2 Lecture 2: Logical statements and proof by contradiction Lecture 10: More on Permutations, Group Homomorphisms 31


 Georgina Norris
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1 Contents 1 Lecture 1: Introduction 2 2 Lecture 2: Logical statements and proof by contradiction 7 3 Lecture 3: Induction and WellOrdering Principle 11 4 Lecture 4: Definition of a Group and examples 15 5 Lecture 5: Basic properties of Groups and Group structure on bijective functions 18 6 Lecture 6: More examples of groups (S n, Z n ), Multiplication Tables 21 7 Lecture 7: Subgroups (definition and examples) 23 8 Lecture 8: Cyclic Groups 26 9 Lecture 9: More on Cyclic groups, order of elements in S n Lecture 10: More on Permutations, Group Homomorphisms Lecture 11: More on Group homomorphisms Lecture 12: Review for Midterm Lecture 13: More on the Symmetric group: Transpositions Lecture 14: Equivalence relations and cosets Lecture 15: Lagrange s Theorem Lecture 16: Applications of Lagrange s Theorem Lecture 17: Normal Subgroups and quotient groups Lecture 18: Proof of FTOA Lecture 19: Review for Midterm Lecture 20: Group actions Lecture 21: Group actions Lecture 22: Rings and Fields Lecture 23: Ideals and Homomorphisms Lecture 24: Integral domains and Factorization 60
2 25 Lecture 26: The GCD and irreducibility over Q[x] Lecture 26: Eisenstein s Criterion 65 1 Lecture 1: Introduction This course is meant to serve as a first introduction to Abstract Algebra. To begin with, one might ask: What is abstract algebra? According to wikipedia, Abstract Algebra is the study of algebraic structures such as groups, rings, fields, modules, vector spaces etc. Some of you might have already seen some of these objects in your other courses. For example, most of you have probably seen vector spaces and linear transformations in a Linear Algebra course. Historically, such algebraic structures arose when studying solutions to systems of equations. For example, the study of linear systems of equations leads naturally to the concepts of vector spaces and linear transformations. In general, one might be interested in possibly nonlinear equations (for example polynomials of higher degree). The study of solutions to such equations leads naturally to the concepts of groups, rings and fields. These are examples of algebraic structures which we shall study in this course. At this point, I ll refer you to Chapter 1 of the textbook for a more leisurely introduction to the concepts we will study in this course. This chapter also has a nice historical survey as well good motivating examples. Groups will be the first algebraic structures we shall study. Historically, attempts at finding solutions to polynomial equations lead naturally to the notion of groups as a certain set of symmetries. These symmetries come equipped with natural operations satisfying some natural conditions, and this data was then axiomatized into the definition of a Group. Before we start defining these notions rigorously, lets work out some motivating examples. As I mentioned above, groups arise when studying symmetries of certain objects. For example, we could consider a square, which has many symmetries or a circle (which has even more symmetries). Before we go further, we should first understand what we mean by symmetry of a circle. One way of saying this is that we can rotate the circle by say an angle θ. This is a symmetry since it sends the circle to itself (though it moves around the points on the circle). For a square, we can also try rotating. But if we are to send the square to itself, then we have to send a corner of the square to another corner. We will consider these examples in more detail later. For now, lets analyze the case of the real number line (which can be thought of as the circle of infinite radius). Example 1.1. The symbol R will denote the set of real numbers (i.e. the real number line). Rotation in this case is the operation which takes a point on the line (i.e. a real number x) and shifts it to the right or left by a certain fixed distance y. This amounts to sending the point x to x + y or x y (depending on whether we are shifting to the left or right). In particular, each real number y R gives rise to a symmetry of R. It follows that the set of symmetries itself can be identified with R. Moreover, the basic laws of addition have a natural interpretation in terms of symmetries. For example, shifting by 0 does nothing since x + 0 = x.
3 Suppose, we shift by y and then by z. This is the same as shifting by z+y. Mathematically, shifting by y and then by z sends a point x R to (x + y) + z. But, the associative law for addition tell us that (x + y) + z = x + (y + z). The right hand side of the equation describes the symmetry which shifts x by (y + z). We also note that shifting x by y is the same as shifting y by x since by the commutativity property of addition x + y = y + x. Finally, we can always shift back to where we started. This is expressed algebraically by the fact that given any x there is a y such that x + y = 0 (i.e. existence of additive inverses). In particular, the geometry of our symmetries is captured through various properties of the operation of addition on the real numbers. In fact, R with the operation of addition is an example of a group. In order to define the notion of a group and discuss symmetries in a rigorous manner, we have to use the language of set theory. Therefore, we shall begin by reviewing some of the basic concepts of set theory. Definition 1.2. A set is a collection of objects. We will usually use upper case letters (such as A, B, C etc.) to denote a set. There will be some sets for which we reserve special notation. For example, the set of real numbers is denoted by R. It is by definition the collection of all real numbers. Similarly, Z will denote the set of all integers, N denotes the set of all positive (including 0) integers, and Q denotes the set of all rational numbers. We define the empty set, denoted by the symbol, to be the set which has no elements. We will often use braces to define sets. For example, the notation A := {x x Z and x > 2} should be understood as follows: A is defined to be the set of x such that x is an integer and is greater than 2. The symbol := is synonymous to the word definition. Moreover, the objects x within the braces are the elements that are being defined to be in the set A. Finally, the notation is synonymous to the phrase such that. Remark 1.3. (Notation) We will be learning a lot of mathematical notation in the course. This is necessary to be able to concisely (and precisely) describe various abstract concepts. Let me recall that the symbol is used to denote the phrase for all, and the symbol is used to denote the phrase there exists. The symbol denotes the word in, and therefore the notation x A means that x is an element or object of A. Lower case letters such as x will be reserved to denote elements of a set. There are several natural operations one can perform on sets to create new sets. Definition 1.4. A set B is a subset of A if every element of B is also an element of A. The notation B A is used to denote that B is a subset of A.
