induction - know this proof.

Theorem 1.13 (Fundamental Theorem of Arithmetic). Every integer n ≥ 2 can be factoreduniquely into a product of primes.

Proof. Existence: already did.Uniqueness: uniqueness means up to the order of the factors.

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Theorem 1.14 (Euclid’s Theorem). There are infinitely many primes.

Proof. .

The gcd(m,n) is the product of the smallest prime powers occurring in m and n (incl 0).The lcm(m,n) is the product of the largest prime powers occurring in m and n.

Example 1.4. Find gcd(252, 15) and lcm(252, 15).

Proposition 1.15. gcd(m,n)lcm(m,n)=

1.3 Integers Modulo n

Definition (congruent modulo n). Let a, b, n ∈ Z with n ≥ 2. a and b are congruent modulon if n|(a− b).

• a(mod n) is the remainder of dividing a by n. So, if a = qn+ r, then a ≡ r(modn).

Theorem 1.16. Congruence modulo n is an equivalence relation.

Question: How many equivalence classes are there? What are they?

Definition (integers modulo n). The quotient set mod n (set of equivalence classes) is calledthe integers modulo n and are denoted by Zn. Zn = {[0], [1], ..., [n− 1]}.

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• Recall that equivalence relations partition the set, so modulo n partitions the integers.They are partitioned by what their remainder is after being divided by n. Thus, [0]contains those numbers divisible by n.

• We think of modular arithmetic as “clock” arithmetic.

Example 1.5. Practice calculations.

Question: What is [a] + [b]? [a][b]?

Proposition 1.17. [a]n+[b]n = [a+b]n and [a]n[b]n = [ab]n are well-defined. In other words,if a ≡ a′ (mod n) and b ≡ b′ (mod n), then a+ b ≡ a′ + b′ (mod n) and ab ≡ a′b′ (mod n).

Proof. We are checking that addition and multiplication do not depend on the representativethat we use.

Example 1.6. Here is a case where it would not be well-defined.

What are properties of addition and multiplication of the integers which also hold in Zn?Which are some properties which are different?

Theorem 1.18. The following hold in Zn:

• addition and multiplication are commutative

• addition and multiplication are associative

• there is an additive identity ([0]) and multiplicative identity ([1])

• every element has an additive inverse ([n-a])

• distributive

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Example 1.7. Make a multiplication table for Z6. What do you notice?

Properties which are different: do not have the cancelation law in general, have zerodivisors, some elements other than [1] have a multiplicative inverse.

Definition (zero divisor). A nonzero element [a] is a zero divisor if there exists a nonzeroelement [b] such that [a][b] = [0]. (In general, ab=additive identity.)

Question: Which elements are invertible? Which are zero divisors?

Theorem 1.19. Let a and n be integers with n ≥ 2. Then [a]n is invertible if and only if aand n are relatively prime.

Proof. .

If [a] 6= [0] and a and n are not relatively prime, then [a] is a zero divisor.

Example 1.8. What is 4123 (mod 5)?

Example 1.9. Prove that for all a ∈ Z, a7 ≡ a (mod 7).

Example 1.10. Prove that a positive integer is divisible by 9 if and only if the sum of itsdigits is divisible by 9.

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Theorem 1.20 (Chinese Remainder Theorem). Let m,n be relatively prime. Let s, t ∈ Z.Then there exists a solution x ∈ Z to the system x ≡ s (mod m) and x ≡ t (mod n).

Proof. Use the Euclidean Algorithm. There exist p,q such that 1 = mp + nq. Then x =mpt+ nqs.

Question: Which elements are invertible in Zp for p prime? What does this lead you tothink about the role Zp will play for us?

Theorem 1.21 (Fermat’s Theorem). If p is prime, then ap ≡ a (mod p) for all integers a.Furthermore, ap−1 ≡ 1 (mod p) if a is relatively prime to p.

Proof. .

Note - this is a very important theorem in Number Theory.

1.4 Permutations

Questions:

• Think about the cup game. How do you keep track of where the ball is hidden?

• How many ways are there to rearrange three identical objects?

• What are the symmetries of a triangle? How can you label what each does to thevertices?

Definition (permutation). Let Xn = {1, ..., n}. A permutation is a bijective function σ :Xn → Xn.

It is a symmetry of a configuration of identical objects.

Definition (Sn). The set of all permutations of Xn is Sn, called the symmetric group ofdegree n.

Example 1.11. What is S3?

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Notation: matrix (2-line) and cycle notation

Proposition 1.22. |Sn| = n!

Example 1.12. Composition of permutations:

Questions:

• Does order matter?

• What is the identity permutation?

• What is the inverse of a given permutation?

Note: Because permutations are invertible στ = σµ ⇒ τ = µ. However, because Sn isnot commutative for n ≥ 3, στ = µσ does NOT imply τ = µ!

Properties: Notice similarities and differences with Zn.

1. The composition of permutations is a permutation.

2. There is an identity permutation.

3. Every permutation has an inverse.

4. Composition of permutations is associative (note: composition of functions is alwaysassociative).

Definition (disjoint cycles). Two cycles are disjoint if each leaves fixed the numbers movedby the other. In other words, no number appears in more than 1 cycle.

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NB: We will write all permutations as products of disjoint cycles. The benefit is thatdisjoint cycles commute, and it is in simplified form.

Definition (length). The length of a cycle (a1 a2 ... ar) is r.

Definition (order). A cycle σ has order k if σk = (1) and σl 6= (1) for all l < k.

Example 1.13. What is the order of (1 2 3 4)?What is the order of (1 2)(3 4 5)(6 7 8 9)?

Theorem 1.23 (Cycle Decomposition Theorem). Every permutation σ ∈ Sn can be writtenuniquely (up to order) as a product of disjoint cycles.

Proof. Use strong induction on the cardinality of the set.

Definition (transposition). A cycle of length 2.

Theorem 1.24. Every cycle of length r > 1 is a product of r − 1 transpositions.

Proof. Homework

NB: This is NOT a unique product (existence, but not uniqueness). Eg (1)=(12)(12) or(13)(13) or... in S3.

However,

Theorem 1.25. Any factorization of a permutation contains either an even number oftranspositions or an odd number (not both). In other words, if σ = τ1...τn = µ1 · · ·µm,then either m and n are both even or both odd. This is called the parity of the permutation.

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Definition (alternating group). The set of all even permutations in Sn is denoted An andis called the alternating group. |An| = n!

2. Note the identity permutation is in An.

Example 1.14. Write each permutation as a product of disjoint cycles. And, write each asa product of transpositions.

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2 Groups

2.1 Binary Operations

Definition. A binary operation assigns to each ordered pair in a set an element a ∗ b. Aset M is closed under a binary operation * if for all (a, b) ∈M , a ∗ b ∈M .

Theorem 2.1. If a binary operation on a set M has an identity, it is unique.

Proof. Assume e, e′ are identities.

Definition (monoid). A set M with binary operation * is a monoid if it is closed, associative,and has an identity.

Example 2.1. Examples:

Non-examples

Definition (Cayley table). Table showing the result of the binary operations (like a multi-plication table) - vertical times horizontal.

Example 2.2. Make the Cayley table for Z3.

Example 2.3. Fill in a 4 by 4 Cayley table with random products. Have a friend decidewhether it is a Cayley table for a monoid.

Theorem 2.2. Let a, b ∈M , monoid. Let m,n ∈ Z

1. aman = am+n.

2. (am)n = amn

3. If ab = ba, then (ab)n = anbn.

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Definition (unit). A element which has an inverse. An element a ∈ M such that thereexists b ∈M where ab = e = ba.

Theorem 2.3. Inverses are unique: If a has an inverse, that inverse is unique.

Proof. Assume b, b′ are inverses of a.

Example 2.4. What elements are units in the following monoids?

Theorem 2.4. Let M be a monoid.

