Abstract Algebra

Abstract Algebra Maths 113

Alexander Paulin

December 8, 2010

Lecture 1

The basics:

e-mail: [email protected]

My website: math.berkeley.edu/apaulin/

Im Scottish Math = Maths.

Office hours: Monday 11am -12.30pm and Wednesday 11am - 12.30pm

Grading:

1 Homework per week : Total worth 10 percent

2 midterms each worth 15 percent

1 Final Exam worth 60 percent

Make sure now that you can make the final - its on the 16th of December at 3pm. Ifyou cant then you cant take the course. We wont be using a specific text. Ill give outcourse notes each lecture. Books you may want to look at include:

Classic Algebra by P.M.Cohn - This is hard but is the gold standard in my opinion.

Algebra by Michael Artin - fantastic and easier to digest.

Frankly there are loads of good basic books. These notes will be enough to understandeverything, but going to the library to investigate concepts in more depth is a very goodskill to learn.

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What is Algebra

If you ask someone on the street this question the most likely response is something horribleto do with Xs and Y s. If youre lucky enough to bump into a mathematician then youmight get something along the lines of Algebra is the abstract encapsulation of our intuitionfor how arithmetic behaves.

Algebra is deep. Deep means that it permeates all of our mathematical intuitions. Infact the first mathematical concepts we ever encounter are the foundation of the subject.Let me summarize the first six years of your mathematical education:

The concept of unity: The number 1.You always understood this, even as a little baby!

N := {1, 2, 3...}. The natural numbers.

Comes equipped with two natural operations + and .

Z := {... 2,1, 0, 1, 2, ...}. The integers.We form these by using geometric intuition thinking of N as sitting on a line. Z also comes

with + and . Addition is particularly nice on Z, e.g. we have additive inverses.

Q := {ab |a, b Z, b (= 0}. The rationals.We form these by formally dividing through by non-negative integers. We again use

geometric insight picture Q. Q comes with + and . This time multiplication is nice onQ\{0}, e.g multiplicative inverses exist.

Notice that at each stage the operations of + and become better behaved. These ideasare very simple, but also profound. We spend years understanding how + and behave onthese objects. e.g.

a+ b = b+ a a, b Z,or

a (b+ c) = a b+ a c a, b, c Z.The central idea behind modern Algebra is to define a larger class of objects (sets with extrastructure), of which Z and Q are canonical members. Canonical means definitive.

(Z,+) Groups(Z,+,) Rings(Q,+,) Fields

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If youve done Linear algebra before the analogous idea is

(Rn,+) V ector Spaces over RThe amazing thing is that these vague ideas mean something very precise and have far

far more depth than one could ever imagine.

Sets and Mappings

A set is any collection of objects, e.g. six dogs, all the protons on Earth, every thoughtyouve ever had, N, Z, Q, R. The objects we will study in this course will all be sets.

Notation

Let S and T be two sets.

1. S T - S is contained in T. Note that they may be equal.2. S T - The intersection of S and T.3. S T - The union of S and T .4. S T = {(a, b)|a S, b T}. The product of S and T.5. - The empty set.

Definition. We say that S and T are disjoint if S T = .Definition. A map f (or function) from S to T is a rule sending elements of S to T.

f : S Tx f(x)

A map f : S T can have various nice propertiesDefinition. 1. f : S T is injective if f(x) = f(y) x = yx, y S.

2. f : S T is surjective if given y T there exists x S such that f(x) = y.3. If f : S T is both injective and surjective we say it is bijective. Intuitively this

means there is a one to one correspondence between elements of S and T.

Some examples of maps:

1. S =T = N,f : N N

x x2

2. S= Z Z, T = Z,f : Z Z Z

(a, b) a+ b

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Equivalence Relations

Within a set it is sometimes natural to talk about different elements being related to eachother. For example, on Z we could have

a, b Z, a b a b is divisible by 2.Definition. An equivalence relation on a set S is a subset U S S satisfying:

1. (symmetric) (x, y) U (y, x) U .2. (reflexive) x S, (x, x) U .3. (transitive) Given x, y, z S if (x, y) U and (y, z) U (x, z) U .In more convenient notation we write x y to mean (x, y) . An equivalence class is

a maximal subset of elements of S which are equivalent to each other. This makes sensebecause of the above axioms. I leave it as an exercise to show that the union of all theequivalence class is equal to S and they are all disjoint. We say they form a partition of S.

These are all the basic set theoretic concepts well need

Back to ZWe may naturally express + and in the following set theoretic way:

+ : Z Z Z(a, b) a+ b

: Z Z Z(a, b) a b

Here are 4 properties that + satisfies:

1. (associativity): a+ (b+ c) = (a+ b) + c a, b, c Z2. (Existence of additive identity) a+ 0 = 0 + a = a a Z.3. (Existence of additive inverses) a+ (a) = (a) + a = 0 a Z4. (Commutativity) a+ b = b+ a a, b Z.

Here are 3 properties that satisfy:1. (associativity): a (b c) = (a b) c a, b, c Z2. (Existence of multiplicative identity) a 1 = 1 a = a a Z.3. (Commutativity) a b = b a a, b Z.

+ and interact by the following:1. (Distributivity) a (b+ c) = (a b) + (a c) a, b, c Z.

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Remarks

1. each of these properties is totally obvious but will for the foundations of future defini-tions: Groups and Rings.

2. All of the above hold for Q. There is an extra property that non-zero elements havemultiplicative inverses e.g

Given a Q\{0} b Q such that a b = b a = 1.This extra property will motivate the definition of a field.

3. The significance of the associativity laws is that summing and multiplying a finitecollection of integers makes sense, i.e. is independent of how we do it.

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Alexander Paulin

September 2, 2009

Lecture 2

Last time we introduced the basic language of set theory we required. We then expressed+ and on Z in this language. We wrote down eight elementary properties they satisfied.Lets look at some of their deeper properties on Z. Again, these will motivate definitions ina more abstract setting.

It is an important property of Z (and Q) that the product of two non-zero elements isagain non-zero. More precisiely: if a, b Z such that ab = 0 either a = 0 or b = 0.Later this property will mean that Z is an integral domain. This has the following usefulconsequence:

Cancellation Law: For a, b, c Z if ca = cb and c $= 0 a = b.

A fundamental concept is that of divisibility.

Definition. Let a, b Z. We say that a divides b c Z such that b = ca. We denotethis by a|b and say that a is a divisor of b.

Any way of expressing an integer as the product of a finite collection of integers is calleda factorization. Observe that 0 is divisible by every integer. The only integers which divide1 are 1 and -1.

Definition. A prime number p is an integer greater than 1 whose only positive divisors arep and 1.

Remarks. Z is generated by 1 under addition. By this I mean that every integer can beattained by successively adding 1 (or -1) to itself. Under multiplication the situation is muchmore complicated. There is clearly no generator of Z under multiplication in the above sense.

Definition. Two non-zero integers a, b Z are coprime if there is no integer other than 1dividing both a and b. E.g. a=12, b=23.

Here are four important elementary properties of divisibility dating back to Euclid (300BC),which Ill state without proof. It would be a good exercise to try and prove them yourselves:

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Proposition. (Division Algorithm) Given a, b Z, if b > 0 then ! q, r Z such thata = bq + r with 0 r < b.Lemma. Given a, b Z, both non-zero, u, v Z such that au + bv = 1 a and b arecoprime.

Lemma. Given a, b1, ..., br Z all non-zero, if a is coprime to each bi a is coprime tob1b2...br.

Proposition. Let a Z1. Let b1, ...br be pairwise coprime integers. If bi|a i b1b2...br|a.2. Let p Z be a prime. If b1, ...br Z such that p|b1b2...br p|bi for some i.Using these we can prove the following very important property of Z.

The Fundamental Theorem of Arithmetic. Every positive integer a can be written asa product of primes:

a = p1p2...pr.

Such a factorization is unique up to ordering.

Proof. The number 1 can be represented as an empty product. If there is a positive integernot expressible as a product of primes , let c N be the least such element. c is not 1 or aprime, hence c = c1c2 where c1, cc N, c1 < c and c2 < c. By our choice of c we know thatboth c1 and c2 are the product of primes. Hence c much be expressible as the product ofprimes. This is a contradiction. Hence all positive integers can be written as the product ofprimes.

We must prove the uniqueness (up to ordering) of any such decomposition. Let

a = p1p2...pr = q1q2...qs

be two factorizations of a into a product of primes. Then p1|q1q2...qs. By the second partof the above proposition we know that p1|qi for some i. After renumbering we may assumei = 1. However q1 is a prime, so p1 = q1. Applying the cancellation law we obtain

p2...pr = q2...qs.

Assume, without loss of generality, that r < s. We can continue this process until we have:

1 = qr+1..qs.

This is a contradiction as 1 is not divisible by any prime. Hence r = s and after renumberingpi = qi i.

Using this we can prove the following beautiful fact:

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Theorem. There are infinitely many distinct prime numbers.

Proof. Suppose that there are finitely many distinct primes p1, p2.....pr. Consider c =p1p2...pr + 1. Clearly c > 1. By the Fundamental Theorem of Arithmetic, c is divisibleby at least one prime, say p1. Then c = p1d for some d Z. Hence we have

p1(d p2...pr) = c p1p2..pr = 1.This is a contradiction as no prime divides 1. Hence there are infinitely many distinctprimes.

The Fundamental Theorem of Arithmetic also tells us that every positive element of Qcan be written uniquely (up to reordering) in the form:

a = p11 pnn , pi prime and i ZThe Fundamental Theorem also tells us that two positive integers are coprime if and only ifthey have no common prime divisor. This immediately shows that every positive element ofQ can be written uniquely in the form

a = ,, N and coprime.As previously mentioned, Q has the property that non-zero elements have multiplicativeinverses, i.e. Given a Q \ {0}, b Q such that a b = 1. This extra property (added tothe other 8 it shares with Z, given in 1st lecture) makes Q an example of an object called afield (This will be defined properly later).So both Z and Q are examples of sets with two concepts of composition (+ and ) whichsatisfy a collection of abstract conditions. Can we think of other examples of sets with aconcept of + and which satisfy these rules?

Congruences

Fixm N. Division Algorithm Given a Z, ! q, r Z such that a = qm+r, 0 r < m.We call r the remainder of a modulo m. This gives the natural equivalence relation on Z :

a b a and b have the same remainder modulo m m|(a b)Exercise: Check this really is an equivalence relation!