4 Two sets A and B are equal if B A and A B. The notation A = B is used to denote that the two sets A and B are equal. Example 1.5. By defintion, we have N Z Q R. Moreover, if we let C denote the set of complex nnumbers, then R C. Exercise 1.6. Let A, B, and C be sets. Show that if A B and B C, then A C. Example 1.7. Let Mat(n n, R) denote the set whose elements are n n matrices with entries that are real numbers. Let GL(n, R) denote the set of such matrices which are invertible (recall, this is equivalent to saying that the determinant of such a matrix is nonzero). Then GL(n, R) Mat(n n, R). Definition 1.8. The union of two sets A and B, denoted by A B, is defined as follows: A B := {x x A or x B}. Definition 1.9. The intersection of two sets A and B, denoted by A B, is defined as follows: A B := {x x A and x B}. Example Suppose A is the set of integers > 3 and B is the set of integers < 7. Then A B = {4, 5, 6}. On the other hand, A B = Z. Definition Suppose B A. Then the complement of B in A, denoted by B c, is the set of all elements in A which are not in B. Example The complement of GL(n, R) in Mat(n n, R) is the set of all n n matrices with determinant 0. Definition The difference of A and B, denoted by A \ B, is the set of elements which are in A and not in B: A \ B := {x x A and x / B}. Note that this is not the same as the difference of B and A. The latter is the set B \ A, which consists of elements in B which are not in A. In particular, this notion of difference is not symmetric. Example Note that R \ Z is the set of a real numbers which are not integers. On the other hand, Z \ R is the empty set. These constructions on sets satisfy some very natural properties. One of the goals of this course is also to learn how to prove mathematical statements in a rigorous manner. So we will take this opportunity to prove a simple property of the previously defined operations on sets. You are asked to prove many more such properties in your homework. Lemma (Distributive Law) Suppose we are given sets A, B, and C. Then one has A (B C) = (A B) (A C).
5 Proof. By Definition 1.4, two sets are equal if each is contained in the other. Therefore, to verify that the statement of the lemma is true, we must verify the following two statements: (1) A (B C) (A B) (A C) (2) (A B) (A C) A (B C) We will verify (1) and leave (2) as an exercise. By Definition 1.4, we must show that if x A (B C), then x (A B) (A C). Let x A (B C). Then, by Definition 1.9, x A and x B C. On the other hand, if x B C, then by Definition 1.8, x B or x C. Therefore, we have shown that x A and x B, or x A and x C. But, again by Definitions 1.8 and 1.9, this means that x A B or x A C. In particular, x (A B) (A C). Before proceeding, I want to describe another way to write the proof of the previous Lemma. We could have written the whole proof using set theoretic notation: A (B C) = {x x A and x (B C)} (by Definition 1.9) = {x x A and ((x B) or (x C))} (by Definition 1.8) = {x (x A and x B) or (x A and x C)} (just rearranging terms) = {x x (A B) and x (A C)} (by Definition 1.9) = (A B) (A C) (by Definition 1.8) In order to write the proof this way, we have to verify each of the equalities appearing in this chain of equalities. The point is that we broke up the equality we wanted to prove into a chain of equalities that are obvious in the sense that each such equality is easily justified using the definitions of union and intersection. We end with an important construction on sets which will be very useful later in the course. Definition Given two sets A and B, the cartesian product of A and B, denoted A B, is defined as follows: A B is the set of symbols (a, b) where a A and b B. We can also write this as A B := {(a, b) a A and b B}. Remark Note that two elements (a, b) and (c, d) of A B are equal (i.e. give the same element of the set A B) if and only if a = b and c = d. It follows that the element (a, b) is not equal to (b, a) unless a = b. Example The real number plane, denoted R 2, is the cartesian product of R with itself. Lets prove another set theoretic identity (like in the previous Lemma), but involving the cartesian product. Lemma A (B C) = (A B) (A C)
6 Proof. Recall, we need to prove the following statements: (1) A (B C) (A B) (A C) and (2) (A B) (A C) A (B C) We will prove (1), and leave (2) as an exercise. Let x A (B C). Then x = (a, r) where a A and r B C. It follows that (a, r) A B and (a, r) A C. In particular, x = (a, r) (A B) (A C). The previous lemma can also be proved completely using a chain of set theoretic equalities as we did after the first Lemma. We leave it to the reader to work this out as an exercise. In practice, we also want to be able to compare two different sets. One way to formalize this mathematically is to introduce the concept of a function. All of you will have seen functions before in Calculus. In Calculus, functions had domain some subset of R and range also some subset of R. Here we will simly generalize that concept to allow the domain to be an arbitrary set A and range to be some set B. Definition Let A and B be two sets. Then a function from A to B, denoted by f : A B, is a rule which associates to an element a A a unique element b B. We say that A is the domain of f, and B is the range of f. We note that uniqueness here means that there is only one element in B associated to a. We usually denote this element by f(a). Often one also says that f is a function from A to B or that f maps A to B or that f is a mapping from A to B. All of these mean the same thing. I won t say much about functions at the moment. We will study them in much more detail later in the course.