1. the identity is a unit: e−1 = e

2. If a is a unit, (a−1)−1 = a

3. If a, b are units, (ab) = b−1a−1

4. If a1, ..., an are units, (a1 · · · an)−1 = a−1n · · · a−11

5. If a is a unit, (an)−1 = (a−1)n

Proof. Had same results for invertible matrices.

2.2 Groups

Definition (Group). A nonempty set G with binary operation * is a group if

1. G is closed

2. * is associative

3. there is an identity

4. every element has an inverse in G

In other words, a group is a monoid in which every element has an inverse.If * is commutative, we call G an abelian group.

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Definition. If G is finite, |G|, the number of elements in G, is called the order of the group.If G has infinitely many elements, we say G is an infinite group.

NB We also have a definition of an order of an element; they are related, but not thesame thing!

Example 2.5. Examples of groups:

Non-examples of groups:

Definition (direct product). Let G1, ..., Gn be groups. The direct product G1×· · ·×Gn is theset of all ordered n-tuples with multiplication defined by (g1, ..., gn)(g′1, ..., g

′n) = (g1g

′1, ..., gng

′n).

Theorem 2.5. The direct product of groups is a group.

Proof. Closed:Identity:Inverses:Associativity:

Because groups are monoids, all of the theorems about monoids are true for groups. Namely,we have the same laws of exponents and inverse properties.

Theorem 2.6 (Cancellation Laws). Let f,g,h be in a group G. Then gh = gf ⇒ h = f andhg = fg ⇒ h = f .

Proposition 2.7. If G is a finite group and g ∈ G, then gn = e for some n.

Proof. Keep taking powers, eventually have to repeat an element.

Proposition 2.8. Every row (and column) of the Cayley table of a group G contains everyelement of G exactly once.

Proof. The equations gx = h and xg = h have a unique solution in G.

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Example 2.6. How many different Cayley tables are there for groups of size 1, 2, 3, and 4?

Definition (cyclic group). The cyclic group generated by g is the set of all powers ofg: Cn = {e, g, g2, ..., gn−1}. G is a cyclic group if every element in G is a power of anelement g ∈ G, called the generator of G.

We can have an infinite cyclic group: {e, g, g2, ...} where all elements are distinct.

For a finite cyclic group, exponents are reduced modulo n.

Proposition 2.9. The cyclic group generated by g: Cn = {e, g, g2, ..., gn−1} is a group.

Proof. .

Example 2.7. Examples of cyclic groups:

Non-examples

2.3 Subgroups

Definition (subgroup). H is a subgroup of a group G if H is a subset of G and H is agroup with the same operation as G.

Theorem 2.10 (subgroup test). A nonempty subset H of a group G is a subgroup of Giff 1) ∀g, h ∈ H, gh ∈ H 2) ∀h ∈ H, h−1 ∈ H.

These two can be combined to say that a nonempty subset H is a subgroup iff ∀g, h ∈H, gh−1 ∈ H.

Proof. Need to check associativity and identity:

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Every group has at least two subgroups: the identity (trivial subgroup) and the whole groupG.

Example 2.8. Which of the following are subgroups? which are not?

• An < Sn

• nZ < Z

• {0, 2} < Z4

• Z2 < Z4

• Sk < Sn when k ≤ n.

• GLn(R) < Mn(R)

Other examples?

Proposition 2.11. If H is a finite set, then we only need to check nonempty and closed.

Proof. Follows from subgroup test and proposition from previous section that ∀h ∈ H, hn = efor some n.

Proposition 2.12. The intersection of subgroups is a subgroup.

Proof. .

NB: The union of subgroups is usually NOT a subgroup.

The subgroups of G are partially ordered by set inclusion. The lattice diagram for a groupG shows all the subgroups of G and their containments.

Example 2.9. Draw the subgroup lattices for S3, Z6

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Definition (center of G). The center of G is Z(G) = {z ∈ G : zg = gz∀g ∈ G}.

Theorem 2.13. If G is a group, Z(G) is an abelian subgroup of G.

Proof.

Example 2.10. What is Z(S3)? Z(Zn)?

Proposition 2.14. Let H be a subgroup of a group G and let g ∈ G. Then gHg−1 ={ghg−1 : h ∈ H} is a subgroup of G. These subgroups are called the conjugates of H in G.

Proof. e = geg−1 Check closure:

Note: These subgroups will play an important role later this chapter.

2.4 Cyclic Groups

Definition. For any group G and subset S ⊆ G, the smallest subgroup of G that contains S iscalled the subgroup generated by S, < S >. In the case where |S| = 1, we say < S >=< g >is a cyclic subgroup.

If there exists a set S such that G =< S >, we say that G is generated by S and theelements of S are called the generators of G. G is a cyclic group if there exists a singleelement which generates G.

Proposition 2.15. Let S be a nonempty subset of a group G. Then < S >= {xk11 · · · xkmm :xi ∈ S, ki ∈ Z,m ≥ 1} is the smallest subgroup of G containing S.

Proof. Homework problem.

Example 2.11. • Z =< 1 >=< −1 >

• < d >= dZ = {nd : n ∈ Z}

• rotations of a regular polygon are generated by the smallest degree rotation

• S3 =< (123), (12) >

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• Zn = ?

• Sn = ?

Proposition 2.16. Let < g > be the cyclic group generated by g.

1. If all gk are distinct, < g > is an infinite group and we say g has infinite order.

(a) gk = e iff k = 0

(b) gk = gm iff k = m

(c) G = {..., g−1, e, g, g2, ...}

2. If gk = gm for some k 6= m, then < g > is a finite group. Let n be the smallest positiveinteger such that gn = e. n is called the order of g.

(a) gk = e iff n|k(b) gk = gm iff k ≡ m (mod n)

(c) G = {e, g, g2, ..., gn−1}.

Proof. Idea: 2) Use Euclidean algorithm.

The order of an element g ∈ G, |g|, equals the size of the cyclic subgroup generated by g,| < g > |.

Example 2.12. Calculate the orders of the following elements:

• [6]15

• [6]16

• [4] ∈ Z∗15

• 90 rotation of the square

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• (1 3 5 9 8)(7 4)(2 6 10)

Question: Find the orders of all elements in Z10.What conjecture do you have for possible orders of elements in Zn?

Proposition 2.17. Every cyclic group is abelian.

Proof. Note: the converse does not hold.Use laws of exponents:

Theorem 2.18. Every subgroup of a cyclic group is cyclic.

Proof. .

Theorem 2.19. Let G =< g > and |g| = n. Then G =< gk > if and only if gcd(k,n)=1.

Proof. A group can have more than one generator.

Theorem 2.20 (Fundamental Theorem of Finite Cyclic Groups). Let G =< g > be a cyclicgroup of order n.

1. If H is a subgroup of G, then H =< gd > for some d|n. Hence |H| divides n.2. If k|n, then < gn/k > is the unique subgroup of G of order k.

Note: this says that the order of a subgroup divides the order of a cyclic group (true forany finite group) AND there is a subgroup of every order dividing the order of the group.

Proof. .

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Example 2.13. Draw the subgroup lattice for Z10.Draw the subgroup lattice for C10.Draw the subgroup lattice for C18.

2.5 Homomorphisms and Isomorphisms

The quote on page 95 is a very good summary of this section: “Mathematicians do not dealin objects, but in relations among objects; they are free to replace some objects by others solong as the relations remain unchanged. Content to them is irrelevant: they are interestedin form only.” Henri Poincare

Definition (homomorphism). A mapping from group G to group H which preserves theoperation. A function α : G→ H is a homomorphism if α(a ∗G b) = α(a) ∗H α(b).

Example 2.14. Standard examples of group homomorphisms.

• Identity map: 1G : G→ G defined by 1G(g) = g.

• Trivial map: α : G→ H defined by α(g) = eH for all g.

• Projection map: π1 : G1 ×G2 → G1 defined by π1(g1, g2) = g1.

• Injection map: ι1 : G1 → G1 ×G2 defined by ι1(g1) = (g1, e).

Example 2.15. Which of the following maps are homomorphisms?

• f : Z→ Z defined by f(x) = ax for a fixed integer a.