Definition. a, b Z are congruent modulo m m|(a b). This can also be written:a b mod m.

Remarks. 1. The equivalence classes of Z under this relation are indexed by the possibleremainder modulo m. Hence, there are m distinct equivalence classes which we callresidue classes. We denote the set of all residue classes Z/mZ.

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2. There is a natural surjective map

[ ] : Z Z/mZa a mod m

Note that this is clearly not injective. Also observe thatZ/mZ = {[0], [1], ...[m 1]}.The following result allows us to define + and on Z/mZ.

Propostion. Let m N. Then,a, b, a, b Z :

[a] = [a] and [b] = [b] [a+ b] = [a + b] and [a b] = [a b].

where [ ] : a a mod m.Proof. This is a great exercise.

Definition. We define addition and multiplication on Z/mZ by

[a] [b] = [a b] a, b Z [a] + [b] = [a+ b] a, b ZRemarks. 1. The proposition shows us that this is well defined, i.e. it doesnt matter

which members of the equivalence class we choose.

2. Exactly the same idea as showing that addition and scalar multiplication is well definedon the set of cosets of a subspace of a vector space. Dont worry if you havent donemuch linear algebra well see this idea over and over.

Our construction of + and on Z/mZ is basically stolen from Z, hence they satisfy all theproperties given in the first lecture.e.g.

[0] Z/mZ behaves like 0 Z : [0] + [a] = [a] + [0] = [a] [a] Z/mZ.[1] Z/mZ behaves like 1 Z : [1] [a] = [a] [1] = [a] [a] Z/mZ.

Notice that Z/mZ has the following property, quite different than Z:

[1] + [1] + [1] + + [1](m times)= [m] = [0]Hence we can add 1 (in Z/mZ) to itself and eventually get 0 (in Z/mZ). Also observe thatif m is composite with m = rs, where r < m and s < m then [r] and [s] are both non-zero('= [0]) in Z/mZ, but [r] [s] = [rs] = [m] = [0] Z/mZ. Hence we can have two non-zeroelements multiplying together to give zero. This shows that even though +and onZ/mZ have the same elementary properties as + and on Z, they behave very differently.Proposition. For every m N, a Z the congruence

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ax 1 mod (m)has a solution (in Z) iff a and m are coprime.

Proof. This is just a restatement of the fact that a and m coprime u, v Z suchthat axu+mxv = 1.

Observe that the congruence above can be rewritten as [a] [x] = [1] in Z/mZ. We say that[a] Z/mZ has a multiplicative inverse if [x] Z/mZ such that [a] [x] = [1]. Hencewe deduce that the only elements of Z/m with muliplicative inverse are those given by [a],where a is coprime to m.Remember that on Q had the extra property that all non-zero elements had multiplicativeinverses. When does this happen in Z/mZ?, By the above we see that this can happen {2, ,m 1} are all coprime to m. This can only happen if m is prime. We have thusproven the following:

Corollary. All non-zero elements of Z/mZ have a multiplicative inverse m is prime.Later this will be restated as Z/mZ is a field m is a prime. These are examples

of finite fields. What this shows is that it is possible to have sets with a good concept of +and , but they may have very different properties from Z or Q.

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Alexander Paulin

September 7, 2009

Lecture 3

Recall the construction introduced in lecture 2:

Fix m N. We defined the equivalence relation on Z: a b m|(a b), alsowritten a bmodm. We denote the equivalence classes by Z/mZ. And observe that therewere exactly m such classes, indexed by the possible remainder modulo m. We also defined+ and on Z/mZ:

[a] [b] = [a b] a, b Z [a] + [b] = [a+ b] a, b Z

Proposition. For every m N, a Z the congruenceax 1 mod m

has a solution (in Z) a and m are coprime.Proof. This is just a restatement of the fact that a and m coprime u, v Z suchthat au+mv = 1.

Observe that the congruence above can be rewritten as [a] [x] = [1] in Z/mZ. We say that[a] Z/mZ has a multiplicative inverse if [x] Z/mZ such that [a] [x] = [1]. Hence wededuce that the only elements of Z/m with muliplicative inverse are those given by [a], wherea is coprime to m. What this is really saying is that an element of Z/mZ has a multiplicativeinverse every member of it is coprime to m. HOwever if one is then they all are.Remember that on Q had the extra property that all non-zero elements had multiplicativeinverses. When does this happen in Z/mZ?, By the above we see that this can happen {2, ,m 1} are all coprime to m. This can only happen if m is prime. We have thusproven the following:

Corollary. All non-zero elements of Z/mZ have a multiplicative inverse m is prime.

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Later this will be restated as Z/mZ is a field m is a prime. These are examples offinite fields.

Ive shown you this example to show that there are natural examples of sets with goodconcepts of addition and multiplication but which behave very differently than Z and Q.By good I mean they share the same elementary properties I wrote down in lecture 1. Thisgives a hint that perhaps there is some kind of unifying underlying language uniting theseideas.

Groups

Definition. Let G be a set. A binary operation is a map of sets:

: GG G.For ease of notation we write (a, b) = a ba, b G.Any binary operation on G gives a way of combining elements. As we have seen if G = Zthen + and are examples.Fundamental Definition. A group is a set G, together with a binary operation , suchthat the following hold:

1. (Associativity): (a b) c = a (b c) a, b, c G.2. (Existence of identity): e G suhc that a e = e a = a a G.3. (Existence of inverses): Given a G, b G such that a b = b a = e.

Remarks. 1. The we have seen three different examples thus far: (Z,+), (Q\{0},) and(Z/mZ,+). Another example is that of a real vector space under addition. Note that(Z,) is not a group. Also note that this gives examples of groups which are bothfinite and infinite. The more mathematics you learn the more youll see that groupsare everywhere.

2. A set with a binary operation is called a monoid if only the first two properties hold.From this point of view, a group is a monoid in which every element is invertible. Z,)is a monoid.

3. Observe that in all of the examples Ive given the binary operation is commutative, i.e.a b = b a a, b G. We do not include this in our definition as this would be toorestrictive as well see later. For example the set of invertible n n matrices with realcoefficients forms a group under matrix multiplication. However we know that matrixmultiplication does not commute in general.

So a group is a set with extra structure. In set theory we have the natural concept of a mapbetween sets. The following is the analogous concept for groups.

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Definition. Let (G, ) and (H, ) be two groups. A homomorphism f , from G to H is amap of sets f : G H, such that f(x y) = f(x) f(y) x, y G.Remarks. 1. Intuitively one should thing about a homomorphism as a map of sets which

preserves the underlying group structure. Its the same idea as a linear map betweenvector spaces.

2. A homomorphism f : G H which is bijective is called an isomorphism. Twogroups are said to be isomorphic if there exists an isomorphism between them. Intu-itively two groups being isomorphic means that they are the same group, but viewedfrom different points of view.

3. A homomorphism from a group to itself (e.g. f : G G) is called an endomorphism.An endomorphism which is also an isomorphism is called an automorphism.

Proposition. Let (G, ), (H, ) and (M,!) be three groups. Let f : G H and g : H Mbe homomorphism. Then the composition gf : GM is a homomorphism.Proof. Let x, y G. gf(x y) = g(f(x) f(y)) = gf(x)!gf(y).Remarks. Composition of homomorphism gives the collection of endomorphisms of a groupthe structure of a monoid. The subset of automorphisms has the stucture of a group undercomposition. We denote it by Aut(G).

Simple consequences of axioms:

Proposition. Let (G, ) be a group. The identity element is unique.Proof. Assume e, e G both behave like the identity. Therefore e = e e = e.Proposition. Let (G, ) be a group. For a G there is only one element which behaves likethe inverse of a.

Proof. Assume a G has 2 inverses, b, c G. Then:(a b) = e

c (a b) = c e(c a) b = c (associativity and identity)

e b = cb = c

The first tells us that we can write e G for the identity and it is well defined. Similarlythe second tells us that for a G we can write a1 G for the inverse in a well defined way.

The proof of the second result gives a good example of how we prove results for abstractgroups. We can only use the axioms, nothing else.

We also have an analogue of the cancellation law:

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Proposition. Let a, b, c G a group. Thena c = a b c = b and c a = b a c = b

Proof. Compose on left or right by a1 G, then apply the associativity and inverses andidentity axioms.

Definition. A group (G, ) is called abelian if it also satisfiesa b = b a a, b G

In linear algebra, we can talk about subspaces of vector spaces. We have an analogousconcept in group theory.

Definition. Let (G, ) be a group. A subgroup of G is a subset H G such that1. e H2. x, y H x y H3. x H x1 H

Remarks. A subgroup is naturally a group under the induced binary operation. It clearlyhas the same identity element.

Proposition. If H,K G are subgroups H K G is a subgroup.Proof. 1. As H,K subgroups, e Hande K e H K.

2. Let x, y H K x y H and x y K x y H K.3. Let x H K x1 H and x1 K x1 H K.

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Alexander Paulin

December 17, 2010

Lecture 4

Recall the fundamental concepts we introduced in lecture 3:

Definition. Let G be a set. A binary operation is a map of sets:

: GG G.Definition. A group is a set G, together with a binary operation , such that the followinghold:

1. (Associativity): (a b) c = a (b c) a, b, c G.2. (Existence of identity): e G suhc that a e = e a = a a G.3. (Existence of inverses): Given a G, b G such that a b = b a = e.Recall that we also introduced the concept of a subgroup:

Definition. Let (G, ) be a group. A subgroup of G is a subset H G such that1. e H2. x, y H x y H3. x H x1 H

Recall that a subgroup is naturally a group under the induced binary operation and theintersection of subgroups is a subgroup.

Definition. Let X G be a subset. Then we define the subgroup generated by X to bethe intersection of all subgroups containing X. We denote it by gp(X) G.Remarks. 1. gp(X) is the minimal (smallest) subgroup containing X.

2. A more constructive way of defining gp(X) is as all possible finite compositions ofelements of X and their inverses. I leave it as an exercise to check that this subset isa subgroup.

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3. Let us consider the group (Z,+) and X = {1} Z. Then gp(X)=Z. This is theprecise sense in which Z is generated by 1 under addition.

Definition. We say a group (G, ) is finitely generated if X G finite such thatgp(X)=G.

Remarks. 1. All finite groups are finitely generated

2. Infinitely many primes (Q\{0},) is not finitely generated.Definition. A group (G, ) is said to be cyclic if x G such that gp({x}) = G, i.e. G canbe generated by a single element.