7 2 Lecture 2: Logical statements and proof by contradiction Last time we discussed some basic set theory and proved some set theoretic identities. Some of you may have seen proofs before, or even learned how to write proofs. But, what does it mean to prove a statement? How does one prove something? In this lecture, I want to explain what it means to prove a mathematical statement, and give some techniques for proving such statements. We have already seen some examples of a proofs in the context of set theory last time. Before we discuss proofs of statements, we should discuss what we mean by a statement. So we will begin by discussing the vocabulary for mathematical statements. Abstractly a statement is simply a declarative sentence. amounts to verifying its veracity (i.e. is it true or false). Proving a statement then Example 2.1. I am wearing black jeans today is an example of such a statement. It would be true if I was indeed wearing black jeans today. Example 2.2. (1) 2 is an even number is an example of a mathematical statement. It is true since by definition a number is even if it is divisible by 2. (2) 7 is an even number is a also a statement. Explain why A mathematical statement is simply a statement which uses the language or vocabulary of mathematics. It is a special type of statement. For example, the set theoretic identity Given any two sets A and B, one has A B = B A. is a mathematical statement. Sometimes, as a short hand we write simply A B = B A. The statement asserts that the two sets A B and B A are equal. Proving such a statement amounts to showing that the statement is true. In mathematics, it is common to form statements using the language of logic. In particular, there are standard ways to create more statements form given statements. For example, suppose p and q are some statements. For example, p could be the statement from the first example. Then we can create new statements using the following operations: (1) Not: We can consider the statement not p. This is also denoted by p. This is also known as the negation of p. It is the statement which is defined to be true when p is false and it s defined to be false when p is true. (2) And: Given two statements as above, we can form the statement p and q (also denoted by p q). By definition, this statement is true when both p and q are true, and false if either p or q is false. (3) Or: Given two statements as above, we can form the statement p or q (also denoted by p q). By definition, this statement is true when p or q or both of them are true (i.e. any one of these is true), and false if both p and q are false. Note that in everyday English, the statement p or q would be reserved to mean that one of p or q is true but
8 not both. However, in mathematics we allow for the possibility that both are true. (4) if..., then... : Given two statements p and q, we can form the statement If p then q. By definition, the validity of this statement amounts to saying that if p is true then q is true, or p is false. On the other hand, If p then q is false when it happens that p is true but q is false. (5) if and only if: Given two statements p and q, we can form the statement p if and only if q. By definition, this statement is true if the statements p q (see below for notation) and q p are both true. Otherwise, the statement is false. Explicitly, this means that the statement is true when p and q are both false or both are true. We say that p is equivalent to q. Notation An ifthen statement is sometimes denoted as p q. An if and only if statement is sometimes denoted by p q. We may also combine the previous operations. Example 2.3. Consider the statement p q. If we negate this statement, we get formally the statement (p q). We note that this statement is equivalent to the statement ( p) ( q) i.e (p q) ( p) ( q). How dow we show this? We must show that the statement (p q) ( p) ( q) is true. Recall, this means that we have to show two things: (1) if (p q) is true, then so is ( p) ( q), and (2) we must show the reverse implication (this is sometimes also called the converse statement) i.e. ( p) ( q) (p q). We will show (1) and leave (2) as an exercise (the proof is similar). So we assume (p q) is true. Then (p q) must be false. It follows that both p and q are false. Then both p and q must be true. But, this is means that ( p) ( q) is true. To recap, in this course we will be interested in proving various mathematical statements. Proving such a statement amounts to verifying that the statement is in fact true. We want to now develop some techniques for verifying mathematical statements or proving them. Instead of developing these techniques abstractly, we shall work out some instructive examples. Before I state the next example, let me recall the definition of even and odd integers. Definition 2.4. (1) An integer x is even if there is an integer y such that x = 2y. (2)An integer x is odd if there is an integer y such that x = 2y + 1. Example 2.5. Consider the following mathematical statement: If x is an even integer, then x 2 is an even integer. This is a statement of type (4) in our previous list of statements. Here x is an even integer plays the role of p and x 2 is an even integer plays the role of q. So prove the statement we must show that if x is even, then the same is true for x 2. So we suppose that x is even.
9 Then we can write x = 2y for some integer y. It follows that x 2 = (2y) 2 = 4y 2 = 2(2y 2 ). So we have written x 2 and 2 times the number 2y 2. To complete the proof we only need to note that 2y 2 is also an integer. To see this, note that a product of integers is also an integer. Example 2.6. Consider the statement: Let x be an integer. Then x is even x + 1 is odd. To prove this statement, we must verify the following two statements: (1) If x is even then x + 1 is odd and, (2) If x + 1 is odd then x is even. Lets show (1), and again we will leave (2) as an exercise. So we suppose x is even. We want to show that x + 1 is odd. If x is even, then there exists a y such that x = 2y. It follows that x + 1 = 2y + 1 must be odd since 2y is also an integer. Sometimes to prove a statement, we have to split it up into several cases. For example, consider the statement: For every integer x, x(x + 1) is even. Note that the statement is an example of an ifthen statement. For example, it can be restated in the form: If x is an integer, then x(x + 1) is even. Proof. In order to prove this statement, it is helpful to consider two different cases. Namely, we first note that any integer is either even or odd (this is not an obvious statement, and we will prove it rigorously later when we discuss the euclidean algorithm). So we can split the claim into the following two cases: (1) If x is even, then x(x + 1) is even, (2) If x is odd, then x(x + 1) is even. Lets prove (1). Then x is even. It follows that x = 2y, and, therefore, x(x+1) = 2yx(x+1). Since yx(x + 1) in an integer, it follows that we have written x(x + 1) as 2 times and integer. Therefore, x(x + 1) is even. Now lets prove the second case (2). Then x is odd. It follows that x = 2y + 1. Therefore, x(x + 1) = x(2y ) = x(2y + 2) = 2x(y + 1). Again, it follows that x(x + 1) must be even. Another useful technique for proving statements is to try deduce a contradiction if the statement were false. In particular, suppose we want to prove that a statement p is true by contradiction. Then we first assume that p is false and show that this implies some other statement q is true, which we know to be false. This then leads to a contradiction, and therefore p could not have been false. Example 2.7. A real number is said to be irrational if it s not rational. Consider the following claim: If x is rational and y is irrational, then x + y is irrational.