• h : Z6 → Z4 defined by h([a]6) = [a]4.

• Let a ∈ G. g : Z→ 〈a〉 defined by g(k) = ak.

• The determinant of invertible matrices.

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• G→ G, fix natural number n, g 7→ gn.

Proposition 2.21. The composition of homomorphisms is a homomorphism.

Proof. Let α : G→ H and β : H → K be group homomorphisms.

Theorem 2.22. Let α : G→ H be a homomorphism. Then1) α(eG) = eH2) α(g−1) = (α(g))−1

3) α(gk) = (α(g))k

Proof. .

Corollary 2.23. Let α : G→ H be a group homomorphism. If g ∈ G has finite order, thenα(g) has finite order which divides the order of g.

Proof. If |g| = n, then α(g)n = α(gn) = α(eG) = eH .

Theorem 2.24. Let G = 〈X〉 and α : G → H and β : G → H be homomorphisms. Thenα = β iff α(x) = β(x)∀x ∈ X. In other words, a homomorphism is completely determinedby what it does to the generators of a group.

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Proof. .

Example 2.16. Find all the homomorphisms from S3 → Z6.

Definition (isomorphism). Bijective homomorphism. A bijective function f : G→ H suchthat f(ab) = f(a)f(b) is an isomorphism. We say G is isomorphic to H, G ∼= H.

Example 2.17. Which of the homomorphisms above are isomorphisms?

Example 2.18. Determine whether the following are isomorphic.

• Z, nZ

• R, R+

• Q, Q∗

• Z9, Z3 × Z3

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Proposition 2.25. Let G = 〈g〉 be a cyclic group.1. If |G| = n <∞, then G ∼= Zn2. If |G| =∞, then G ∼= Z

Proof. .

Theorem 2.26. Isomorphic groups have the same structural properties. If G ∼= H and Ghas property (fill in the blank), then H does too.

Structural properties (not a complete list): order n, finite, abelian, cyclic, have samenumber of elements of every order, generators mapped to generators, ...

One very important result of this theorem is that you can use these to prove two groupsare NOT isomorphic. For example, if one group has an element of order 2 and the otherdoes not, they are not isomorphic.

Theorem 2.27. 1. The identity map is an isomorphism.2. The inverse of an isomorphism is an isomorphism.3. The composition of isomorphisms is an isomorphism.

Proof. 1. obvious 3. proved composition of homomorphisms is a homomorphism; and thecomposition of invertible functions is invertible.

2.

Corollary 2.28. ∼= is an equivalence relation for groups.

Definition (automorphism). An isomorphism from G to G.

Theorem 2.29. The set of all automorphisms of a group G is a group, called the auto-morphism group of G.

Example 2.19. Some common automorphisms are: identity, inverse mapping (g 7→ g−1),and inner automorphism (g 7→ aga−1 for a fixed a)

27

Theorem 2.30. Let σ : G→ H be an isomorphism. Then |σ(g)| = |g| for all g ∈ G.

This is very helpful in trying to find an isomorphism because all the orders have to bethe same and they have to line up.

Theorem 2.31 (Cayley’s Theorem). Every group of order n is isomorphic to a subgroup ofSn.

Proof. .

2.6 Cosets and Lagrange’s Theorem

Definition (coset). Let H < G and let g ∈ G. Then Ha = {ha : h ∈ H} is the right cosetof H generated by a and aH = {ah : h ∈ H} is the left coset of H generated by a.

Ha 6= aH in general. It is a special case when it does, which we will discuss in 2.8.Cosets partition the group (and so is an equivalence relation).

Theorem 2.32. Let H < G and let a, b ∈ G. (same for left or right coset)1. H = eH2. aH = H iff a ∈ H3. aH = bH iff b−1a ∈ H iff a−1b ∈ H4. If a ∈ bH, then aH = bH5. Either aH = bH or aH ∩ bH = ∅6. The set of cosets form a partition of G.

Proof. Notice this is the same as an equivalence relation partitions the set.

Example 2.20. Find all left and right cosets of the following groups and subgroups.

28

1. G = K4 = {e, a, b, ab}, H = {e, a}

2. G = Z8, H =< [2] >, K =< [4] >

3. G = Z, H = 8Z

4. G = S4, H =< (1234) >

Lemma 2.33. Let H be a finite subgroup of a group G. Then |H| = |aH| = |Ha| for alla ∈ G.

Proof. Define σ : H → aH by h 7→ ah. This is a bijection: σ−1 maps ah 7→ a−1ah.

Definition. The number of cosets of H in G is called the index of H in G, [G : H].

Theorem 2.34 (Lagrange’s Theorem). Let H be a subgroup of a group G. Then |H| divides|G|.

Proof. G is a union of disjoint cosets, all having the same size. And, H is one of thosecosets.

This theorem is important because it gives us the possible sizes of subgroups. Note thatunlike for cyclic groups, there does not need to be a subgroup of every size dividing the orderof the group. The theorem also gives us a lot of important consequences.

Example 2.21. Find all subgroups of A4

Corollary 2.35. If g ∈ G, a finite group, then |g| divides |G|.

Corollary 2.36. If G is a group and |G| = n, then gn = e for all g.

29

Proof. .

Corollary 2.37. If p is prime, then every group G of order p is cyclic. G =< g > for everynon-identity element in G. The only subgroups of G are {e} and G.

Proof. .

Corollary 2.38. Let H and K be finite subgroups of a group G. If gcd(|H|, |K|) = 1, thenH ∩K = {e}.

Proof. .

Corollary 2.39. If H is a subgroup of a finite group G, then [G : H] = |G||H| .

Recall Z∗n = {k : 1 ≤ k < n, gcd(k,n) = 1}.

Definition (Euler φ-function). For n ≥ 2, the Euler φ-function, φ(n), is the number ofnatural numbers k, 1 ≤ k < n, such that k is relatively prime to n. Thus, φ(n) = |Z∗n|.

Define φ(1) = 1 and note that φ(p) = p− 1 for a prime p.

Theorem 2.40 (Euler’s Theorem). If a and n ≥ 2 are relatively prime integers, then aφ(n) ≡1 (mod n).

Proof. Use Lagrange, Corollary 2.

We can use this theorem to give an easier proof of Fermat’s Theorem: ap ≡ a (mod p).

30

2.7 Symmetry

Note: this book uses motion and symmetry as having two different meanings. Most peoplewhen they are talking about a symmetry of a geometric figure mean a motion of a figure asthis book is using those terms.

The symmetries of geometric figures are important examples of groups. They are sub-groups of Sn. A motion of a geometric figure is a permutation of its vertices that can berealized by a rigid motion; i.e. reflection or rotation.

The set of motions of a figure with n vertices is a subgroup of Sn.

Example 2.22. What are the symmetries of an equilateral triangle? square? Write themas permutations.

The symmetries of a regular n-gon form a very important set of groups.

Definition (dihedral group). For n ≥ 2, the dihedral group Dn is the group of order 2ngenerated by a, b with relations an = e, b2 = e, and aba = b.

Example 2.23. Show that D3 and D4 are the symmetries of the equilateral triangle andsquare as found above. (Notice D3

∼= S3.)

Theorem 2.41. The group of symmetries of a regular n-gon is isomorphic Dn with a therotation of 360/n and b being any reflection.

Think about the group of motions for other figures.

Example 2.24. The group of motions of the tetrahedron is A4.

31

2.8 Normal Subgroups

Definition (normal subgroup). A subgroup H of G is a normal subgroup of G if gH = Hgfor all g ∈ G. Write H �G.

All the right cosets equal the left cosets. Note that it is not true that gh = hg; in otherwords, it does not have to be abelian. However, gh1 = h2g for some h2 ∈ H. Also, note thatall subgroups of abelian groups are normal.

Example 2.25. • {e}�G

• G�G

• Theorem: If G is abelian, H �G for all subgroups H.

• Theorem: Z(G) �G. Also, any subgroup of the center is normal in G.

