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Algebra & Number Theory [09/01/2018] BY: Andrew Baker (2009) Andrew Baker Department of Mathematics, University of Glasgow. Email address: a.baker@maths.gla.ac.uk URL: http://www.maths.gla.ac.uk/∼ajb

Contents Chapter 1. Basic Number Theory 1. The natural numbers 2. The integers 3. The Euclidean Algorithm and the method of back-substitution 4. The tabular method 5. Congruences 6. Primes and factorization 7. Congruences modulo a prime 8. Finite continued fractions 9. Infinite continued fractions 10. Diophantine equations 11. Pell’s equation Problem Set 1 1 1 3 4 7 9 12 14 17 19 24 25 28 Chapter 2. Groups and group actions 1. Groups 2. Permutation groups 3. The sign of a permutation 4. The cycle type of a permutation 5. Symmetry groups 6. Subgroups and Lagrange’s Theorem 7. Group actions Problem Set 2 31 31 32 33 34 36 38 41 47 Chapter 3. Arithmetic functions 1. Definition and examples of arithmetic functions 2. Convolution and M¨ obius Inversion Problem Set 3 51 51 52 56 Chapter 4. Finite and infinite sets, cardinality and countability 1. Finite sets and cardinality 2. Infinite sets 3. Countable sets 4. Power sets and their cardinality 5. The real numbers are uncountable Problem Set 4 57 57 59 60 62 64 65 Index 67 1

CHAPTER 1 Basic Number Theory 1. The natural numbers The natural numbers 0, 1, 2, . . . form the most basic type of number and arise when counting elements of finite sets. We denote the set of all natural numbers by N0 = {0, 1, 2, 3, 4, . . .} and nowadays this is very standard notation. It is perhaps worth remarking that some people exclude 0 from the natural numbers but we will include it since the empty set ∅ has 0 elements! We will use the notation Z+ for the set of all positive natural numbers Z+ = {n ∈ N0 : n 6= 0} = {1, 2, 3, 4, . . .}, which is also often denoted N, although some authors also use this to denote our N0 . We can add and multiply natural numbers to obtain new ones, i.e., if a, b ∈ N0 , then a + b ∈ N0 and ab ∈ N0 . Of course we have the familiar properties of these operations such as a + b = b + a, ab = ba, a + 0 = a = 0 + a, a1 = a = 1a, a0 = 0 = 0a, etc. We can also compare natural numbers using inequalities. Given x, y ∈ N0 exactly one of the following must be true: x = y, x < y, y < x. As usual, if one of x = y or x < y holds then we write x 6 y or y > x. Inequality is transitive in the sense that x < y and y < z =⇒ x < z. The most subtle aspect of the natural numbers to deal with is the fact that they form an infinite set. We can and usually do list the elements of N0 in the sequence 0, 1, 2, 3, 4, . . . which never ends. One of the most important properties of N0 is The Well Ordering Principle (WOP): Every non-empty subset S ⊆ N0 contains a least element. A least or minimal element of a subset S ⊆ N0 is an element s0 ∈ S for which s0 6 s for all s ∈ S. Similarly, a greatest or maximal element of S is one for which s 6 s0 for all s ∈ S. Notice that N0 has a least element 0, but has no greatest element since for each n ∈ N0 , n + 1 ∈ N0 and n < n + 1. It is easy to see that least and greatest elements (if they exist) are always unique. In fact, WOP is logically equivalent to each of the two following statements. The Principle of Mathematical Induction (PMI): Suppose that for each n ∈ N0 the statement P (n) is defined and also the following conditions hold: 1

• P (0) is true; • whenever P (k) is true then P (k + 1) is true. Then P (n) is true for all n ∈ N0 . The Maximal Principle (MP): Let T ⊆ N0 be a non-empty subset which is bounded above, i.e., there exists a b ∈ N0 such that for all t ∈ T , t 6 b. Then T contains a greatest element. It is easily seen that two greatest elements must agree and we therefore refer to the greatest element. Theorem 1.1. The following chain of implications holds PMI =⇒ WOP =⇒ MP =⇒ PMI. Hence these three statements are logically equivalent. Proof. PMI =⇒ WOP: Let S ⊆ N0 and suppose that S has no least element. We will show that S = ∅. Let P (n) be the statement P (n): k ∈ / S for all natural numbers k such that 0 6 k 6 n. Notice that 0 ∈ / S since it would be a least element of S. Hence P (0) is true. Now suppose that P (n) is true. If n + 1 ∈ S, then since k ∈ / S for 0 6 k 6 n, n + 1 would be the least element of S, contradicting our assumption. Hence, n + 1 ∈ / S and so P (n + 1) is true. By the PMI, P (n) is true for all n ∈ N0 . In particular, this means that n ∈ / S for all n and so S = ∅. WOP =⇒ MP: Let T ⊆ N0 have upper bound b and set S = {s ∈ N0 : t < s for all t ∈ T }. Then S is non-empty since for t ∈ T , t 6 b < b + 1, so b + 1 ∈ S. If s0 is a least element of S, then there must be an element t0 ∈ T such that s0 − 1 6 t0 ; but we also have t0 < s0 . Combining these we see that s0 − 1 = t0 ∈ T . Notice also that for every t ∈ T , t < s0 , hence t 6 s0 − 1. Thus t0 is the desired greatest element. MP =⇒ PMI: Let P (n) be a statement for each n ∈ N0 . Suppose that P (0) is true and for n ∈ N0 , P (n) =⇒ P (n + 1). Suppose that there is an m ∈ N0 for which P (m) is false. Consider the set T = {t ∈ N0 : P (n) is true for all natural numbers n satisfying 0 6 n 6 t}. Notice that T is bounded above by m, since if m 6 k, k ∈ / T . Let t0 be the greatest element of T , which exists thanks to the MP. Then P (t0 ) is true by definition of T , hence by assumption P (t0 + 1) is also true. But then P (n) is true whenever 0 6 n 6 t0 + 1, hence t0 + 1 ∈ T , contradicting the fact that t0 was the greatest element of T . Hence, P (n) must be true for all n ∈ N0 . An important application of these equivalent results is to proving the following property of the natural numbers. 2

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