10 Let s prove it using proof by contradiction. So we suppose the x + y is rational. Then we note that the sum or difference of two rational numbers is also a rational number. It follows that x + y x = y is rational. But, this leads to a contradiction since by our hypothesis y is irrational. Therefore, we conclude that x + y must be irrational.
11 3 Lecture 3: Induction and WellOrdering Principle Today we are going to learn one of the most important techniques for proving mathematical statements: mathematical induction. Suppose that for each positive integer n we are given a statement P (n). Mathematical induction is a principle which asserts that under some natural hypotheses, we can show that P (n) is true for all positive integers n once we show that P (1) is true. As you can imagine, this can sometimes be a very powerful tool when one wants to prove an infinite number of statements indexed by the positive integers. Example 3.1. For any positive integer n, consider the statement P(n): n = n(n+1) 2. You ve probably seen this statement in your Calculus courses, and perhaps even given a proof using induction. It is easy to verify that this statement is true for n = 1. Moreover, if we assume that the statement P(n) is true for some fixed n, then we can show that P(n+1) is also true. In this setting, the principle of mathematical induction will allow us to conclude that P(n) is true for all positive integers n. Induction is essentially equivalent to the following fundamental principle (or axiom) of set theory. WellOrdering Principle: Every nonempty set of positive integers has a least element. Remark 3.2. Note that the Well Ordering Principle implies the analogous assertion with positive integers replaced by the set N. Theorem 3.3. (Induction) Let A Z >0. Consider the following two statements: (1) 1 A. (2) For any positive integer k, if k A, then k + 1 A. If A satisfies these two conditions, then A = Z >0. Proof. We need to show that under the assumptions (1) and (2), Z >0 A. We shall prove this by contradiction. Consider the complement A c of A in Z >0. By the WellOrdering principle, it has a least element. Lets denote this element by b. Then b > 1, since 1 A. It follows that b 1 Z >0 and by construction it must be in A. Then, by property (2), b = b A. This is a contradiction since b A c and therefore cannot be in A. It follows that A c must be the empty set. Remark 3.4. The empty set, denoted, is the set consisting of no elements. By definition it is a subset of every set. Then it is easy to see that A A c =. Lets apply the principle to prove the statement from Example 3.1. In that case, we set A Z >0 to be the set of those positive integers such that P(n) is true. Note P(1) is true. Now suppose P(k) is true. This means that one has k = k(k + 1). 2 To apply the theorem, we must show that P(k+1) is also true. We have
12 k + (k + 1) = k(k+1) 2 + k + 1 (since P(k) is true) = k(k+1)+2(k+1) 2 = k2 +k+2k+2 2 = (k+1)(k+2) 2. Then, by induction, A = Z i.e. P(n) is true for all n. When applying induction, we have to verify the statements (1) and (2) in Theorem 3.3. In practice, one refers to (1) as the base case and (2) as the induction hypothesis. There are many other forms of the principle of mathematical induction. Some of these may look stronger, but they are in fact equivalent to Theorem 3.3. Theorem 3.5. (Strong induction) Let A Z >0. Suppose the following is true for every positive integer n: Then A = Z >0 If k A for all 0 < k < n, then n A. Before I prove the theorem, note that Theorem 3.3 follows from Theorem 3.5. Suppose A satisfies the hypothesis (1) and (2) of Theorem 3.3. Note that these hypotheses imply the hypothesis of Theorem 3.5. Namely, if k A for all k < n it follows that n 1 A, and by hypothesis (2) of Theorem 3.3 we must have that n A. Note here that we have to assume n 1 > 0. If it is 0, then n = 1 and therefore (1) will imply that 1 A. It follows that the hypotheses (1) and (2) of Theorem 3.3 imply the hypothesis in Theorem 3.5. Therefore, we can again conclude that A = Z >0. In particular, Theorem 3.3 can be deduced from Theorem 3.5. Proof. (Theorem 3.5) We can again apply the wellordering principle to A c. Note that 1 A. Let b denote the least element of A c. Then b > 1. It follows that 1,..., b 1 are in A. Therefore, b A. But this is a contradiction. It follows that A c =. Here is another form of induction which is easily deduced from Theorem 3.5. Theorem 3.6. Let A Z >0 and r Z >0. Suppose that the statement If k A for all r k < n, then n A. is true for all n r. Then A = {x x Z >0 and x r}. Note that if r = 1, then we recover Theorem 3.5. The idea is here is that when applying induction we don t have to start at n = 1. The base case can be some other positive integer r. Here is an application. Before stating it, let me recall a definition. Definition 3.7. A natural number p > 1 said to be prime if it is only divisible by 1 and itself.