• 〈(123)〉� S3, 〈(12)〉 6 S3

Proposition 2.42. If H is a subgroup of index 2 in G, then H is normal in G.

Proof. There are only 2 cosets, H and everything else.

Theorem 2.43 (Normality Test). The following are equivalent for H < G:

1. H �G

2. gHg−1 ⊆ H∀g ∈ G

3. gHg−1 = H∀g ∈ G

Proof. .

Corollary 2.44. If G = 〈X〉, a subgroup H is normal in G iff xHx−1 ⊆ H∀x ∈ X.

Proof. It is saying that it is enough to check the generators of the group since every elementcan be written as a product of the generators.

32

Corollary 2.45. If H < G and if G has no other subgroups isomorphic to H, then H �G.

Proof. gHg−1 is a subgroup of G which is isomorphic to H for all g.

Theorem 2.46. If H,K �G and H ∩K = {e}, then hk = kh for all h ∈ H, k ∈ K.

Proof. .

Theorem 2.47. Let G be a group. If H,K �G, H ∩K = {e}, and HK = {hk : h ∈ H, k ∈K} = G, then G ∼= H ×K (direct product).

In the special case of G is a finite group, we need H,K�G, H ∩K = {e}, and |H||K| =|G|. Then G ∼= H ×K.

Proof. Breaking a group up into a direct product is a great way to better understand agroup’s structure.

For all g ∈ G, g can be written uniquely as hk for some h ∈ H, k ∈ K.

Example 2.26. Let m, n be relatively prime. Then Zmn ∼= Zm × Zn.We will use this to classify all finitely generated abelian groups in Chap 7.

What if I have one normal subgroup and the other conditions above? i.e. H � G,K 6 G,G = HK, H ∩K = {e}. Then every g can still be written uniquely as hk, however, hk 6= khfor all h, k.

33

So, g1 = h1k1, g2 = h2k2, what is g1g2? g1g2 = h1k1h2k2 = (h1k1h2k−11 )(k1k2); because H

is normal, k1h2k−11 is in H, so we have written g1g2 has a product hk. This can be generalized

as follows:

Theorem 2.48. Let H, K be groups and let θ : K → autH be a homomorphism definedby θ(k) = σk. If G = H ×K is the cartesian product, define (h, k)(h′, k′) = (hσk(h

′), kk′).Then 1. G is a group with identity (eH , eK) and inverses (h, k)−1 = (σk−1(h−1), k−1) 2.H1 = H × {e} and K1 = {e} × K are subgroups of G 3. G = H1K1, H1 � G, andH1 ∩K1 = {e}.

Definition (semi-direct product). Let G be a group. G is a semi-direct product if and onlyif it has subgroups H and K with G = HK,H �G,H ∩K = {e}.

Then G ∼= H ×θ K for some homomorphism θ : K → aut H.

Note that a direct product is a semi-direct product where θ is the trivial homomorphismθ(k) = 1H .

Example 2.27. Dn∼=< a > ×θ < b >.

Definition (simple group). A group G 6= {e} is a simple group iff its only normal subgroupsare {e} and G.

It is really hard to classify finite simple groups. Read the last paragraph on p 125.

Example 2.28. An abelian group is simple iff it is cyclic of prime order.

Example 2.29. For n ≥ 5, An is a simple group.See proof in book.

34

2.9 Factor Groups

Theorem 2.49. K is a normal subgroup of G if and only if aKbK = abK is a well-definedmultiplication of cosets.

Proof. .

Theorem 2.50. Let N �G of a group G and denote the set of cosets by G/N . Then G/Nis a group, called the factor group or the quotient group. The order of the group is [G : N ].

Proof. The identity is N , inverse of aN is a−1N , closed because operation is well defined:aNbN = abN , which is a coset, associativity follows from associativity in G.

Theorem 2.51. Let N �G. The function π : G→ G/N given by π(g) = Ng is a surjectivehomomorphism, called the quotient map.

Proof. .

Proposition 2.52. Let N �G.1. If G is abelian, then G/N is abelian.2. If G =< a >, then G/N =< Na >.

Proof. .

Example 2.30. Find the quotient group and make a Cayley table. To what group is thequotient group isomorphic?

• Z16/ < [4] >

35

• K = {(1), (12)(34), (13)(24), (14)(23)}, A4/K and S4/K

• Dn/Z(Dn)

• Z/nZ

• R/Z

When is the quotient group abelian? aNbN = bNaN ⇒ abN = baN ⇒ a−1b−1abN =N ⇒ a−1b−1ab ∈ N (or aba−1b−1 ∈ N). This form is important (for eg, in Lie Alg):

Definition (commutator). [a, b] = aba−1b−1 is the commutator of a and b.

Definition (commutator subgroup/derived subgroup). The commutator subgroup of a groupG is G′ = {all finite products of commutators}.

Theorem 2.53. Let H < G. 1. G′ is a normal subgroup of G and G/G′ is abelian.2. G′ ⊆ H iff H �G and G/H is abelian.

Proof. .

Example 2.31. Find

• Q′

• D′6

• S ′n

36

2.10 Isomorphism Theorems

Definition. Let α : G → H be a group homomorphism. kerα = {g ∈ G : α(g) = eH}.imα = {α(g) : g ∈ G}

Theorem 2.54. Let α : G→ H be a group homomorphism. Then kerα is a normal subgroupof G and imα is a subgroup of H.

Proof. Note that kernel and image are important concepts for many functions (eg vectorspaces in Linear).

Theorem 2.55. If N �G, then N = ker π where π is the quotient map G→ G/N .

These two theorems say that the kernel is a normal subgroup, and every normal subgroupis a kernel of some map (namely the quotient map).

Proof. N is the identity of G/N .

Example 2.32. Find the kernel of the following maps, or find a map for which the subgroupgiven is the kernel.

• φ : GLn(R)→ R∗ given by A 7→ det(A).

• ε : Sn → {1,−1} given by σ 7→ sgn (σ)

• K = {(1), (12)(34), (13)(24), (14)(23)}� A4

• G�G

• {e}�G

37

Theorem 2.56 (injective test). If φ : G→ H is a homomorphism, then φ is injective if andonly if ker φ = {e}.

Proof. If φ is injective, then the kernel is clearly trivial since the identity always maps tothe identity.

Theorem 2.57 (1st Isomorphism Theorem). Let φ : G → H is a group homomorphism.Then φ(G) ∼= G/ker φ. Let ker φ=N. φ induces the isomorphism φ̃ : G/N → φ(G) given byφ̃(Ng) = φ(g). Thus, φ = φ̃π where π is the quotient map, and φ̃ is the unique homomor-phism satisfying this equation.

Proof. .

Example 2.33. Apply this theorem to the examples above and the quotient groups in theprevious section.

38

Proposition 2.58. Let N �G and K�H. Then N ×K�G×H and (G×H)/(N ×K) ∼=(G/N)× (H/K).

Proof. Define φ(g, h) = (Ng,Kh)

Proposition 2.59. G/Z(G) ∼= innG.

Proof. Define φ(a) = σa.

Proposition 2.60. Let φ : G→ φ(G) be a surjective homomorphism with kernel N. Then1. there is a bijection between the subgroups of the image and the subgroups of G con-

taining N. Map the subgroup of φ(G) to its preimage.2. B � φ(G) iff φ−1(B) �G.

Proof. .

Theorem 2.61 (2nd Isomorphism Theorem). Let H < G, N�G. Then NH < G, N�NH,

N ∩H �H, andNH

N∼=

H

N ∩H.

Proof. Define φ : H → NH/N by φ(h) = Nh. Apply the 1st Iso Thm.

Theorem 2.62 (3rd Isomorphism Theorem). Let K < H < G where K,H � G. Then

H/K �G/K andG/K

H/K∼= G/H.

Proof. Define φ : G/K → G/H by φ(Kg) = Hg.