We will study such groups in detail in future lectures. By the above observations (Z,+)is an example.

We now introduce the fundamental class of group.

Definition. Let S be a set. We define the group of permutations of S to be the set ofbijections from S to itself, denoted (S), where the group binary operation is compositionof functions.

Remarks. 1. By composition of functions I always mean on the left, i.e. f, g (S)and s S (f g)(s) = f(g(s)).

2. Associativity clearly has to hold. The identity element e of this group is the identityfunction on S, i.e. x S, e(s) = s. Inverses exist because any bijective map from aset to itself has an inverse map.

3. Let n N. We write Symn := ({1, 2, ..., n}). If S is any set of cardinality n then(S) is isomorphic to Symn, the isomorphism being induced by writing a bijection fromS to {1, 2, ..., n}. These groups will prove critical in understanding finite groups.

4. Observe that given (S) we can think about as moving S around. In thissense the group (S) naturally acts on S. Lets may this precise.

Definition. Let (G, ) be a group and S a set. By a group action of (G, ) on S we under-stand a map:

: G S Ssuch that

1. x, y G, s S, (x y, s) = (x, (y, s))2. (e, s) = s

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If the action of the group is understood we will write x(s) = (x, s)x G, s SThis notation makes the axioms clearer: (1) becomes (x y)(s) = x(y(s))x, y G, s Sand (2) becomes e(s) = s s S.Remarks. 1. Notice that there is a natural action of (S) on S:

: (S) S S(f, s) f(s)

In a sense this is where the definition comes from.

2. Let (G, ) be a group. There is a natural action of G on itself:

: GG G(x, y) x y

Property (1) holds as is associative. Property (2) holds because e x = x x G.This is called the left regular representation of G.

3. We define the trivial action of G on S by

: G S S(g, s) s s S, g G

4. There is another natural action of G on itself:

: GG G(x, y) x y x1

Property (1) holds because of associativity of and that (g h)1 = h1 g1. Property(2) is obvious. This action is called conjugation.

Let : G S S an action of a group G on a set S. Any g G naturally gives rise to amap:

g : S Ss g(s)

Observe that property (1) of being an action implies that gh = gh g, h G. Herethe second term is composition of functions. Similarly property (2) tell is that e = identitymap on S.

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Proposition. g is a bijection.

Proof. Given g, if we can find an inverse function, then we will have shown bijectivity. Bythe above two observations it is clear that g1 is inverse to g.

Hence we now have a map of sets

: G (S)g g

Proposition. is a homomorphism.

Proof. Property (1) of being an action h g = hgh, g G. This is precisely thestatement that is a homomorphism.

So an action of a group G on a set S gives a homorphism : G (S). It is true thatany such homorphism gives comes from a group action. In this sense an action of G on S isjust a homorphism as above.

Definition. An action of G on S is called faithful if

: G (S)g g

is injective.

Notice that if f : G H is an injective homomorphism then we may view G as a subgroupof H by identifying it with its image in H. Hence if G acts faithfully on S then G is naturallyisomorphic to a subgroup of (S).

Claim. The left regular representation of G on itself is faithful.

Proof. We need to show that is injective. For h, g G, suppose h = g. Then h x =g xx G h = g.Hence G is isomorphic to a subgroup of (G). Hence the entire concept of a group iscontained in the concept of bijective maps from a set to itself.We have proven the following:

Proposition. Let G be a finite group of order n N. Then G is isomorphic to a subgroupof Symn.

Proof. We can use the Left regular representation of G on itself to give an injective homo-morphism:

: G (G) = Symn

This result is often called Cayleys Theorem.

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Alexander Paulin

December 30, 2010

Lecture 5

Last time we introduced the important concept of a group (G, ) acting on a set S. Thedefinition we gave turned out to be equivalent to the following:

Definition. An action (G, ) on S is a group homomorphism

: G (S),where (S) is the permutation group of S.

Recall that if the action is understood then given g G and s S we write g(s) := (g)(s).Definition. Let (G, ) be a group, together with an action on a set S. We can define anequivalence relation on S by

s t g G such that g(s) = tRemarks. This is an equivalence relation as a consequence of the group axioms, togetherwith the definition of an action.

Definition. Let (G, ) be a group, together with an action on a set S. Under the aboveequivalence relation we call the equivalence classes orbits, and we write

orb(s) := {t S|g G such that g(s) = t} Sfor the equivalence class containing s S. We call it the orbit of s.It is important to observe that orb(s) is a subset of S and hence is merely a set with noextra structure.

Definition. Let (G, ) be a group, together with an action on a set S. We say that G actstransitively on S is there is only one orbit. Equivalently, is transitive if given s, t S,g G such that g(s) = t.

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An example of a transitive action is the natural action of (S) on S. This is clear becausegiven any two points in a set S there is always a bijection which maps one to the other. IfG is not the trivial group (the group with one element) then conjugation is never transitive.To see this observe that under this action orb(e) = {e}.Definition. Let (G, ) be a group, together with an action on a set S. Let s S. Wedefine the stabiliser subgroup of s to be all elements of G which fix s under the action. Moreprecisely

Stab(s) = {g G|g(s) = s} G

For this definition to make sense we must prove that stab(s) is genuinely a subgroup.

Proposition. Stab(s) is a subgroup of G.

Proof. 1. e(s) = s e Stab(s)2. x, y Stab(s) (x y)(s) = x(y(s)) = x(s) = s x y Stab(s).3. x Stab(s) x1(s) = x1(x(s)) = (x1 x)(s) = e(s) = s x1 Stab(s)

So given any a group (G, ) acting on a set S then for every s S we have two naturalobjects: orb(s) S and stab(s) G. We will study these in the next lecture but first weprove an important result about subgroups.

Lagranges Theorem

Let (G, ) be a group and let H G. Let us define a relation on G using H as follows:

Given x, y G, x y x1 y H

Proposition. This gives an equivalence relation on G.

We need to check the three properties of an equivalence relation:

1. (Reflexive )e H x1 x H x G x x2. (Symmetric) x y x1 y H (x1 y)1 H y1 x H y x3. (Transitive) x y, y z x1 y, y1 z H (x1 y) (y1 z) H x1 z

H x zDefinition. We call the equivalence classes of the above equivalence relation right cosetsof H in G.

Proposition. For x G the equivalence class containing x equals

2

xH := {x h|h H} GProof. The easiest way to show that two subsets of G are equal is to prove containment inboth directions.

x y x1 y H x1 y = h for some h H y = x h xH. Therefore{Equivalence class containing x} xH.

y xH y = x h for some h H x1 y H y x. Therefore xH {Equivalence class containing x}.

This has the following very important consequence:

Corollary. Hence for x, y G, xH = yH x1 y H.Proof. By the above proposition we know that xH = yH x y x1y H.It is very very important you understand and remember this fact. An immediate conse-quence is that y xH yH = xH. Hence cosets can in general be written with differentrepresentatives at the front. This is very important. Make sure you properly understandthis remark. Also observe that the equivalence class containing e G is just H. Hence theonly equivalence class which is a subgroup H, as no other contains the identity. If H = {e}then the cosets are singleton sets.

Definition 1. For and finite set S we denote its order by |S|.Definition. Let (G, ) be a group and H G a subgroup. We denote by G/H the set ofright cosets of H on G. If the size of this set is finite then we say that H has finite indexin G. In this case we write

(G : H) = |G/H|,and call it the index of H in G.

The subset of (Z,+) consisting of even integers is of index 2. If G is infinite and H = {e}then H is not of finite index in G.

Proposition. Let x G. The map (of sets)

: H xHh x h

is a bijection.

Proof. We need to check that is both injective and surjective. For injectivity observe thatfor g, h H, (h) = (g) x h = x g h = g. Hence is injective. For surjectivityobserve that g xH h H such tat g = x h g = (h).Now lets restrict to the case where G is a finite group.

3

Proposition. Let (G, ) be a finite group and H G a subgroup. Then x G , |xH| = |H|.Proof. We know that there is a bijection between H and xH. Both must be finite becausethey are contained in a finite set. A bijection exists between two finite sets if and only ifthey have the same order.

We can now prove the following important result.

Lagranges Theorem. Let (G, ) be a finite group and H G a subgroup. Then |H|divides |G|.Proof. We can use H to define the above equivalence relation on G. Because it is an equiv-alence relation, its equivalence classes cover G and are all disjoint. Recall that this is calleda partition of G.We know that each equivalence class is of the form xH for some (clearly non-unique ingeneral) x G. We know that any right coset of H has size equal to |H|. Hence we havepartitioned G into subsets all of size |H|| H| divides |G|.This is a powerful result. It gives tight control on the behavior of subgroups of a finite group.For example:

Corollary. Let p N be a prime number. Let (G, ) be a finite group of order p. Then theonly subgroups of G are G and {e}.Proof. Let H be a subgroup of G. By Lagrange |H| divides p. But p is prime so either|H| = 1 or |H| = p. In the first case H = {e}. In the second case H = G.

4


Alexander Paulin

December 30, 2010

Lecture 6

Let (G, ) be a group together with an action, , on a set S. Recall that : G (S) isa group homomorphism. Fix s S. Recall that we associated to s two important objects:

orb(s) := {t S|g G such that g(s) = t} Sand

stab(s) = {g G|g(s) = s} G.recall that stab(s) G is a subgroup. Hence we may form the right cosets of stab(s) in G:

G/stab(s) := {x(stab(s))|x G}.Recall that these subsets of G are the equivalence classes for the equivalence relation:

Given x, y G, x y x1 y stab(s),

hence they partition G into disjoint subsets.

Lemma. Let x, y G be in the same right coset of stab(s) in G. Then x(s) = y(s) S.Proof. Recall that x and y are in the same right coset x1y stab(s). Hencex1y(s) = s. Composing both sides with x and simplifying by the axioms for a group actionimplies that x(s) = y(s).

Hence there is a well defined map (of sets)

: G/stab(s) orb(s)x(stab(s)) x(s)

Proposition. is a bijection.

1

Proof. By definition, orb(s) := {x(s) S|x G}. Hence is trivially surjective.Assume (x(stab(s)) = (y(stab(s)) for some x, y G. This implies the following:

x(s) = y(s) x1(y(s)) = s (x1 y)(s) = s x1 y stab(s) x(stab(s)) = y(stab(s))

Therefore is injective.