13 Notation: The notation m n is used to denote the fact that m divides n. Claim: Any integer n > 1 can be expressed by a product of prime numbers. Proof: Let A denote the set of integers greater than 1 which can be expressed as a product of prime numbers. We want to show that A is the set of all integers greater than 1. We will apply Theorem 3.6 with r = 2. In particular, we need to verify the hypothesis of Theorem 3.6. Therefore, we have to show that the following statement is true for all n > 1: P(n): If any integer 1 < k < n can be factored as a product of prime numbers, then n can also be factored as a product of prime numbers. So we fix an integer n and suppose that every integer 1 < k < n can be factored into primes. We have to show that under these hypothesis n can also be factored into primes. Now we have two cases. First, n could be prime, then we are done. If n is not prime then we can write n = ab where a and b are integers such that 1 < a, b < n. Then our induction hypothesis says that a can be factored into primes, and so can b. It follows that n can also be written as a product of primes. I want to now discuss some applications of the wellordering principle. First, we shall prove the division algorithm or euclidean algorithm. Theorem 3.8. (Division Algorithm) Let m be an integer and n > 0 be a positive integer. Then there exist unique integers q and r such that where 0 r < n. m = qn + r Proof. Consider the set S = {m nb 0 b Z}. First, note that S. If m 0, then we can simply choose b = 0. We leave the other case for the reader to verify. Therefore, we can apply the wellordering principle (or rather Remark 3.2) and conclude that S has a least element. Lets denote this element by r, and set r := m nq. Note that by definition r 0. On the other hand, r < n. To see this, we use proof by contradiction. Namely, suppose that r n. Then 0 r n = m n(q + 1). It follows that m n(q + 1) S. Moreover, this element is smaller than r. But, this is a contradiction since r was the least element of S. Therefore, we have r < n. Therefore, we have shown that there exist q and r satisfying the desired properties. It remains to show uniqueness. So suppose that there is another pair of integers q, r as in the theorem. We have to show that q = q and r = r. But, we have qn + r = q n + r. It follows that (q q )n = (r r). Since both r and r are in the set {0,, n 1}, their difference cannot be a nonzero multiple of n. It follows that q q = r r = 0. Here is another application. Claim: 2 is irrational. Proof: We shall use proof by contradiction to prove the statement. So suppose that
14 2 is rational. Then we can write 2 = a b where b > 0. There might be more than one way of representing 2 in such a manner. But, lets choose the representation such that b is positive and the smallest possible (we can do this by the wellordering principle). It follows that 2b 2 = a 2. Since a 2 is even, it follows that a is even. Therefore, a = 2k for some integer k and so a 2 = 4k 2. It follows that 2 divides b 2. In particular, b 2 /2 is an integer and 2 = a/2 b/2. But, then we have written 2 as a fraction where the denominator is less than b. But, by our assumption b was the least such integer, and therefore we are led to a contradiction. It follows that 2 is irrational. Exercise: If p and p + 2 are primes such that p > 3, then p + 1 is divisible by 6.
15 4 Lecture 4: Definition of a Group and examples During the last couple of lectures we discussed some set theory and techniques for proving mathematical statements. We are now ready to go back to our discussion of symmetries. Lets begin with the circle. We have already noted that the circle (say of radius 1 and centered at the origin in the plane) has symmetries given by rotating by an angle θ counterclockwise. In principal, θ could be any positive real number. But, many of these will give the same rotation. For example, rotating by θ or θ + 2π are the same. And therefore we see that the distinct rotations are parametrized by the set [0, 2π). Any element of this halfopen interval gives a unique rotation of the circle, and all rotations are obtained in this manner. Remark 4.1. There are other symmetries of the circle. For example, reflections along lines through the origin will also give symmetries of the circle. But, for the moment, we shall concentrate only on the rotational symmetries. When we analyzed the symmetries of the line we saw that the geometry of the symmetries was reflected in the additive structure on the real numbers. There is a similar story for the circle. Rotating by θ and then again by θ is the same as rotating by θ+θ. Unfortunately, if we identify our symmetries with the set [0, 2π) then θ + θ might not be in the set. This happens precisely when θ +θ 2π. However, note that if θ, θ [0, 2π), then θ +θ < 4π. Therefore, if θ + θ 2π, then θ + θ 2π [0, 2π). This leads us the the following definition: Definition 4.2. Given θ, θ [0, 2π), we set θ θ := θ + θ if θ + θ < 2π and θ θ := θ + θ 2π if θ θ 2π. Remark 4.3. We can think of as a function : [0, 2π) [0, 2π) [0, 2π). However, we shall see below that it is a special type of function which satisfies many more properties which are formally analogous to the usual addition operation + : Z Z Z on the integers. We see that the symmetry of the circle given by first rotating by θ and then by θ is the same as θ θ. So the geometry of the symmetries is reflected in a sort of new addition operation on the set [0, 2π). The following Lemma shows that this operation satisfies many of the properties that usual addition satisfies. Lemma If θ [0, 2π), then θ 0 = θ. 2. If θ, θ [0, 2π), then θ θ = θ θ. 3. Given θ [0, 2π), there is a φ [0, 2π) such that θ φ = If θ, φ, ψ [0, 2π), then (θ φ) ψ = θ (φ ψ). Proof. (1) Note that θ + 0 = θ < 2π. Therefore, by definition, θ 0 = θ. (2) Left as an exercise. (3) If θ = 0, then one can let ψ = 0. If θ > 0, then ψ := 2π θ < 2π and one can check