39

3 Rings

3.1 Basic properties

Definition (ring). A ring is a nonempty set R with two binary operations (+ and *) satis-fying:

1. R is an abelian group under + (will usually use 0 as notation for identity)2. * is associative and3. distributive property a ∗ (b+ c) = a ∗ b+ a ∗ c and (a+ b) ∗ c = a ∗ c+ b ∗ cIf there exists a multiplicative identity (unity): 1*r=r*1=r for all r, then we say that R

is a ring with identity or ring with unity.If multiplication is commutative, say R is a commutative ring.

Note: the book means a ring with identity when it says ring. Most of the examples welook at will have an identity, but we will not assume it must have one.

Example 3.1. Examples of rings:Z, Q, R, Mn(R), Zn, K[x], continuous functions, direct product of rings (both operations

are component-wise)

Theorem 3.1. Let r, s be in a ring R; m,n ∈ Z.1. (−r)s = r(−s) = −(rs)2. (−r)(−s) = rs3. (mr)(ns) = mn(rs)

Theorem 3.2. Let kR = {kr : r ∈ R} for a ring R and integer k. If a positive integer nexists such that nR = 0R (equivalently n1R = 0R), then the characteristic of R, charR, is the smallest such natural number. If no such natural number exists, then char R=0.Moreover, if char R = n > 0, kR = 0R iff n|k and if char R=0, then kR = 0R iff k = 0.

Example 3.2. What are the characteristics of the examples above?

Definition (subring). If R is a ring, a subset S ⊆ R is a subring of R if S is a ring withthe same operations as R.

40

Theorem 3.3 (Subring Test). A nonempty subset S of a ring R is a subring iff for alls, t ∈ S, s + t, st and −s are in S; ie. closed under both operations and under additiveinverses.

Note: the additive identity must be the same in S as in R. And, if R has unity, S musthave the same one.

Example 3.3. • Triangular matrices< Mn

• Q < R

• continuous functions<functions

• Z(R) < R (center is subring)

• {0} and R are subrings of R

Definition (idempotent). r ∈ R is idempotent if r2 = r

Proposition 3.4. If s2 = s in a ring R with unity, sRs is a ring with unity s.

Definition (unit). u ∈ R, R ring with unity, is called a unit if u has a multiplicativeinverse. R∗ = {u : u is a unit} is the group of units of R.

Example 3.4. Mn(R)∗ = GLn(R)

Note: we have the cancelation laws for units only in a ring.

Definition (division ring). A ring in which every nonzero (non-additive identity) is a unit.(No zero divisors.)

Definition (field). Commutative division ring.

Example 3.5. Ring or division ring or field?

• Z

• Q

• R

• Mn(R)

• Zn if n is not prime, if n is prime

41

• quaternions

Definition (nilpotent). An element r ∈ R is nilpotent if there exists k ∈ N such thatrk = 0R. R is a nilpotent ring if there exists k ∈ N such that rk = 0R for all r ∈ R.

Proposition 3.5. If r is nilpotent, then 1− r is a unit.

Proof. (1− r)(1 + r + r2 + rk−1) = 1

Definition (ring isomorphism). Two rings R, S are isomorphic if there exists an isomor-phism φ : R→ S such that

1. φ is a bijection

2. φ(r + r′) = φ(r) + φ(r′) preserves addition

3. φ(rr′) = φ(r)φ(r′) preserves multiplication

Example 3.6. Are the following isomorphisms group or ring isomorphisms?

• R =

{[a −bb a

]}∼= C

• |R| = p, R ∼= Zp for p prime

• R =

{[a b0 a

]: a, b ∈ Z2

}∼= Z2 × Z2

• Z ∼= 2Z

Example 3.7. As we know, there are two groups of order 4. How many rings are there oforder 4?

42

3.2 Integral Domains and Fields

Definition (domain). A ring in which for all a, b ∈ R, if ab = 0R, then a = 0R or b = 0R.In other words, there are no zero divisors. A commutative domain is called an integraldomain.

Proposition 3.6. TFAE for a ring R:1. If ab = 0, then a = 0 or b = 0.2. If ab = ac and a 6= 0, then b = c - left cancelation.3. If ba = ca and a 6= 0, then b = c - right cancelation.

Proof. .

Note: every division ring is a domain. We have field ⊂ division ring ⊂ domain ⊂ ring ⊂group. (There is more than one type of domain - 3 main types of integral domains.)

Example 3.8. Domains:

• Z

• field, division ring

• Gaussian integers Z[i] - a+ bi where a,b are integers

• rings with division algorithms: integers, polynomial rings

Example 3.9. Show that Q[√

3] is a field.

Proposition 3.7. The characteristic of any domain is either zero or prime.

Proof. .

43

Proposition 3.8. Every finite integral domain is a field.

Proof. .

Theorem 3.9 (Wedderburn’s Theorem). Every finite division ring is a field.

Definition (quaternion). H = {a+ bi+ cj + dk : a, b, c, d ∈ R}, where i2 = j2 = k2 = ijk =−1. Note that ij = k = −ji, jk = i = −kj, ki = j = −ik. So, it is a noncommutativedivision ring.

Definition. The conjugate of a quaternion q = a+ bi+ cj + dk is q∗ = a− bi− cj − dk andthe norm is N(q) = a2 + b2 + c2 + d2.

Theorem 3.10. H is a noncommutative division ring. Moreover, for q 6= 0, q−1 = 1N(q)

q∗.

Fields of fractions generalize the relationship between integers and rational numbers.

Definition (field of fractions). Let R be an integral domain. Let X = {(r, u) : r, u ∈ R, u 6=0R}. Define an equivalence relation on X by (r, u) ≡ (s, v) iff rv = su. The set of equivalenceclasses, [(r, u)] = r

u, form a field, called the field of fractions (or field of quotients).

Theorem 3.11 (Embedding Theorem). If R is an integral domain, there is a field Q con-sisting of quotients r

u, where r, u ∈ R and u 6= 0R. By identifying r = r

1, R is a subring of

Q.

3.3 Ideals and Factor Rings

Definition (ideal). A subring I of a ring R is a (two-sided) ideal if ∀a ∈ I and ∀r ∈ R,ar, ra ∈ I. In other words, Ra ⊆ I and aR ⊆ I. I is a left-ideal if Ra ⊆ I and a right-idealif aR ⊆ I.

NOTE! Because the book requires the unity to be in a ring, an ideal is defined as anadditive subgroup such that the property above holds. It is not a subring in general (with thebook’s definition) because an ideal does not generally have unity. If it has the multiplicativeidentity, it must be the whole ring - see a few theorems below. The Ideal Test below workswith my and the book’s definition.

Theorem 3.12 (Ideal Test). A nonempty subset I of a ring R is an ideal of R if 1) a−b ∈ Ifor all a, b ∈ I and 2) ra, ar ∈ I for all a ∈ I, r ∈ R.

Proof. 1) check closed under addition and additive inverses. 2) checks closed under multi-plication (since I is a subset of R) and ideal.

44

Example 3.10. Every ring has two ideals: {0} and R itself.R[x] - polynomial ring. The polynomials with constant term 0 is an ideal.The subset of differentiable functions of real-valued functions is a subring, but not an

ideal.

Definition (principal ideal). Let R be a commutative ring with unity. Let a ∈ R. Then theprincipal ideal generated by a is 〈a〉 = {ra : r ∈ R}. Note that this is analogous to cyclicsubgroup.

If R is not commutative, you can have the left principal ideal Ra and the right principalideal aR.

We will now use additive notation for cosets; ie. G/H = {g + H}. Because a ring is anabelian group, all subgroups are normal (with addition). Thus, we can form the quotientgroup. The question becomes whether it is actually a ring.

Theorem 3.13. Let I be an ideal in a ring R. Then R/I is a ring, called the quotient orfactor ring, and the quotient map a 7→ a + I is a surjective ring homomorphism from Rto R/I with kernel I. If R has unity, then R/I does.

Proof. The multiplication in the ring is (r + I)(s + I) = rs + I. We need to prove this iswell-defined. It satisfies the group properties, because it is a quotient group under addition.Associativity and distributivity follow from those properties in R.

well-defined:

Example 3.11. • R is the ring of 2 by 2 matrices with integer entries. I is the subsetof even entries. Is I an ideal? If yes, what is R/I?

