Notes for Math 290 using Introduction to Mathematical Proofs by Charles E. Roberts, Jr.

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1 Notes for Math 290 using Introduction to Mathematical Proofs by Charles E. Roberts, Jr. Chapter : Logic Topics:. Statements, Negation, and Compound Statements.2 Truth Tables and Logical Equivalences.3 Conditional and Biconditional Statements.4 Logical Arguments.5 Open Statements and Quantifiers Notes: Statement: a sentence that can be assigned a truth value. Example: 5 is an odd number. Example: 6 is an odd number. Negation: If P represents a statement, P represents the statement with the opposite truth value. Compound statements can be formed with the logical connectives and, or, implies, if and only if. The truth value of a compound statement is determined by truth tables. These are arbitrary rules which everyone agrees on. In P = Q, P is called the hypothesis and Q is called the conclusion. Equivalences: for example, DeMorgan s Laws. Statements derived from P = Q: Converse, inverse, contrapositive. Necessary versus sufficient. Avoid using vague terms subject to misinterpretation. Logical argument: Given premises P through P n, deduce conclusion C. The argument is valid if and only if the statement P P n = C is true for all truth values assigned to the variables. Illustrate by proving that the sum of two odd numbers is an even number. Premises: m and n are odd. Conclusion: m + n is even. A quantified statement P (x, y, z,... ) whose truth depends on the value of the variables x, y, z,.... Associated ideas: universe of variable values, universal and existential statements and their negation.

2 Chapter 2: Deductive Mathematical Systems and Proofs Topics: 2. Deductive Mathematical Systems 2.2 Mathematical Proofs 2.2. Techniques for proving P = Q Additional Proof Techniques Conjectures, Proofs, and Disproofs The System of Rational Numbers and the System of Real Numbers Notes: Many mathematical theorems are stated in the following form: Theorem: n : P (n) = Q(n). Proof: Let n be an arbitrary element of the universe. Assume P (n) is true. Then argue that Q(n) is true. Define even numbers as natural numbers ending in a digit in {0, 2, 4, 6, 8}. Define odd numbers as natural numbers ending in a digit in {, 3, 5, 7, 9}. Every number is either even or odd, and cannot be both. Here are some typical theorems about natural numbers: Theorem: Every even natural number can be written in the form 2k for some k N. Logical Form: n : n is even = k N : n = 2k. Proof: Let n be given. If n is odd, the implication is true. If n is even, then by definition n = 0a + b where b {0, 2, 4, 6, 8}, and in every case n = 2k for some k N. Theorem: Every odd natural number can be written in the form 2k + for some k N. Logical Form: n : n is odd = k N : n = 2k +. Proof: Similar. Theorem: The square of an even number is even. Logical form: n Z : n is even = n 2 is even. Proof: Let n be given. If n is odd, the implication is true. If n is even, we must show that n 2 is also even. It is: n 2 = (2k) 2 = 2(2k 2 ). 2

3 This is a direct proof. Theorem: If the square of a number is even, the number must be even. Logical Form: n : n 2 even = n even. Proof: Let n be given. Assume n 2 is even. If n is odd then n 2 = (2k + ) 2 = 4k 2 + 4k + is odd, a contradiction, therefore n is even. The idea behind proof by contradiction is that a statement must be true or false. If a choice of truth value forces another statement to have both truth values, the choice is wrong. Another way to prove this theorem is to note that it is logically equivalent to its contrapositive, then to prove the contrapositive statement. Rational numbers are quotients of integers. Real numbers are numbers with decimal expansions. Theorem: 2 is not a rational number. Logical Form: a : b 0 : 2 a b. Proof: Let a and b be given. Suppose 2 = a. Let k be the largest common b factor of a and b, so that a = ka and b = kb where A and B have no common factor. Then 2 = A, and after cross-multiplying and examining the results, B we can see that A and B have common factor 2, a contradiction. Therefore 2 a. b 3

4 Chapter 3: Set Theory Topics: 3. Sets and Subsets 3.2 Set Operations 3.3 Additional Set Operations 3.4 Generalized Set Union and Intersection Terminology: set, element. List elements or give rule for belonging. When x is a variable and P (x) is a statement about x, then by {x : P (x)} we mean the set of all x for which P (x) is true. Empty set. Russell s Paradox: The set A = {} satisfies A A. Now consider the expression S = {A A A}. Then either S S or S S, both of which lead to a contradiction. So our idea of what constitutes a set is not well-defined. Interval notation. Union, intersection, set difference, complement. Set Inclusion: Quantified definitions of A B and A = B. Logical proofs of set statements, for example DeMorgan s Laws and others stated in review section. Indexed sets and their use in forming unions and intersections. Ordered pairs and the cartesian product of two sets. Ordered pair definition is (a, b) = {{a}, {a, b}}. Power set. 4

5 Chapter 4: Relations Topics: 4. Relations 4.2 The Order Relations <,, >, 4.3 Reflexive, Symmetric, Transitive, and Equivalence Relations Supplementary: the ring Z n. Notes: We will skip most of Section 4. and Section 4.2, and introduce some supplementary material. Let A and B be sets. A relation from A to B is any subset R of A B. Domain, codomain, range. Let A and B be sets. A function from A to B is relation f with the additional property that if (a, b) f and (a, c) f then b = c. In this case, we can use the notation f(a) to denote the second coordinate. We will study functions in more detail in Chapter 5. Relation on a set S: a subset R of S S. arb if and only (a, b) R. Often written a b. Reflexivity, symmetry, transitivity. Examples of relations: <,, a b, etc. Equivalence relation: reflexive, symmetric, transitive. How to remember these properties: () Alphabetical order R,S,T. (2) Number of elements involved,2,3. An equivelence relation: a b iff 3 (a b). Not an equivalence relation: A B iff A B. When is an equivalence relation, we can define equivalence classes. Theorem: a b [a] = [b]. Corollary: The set of distinct equivalence classes partitions the set. Example: Congruence modulo n partitions all integers into n equivalence classes. To decide which class a number falls in, find the remainder after division by n. In the case of a negative number, add n. The Ring Z n : Given [a] and [b] where 0 a, b {0,,..., n }, define [a] + [b] = [a + b] and [a][b] = [ab], where 0 a, b < n. 5