16 that θ ψ = 0. (4) Left as an exercise. We say that is an operation on [0, 2π) which makes it into an abelian group. In general, we have the following definition: Definition 4.5. A group G is a set which comes equipped with an operation m : G G G (i.e. a function) satisfying the following properties: 1. There is an element e G such that m(e, g) = m(g, e) = g. 2. For every g G, there is a g G such that m(g, g ) = m(g, g) = e. 3. Given g, h, i G, m(m(g, h), i) = m(g, m(h, i)). Given a group G, we usually write m(g, h) simply as gh or sometimes we will write g + h. The idea is that the function m is allowing one to add or multiply elements of the set G. With this notation, properties (1), (2) and (3) of 4.5 can then be written as follows: 1. There is an element e g such that ge = eg = e. 2. For every g G, there is a g G such that gg = g g = e. 3. Given g, h, i G, (gh)i = g(hi). The first property states that we have an element, denoted e, that functions like the identity element. The second tells us that every element has an inverse with respect to the given operation. Finally the third property is telling us that the given operation is associative. Note that we do not require the commutativity property. Definition 4.6. A group G is abelian if it also satisfies the commutativity property: For all g, h G, gh = hg. Lemma 4.4 shows that the set [0, 2π) with the operation is an abelian group. Example 4.7. (1) (Z, +). (2) (Q, +), (Q, ). (3) (R, +), (R, ). (4) (C, +), (C, ). (5) GL n (R). We will end this section with another example of a group coming from symmetries. Consider an equilateral triangle with vertices 1, 2, 3 (say in the plane centered at the origin). Lets consider the set of symmetries, denoted by G, given by rotations and flips. In particular, any rotation or flip or any combination of flips and rotations which preserves our triangle gives an element of G. Lets write down all the symmetries. First, note that a symmetry is completely determined by what it does to the vertices. Therefore, to write down a symmetry we only need to say where it sends the vertices. Note that we can do nothing. We denote this symmetry by I (for identity). In terms of vertices it sends
17 1 1, 2 2 and 3 3. We note that there are two rotations R + and R where we rotate counterclockwise and clockwise. Similarly there are three flip F 12, F 13, F 23. Here F ij sends vertex i to vertex j, vertex j to vertex i and fixes the other vertex. So G has six elements. Now, multiplying two elements means that we follow one motion by another. So for example R + R + = R 2 + is the motion which sends 1 3, 2 1, and 3 2. Note that this is just R. Therefore, R 2 + = R. Lets compute F 12 followed by F 13. We will denote this as F 12 F 13. This symmetry will send 1 2, 2 3, and 3 1. It follows that F 12 F 13 = R +. Lets compute F 13 F 12. This will send 1 3, 2 1, and 3 2. Note that this is precisely R. Note that F 12 F 13 F 13 F 12.
18 5 Lecture 5: Basic properties of Groups and Group structure on bijective functions Last time we defined the notion of a group and discussed several examples. In particular, a group was defined to be a set G equipped with an operation (i.e. a function m : G G G ), which was required to satisfy some properties (existence of identity, existence of inverses, and the associativity property). Sometimes it s useful to consider operations on sets which satisfy some but not all of these properties. This leads to the following definition. Definition An operation on a set K is a function m : K K K. 2. An operation m on a set K is said to have an identify if there is an element e K such that m(e, k) = m(k, e) = k for all k K. 3. An operation m on a set K is said to be associative if for all g, h, k K, one has m(m(g, h), k) = m(g, m(h, k)). 4. An operation m on a set K is said to have inverses if it has an identity e and for all k K, there is a k such that m(k, k ) = m(k, k) = e. Example 5.2. Consider the operation of division on R. In particular, it associates to two real numbers a and b, the real number a/b. This operation satisfies all the properties required for it to be a group except for the associativity property! Verify this. We would like to be able to do algebra in a group. Namely, suppose we have an equation ac = bc in a group G. We would like to conclude that a = b, as we would for usual addition or multiplication on the real numbers. It turns out that this property does hold in an arbitrary group. Before we prove this, we prove some uniqueness lemmas. More precisely, note that while the definition of a group posits the existence of an identity and the existence of inverses, it does not require the uniqueness of such elements. However, it turns out that this is a formal consequence of the definition of a group. Lemma 5.3. Any group G has a unique identity element. Proof. Suppose e and e are two identity elements. Then ee = e since e is an identity of G. On the other hand, by the same argument, ee = e. It follows that e = e. Lemma 5.4. Let G be a group, and g G. Then g has a unique inverse. We shall denote the inverse of g by g 1. Proof. Suppose g 1 and g 2 are both inverses for g. It follows that gg 1 = e = gg 2. We can now multiple both sides by g 1 on the left, and we get g 1 gg 1 = g 1 gg 2. But, g 1 gg 1 = eg 1 = g 1 and similarly g 1 gg 2 = eg 2 = g 2. It follows that g 1 = g 2. Since inverses of elements in a group G are unique, we shall use the notation g 1 to denote the inverse of g G. Similarly, we shall use the notation e to denote the identity element of a group. Finally, we prove the cancellation law we alluded to above. Proposition 5.5. Let G be a group and suppose a, b, c G satisfy ab = ac. Then b = c.