• R = R[x], I = 〈x2 + 1〉. What is R/I?

Definition (annihilator). Let a ∈ Z(R). ann(a) = {r ∈ R : ra = 0R} is an ideal of R,called the annihilator of a.

45

Definition (prime ideal). Proper ideal P of a commutative ring R such that if rs ∈ P , thenr ∈ P or s ∈ P .

These fill the role of prime numbers. In other words, the principal ideal generated by aprime number in Z is a prime ideal.

Theorem 3.14. If R is a commutative ring with unity, then P is a prime ideal of R if andonly if R/P is an integral domain.

Proof. If R/P is an integral domain and rs ∈ P , then (r + P )(s+ P )...

Definition (maximal ideal). Proper ideal M of a ring R such that whenever B is an idealof R and M ⊆ B ⊆ R, then B = M or B = R. In other words, there are no ideals strictlybetween M and R.

Example 3.12. • nZ is prime ideal iff n is prime.

• The max ideals in Z36 are < 2 >, < 3 >.

• < x2 + 1 > is not prime in Z2 since (x+ 1)2 = x2 + 1 ∈ I but x+ 1 6∈ I.

Definition (simple ring). A nonzero ring such that its only ideals are 0 and R.

Theorem 3.15. If R is commutative with unity, then R is simple iff it is a field.

Proof. .

Note: every division ring is simple.

Theorem 3.16. A proper ideal M is a maximal ideal of R iff R/M is a simple ring.

46

Proof. .

Corollary 3.17. An ideal M in a commutative ring R with unity is maximal iff R/M is afield.

Corollary 3.18. Every maximal ideal of a commutative ring with unity is a prime ideal.

Lemma 3.19. Let I be an ideal of a ring R. Then Mn(I) is an ideal of Mn(R).Every ideal of Mn(R) has the form Mn(I) for some ideal I of R.

Theorem 3.20. If R is a ring, then Mn(R) is simple iff R is simple.

Proof. It follows from the lemma.

Corollary 3.21. If R is a division ring, then Mn(R) is simple.

3.4 Ring Homomorphisms

Example 3.13. Examples of Ring Homomorphisms:

• Like for groups, the coset map is a surjective ring homomorphism: I ideal of R, R→R/I by r 7→ r + I.

• projection maps: π1 : R1 ×R2 → R1 and π2 : R1 ×R2 → R2

• θ :

[R R0 R

]→ R×R by θ(A) = (a11, a22)

• If θ : R → S is a ring homomorphism, then θ̄ : M2(R) → M2(S) is a ring homomor-phism.

Ring homomorphisms preserve rational expressions. This can be used as in the aboveexample, but also can be used to show an equation has no solution in the ring.

Example 3.14. Show x3 − 8x2 + 5x+ 3 = 0 has no solution in the integers.

47

Theorem 3.22. Let φ : R→ S be a ring homomorphism.1. φ(R) is a subring of S.2. kerφ is an ideal of R.

Proof. .

Again, every ideal is the kernel of the coset map.

Theorem 3.23 (1st Isomorphism Theorem for Rings). Let φ : R → S be a ring homomor-phism. Let I = ker φ. Then φ induces the ring isomorphism φ̄ : R/I → φ(R) given byφ̄(r + I) = φ(r).

Proof. .

Proposition 3.24. If I is an ideal of R, then Mn(I) is an ideal of Mn(R) andMn(R)

Mn(I)∼=

Mn(R

I).

Proof. .

Theorem 3.25. If R is a ring, then Z1R is a subring of R (all multiples of the unity).1. If R has characteristic n > 0, then Z1R ∼= Zn2. If R has characteristic 0, then Z1R ∼= Z

Proof. .

48

Corollary 3.26. Every field R contains a subfield isomorphic to Zp or Q, called prime fields.

Definition. A + B = {a + b : a ∈ A, b ∈ B}. If A and B are ideals, then their sum is aswell.

Theorem 3.27. Let R be a ring with ideals A and B such that R = A+B and A∩B = {0R}.Then R ∼= A×B as rings.

Every element in R can be written as a unique combination (a, b).

Proof. .

Corollary 3.28. If e2 = e ∈ R is a central idempotent in a ring with unity, then R ∼=eRe× (1− e)R(1− e).

Theorem 3.29 (Chinese Remainder Theorem). Let A, B be ideals of R with unity.1. If A+B = R, then R/(A ∩B) ∼= R/A×R/B.2. If A+B = R and A ∩B = {0R}, then R ∼= R/A×R/B

Proof. .

Corollary 3.30. If m and n are relatively prime, then Zmn ∼= Zm × Zn

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4 Polynomials

4.1 Polynomials

We will take the coefficients in our polynomial from a ring with unity: R[x]

Theorem 4.1. Let R be a ring and x an indeterminate over R. Then, 1. R[x] is a ring 2.x ∈ Z(R[x]) and 3. if R is commutative, R[x] is commutative

Proposition 4.2. R[x]/〈x〉 ∼= R

Proof. Define θ : R[x]→ R by θ(a0 + a1x+ · · · anxn) = a0.

Theorem 4.3. Let R be a domain. Then, 1. R[x] is a domain 2. deg(f(x)g(x)) =degf(x)+deg g(x) and 3. the units in R[x] are the units in R

Theorem 4.4 (Division Algorithm). Let R be a ring and f(x), g(x) ∈ R[x]. Assume f(x) is a(nonzero) monic polynomial. Then there exist unique q(x), r(x) such that g(x) = q(x)f(x) +r(x), deg r(x) <deg f(x).

Note: you do not need a monic polynomial as long as the leading coefficient is a unit;essentially factor out the unit and then you have the theorem.

Proof. Same as for integers, essentially.

Example 4.1. Use the division algorithm on g(x) = x5 + 4x4 + x3 + 5x2 + x + 2 andf(x) = x2 + x+ 1 in Z6[x].

Also, g(x) = 3x3 + 2x2 − 8x+ 1 and f(x) = x2 + 2 in Q[x].

Definition. Let f(x) ∈ R[x]. Let a ∈ Z(R). Then f(a) = a0 + a1a + · · · anan is theevaluation of f(x) at a.

Note: the book requires a to be in the center so that the evaluation makes sense when youfactor the polynomial (i.e. finding roots). There is a whole other, very interesting, theory ofroots of noncommutative polynomials.

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Theorem 4.5 (Evaluation Theorem). Let R be a ring and let a ∈ Z(R). Define the evalu-ation mapping φa : R[x] → R by φa(f(x)) = f(a). Then φa is a surjective ring homomor-phism.

Theorem 4.6 (Factor Theorem). Let R be a commutative ring and let f(x) ∈ R[x]. Ifa ∈ R, then kerφa = 〈x− a〉. In other words, f(a) = 0 iff f(x) = (x− a)q(x).

Theorem 4.7 (Remainder Theorem). If R is a commutative ring and f(x) ∈ R[x] is dividedby x− a, the remainder is f(a).

Example 4.2. Use the polynomials g(x) above with a = 1.

Definition. Let R be a commutative ring. a ∈ R is a root of f(x) ∈ R[x] if f(a) = 0.

NB: A polynomial of degree n has at most n roots if R is an integral domain. It is nottrue in general. For example, over matrices.

NB: You must pay attention to the coefficient ring. You could have roots over one ringbut not another. For example, x2 + 1 does not have any roots in R, but it does in C.

Theorem 4.8 (Rational Roots Theorem). Let f(x) ∈ Z[x], where a0, an 6= 0. Then everyrational root of f(x) has the form c/d where c|a0 and d|an.

Example 4.3. Find the roots of (x− 4)(x− 5) ∈ Z6.

Example 4.4. Factor f(x) = x4 + 12 in Z13

Example 4.5. Find all rational roots of f(x) = 4x4 + x3− 3x2 + 4x− 3 and factor as muchas possible over Q.