This immediately gives the following key result:

Orbit-Stabiliser Theorem. Let (G, ) be a group together with an action, , on a set S.Let s S such that the orbit of s is finite (|orb(s)| < ). Then stab(s) G is of finiteindex and

(G : Stab(s)) = |orb(s)|Proof. Immediate from above.

We have the following corollary:

Corollary 1. If (G, ) is a finite group acting on a set S and s S then |G| = |stab(s)||orb(s)|.Proof. In this case (G : Stab(s)) = |G|/|stab(s)|. Applying the above theorem yields theresult.

Remarks. 1. Observe that if G acts transitively on S then |S| divides |G|.2. Let G be the symmetry group of a cube in R3. More precisely G is the collection of

combinations of rotations and reflections in R3 which preserve the a fixed cube. Thegroup law is given by composition of functions.

How big is G?

Any G naturally permutes the vertices of the cube (of which there are 8). Any, G which give the same permutation are clearly equal. This gives a naturalaction of G on the vertices. Fix s {vertices of cube}. Clearly the action is transitive,so |orb(s)| = 8. What about the stabilizer of s? stab(s) = symmetry groups of the cubecorners in R3. This is the same as the symmetry group of an equilateral triangle in R2.There are 6 possible symmetries. Hence |stab(s)| = 6 and therefore |G| = 6 8 = 48.Very clever! What Ive said is vague. Well return to this idea of symmetry later.

2

3. If |G|

Theorem. Let (G, ) be a group and H,K G two subgroups of finite index. Assume that(G : H) = r and (G : K) = s. Then H K is of finite index in G and (G : H K) rs.Proof. By the above theorem let S be a set on which G acts transitively such that for s S,stab(s) = H. Similarly, let T be a set on which G acts transitively such that for t T ,stab(t) = K. By the orbit stabiliser theorem we know that |S| = r and |T | = s. Thenthere is a natural action of G on S T given by g(x, y) = (g(x), g(y)) fora all g G, x Sand y T . Observe that |S T | = rs and by construction H K = stab(s, t). By theorbits-stabiliser the we deduce that

(G : H K) = |orb(s, t)| rs.

4


Alexander Paulin

September 21, 2009

Lecture 7

Recall that if (G, ) is a group and X G is a subset (not necessarily a subgroup) thengp(X) G, the group generated by X, is the minimal subgroup of G containing X. Con-cretely, gp(X) is all possible combinations of elements of X and their inverses. Recall thefollowing definition:

Definition. A group G is said to be cyclic if it can be generated by a single element, i.e.x G such that gp({x}) = G.Remarks. 1. Intuitively, gp({x}) is the set { x2, x1, e, x, x2, } G. Note that we

are using multiplicative notation here, i.e x x = x2. Observe that this set may not ingeneral be infinite. For example, if G is finite then it is forced to be finite. From thisperspective associativity implies it is Abelian.

2. (Z,+) is an infinite cyclic group generated by {1} (or {1}). This shows that genera-tors are not unique. For m N, (Z/mZ,+) is cyclic with generator [a] for any a Zcoprime to m.

Theorem. Let G be a cyclic group. Then

1. If G is infinite, G = (Z,+)2. If |G| = m N, then G = (Z/mZ,+)

Proof. We have two cases to consider.

1. If G = gp({x}), then G = { x2, x1, e, x, x2 }. Assume all elements in this setare distinct, then we can define a map of sets:

: G Zxn n

Then,a, b Z,(xa xb) = (xa+b) = a+ b = (xa) + (xb) so is a homomorphismwhich by assumption was bijective. Thus, (G, ) is isomorphic to (Z,+).

1

2. Now assume a, b Z, b > a such that xa = xb. Then x(ba) = e x1 = x(ba1) G = {e, , xna1}. In particular G is finite. Choose minimal m N such thatxm = e. Then G = {e, x, , xm1} and all its elements are distinct by minimality ofm. Hence |G| = m.

Define the map

: G Z/mZxn [n] for n {0, ...m 1}

This is clearly a surjection, hence a bijection because |G| = |Z/mZ| = m. Again a, b {0, ...m 1} we know (xa xb) = (xa+b) = [a + b] = [a] + [b] = (xa) + (xb) is ahomomorphism. Hence (G, ) is isomorphic to (Z/mZ,+).

Hence two finite cyclic groups of the same size are isomorphic. What are the possiblesubgroups of a cyclic group?

Proposition. A subgroup of a cyclic group is cyclic.

Proof. If H is trivial we are done. Hence assume that H is non-trivial. By the above weneed to check two cases.

1. (G, ) = (Z,+). Let H Z be a non-trivial subgroup. Choose m N minimal suchthat m H(m *= 0). Hence mZ = {ma|a Z} H. Assume n H such thatn / mZ. By the remainder theorem, n = qm + r, r, q Z and 0 < r < m r H.This is a contradiction by the minimality of m. Therefore mZ = H. Observe thatgp({m}) = mZ Z. Hence H is cylic.

2. (G, ) = (Z/mZ,+). Let H Z/mZ be a non-trivial subgroup. Again, choose n Nminimal and positive such that [n] H. The same argument as above shows that thecontainment gp({[n]}) H is actually equality. Hence H is cyclic.

Proposition. Let (G, ) be a finite cyclic group of order d. Let m N such that m divides|G|. Then there is a unique cyclic subgroup of order m.Proof. Because |G| = d we know that G = (Z/dZ,+). Hence we need only answer thequestion for this latter group. Let m be a divisor of d. Then if n = d/m then gp({[n]}) Z/dZ is cyclic of order m by construction. If H Z/dZ is a second subgroup of orderm then by the above proof we know that the minimal n N such that [n] H must ben = d/m. Hence H = gp({[n]}).

2

Let (G, ) be a group (not necessarily cyclic) and x G. We call gp({x}) G the subgroupgenerated by x. By definition it is cyclic.

Definition. If |gp({x})| < we say that x is of finite order and its order, written ord(x)equals |gp({x})|. If not we say that x is of infinite order.Remarks. 1. Observe that by the above we know that if x G is of finite order, then

ord(x) = minimal m N such that xm = e

2. e G is the only element of G of order 1.3. The only element with finite order in Z is 0.

Proposition. Let (G, ) be a finite group and x G. Then ord(x) divides |G| and x|G| = e.Proof. By definition ord(x) = |gp({x})|. Therefore, by Lagranges theorem, ord(x) mustdivide |G|. Also note that by definition xord(x) = e. Hence

x|G| = x(ord(x)|G|

ord(x) ) = e|G|

ord(x) = e.

Symmetric groups

As we have proven, if (G, ) is a finite group of order n. Then G is isomorphic to a subgroupof Symn, the symmetric group on {1, 2...n}. Hence to properly understand finite groups wemust under stand these finite symmetric groups.

Proposition. For n N, |Symn| = n!.Proof. Any permutation of {1, 2...n} is totally determined by a choice of (1), then of (2)and so on. At each stage the possibilities drop by one. Hence the number of permutationsis n!.

We need to think of a way of elegantly representing elements of Symn. For a {1, 2...n}and Symn we represent the action of on a by a cycle:

(abc...f) where b = (a), c = (b)...(f) = a.

We know that eventually get back to a because has finite order. In this way every Symncan be written as a product of disjoint cycles:

= (a1...ar)(ar+1...as)...(at+1...an).

3

This representation is unique up to ordering the cycles.

E.g. Let n = 5 then = (123)(45) corresponds to

1 22 3

: 3 14 55 4

If an element is fixed by we omit it from the notation.

E.g. Let n = 5 then = (523) corresponds to

1 12 3

: 3 54 45 2

This notation makes it clear how to compose two permutations: E.g. Let n = 5 and = (23), = (241), then = (241)(23) = (1234) and = (23)(241) = (1324). Observethat we are keeping our composition on the left notation when composing permutations.This example also shows that in general Symn is not Abelian.

4


Alexander Paulin

December 30, 2010

Lecture 8

Let n N. Recall that Symn is the permutation group of the set {1, , n}. We observedthat every Symn could be written as the product of disjoint cycles.

= (a1, , ar)(ar+1, , as) (at+1, , an)

This representation is unique up to reordering the cycles, i.e. (123)(45) = (45)(123). Hencegiven Symn, we naturally get a well defined partition of n, taking the lengths of thedisjoint cycles appearing in .

Proposition. Let Symn decompose as the disjoint product of cycles of length n1, ..nm(so

ni = n). Then ord() = LCM(n1, ...nm), where LCM denotes the lowest commonmultiple.

Proof. Let = (a1, , ar)(ar+1, , as) (at+1, , an), be a representation of as thedisjoint product of cycles. We may assume that r = n1, etc, without any loss of generality.Observe that a cycle of length d N must have order d in Symn. Also recall that if G isa finite group then for any d N, x G, xd = e ord(x)|d. Also observe that forall d N, d = (a1, , ar)d(ar+1, , as)d (at+1, , an)d. Thus we know that d = e ni|d i. The smallest value d can take with this property is LCM(n1, ...nm).Definition. Suppose that Symn can be represented as the disjoint product of 1 cyclesof length 1, 2 cycles of length 2, ....., and r cycles of length r. In this case we say that has cycle structure equal to 11 22 ... rr . Note that this is not a number, it is a formalsymbol.

Theorem. Two permutations are conjugate in Symn they have the same cycle struc-ture.

Proof. Let , Symn have the same cycle structure. Hence we may represent both in theform:

= (a1, , ar)(ar+1, , as) (at+1, , an),

1

= (b1, , br)(br+1, , bs) (bt+1, , bn).Define Symn such that (ai) = bi i. By construction 1 = . Going through theabove process in reverse, the converse is clear.

Corollary. Conjugacy classes in Symn are indexed by cycle structure.

Proof. Immediate from the above.

Definition. A transposition is a cycle of length 2.

Observe that we can write any cycle as a product of transpositions:

(a1 as) = (a1as) (a1a3)(a1a2)Hence any permutation Symn may be written as the (not necessarily disjoint) productof transpositions. This representation is non-unique as the following shows:e.g. n=6, =(1 2 3)=(1 3) (1 2)=(4 5)(1 3)(4 5)(1 2)Notice that both expressions involve an even number of transpositions.