19 Proof. We can multiply both sides of this equation by the inverse of a. Then a 1 ab = a 1 ac. Therefore, b = c. In the previous proofs, and examples, we have written expressions of the form abc for elements a, b, c G. Strictly speaking this notation should be justified. Namely, a law of composition allows us to combine two elements in a group. So for example we can consider ab or bc. We can then iterate this procedure and consider the two elements (ab)c and a(bc). Since the law of composition on a group is associative, we know that both of these are in fact the same element, and we may denote it simply by abc. More generally, if we are given n elements a 1,..., a n G, then we can define a unique element a 1 a 2 a n G given that a 1 a 2 is just the law of composition on the group. Moreover, it satisfies the property that for any i, a 1 a n = (a 1 a i )(a i+1 a n ). To define this element, we proceed by induction on n. So suppose we have defined this for any product of n 1 elements. Then we define a 1 a n := (a 1 a n 1 )a n. We can check this satisfies the required property by using the induction hypothesis. We leave the details to the reader. I want to now discuss an important example of a group. In fact, abstractly all groups are subsets of such groups. Example 5.6. (Functions on a Set) Let K and M denote two sets. Then we can consider the set of functions f : K M, denoted Fun(K, M). If K = M, then this set comes equipped with a natural operation given by the composition of functions: : Fun(K, K) Fun(K, K) Fun(K, K) where f g(x) := f(g(x)). Note that this operation satisfies the associative property. We can consider the identity function I(x) = x. This function is an identity in Fun(K, K) under the operation of composition of functions. Unfortunately, not all functions have an inverse under this operation. Definition A function f : K M is injective if given any x, y K, f(x) = f(y) if and only if x = y. 2. A function f : K M is surjective if for any m M, there exists a k K such that f(k) = m. 3. A function f : K M is bijective if it is both surjective and injective. Proposition 5.8. The set of bijective functions f : K K, denoted by Bij(K, K), form a group under the operation of composition of functions. Proof. We must first show that composition of functions is an operation on Bij(K, K). In particular, we must show that if f and g are bijective functions, then so is f g. Lets show that f g is injective. Suppose x, y K such that f g(x) = f g(y). Then, by definition of composition, f(g(x)) = f(g(y)). Since f is injective, it follows that g(x) = g(y).
20 Finally, since g is injective, it follows that x = y. Therefore, f g is injective. We leave it as an exercise to show that f g is surjective. Note that the identity function I is a bijection, and therefore an element of Bij(K, K). In particular, Bij(K, K) has an identity under the operation given by composition of functions. We ve already remarked in Example 5.6 that composition of functions is an associative operation. It remains to show that every bijective function has an inverse under composition. Let f Bij(K, K). Then for x K, we define f 1 (x) := y where y is an element such that f(y) = x. Since f is surjective, such a y always exists. Moreover, there is a unique such y since f is injective. It follows that f 1 is a function on K. We leave it as an exercise to check that this is also a bijective function, and that f 1 is the inverse of f under the operation of composition. For the last part, you should check that f f 1 = f 1 f = I.
21 6 Lecture 6: More examples of groups (S n, Z n ), Multiplication Tables Recall, last time we proved the following proposition: Proposition 6.1. The set of bijections Bij(K, K) forms a group under the operation of composition of functions. In the proof of the previous proposition, we proved a special case (where K = M) of the following lemma which is important enough be stated separately: Lemma 6.2. Given sets K, M, and L, and functions f : K M, and g : M L: 1. If f and g are injective, then g f is injective. 2. If f and g are surjective, then g f is surjective. 3. If f and g are bijective, then g f is bijective. Proof. Exercise Example 6.3. Let K = {1,..., n} denote the set consisting of the integers between 1 and n. Then S n := Bij(K, K) is a group by the previous proposition. This is usually called the symmetric group on nletters or nelements. Note that an element of S n is exactly a permutation of the integers between 1 and n. We have already seen that S 3 is exactly the symmetry group of the triangle. Definition 6.4. We say that a group is finite if the underlying set is finite. In this case, the number of elements in this set is called the order of the group. It is denoted by the symbol G. Lemma 6.5. The order of S n is n!. Proof. By definition any f S n is a bijective function f : {1,..., n} {1,..., n}. Then there are n choices for f(1), since this can take any value between 1 and n. Since f is one to one f(1) f(2), and so there are n 1 choices for f(2) i.e. any integer not equal to f(1). Continuing in this way, one sees that there are n (i 1) choices for f(i). Therefore, the total number of such f s is given by n(n 1)(n 2) 1 = n!. The symmetric groups are an important class of groups, and we shall study their structure in much more detail later in the course. Another important class of groups are the abelian groups Z n. As a set Z n = {0, 1,, n 1}. We define an operation of addition on Z n as follows. Given two integers a and b we can consider the remainder of a + b modulo n (recall, modulo simply means when divided by ). By the division algorithm, this is a unique number in the set Z n. This defines an operation : Z n Z n Z n. In practice, we will usually use the notation + to denote this operation. This should cause no confusion with the usual operation of addition on the integers. Lemma 6.6. Z n is an abelian group with respect to the operation defined above.
22 Proof. We have to verify that (Z n, +) satisfies the three defining properties of a group, and show that the operation is commutative. (1) For any a Z n, a + 0 = 0 = 0 + a. Since the remainder of 0 is 0 modulo n, this shows that a 0 = 0 = 0 a, and therefore 0 is the identity. (2) For any number n > a > 0, note that 0 < n a < n and therefore in Z n. Moreover, a + (n a) = n and this has remainder 0 modulo n. Therefore, a (n a) = 0. (3) Given any a, b Z n, the remainder of a + b modulo n is the same as that of b + a modulo n since a+b = b+a. It follows that a b = b a and the operation is commutative. (4) Finally, we leave it as an exercise to check the associative property. Let Z n := {1, n 1}. We can also try to define an operation of multiplication : Z n Z n Z n in a similar manner. Namely, we define a b to be the remainder of ab modulo n. Unfortunately, this is not welldefined since there are nonzero integers a, b such that the product ab is divisible by n, and therefore has remainder 0. For example, consider Z 6. Then 3 2 = 6 will be 0 modulo 6. It turns out that this can never happen if n is a prime. In particular, we have the following remarkable proposition. We shall prove it later. Proposition 6.7. The operation induces an abelian group structure on Z p prime p. for any Notation: Given a group G, an element g G and n N, we will denote by g n the element given by multiplying g with itself ntimes. In particular, g 2 = gg etc. If n = 0, we set g 0 := e, where e is the identity. Finally, if n < 0, we define g n := (g 1 ) n i.e. we multiply the inverse g 1 with itself n > 0 times. We note that the usual rules for exponents hold in a group. In particular, (g n ) m = g nm and g n g m = g n+m. Once we have a group, we would like to understand how to multiply elements in the given group. For finite groups, we can use a multiplication table to understand how to multiply elements in the group. The method for constructing such tables was discussed in class.