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4.2 Factorization of Polynomials over a Field

Polynomials over a field have the same structure as the integers. In other words, for everyresult we had in section 1.2, we have the analogous result in this section. As such, we willnot go over the proofs, but you should try to prove each of the theorems (mimic the proofsin section 1.2).

Recall that the main result about natural numbers is that they can be factored uniquelyinto a product of primes. Polynomials can be factored uniquely into a product of irreducibles.Note that this will depend on what field we take the polynomial over.

Theorem 4.9 (Unique Factorization Theorem). Let F be a field and f(x) ∈ F [x] be anonconstant polynomial. Then f(x) can be factored uniquely into a product of irreduciblemonic polynomials, up to a constant and order of the factors.

This theorem is the goal of the section. To achieve that, we have to now what polynomialsare irreducible.

Definition (irreducible polynomial). A polynomial p(x) ∈ F [x] is irreducible if deg p(x) ≥ 1and if p(x) = f(x)g(x), then either f or g is a unit in F.

Example 4.6. • Every linear polynomial is irreducible.

• Every constant multiple of an irreducible is irreducible.

• If f(x) has a root in the field, it is reducible; if f(x) has degree 2 or 3, then if it doesnot have a root, it is irreducible.

• Over R, irreducibles are linear or quadratic. Over C, irreducibles are linear.

Theorem 4.10 (Fundamental Theorem of Algebra). If f(x) is a nonconstant polynomialsin C[x], then f(x) has a root in C.

Corollary 4.11. f(x) ∈ C[x] factors as u(x − r1) · · · (x − rn), u ∈ C. Thus, the onlyirreducibles are linear and every polynomial completely factors as a product of linear polyno-mials.

It is more difficult to determine whether a polynomial is irreducible over other fields (andrings). Because we can multiply a polynomial by a constant to clear denominators, questionsabout polynomials over Q can be answered by looking at polynomials over Z.

Lemma 4.12 (Gauss’ Lemma). Let f(x) = g(x)h(x) ∈ Z[x]. If a prime p divides everycoefficient of f(x), then either p divides every coefficient of g(x) or h(x).

Proof. Idea: reduce all the coeffs. modulo p. Z[x]→ Zp[x] defined by f 7→ f̄ is a surjectivering homomorphism. Use the fact that Z is an integral domain.

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Theorem 4.13. Let f(x) 6= c ∈ Z[x]. f(x) is irreducible in Q[x] if and only if it has noproper factorization in Z[x].

Note: we say no proper factorization because a constant is not a unit in Z so it could beirred. but not a proper factorization.

The question then becomes, how do we know if a polynomial has a proper factorization?You can use the Rational Roots Theorem, but it only tells you if there is/is not a linearfactor.

Theorem 4.14 (Modular Irreducibility Test). Let 0 6= f(x) ∈ Z[x] and suppose a prime psatisfies 1. p does not divide the leading coefficient and 2. f(x) modulo p is irreducible inZp[x]. Then f(x) is irreducible over Q.

Proof. .

Sometimes using this test requires trying many primes.

Theorem 4.15 (Eisenstein Criterion). Consider f(x) = a0 + a1x + · · · anxn ∈ Z[x], wherean 6= 0, n ≥ 1. Suppose a prime p satisfies

1. p|ai for 0 ≤ i ≤ n− 12. p - an3. p2 - a0Then f(x) is irreducible over Q.

Example 4.7. Show the following are irreducible over Q.

• f(x) = 21x3 − 3x2 + 2x+ 8

• f(x) = −17x4 − 87x3 − 99x2 + 82x+ 83

• f(x) = 26x4 + 15x3 − 9x2 − 2x+ 6

• Nonexample: f(x) = x4 + 1

• f(x) = x3 + 14x+ 7

• f(x) = x5 + 9x4 + 3x3 − 27x2 + 81x− 3

• xp−1 + xp−2 + · · ·+ x+ 1, cyclotomoic polynomial

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Proposition 4.16. If f(x), g(x) are monic polynomials such that f |g and g|f , then f = g.

Theorem 4.17. Let 0 6= f(x), g(x) ∈ F [x], F field. Then there exists a unique d(x) ∈ F [x],called the greatest common divisor, such that

1. d(x) is monic2. d(x)|f(x), g(x)3. If h(x)|f, g, then h|d.4. d(x) = u(x)f(x) + v(x)g(x)

Two polynomials are relatively prime if there gcd is 1.

Theorem 4.18. Let p(x) ∈ F [x] be irred. If p(x)|f1(x) · · · fn(x), then p|fi(x) some i.

This leads to the Unique factorization theorem from the beginning of the section.

4.3 Factor Rings of Polynomials over a Field

This is along the lines of our earlier example where we mod out by a polynomial; we treatthe polynomial as equaling 0.

Theorem 4.19. Let F be a field and I a nonzero ideal of F [x]. Then there exists a uniquelydetermined monic polynomial h(x) such that I =< h >.

Note: this says that all ideals of F [x] are principal. And every monic polynomial generatesan ideal (1-1 correspondence).

Note 2: h(x) is the gcd of the polynomials in the ideal.

Definition. An integral domain where every ideal is principal is a principal ideal domain(PID).

Theorem 4.20. Let F be a field and let h be a monic polynomial in F [x] of degree m ≥ 1.Then the factor ring F [x]/ < h > is given by F [x]/ < h >= {a0 + a1t + · · · am−1tm−1 : ai ∈F, h(t) = 0}.

Example 4.8. Describe the following quotient rings.

• F [x]/ < x3 >

• Z2[x]/ < x4 − 1 >

• Q[x]/ < x2 − 3 >

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Theorem 4.21. Let h be a monic polynomial of degree m ≥ 1 in F [x], F a field. TFAE

1. F [x]/ < h > is a field

2. F [x]/ < h > is an integral domain

3. h is irreducible over F

Proof. Idea. 1 → 2 obvious. It is a field if < h > is a max ideal. This is going to happenwhen h is irred. If it were reducible: h = fg, then < h >⊂< f >.

How many elements does Zp[x]/ < h(x) > have if h has degree m?p choices for each coefficient in a0 + a1t+ · · ·+ am−1t

m−1, so pm elements.In Zp[x], there exists an irreducible monic polynomial for every degree m ≥ 1.

What does this mean for us? It means that we can construct a field of order pm for everyprime p and natural number m. In abstract algebra 2, you learn that these are all the finitefields; ie. every finite field has prime power order.

Example 4.9. Construct a field of order 8.f(x) = x3 + x+ 1 over Z2

Construct a field of order 121.g(x) = x2 + x+ 1 over Z11

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5 Factorization in Integral Domains

5.1 Irreducibles and unique factorization - UFDs

Recall that natural numbers and polynomials can be written uniquely as a product of irre-ducibles. We are going to generalize that concept.

Definition (irreducible). Let a be a nonzero, non-unit element of a commutative ring withunity. A proper factorization of a is an equality a = bc where b and c are not units. a isirreducible if it has no proper factorizations.

Definition (associates). Two elements a and b in an integral domain are associates if a|band b|a.

If a and b are associates, we write a ∼ b. This forms an equivalence relation, a = ub forsome unit u, and 〈a〉 = 〈b〉.

Theorem 5.1. The following are equivalent for a nonzero, non-unit element p in an integraldomain R:

1. p is irreducible

2. If d|p, then either d ∼ 1 or d ∼ p

3. If p ∼ ab, then p ∼ a or p ∼ b.

4. If p = ab, then p ∼ a or p ∼ b.

Definition (Ascending Chain Condition on Principal Ideals (ACCP)). An integral domainsatisfies the ACCP if R contains no strictly increasing infinite chain 〈a1〉 ⊂ 〈a2〉 ⊂ 〈a3〉 ⊂ · · ·of principal ideals.

Theorem 5.2. Let R be an integral domain that satisfies the ACCP. Then every nonzerononunit element in R is a product of irreducibles (not necessarily unique).