Theorem. Let Symn be expressed as the product of transpositions in two potentiallydifferent ways. If the first has m transpositions and the second has n transpositions then2|(m n).Proof. First notice that a cycle of length r can be written as the product of r1 transpositionsby the above. Let us call even if there are an even number of even length cycles (onceexpressed as a disjoint product); let us call odd if there are an odd number of even lengthcycles. We also define the sign of , denoted sgn(), to be +1 or 1 depending on whether is even or odd.Consider how sign changes when we multiply by a transposition (1 i). We have two cases:

1. 1 and i occur in the same cycle in . Without loss of generality we consider (1 2 i r) as being in .

(1 i)(1 2 i r)=(1 2 i 1)(i i+ 1 r)

If r is even then either we get two odd length cycles or two even length cycles. Ifr is odd then exactly one of the cycles on the right is even length. In either case,sgn((1 i)) = sgn().

2. 1 and i occur in distinct cycles. Again, without loss of generality we may assume that(1 i 1)(i r) occurs in . In this case

(1 i)(1 2 i 1)(i r)=(1 r).

In either of the cases r even or odd, we see that the number of even length cycles mustdrop or go up by one. Hence sgn((1 i)) = sgn() as in case 1.

2

We deduce that multiplying on the left by a transposition changes the sign of our permuta-tion. The identity must have sign 1, hence by induction we see that the product of an oddnumber of transpositions has sign 1, and the product of an even number of transpositionshas sign 1.

Note that if we write any product of transpositions then we can immediately write downan inverse by reversing their order. Let us assume that we can express as the product oftranspositions in two different ways, one with an odd number and one with an even number.Hence we can write down as the product of evenly many transpositions and 1 as aproduct of an odd number of transpositions. Thus we can write e = 1 as a product ofan odd number of transpositions. This is a contradiction as sgn(e) = 1.

We should observe that from the proof of the above we see that , Symn, sgn() =sgn()sgn(). Observe that we may view the set {1,1} as a group under multiplication.It is cyclic of order 2, generated by -1. Let us denote it by C2. What this shows is that wemay view sgn as a group homomorphism:

sgn : Symn C2 sgn()

In particular this shows that the set of even elements of Symn is closed under compositionand taking inverse. Hence we have the following:

Definition. The subgroup of Altn Symn consisting of even elements is called the Alter-nating group of rank n.

Observe that Altn contains all 3-cycles.

Proposition. Altn is generated by 3-cylces.

Proof. By generate we mean that any element of Altn can be expressed as the product ofthree cycles. As any element of Altn can be written as the product of three cycles we onlyhave to do it for the product of two transpositions. There are two cases:

1. (i j)(k l) = (k i l)(i j k).

2. (i j)(i k) = (i k j).

3


Alexander Paulin

September 30, 2010

Lecture 9

Recall that last time we defined the alternating subgroup Altn Symn to be the collectionof all even permutations. An even permutation is one that can be written (in any way) asthe product of evenly many transpositions.

Proposition. |Altn| = n!2 .Proof. Recall that |Symn| = n!, hence we just need to show that (Symn : Altn) = 2. Let, Symn. Recall that

Altn = Altn 1 Altn.But sgn(1) = sgn()sgn(), hence

Altn = Altn sgn() = sgn().Hence Altn has two right cosets in Symn, one containing even permutations and one oddpermutations.

Symmetry of Sets with Extra Structure

Let S be a set and (S) its permutation group. The permutation group completely ig-nores the fact that there may be extra structure on S. Today we are going to look at thepermutations which preserve certain geometric structures.

Within mathematics there are objects which have internal structure. For example, a cubesitting in R3 is not just a set of points, it has some sort of geometric structure. So far in thiscourse we have already encountered sets with exact structure, namely groups. We definedmaps between groups which preserved their structure - we called these homomorphisms. Let(G, ) be a group. The collection of automorphisms from G to itself form a group (undercomposition), hence they naturally form a subgroup aut(G) (G). Similarly, the collectionof invertible linear maps from Rn to itself form a subgroup of (Rn). Both of these examplesare for abstract algebra, but there is a much more obvious class:

1

Symmetry in Euclidean Space

Observe that in R2 we have a natural concept of the distance between two points : ifx = (x1, x2) and y = (y1, y2) then by Pythagoras theorem the distance from x to y is

d(x,y) =(

2i=1

(xi yi)2).

Similarly in R3 we can again us Pythagoras to show that for x = (x1, x2, x3) and y =(y1, y2, y3),

d(x,y) =(

3i=1

(xi yi)2).

For n N we may extend this concept of distance to Rn as follows: for x = (x1, xn) andy = (y1, yn) we define the distance between x and y to be

d(x,y) =(

ni=1

(xi yi)2).

This gives a good concept of distance on Rn because it shares many good properties withthe usual concept in R2 and R3. For example d(x,y) = 0 x = y. In the language ofanalysis d is a metric on Rn. It is called the Euclidean metric on Rn.

Definition. The set Rn together with the Euclidean metric d is called Euclidean space ofdimension n.

This gives the set Rn some extra structure which is fundamentally geometric. Note thatthis is more structure than the fact that Rn is a vector space over R.

Definition. An isometry on Rn is a map of sets f : Rn Rn (not necessarily linear) suchthat x,y Rn, d(x,y) = d(f(x), f(y)).

Intuitively an isometry preserves distance. In R3 examples include rotations around anyline, or reflections in any plane. In R2 we have rotation around points and reflections in anyline. We can also translate by x Rn i.e.

fx : Rn Rny y+ x.

This example clearly shows that an isometry need not be linear, taking x '= 0.

We denote the collection of all isometries of Rn by Isom(Rn). It is clear that Isom(Rn)is closed under composition and contains the identity map.

We must address wether isometries are bijective. It is clear from the definition that iff Isom(Rn) is injective. Let us restrict for a moment to the case when n = 2. Assume

2

we are given three points x1,x2,x3 R2 which form the corners of a triangle. Observe thatgiven any y R2 it is uniquely determined by its distance from x1,x2 and x3. This tellsus that given f Isom(R2), f(y) is uniquely determined by f(x1), f(x2) and f(x3). Con-versely, because f is an isometry f(x1), f(x2) and f(x3) must form the corners of a trianglein R2 which is congruent to the one with corners x1,x2 and x3. Hence given z R2, it isuniquely determine by its distance from f(x1), f(x2) and f(x3). Because the two trianglesare congruent there exists a unique y R2 with these same distances from x1,x2 and x3.This forces f(y) = z., and hence f is surjective, and thus bijective. Clearly its inverse is anisometry also. This basic argument can be extended to Rn for any n N. We deduce thatIsom(Rn) (Rn) is a subgroup.

Let X Rn be a subset (not necessarily a subspace) .Definition. . We define the Symmetry group of X to be the subgroup Sym(X) Isom(Rn)with the property that f Sym(X) f(X) = X.

There is a natural action of Sym(X) on the set X, coming from the fact that Sym(X) (X). Lets do some examples:

The Dihedral Group

Let m N and n = 2. Let X R2 be the regular m-gon. We call the symmetry group ofX the dihedral group of rank m, and we denote it by Dm. First observe that every elementof Dm must fix the center of X. Also observe that f Dm is totally determined by what itdoes to the vertices of X. By the above we see that it must permute the m vertices (i.e. Dmacts on the m vertices) and the action must be faithful. Hence there is a natural injectivehomomorphism Dm Symm. The subset of rotations about the center forms a subgroupor order m, which we denote by Rot Dm. Note that using rotation we can move from onevertex to any other hence the action is transitive. Also observe that given any vertex s thestabiliser subgroup is {e, } Dm, where is the reflection through s and the center. Thusby orbit-stabiliser we deduce that |Dm| = 2m. So Rot has two right cosets in Dm, Rot itselfand Ref , the subset of reflections.

The Cube

Let X R3 be a solid cube. What potential isometries can preserve X. They clearlyhave to fix the center. They also have to fix the 8 vertices, and by the above permutethem. Again by the above distinct elements of Sym(X) must induce distinct permutationsof the vertices. This induces an injective homomorphism Sym(X) Sym8 Can we think ofany other subsets of X which must be permuted? The central points of each face must bepermuted also. There are six such points and again this induces an injective homomorphismSym(X) Sym6. Clearly both actions are transitive. In the If s is a vertex then stab(s)is naturally isomorphic to D3. It t is the center of a face then stab(t) is isomorphic to D4.Orbit-stabiliser in either case shows that |Sym(X)| = 48.

3

Deep Question

Let (G, ) be an abstract group. When is it true that we can find X Rn, for some n Nsuch that

G = Sym(X)?Heuristically, when can an abstract group be realised in geometry.

4


Alexander Paulin

September 30, 2010

Lecture 9

In Linear Algebra the predominant objects we study are the maps between vector spaces,and not the vector spaces themselves. Somehow the structure preserving maps betweenvector spaces are more interesting than the spaces themselves. This a deep observation - itis true far beyond the confines of linear algebra. Philosophically its saying that an objectin isolation is uninteresting; its how it relates to whats around it that matters. The worldof group theory is no different. Here the objects are groups and the maps between themare homomorphisms. Today well study homomorphisms between abstract groups in moredetail.

Let G and H be two groups. Well suppress the notation as it will always be obviouswhere composition is taking place. Let eG and eH be the respective identity elements.Recall that a homomorphism from G to H is a map of sets f : G H such that x, y G,f(xy) = f(x)f(y).

Lemma. Let f : G H be a homomorphism. Then x G, f(x1) = f(x)1.Proof. Recall that f a homomorphism f(eG) = eH . Hence f(xx1) = f(x)f(x1) = eH =f(x1x) = f(x1)f(x).

Definition. Given f : G H a homomorphism of groups, we define the kernel of f to be:ker(f) := {x G|f(x) = eH}

We define the image of f to be:

Im(f) := {y H|x G such that f(x) = y}Proposition. Given a homomorphism f : G H, ker(f) G and im(f) H are sub-groups.

Proof. First we will show true for ker(f):

1. f(eG) = eH eG ker(f).

1

2. Suppose x, y ker(f). Then f(xy) = f(x)f(y) = eH xy ker(f).3. Given x ker(f), f(x1) = e1H = eH x1 ker(f).

Now we will show that Im(f) is a subgroup:

1. f(eG) = eH so eH Im(f).2. f(xy) = f(x)f(y)x, y G so Im(f) is closed under composition.3. Note that f(x)1 = f(x1) y Im(f) y1 Im(f).