23 7 Lecture 7: Subgroups (definition and examples) One way to try to understand a set K is by trying to understand its subsets. We can try to apply a similar strategy to understand groups. This leads us to the notion of subgroups. Notation: We shall sometimes denote a group by (G,, e) if we want to specify the operation, and the identity e G. Definition 7.1. A subset H G of a group (G,, e) is a subgroup if the following three properties hold: (a) (Closure) For all h, h H, h h H. (b) (Identity) e H. (c) (Inverses) If h H, then h 1 H. If H satisfies property (a) in the definition above, then we say that H is closed under multiplication (or ). Similarly, if it satisfies property (b), we say that it is closed under the identity, and if it satisfies (c), we say that it is closed under taking inverses. If H is a subgroup, then by property (a) the operation on G restricts to an operation : H H H on the subset H. Moreover, property (b) shows that e H, and therefore an identity for this operation. We leave it as an exercise to the reader to show that this gives rise to a group structure on H. In particular, we have the following Lemma. Lemma 7.2. Let G be a group and H G be a subgroup. Then the operation on G restricts to an operation on H, and H is a group under this operation. Example 7.3. Let G be a group, then {e} G is a subgroup. It is called the trivial subgroup. Note, G is also a subgroup of itself. Definition A subgroup H G is a proper subgroup if H G. 2. A subgroup H G is nontrivial if H {e}. Example 7.5. Let T 2 GL 2 (C) denote the set of upper triangular matrices. This is then a subgroup of GL 2 (C). The product of any two upper triangular matrices is upper triangular. The identity matrix is clearly upper triangular. Finally, one can check that the inverse of an upper triangular matrix is also upper triangular. Definition 7.6. A subgroup H G of a group G is abelian if H is abelian as a group. It is not hard to show that any subgroup of an abelian group is also an abelian group. On the other hand, there are many examples of nonabelian groups with abelian subgroups. Example 7.7. The subset of rotations in the group of symmetries of the triangle or square form a subgroup. Moreover, these are abelian subgroups. Example 7.8. The subset of permutations in S n that send n to n also forms a subgroup of S n. We will see later that this subgroup is naturally identified with S n 1. Let b Z. Consider the set bz := {n n Z and n = bk for some k Z}. Note that bz is a subset of Z.
24 Theorem For any integer b, bz is a subgroup of (Z, +, 0). 2. Any subgroup H of (Z, +, 0) is of the form bz for some unique integer b N. Proof. (a) Since 0 = b0, it follows that 0 Z. If n bz, then n = bk for some k. Then n = b( k) bz. It follows that bz is closed under inverses. Finally, if n, m bz, then n = bk and m = bk for some k, k Z. It follows that n + m = bk + bk = b(k + b ) bz. (b) Suppose H Z. By definition, 0 H. If there are no other elements, then H = 0Z. Otherwise, we have a nonzero element a H. Then a H, and so H has a positive element. Let H + denote the set of positive elements in H. Since it is nonempty, by the WOP (Well Ordering Principle), there is a least positive integer b in H. Since b H, it follows that kb H for any integer k. For instance, one can see this by induction. It follows that bz H. Finally, it remains to show that H bz. Suppose there is an element h H which is not in bz. Then we can write h = bk + r where 0 r < b. Since H is a subgroup and bk, h H, it follows that r = h bk H. Since b was the smallest positive element in H, it follows that r = 0. Finally, the uniqueness is clear since either H = {0} (and then b = 0), or b is by construction the smallest (nonzero) positive integer in H. Here is an interesting application of this result. Given two integers a, b, consider the subset az + bz := {n Z n = ar + bs for some integers r, s} of Z. We first note that this is in fact a subgroup. Lemma az + bz is a subgroup of (Z, +, 0). Proof. Since 0 = a0 + b0, it s closed under the identity. The inverse of ar + bs is given by a( r) + b( s), and therefore it is closed under inverses. Suppose h, h az + bz. Then there exist integers r, r, s, s such that h = ar + bs and h = ar + bs. Then h + h = a(r + r ) + b(s + s ) az + bz. It follows from Theorem 7.9 that there is a unique positive integer d Z, such that dz = az + bz. Corollary Let a, b Z such that not both are zero. Let d be the unique positive integer such that az + bz = dz 1. d divides both a and b. 2. If e divides a and b, then e divides d. We say that d is the greatest common divisor of a and b. I want to end by discussing another very important construction on groups which allows us to construct a new group if we are given kgroups G 1,..., G k. Consider the cartesian product G := G 1 G k. We define an operation on this set as follows. Given two elements (a 1,..., a k ) and (b 1,..., b k ) in G, we set (a 1,..., a k )(b 1,..., b k ) = (a 1 b 1,..., a k b k ). Lemma With notation as above, G is a group with identity e = (e 1,..., e k ) where e i is the identity of G i.
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