Example 5.1. Examples of integral domains which satisfy ACCP:Zany fieldF [x] (F any integral domain satisfying ACCP)

Definition (Unique Factorization Domain (UFD)). Every nonzero nonunit element in anintegral domain R can be written uniquely as a product of irreducibles (up to the order theyare written and up to associates).

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Example 5.2. Examples of UFD’s:Zany field (vacuously)F [x], F a fieldZ[x]Noneg: Z[

√−5]: 6 = 2 ∗ 3 = (1 + i

√5)(1− i

√5)

Definition (prime element). An element p in an integral domain R is a prime if p isnonzero and not a unit and If p|ab, then p|a or p|b.

How are primes and irreducibles related? For example, in the integers, they mean thesame thing. However, in Z[

√−5] they are not.

Theorem 5.3. Every prime in an integral domain R is irreducible in R.

Theorem 5.4. In a UFD, every irreducible is prime.

Thus prime=irreducible in a UFD (but not lower).

Theorem 5.5. If R is a UFD, then R[x] is a UFD.

5.2 Principal Ideal Domains - PIDs and EDs

Definition (Principal Ideal Domain (PID)). An integral domain in which every ideal isprincipal.

Example 5.3. Zany fieldF [x]Z[i]Noneg: Z[x]

Theorem 5.6. Every PID is a UFD.

Theorem 5.7. TFAE for every nonzero nonunit element p in a PID R.

1. p is prime

2. R/ < p > is a field

3. R/ < p > is an integral domain

So, every nonzero prime ideal is maximal. This theorem should look familiar.

What other property of polynomials and integers have we seen? They have an EuclideanAlgorithm. This generalizes to a class of integral domains called Euclidean domains. It is“harder” to be have an Euclidean Algorithm than to have unique factorization.

Definition (Euclidean function). d : R/{0} → N with the properties that d(fg) ≥ max {d(f), d(g)}and ∀f, g ∈ R/{0},∃q, r s.t. f = qg + r where r = 0 or d(r) < d(g).

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Example 5.4. Integers: d(n) = |n|polynomials: d(f) = deg(f)Gaussian integers: d(z) = |z|2 (multiply element by its conjugate)

Definition (Euclidean domain (ED)). An integral domain which admits an Euclidean func-tion.

Theorem 5.8. Let R be an ED. Then:

1. 0 6= f, g have a unique gcd (up to associates) which can be written as a linear combi-nation of f and g (gcd ∈ Rf +Rg)

2. f and g are relatively prime iff 1 ∈ Rf +Rg

3. every ideal is principal

4. If p is irreducible, then p is prime.

5. Every nonzero, nonunit element has a unique factorization into irreducibles.

Notice: 1 and 2 are what we know about integers and polynomials. 3 says ED→ PID. 5says ED→UFD. 4 follows from the fact it is true in a UFD (so true in anything higher)

Definition (norm). N(x) = xx∗ - the product of an element with its conjugate

Theorem 5.9. Let w ∈ C satisfy w2 ∈ Z, w 6∈ Q.

1. N(ab) = N(a)N(b)

2. a is a unit iff N(a) = ±1

3. N(a) = 0 iff a=0

4. If N(a) is prime in Z, then a is irreducible in Z(w).

This norm sometimes makes Z(w) into a ED, but not always!

Summary: ED→PID→UFD→ID and no converse is true.

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6 Fields

Take Abstract Algebra II! :)

7 Modules over PIDs

Goal: write finitely generated abelian groups as a direct product of cyclic groups.Note: we will not do everything in this chapter; I will give the highlights of the concepts

and focus on the Main Theorem.

7.1 Modules

A module generalizes the notion of vector space; a vector space is over a field, a module isover a ring. So, the scalars are ring elements instead of field elements.

Definition (left R-module). Let M be an abelian group and R a ring with unity. M is a leftR-module if for all r, s ∈ R, x, y ∈M :

1. rx ∈M

2. r(x+ y) = rx+ ry

3. (r + s)x = rx+ sx

4. r(sx) = (rs)x

5. 1x = x

Write RM .We can define right R-module in a similar fashion (put r,s’s on other side).

Definition (external direct sum). Let M1, ...,Mn be R-modules. Then M1 ⊕ · · · ⊕Mn is amodule called the external direct sum of M1, ...,Mn.

We have analogues for most of the definitions we have been using. I will discuss a few;see your book for others.

Definition. A module homomorphism preserves addition and scalar multiplication (linearmap). So, α :R M →R N is a R-homomorphism (or R-linear) if α(x+ y) = α(x) +α(y) andα(rx) = rα(x).

Definition (submodule). If RM is a R-module, a subset N ⊆ M is a R-submodule if 1. Nis a subgroup of M and 2. rx ∈ N ∀r ∈ R, x ∈ N .

Example 7.1. Z-submodules of a group are subgroups.If R is a field, they are subspaces of the vector space.0 and M are submodules.

RR submodules are the left ideals of R.kernel and image

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We can make a quotient module M/N for N a submodule of M. We also have the 1stisomorphism theorem.

Recall that ann(x)= {r ∈ R : rx = 0}.

Definition (torsion-free). x ∈R M is torsion-free if ann(x)=0; ie. no element except theadditive identity annihilates x. M is a torsion-free module if every nonzero element is torsion-free.

Definition (internal direct sum). Let N1, ..., Nm be submodules of RM . N1 + · · · + Nm ={x1 + · · · + xm : xi ∈ Ni} is a submodule. It is the internal direct sum if x1 + · · · + xm = 0implies xi = 0 for all i. Equivalent to every element can be written uniquely as a sum of thisform. (ie. it is a direct sum).

Definition. Let xi ∈ M . Rxi is a submodule of RM ; as above, the sum of submodules is asubmodule. We say Rx1 + · · ·+Rxm is the submodule generated by {x1, ..., xm}. If this is adirect sum, then the xi form a basis for M. If M has a finite basis, it is called a free module.

Note: look over the details in the book.

7.2 Modules over a PID

We are only going to do the special case of finitely generated abelian groups (ie. with Zn’s).Read the book for a general presentation.

Theorem 7.1 (Fundamental Theorem of Finitely Generated Abelian Groups). - If G is afinitely generated abelian group, then G ∼= Zm1⊕· · ·⊕Zms⊕Zr, where mi|mi+1. The naturalnumbers r, s,m1, ...,ms are uniquely determined by G.

Proof. See the proof in the book. It is very enlightening, but long.

I call this way of writing it “way 2”. Way 1 says that G can be written has a direct sumof cyclic groups of prime power order and finitely many copies of the integers.

We will focus our examples on finite groups; ie. no copies of the integers (r=0).This theorem allows us to classify ALL of these types of groups; it is very powerful.

Because we understand Zn well, we understand all fin. gen. abelian groups.Let G be a group of order n. We can write n uniquely has a product of primes: n =

pr11 · · · prkk

Example 7.2. List all distinct abelian groups of the following orders (use way 1):27:

p:p2

p3

60

p4

1176 = 23 ∗ 3 ∗ 72

How do we tell which of the options our group is isomorphic to? compare orders ofelements.

Note: are these external or internal direct products?

Theorem 7.2. Greedy Algorithm - order pn

1. compute the orders of the elements in group G

2. select a1 of maximal order and define G1 = 〈a1〉. Set i := 1

3. If |G| = |Gi|, stop. Else i 7→ i+ 1.

4. Select element ai of max order pk such that pk ≤ |G|/|Gi−1| and none of ai, ..., apk−1

i isin Gi−1. Define Gi = Gi−1 + 〈ai〉.

5. Return to step 3.

Note: is this an external or internal direct product?

Example 7.3. G = {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64} < Z∗65

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Example 7.4. How do we get from Way 1 to Way 2? Remember that if a and b are relativelyprime, Zab ∼= Za ⊕ Zb.

The generalization for the fundamental theorem is:

Theorem 7.3. If R is a PID, then every finitely generated module RM has a unique decom-position as a direct sum of principal submodules Rxi, xi ∈M : M = Rx1 ⊕ · · · ⊕Rxn

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