Prop. f : G H a homomorphism is injective if and only if ker(f) is trivial.Proof. f injective ker(f) = {eG} trivially. Now assume ker(f) = {eG}. Suppose x, y Gsuch that f(x) = f(y).

f(x) = f(y) f(x)f(y)1 = eH f(x)f(y1) = eH f(xy1) = eH xy1 = eG x = y

Thus f is injective.

Recall that the subgroups of (Z,+) were exactly the subsets of the form mZ := {ma|a Z}for some m N. One of the first things we saw was that the collection of right cosets Z/mZinherited the structure of a group from Z. It would be reasonable to expect that this wastrue in the general case, i.e. given G a group and H a subgroup the set G/H naturallyinherits the structure of a group from G. To make this a bit more precise lets think aboutwhat naturally means. Let xH, yH G/H be two right cosets. Recall that x and y arenot necessarily unique. The only obvious way for combining xH and yH would be to form(xy)H. Warning: in general this is not well defined. It will depend on the choice of x andy. Something very special happens in the case G = Z and mZ = H.

Definition. We call a subgroup H G normal if, for all g G, gHg1 = {ghg1|g G, h H} = H.Remarks. 1. Observe that this is not saying that given g G and h H, then ghg1 =

h. It is merely saying that ghg1 H. A normal subgroup is the union of conjugacyclasses of G.

2. If G is Abelian, every subgroup is normal as ghg1 = hg, h G.

2

3. Let G=Sym3, H = {e, (12)}. Then (13)(12)(13) = (23) / HHence H is not normal in Sym3.

Proposition. Let G and H be two groups. Let f : G H a homomorphism. Thenker(f) G is a normal subgroup.Proof. Let h ker(f) and g G. Then f(ghg1) = f(g)f(h)f(g1) = f(g)eHf(g)1 =eH ghg1 ker(f).In general Im(f) H is not normal.Definition. We say a group G is simple if its only normal subgroups are {e} and G.Cyclic groups of prime order are trivially simple by Lagranges theorem. It is in fact truethat for n 5, Altn is simple, although proving this will take us too far afield.

The importance of normal subgroups can be seen in the following:

Proposition. Let H G be a normal subgroup. Then the binary operation:G/H G/H G/H(xH, yH) ( (xy)H

is well defined.

Proof. As usual the problem is that hat coset representatives are not unique and thus wecould have two representatives giving different maps. Thus our goal is to show:

x1, x2, y1, y2 G such that x1H = x2H and y1H = y2H, then (x1y1)H = (x2y2)HBy assumption we know x11 x2, y

11 y2 H. Consider

u = (x1y1)1(x2y2) = y11 x11 x2y2

Hence uy12 y1 = y1(x11 x2)y. Therefore, by the normality of H, uy

12 y1 H u H

(x1x2)H = (y1y2)H.

This shows that ifH G normal, G/H can be endowed with a natural binary operation.Proposition. Proposition Let G be a group; H G a normal subgroup. Then G/H is agroup under the above binary operation.

Proof. Simple check of three axioms of being a group.

1. x, y, z G, (xy)z = x(yz) (xH yH) zH xH (yH zH).2. xH H = xH = H xH H G/H is the identity.3. xH x1H = xx1H = eH = H = x1xH = x1H xH inverses exist.

3


Alexander Paulin

October 7, 2009

Lecture 11

Let G be a group, H G a subgroup. Recall that H is normal if g G, gHg1 = H. Theimportant thing about normal subgroups was that the binary operation:

G/H G/H G/H(xH, yH) xyH

is well defined. This critically needed that H was normal in G. Recall that under this binaryoperation G/H is a group.

Proposition. The natural map

: G G/Hx xH

is a homomorphism with ker() = H.

Proof. Observe that x, y G, (xy) = xyH = xHyH = (x)(y) is a homomorphism.Recall that the identity element in G/H is the coset H. Hence for x ker() (x) = xH = H x H. Hence ker() = H.Observe that this shows that any normal subgroup can be realised as the kernel of a grouphomomorphism.

The First Isomorphism Theorem

Let G and H be groups, with respective identities eG and eH . Let : G H be a homomor-phism. Recall that ker() G is a normal subgroup. Hence we may form the quotient groupG/ker(). Let x, y G such that they are in the same right coset of ker(). Recall thatxker() = yker() x1y ker() (x1y) = eH (x1)(y) = eH (x)1(y) = eH (x) = (y). In summary, (x) = (y) xker() = yker()

1

Hence is constant on each coset of ker().

Observe that this is very similar to the observation that if G acts on a set S and s Swe get a map from G to S by evaluating g at s. We observed that this map is constanton right cosets of stab(s). Formally, this is just the same.

Hence we get a map of sets (well also call it ):

: G/ker() Im()xker() (x)

This is well define precisely because of the above observations.

The First Isomorphism Theorem. Let G and H be two groups. Let : G H be ahomomorphism, then the induced map

: G/ker() Im()xker() (x)

is an isomorphism of groups.

Proof. Firstly we observe that the induced is by definition of Im() surjective. Note thatgiven x, y G, (xker()) = (yker()) (x) = (y) xker() = yker(),hence is injective. Be aware that we are abusing the notation here - means two differentthings depending on the context. Observe that this is pretty much identical to the proof ofthe orbit-stabliser theorem.

It is left for us to show that is a homomorphism. Given x, y G, (xker()yker()) =(xyker()) = (xy) = (x)(y) = (xker())(yker()).Therefore : G/kerIm is a homomorphism, and thus an isomorphism

The Third Isomorphism Theorem

Let G be a group and N a normal subgroup. The third isomorphism theorem concerns theconnection between certain subgroups of G and subgroups of G/N .

Let H be a subgroup of G containing N . Observe that N is automatically normal inH. Hence we may form the quotient group H/N = {hN |h H}. Observe that H/N isnaturally a subset of G/N .

Lemma. H/N G/N is a subgroup.Proof. We need to check the three properties.

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1. Recall that N G/N is the identity in the quotient group. Observe that N H N H/N .

2. Let x, y H. By definition xy H. Thus xNyN = (xy)N H/N .3. Let x H. By definition x1 H. Thus (xN)1 = x1N H/N .

Conversely, let F G/N be a subgroup. Let HM G be the union of the right cosetsof M .

Lemma. H G is a subgroup.Proof. We need to check the three properties.

1. Recall that N G/N is the identity in the quotient group. Hence N M N HM .N is a subgroup hence eG N eG HM .

2. Let x, y HM . This implies that xN, yN M . M is a subgroup, hence xNyN =xyN M . This implies that xy HM .

3. Let x HM . Hence xN M . M is a subgroup, hence (xN)1 = x1N M . Thisimplies that x1 HM .

Hence we have two maps of sets:

: {Subgroups of G containing N} {Subgroups of G/N}H H/N

and

: {Subgroups of G/N} {Subgroups of G containing N}M HM

Proposition. These maps of sets are inverse to each other.

Proof. We need to show that composition in both directions gives the identity function.

1. Let H be a subgroup of G containing N . Then (H) = (H/N) = H. Thus isthe identity map on {Subgroups of G containing N}.

2. Let M be a subgroup of G/N . then (M) = (HM) = M . Thus is the identitymap on {Subgroups of G/N}.

3

We deduce that both and are bijections and we have the following:

The Third Isomorphism Theorem. Let G be a group and N G a normal subgroup.There is a natural bijection between the subgroups of G containing N and subgroups of G/N .

Proof. Either map or exhibits the desired bijection.

4


Alexander Paulin

October 13, 2009

Lecture 11

Let G and H be two groups, with respective identities eG and eH . We may form the directproduct GH = {(x, g)|x G g H}. Let x, y G and g, h H. Observe that there isa natural binary operation on GH given by:

(x, g) (y, h) := (xy, gh).Lemma. GH is a group under the natural binary operation.Proof. 1. Associativity holds for both G and H associativity hold for GH.

2. (eG, eH) is the identity.

3. For g G and h H (g, h)1 = (g1, h1).

There is an obvious generalization of this concept to any finite collection of groups. Let Gbe a group and H,K G two subgroups. Let us furthermore assume that

1. h H k K hk = kh.2. Given g G there exist unique h H, k K such that g = hk.Under these circumstances we say that G is the direct sum of H and K and we write

G = H K. Observe that the second property is equivalent to:

3. H K = {eG} and for g G there exist h H, k K such that g = hk.Proposition. If G is the direct sum of the subgroups H,K G then G = H K.Proof. Define the map

: H K G(h, k) hk

Let x, y H and g, h K. By property one ((x, g)(y, h)) = (xy, gh) = xygh = xgyh =(x, g)(y, h). Hence is a homomorphism. Property two ensures that is bijective.

1

The concept of direct sum has a clear generalization to any finite collection of subsets ofG. Example: (Z/15Z,+) is the direct sum of gp([3]) and gp([5]).

Abelian Groups

Let G be an Abelian group. We shall use additive notation to express composition within G.In particular we will denote the identity by 0 (not to be confused with 0 Z). We do thisbecause we are very familiar with addition on Z being commutative. Assume furthermorethat G is finitely generated. Hence {a1, , an} G such that gp({a1, , an}) = G. Inother words, because G is Abelian, every x G can be written in the form

x = 1a1 + + nan i Z.Observe that iai means ai (or ai) composed with itself |i| times. In general such anexpression is not unique. For example is G is of order m N then (m + 1)a = a for alla G. This is because ma = 0 A reasonable goal would be to find a generating set such thatevery expression of the above form was unique (after possibly restricting 0 1 < ord(ai))for a given x G. Such a generating set is called a basis for G. Observe that it is notclear that such a basis even exists at present. If {a1, , an} G were a basis then lettingAi = gp(ai) G we have the direct sum decomposition:

G = A1 An.Conversely, if G can be represented as the direct sum of cyclic subgroups then choosing agenerator for each gives a basis for G.

Definition. Let G be an Albeian group. x G is torsion is it is of finite order. We denotethe subgroup of torsion elements by tG G, called the torsion subgroup.Lemma. tG G is a subgroup.Proof. This critically requires that G be Abelian. It is not true in general.

1. ord(0) = 1 0 tG2. Let g, h tG n,m N such that ng = mg = 0 nm(g + h) = (mng + nmh) =

m0 + n0 = 0 g + h tG.3. ng = 0 (ng) = n(g) = 0. Hence g tG g tG.

Clearly if G is finite then tG = G.

Definition. If tG = G we say that G is a torsion group. If tG = {0} we say that G istorsion free.

2

Proposition. If G is torsion an finitely generated then G is finite.

Proof. Let {a1, , an} G be a generating set. Each element is of finite order hence everyelement x G can be written in the form

x = 1a1 + + nan, i Z, 0 1 < ord(ai).This is a finite set.

Proposition. G/tGis a torsion free Abelian group.

Proof. Firstly note that tG G is normal as G is Abelian, hence G/tG is naturally anabelian group. Let x G. Assume that x + tG G/tG is torsion. Hence n N suchthat n(x + tG) = nx + tG = tG. Hence nx tG so m N such that mnx = 0. Hencex tG xtG = tG.Definition. An finitely generated abelian group G is said to be free abelian if there exists afinite generating set {a1, , an} G such that every element of G can be uniquely expressedas

1a1 + nan where i Z.In other words, if we can find a basis for G consisting of non-torsion elements.

In this case

G = gp(a1) gp(an) = Z Z Z = Zn.Proposition. Let G be a finitely generated free abelian group. Any two bases must have thesame cardinality.

Proof. Let {a1, , an} G be a basis. Let 2G := {2x|x G}. 2G G is a subgroup.Observe that 2G = {1a1 + nan| 2]z}. Hence (G : 2G) = 2n. But the left hand sideis defined independently of the basis. The result follows.

Definition. Let G be a finitely generated free Abelian group. The rank of G is the size ofa any basis.

Theorem. A finitely generated abelian group is free Abelan it is torsion free.Proof. is trivial.Assume G is torsion-free, let {a1, , an} G generate G. We will prove the result byinduction on n.Base Case: n = 1. G = gp(a) = (Z,+) which is free abelian. Therefore result is true forn = 1.If {a1, , an} G is a basis we have nothing to prove. Suppose that it is not a basis. thenwe have a non-trivial relation:

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1a1 + 2a2 + + nan = 0If d Z such that d|i for all i, then have d(1a1d + 2a2d + + ...) = 0. As G is torsion-free,(1a1d +

2a2d + + ...) = 0. We can therefore assume that the i are collectively coprime.

If 1 = 1, then we can shift terms to get a1 = (2a2 + 3a3 + + nan). Therefore, Gis generated by the {a2, , an} G and the result follows by induction. We will reduceto this cases as follows: Assume |1| |2| > 0. By the remainder theorem we may choose Z such that |1 2| < |2|. Let a2 = a2 + a1 and 1 = 1 2, then

1a1 + 2a2 + + nan = 0.

Also observe that {a1, a2, , an} G is still a generating set and {1, ,n} are stillcollectively coprime. This process must must eventually terminate with one of the coefficientsequal either 1 or 1. In this case we can apply the inductive step as above to conclude thatG is free abelian.

Proposition. Let G be finitely generated and Abelian. Then G/tG is a finitely generatedfree Abelian group.

Proof. G/tG is torsion free. We must show thatG/tG is finitely generated. Let {a1, , an} G generate G. Then {a1 + tG, , an + tG} G/tG forms a generating set. By the abovetheorem G/tG is free Abelian.

Definition. Let G be a finitely generated Abelian group. We define the rank of G to be therank of G/tG.

Let G be finitely generated and Abelian. Let G/tG be of rank n N and let f1, , fnbe a basis for G/tG. Let : G G/tG be the natural quotient homomorphism. Clearly is surjective. Choose {e1, , en} G such that (ei) = fi i {1, , n}. None of the fihave finite order none of the ei have finite order. Moreover

(1e1 + + nen) = 1f1 + + nfn G/tG.Because {f1, , fn} is a free basis for G/tG we deduce that 1e1 + + nen = 0 i = 0i F := gp{e1, , en} G is free abelian with basis {e1, , en} F is torsionfree. Therefore F tG = {0}.Let g G. By definition, 1, ,n Z such that (g) = 1f1 + + nfn. Then wehave:

(g) = 1f1 + + nfn (g) = (1e1 + + nen) (g (1e1 + + nen)) = 0 g (1e1 + + nen) ker = tG h tG s.t. g = (1e1 + + nen) + h

Hence every x may be written uniquely in the form x = f + g where f F and g tG.

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Proposition. Every finitely generated Abelian group can be written as a direct sum of a freegroup and a finite group.

Proof. By the above, we may write

G = F tGDefine the homomorphism :

G = F tG tGf + h h

This is surjective with kernel F , hence by the first isomorphism theorem tG is isomorphic toG/F . The image of any generating set of G is a generating set for G/F under the quotienthomomorphism. Hence tG is finitely generated and torsion, hence finite. F is free abelianby construction.

5


Alexander Paulin

October 17, 2010

Lecture 13

Let G be a finitely generated Abelian group. tG G, the torsion subgroup, is the subsetof all elements of finite order. Recall that because it is isomorphic to a quotient of G itis finitely generated. Hence tG is finite. G/tG is torsion free and finitely generated, hencefree Abelian. Free Abelian means we can find a basis consisting of torsion free elements, i.e.n N such that G/tG = (Zn,+). We call n the rank of G. We also saw that we can finda free subgroup of G, denoted F G of rank equal to the rank of G such that G = F tG.Hence we have reduced the study of finitely generated Abelian groups to understanding finiteAbelian groups.

Finite Abelian Groups

Definition. A finite group G (not necessarily Abelian) is a p-group, with p N a prime,if every element of G has order a power of p.

It is not clear that the order of a p-group is a power of p. This is in fact true but we will notprove it in the general case. From now on let G be a finite Abelian group. Let p N be aprime. We define Gp := {g G|ord(p) is a power of p} G.Theorem 1. Gp G is a subgroup.Proof. 1. ord(0) = 1 = p0 0 Gp.

2. Let g, h Gp r, s N such that prg = psh = 0 pr+s(g+h) = ps(prg)+pr(psh) =0 + 0 = 0 g + h Gp.

3. Let g Gp r N such that prg = 0 prg = pr(g) = 0 g Gp

This critically relies on G being Abelian. By definition Gp is a p-group. Recall that g G,ord(g)||G| by Lagranges Theorem. Therefore Gp = 0 unless possibly if p divides |G|. Atthis point it is in no way obvious that if p divides the order of G, there is an element oforder p. Gp G is the maximal p-subgroup contained in G. The importance of the maximalp-subgroups is the following theorem.

1

Theorem. Let G is a finite Abelian group. Let {p1, , pr} be the primes dividing |G|.Then

G = Gp1 GprMoreover this is the unique way to express as the direct sum of p-subgroups for distinctprimes.

Proof. Let |G| = n = a1a2 ar where ai = pii . Let Pi = n/ai. {P1, , Pr} Z arecollectively coprime Q1, , Qr Z such that

P1Q1 + + PrQr = 1 (Extension of Euclid)Let g G and gi = PiQig. Clearly g = g1 + g2 + + gr and pii gi = Qi(ng) = 0. Hencegi Gpi .We must prove the uniquness of this sum. Assume we had

g = g1 + + gr, gi Gpi .Therefore x = g1 g1 = (g2 g2) + (g3 g3) + + (gr gr). The right hand size has orderdividing P1, the left hand side has order dividing Q1. P1 and Q1 are coprime u, v Zsuch that up1 + vq1 = 1 x = u(p1x) + v(q1x) = 0 + 0 = 0 g1 = g1. Similarly we findgi = gi for all i {1, , r}, hence the sum is unique and we deduce

G = Gp1 Gpr .Let {qi, , qs} be a finite collection of distinct primes. Assume that G can be expressed

as the direct sum

G = H1 Hs = H1 Hswhere Hi is a finite qi-subgroup. Clearly Gqi = Hi and if p is a prime not in {q1, , qs}Gp = {0}. Thus {p1, , pr} = {q1, , qs} and any such representation is unique.Observe that we have still not shown that just because p divides |G| then Gp is non-trivial.We have however reduced the study of finite abelian groups to finite abelian p-groups.

Theorem 1. Every finite Abelian p-group is a direct sum of cyclic groups.

Proof. Let G be a finite Abelian p-group. If G is cyclic, we are done, otherwise take acyclic subgroup B = gp(b) of maximal order, say pn. Our strategy is to show that there is ap-subgroup D G such that G = B D. We apply the following inductive hypothesis: Forany finite Abelian p-group F of size less than |G|, if M F is a maximal cyclic subgroupthen there exists N F such that M N = F . This is clearly true for F trivial.

We claim that there is a subgroup C of order p such that B C = {0}. Recall that becauseG is Abelian G/B is naturally an Abelian p-group. Let c G \ B and suppose cB G/B

2

has order pr for r > 0. Observe that the maximal order of any element in G/B is less thanor equal to pn. Thus we know n r. By definition pr(cB) = B prc B. Thus thereexists s N such that prc = sb. By maximality of the order of b we know 0 = pnc = spnrb.But ord(b) = pn, hence pn|spnr. Therefore we have p|s, say s = ps. Hence c1 = pr1c sbhas order p and is not in B. Therefore C = gp(c1) is the required subgroup.

Let BC = {ab|a B, b C}. We claim that BC G is a subgroup.

1. eG B and eG C eG BC.2. Let a1, a2 B, b1, b2 C. Then (a1b1)(a2b2) = (a1a2)(b1b2) BC. Hence BC is closed

under composition.

3. Let a1 B, b1 C. Then (a1b1)1 = b11 a11 = a11 b11 BC. Hence BC is closedunder taking inverses.

First observe that |G/C| < |G|. Hence the inductive hypothesis applies to G/C. Ob-serve that BC G is a subgroup containing C. Observe that BC/C is cyclic, generated bybC BC/C. Because BC = {0} we also know that |BC/C| = pn. Note that the size of themaximal cyclic subgroup of G must be larger than or equal to the size of the maximal cyclicsubgroup of G/C. However we have constructed a cyclic subgroup BC/C G/C whoseorder equals that of a B. Hence BC/C G/C is a maximal cyclic subgroup. Thus by ourinductive hypothesis N G/C such that BC/C N = G/C. By the third isomorphismtheorem we know that N = D/C for a unique subgroup D G containing C. We claimthat G is the direct sum of B and D.

Let g G. Then gC G/C is uniquely expressible in form g + C = (a + C) + (d + C) =(a+d)+C, where a B and d D. Hence g = a+d+e for some c C. However C D sothis expresses g as a sum of elements of B and D. Let x BD. Hence xC BC/CD/C.Assume that x )= 0. Note that x / C. Hence xC is non-zero on BC/C and D/C. Howeverby construction BC/C D/C = {C}. This is a contraction. Hence B D = {0} and wededuce that G = B D.

Thus we have shown that given any finite Abelian p-group G and a maximal cyclic sub-group B G, there exists a subgroup D G such that G = B D. Observe that D is afinite Abelian p-group, thus we can continue this process until eventually it must terminate.The end result will be an expression of G as a direct sum of cyclic p-groups.

Corollary. For any finite Abelian p group G , there exist a unique decreasing sequence ofnatural numbers {r1, , rn} N such that

G = Z/pr1Z Z/prnZ.

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Proof. By the previous theorem we may express G as the direct sum (hence direct product)of cyclic p-groups. This is done by induction on |G|. It remains to show uniqueness. Let{s1, , sm} N be another decreasing sequence with the above property. The size of amaximal cyclic subgroup of such a G must be pr1 = ps1 . Hence r1 = s1. Let B G be amaximal cyclic subgroup. Hence

G/B = Z/pr2Z Z/prnZ = Z/ps2Z Z/psmZ.By induction on |G| we know that m = n and ri = si for all i.Basis Theorem for Finitely Generated Abelain Groups. Every finitely generatedAbelian group G can be written as a direct sum of cyclic groups:

G = 1 rwhere each i is either infinite or of prime power order, and the orders which occurs areuniquely determined.

Proof. G=F tG. F is free and finitely generated, hence the direct sum of infinite cyclicgroups (Z,+). The number equals the rank of G. tG is finite Abelian, hence the is theunique direct sum of p-groups for distinct primes p. Each p-group is the unique direct sum(up to order) of p-power cyclic groups.

Note that we could have stated this theorem with direct product in place of direct sum.

4


Alexander Paulin

October 19, 2009

Lecture 14

Let p N be a prime. Let G be a finite p-group. We saw last time that G could be writtenas the direct sum (equivalently direct product) of cyclic subgoups. The following shows theuniqueness of this decomposition.

Corollary. For any finite Abelian p-group G , there exist a unique decreasing sequence ofnatural numbers {r1, , rn} N such that

G = Z/pr1Z Z/prnZ.Proof. By the previous theorem we may express G as the direct sum (hence direct product)of cyclic p-groups. . It remains to show uniqueness of the ri. This is done by induction on|G|. We will assume the result holds for groups of size strictly less than |G|. This is clearlytrue for G trivial.

Let {s1, , sm} N be another decreasing sequence with the above property. The sizeof a maximal cyclic subgroup of such a G must be pr1 = ps1 . Hence r1 = s1. Let B G bea maximal cyclic subgroup. Hence

G/B = Z/pr2Z Z/prnZ = Z/ps2Z Z/psmZ.By induction on |G| we know that m = n and ri = si for all i.

Hence if G is a finite Abelain p-group this shows that |G| = pn for some n N.Proposition. Let G is an Abelian group such that p N is a prime dividing |G|. Then Gpis non-trivial.

Proof. Recall that if {p1, , pr} are the primes dividing |G| then

G = Gp1 Gpr .Hence |G| = |Gp1| |Gpr |. By the above corollary pi divides |G| if and only if Gpi is

non-trivial.

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Basis Theorem for Finitely Generated Abelain Groups. Every finitely generatedAbelian group G can be written as a direct sum of cyclic groups:

G = 1 rwhere each i is either infinite or of prime power order, and the orders which occurs areuniquely determined.

Proof. Recall that G may be written in the form G = F tG, where F is free Abelian andtG is finite Abelian. Hence F is the direct sum of infinite cyclic groups (Z,+). The numberequals the rank of G. tG is finite Abelian, hence is the unique direct sum of p-groups fordistinct primes p. Each p-group is the unique direct sum (up to order) of p-power cyclicgroups. The uniqueness of orders in the infinite case is because rank is well defined. In thefinite case by the all of the above.

Note that we could have stated this theorem with direct product in place of direct sum.Hence every finitely generated Abelian group has a basis. As powerful application of thetheorem is for example that all finite abelian groups of order 2 3 5 are isomorphic toZ/2Z Z/3Z Z/5Z.

Rings and Fields

A group (G, ) is a set with a binary operation satisfying three properties. The motivationfor the definition reflected the behavior of (Z,+). Observe that Z also comes naturallyequipped with multiplication . In the first lectures we collected some of the properties of(Z,+,). Motivated by this we make the following fundamental definition:Definition. A ring is a set R with two binary operations, x + y called addition and xycalled multiplication such that:

1. R is an abelian group under addition

2. R is a monoid under multiplication (no inverses necessarily exist)

3. + and are related by the distributive law:(x+ y)z = xz + yz and x(y + z) = xy + xzx, y, z R

The identity for + is zero, denoted 0, and the identity for is one, denoted 1.Distributivity implies that we can multiply together finite sums:

(

xi)(

yj) =

xiyj

in a well defined way. Note that is in general not commutative. If is commutative, i.e.xy = yx x, y R then we say that R is a commutative ring.

2

Definition. Let R and S be two rings. A homomorphism from R to S is a map of sets : R S which preserves both operations, i.e. x, y R

1. (x+ y) = (x) + (y)

2. (xy) = (x)(y)

3. (1R) = 1S

Note that R and S are abelian groups under + so is a group homomorphism with respectto + so (0R) = 0S. We have to include (3) as (R,) is only a monoid so it does not followfrom (2).As before, an isomorphism is a bijective homomorphism, or equivalently one with aninverse homomorphism. A homomorphism from R to itself is called an endomorphism.An endomorphism which is also an isomorphism is called an automorphism. This is ex-actly the same terminology as for groups!

In any ring R we have

x0 = x(0 + 0) = x0 + x0 x0 = 0Similarly, 0x = 0.If R consists of one element, then 1 = 0, conversely if 1 = 0 then x R, x = x1 = x0 = 0,hence R consists of one element. The ring with one element is called the trivial ring. In aring we abbreviate expressions like

a+ a+ a+ + a (n times) = na(n N)

It is clear that we may naturally extend this to all n Z.Similarly,

a a a (n times) = an for n N.By the Distributive Law, we have the identities

1. m(a+ b) = ma+mb

2. (m+ n)a = ma+ na

3. (mn)a = m(na)

a, b R;m,n ZDefinition. Let R be a ring. An element a R is said to be invertible or a unit if ithas a multiplicative inverse, i.e. a R such that aa = aa = 1. We know that such aninverse is unique if it exists, hence we shall write it as a1. Note that if 1 '= 0 then 0 isnever invertible. We denote the set of units in R by R.

3

It is clear that for any ring R, (R,) is a group.Definition. A ring R in which every non-zero element is invertible is called a divisionring (or skew field). If R is a commutative division ring then R is called a field.

(Q,+,) is the canonical example of a field.Remarks. 1. (Q,+,) is the canonical example of a field. Other natural examples in-

clude (R,+) and C,+). It is not clear that there exists division rings which are notfields. The quaternions, as discovered by Hamilton in the 19th century, are an exampleof importance in physic.

2. All of linear algebra (except the issue of eigenvalues existing) can be set up over anarbitrary field. All proofs are exactly the same, we never used anything else about R orC.

Given R and S two rings we say that S is a subring of R if it is a subset and is a ring underthe induced operations (with same 0 and 1). Eg. (Z,+,) (Q,+,)If S is merely isomorphic to S1 R a subring, we say that S can be embedded in R. Thisis because if : S S1 is an isomorphism then we may identify S with S1 via as a subringof R, i.e. the composition

: S1 S R is injective.This is the same as for groups. : H G an injective group homomorphism allows us

to think of H as a subgroup of G.

Some examples of rings

1. Let S be a set and be the set of all subsets. On define + and by

X + Y = (X Y ) (X Y ), XY = X Y

Where X denotes the complement of X in S. Then is a ring with = 0 and S = 1.Applications to mathematical logic.

2. In linear algebra the collection of linear maps from Rn to Rn is the setMnn(Rn). Thishas the structure of a ring under the usual addition and multiplication fo matrices.More generally, if R is any ring, then Mnn(R) is a ring in the same way. Note that ingeneral, Mnn(R) is not commutative, even though R may be.

Let n = 2 and R = R. Let A=(

0 10 0

)and let B=

(0 10 0

).

Clearly AB = 0 but A (= 0 and B (= 0. This shows a fundamental difference betweenthis ring and (Z,+,), where this clearly cannot happen. This motivates the followingdefinitions:

4

Definition. Given a != 0, if there exists b != 0 such that ab or ba is 0, then a is saidto be a zero-divisor; otherwise it is a non-zero divisor. Note that 0 is neither bydefinition.

Definition. A non-trivial ring R with no zero divisors is said to be entire; a commutativeentire ring is called an integral domain.

If R is entire then 1 != 0 and x, y R, xy = 0 x = 0 or y = 0.

(Z,+,) is an integral domain. The above example shows that M2(R) is not entire.

5


Alexander Paulin

October 26, 2009

Lecture 14

Recall that a ring R is a set together with an abstract concept of + and which mimic(Z,+,).

Given R and S two rings we say that S is a subring of R if it is a subset and is a ring underthe induced operations (with same 0 and 1). Eg. (Z,+,) (Q,+,)If S is merely isomorphic to S1 R a subring, we say that S can be embedded in R. Thisis because if : S S1 is an isomorphism then we may identify S with S1 via as a subringof R, i.e. the composition

: S1 S Ris injective.

This is the same as for groups. : H G an injective group homomorphism allows usto think of H as a subgroup of G.

Some exotic examples of rings

1. Let S be a set and be the set of all subsets. On define + and by

X + Y = (X Y ) (X Y ), XY = X Y

Where X denotes the complement of X in S. Then is a ring with = 0 and S = 1.Applications to mathematical logic.

2. In linear algebra the collection of linear maps from Rn to Rn is the setMnn(Rn). Thishas the structure of a ring under the usual addition and multiplication fo matrices.More generally, if R is any ring, then Mnn(R) is a ring in the same way. Note that ingeneral, Mnn(R) is not commutative, even though R may be.

Let n = 2 and R = R. Let A=(

0 10 0

)and let B=

(0 10 0

).

1

Clearly AB = 0 but A != 0 and B != 0. This shows a fundamental difference betweenthis ring and (Z,+,), where this clearly cannot happen. This motivates the followingdefinitions:

Definition. Given a != 0, if there exists b != 0 such

Abstract Algebra

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