6 Theorem: The associative properties hold for addition and multiplication. Proof: () We have [a] + ([b] + [c]) = [a] + [r] = [a + r] and ([a] + [b]) + [c] = [s]+[c] = [s+c] where r, s {0,,..., n }. We must show [a+r] = [s+c]. Write b + c r = nj and a + b s = nk. Then (a + r) (s + c) = (a + b + c nj) (a + b nk + c) = nk nj = n(k j), as desired. (2) We have [a]([b][c]) = [a][bc] = [a][r] = [ar] and ([a][b])[c] = [s][c] = [sc] where r, s {0,,..., n }. We must show [ar] = [sc]. Write bc r = nj and ab s = nk. Then ar sc = a(bc nj) (ab nk)c = anj nkc = n(aj kc), as desired. Theorem: The distributive property holds. Proof: We have [a]([b]+[c]) = [a][b+c] = [a(b+c)] = [ab+ac] = [ab]+[ac] = [a][b] + [a][c]. Additive and multiplicative identities: [0] and []. Zero property: [0]x = [0] for all x Z n. An equation: solve [2]x + [3] = [4] when n = 5. Solution: Adding [2] to both sides, [2]x = [6]. Multiplying both sides by [3], [6]x = [8]. Therefore []x = [8] therefore x = [8] = [3]. An equation with no solution: [2]x = [3] when n = 6. Reason: suppose there is a solution. Multiplying through by [3], [0]x = [9], therefore [0] = [9], a contradiction. 6

7 Chapter 5: Functions Topics: 5. Functions 5.2 Onto Functions, One-to-One Functions, and One-to-One Correspondences 5.3 Inverse of a Function 5.4 Images and Inverse Images Notes: Function was defined in Chapter 4. Review the definition. Domain, codomain, range, image, inverse image. The quantified definitions of injective, surjective, and bijective function. Composition of functions. Associative property. A composition of injections is injective. A composition of surjections is surjective. A composition of bijections is bijective. (This last is useful for proving that two sets are in the same cardinality class.) Cancellation laws. The inverse of a bijective function. 7

8 Chapter 6: Mathematical Induction Topics: 6. Mathematical Induction 6.2 The Well-Ordering Principle and the Fundamental Theorem of Arithmetic Notes: Mathematical Induction: Let the universe be U = {n Z : n a}. The statement n : P (n) is equivalent to the statement P (a) ( n : P (n) = P (n + )). Base case is P (a), induction hypothesis is P (n). Examples: Sum of consecutive integers. Sum of consecutive powers. Number of subsets of a set. Example: For x 0 and n N, ( + x) n + nx. Strong Mathematical Induction: Let the universe be U = {n Z : n a}. The statement n : P (n) is equivalent to the statement P (a) ( n : P () P (n) = P (n + )). Base case is P (a), induction hypothesis is P () P (n). Applications of Mathematical Induction: Some Number Theory Review the definitions of natural number and integer. Define greatest common divisor of two natural numbers. Theorem: When a > b, gcd(a, b) = gcd(a b, b). Proof: Any common divisor of a and b is a common divisor of a b and b, and vice-versa. Theorem: Given a, b N there exist x, y Z such that gcd(a, b) = ax + by. 8

9 Proof: By strong induction on a + b. Base case is trivial using a = b =, x =, y = 0. Assume true whenever a + b < n. Now suppose a + b = n. If a = b we can use x =, y = 0. If a > b, we have gcd(a, b) = gcd(a b, b) = (a b)x + by = ax + b(y x). Use a similar argument if a < b. Prime number: A natural number with exactly two divisors. Corollary: When a, p N and p is prime and p a, there is a solution to ax + pb =. Theorem: If p ab and p is prime then p a or p b. Proof: If p a, we re done. If p a, write ab = pk. Then px + ay = pxb + aby = b pxb + pky = b p b. Corollary: If p a a n then p a i for some i. Proof: Induction on n. Theorem: Every n 2 is prime or a product of primes. Proof: By induction on n. Base case n = 2 is true. Assume for all integers n. If n + is prime, fine. If it is not prime, factor as two integers in {2,..., n}. Each factor prime or a product of primes, hence so is n +. Theorem: There are infinitely many primes. Proof: If not, let the primes be p,..., p n. Then x = p p n + is divisible by one of these. Contradiction. Theorem: If p,..., p m and q,..., q n are primes and p p m = q q n then both lists contain the same distribution of primes. Proof: By induction on m. When m =, p q i for some i, which forces p = q i. Dividing through by p we see that there are no other q j since their product is greater than. 9

10 Now assume true for a given m. Suppose p p m+ = q q n. Then p = q i for some i. Dividing through by p, we can use the induction hypothesis to say that the remaining prime distributions are the same. Theorem: 2 is irrational. Proof: Suppose 2 = p p m q q n. Then 2q 2 q 2 n = p 2 p 2 m. This is impossible because 2 appears an odd number of times on the left and an even number of time on the right. Theorem: (75) 3 is irrational. Proof: At some point we get to q 3 q 3 n = p 3 p 3 m. Impossible because the left product contains + 3j factors of 3 and the right product contains 3k factors of 3, therefore + 3j = 3k, a contradiction. Or: the left product contains 2+3j factors of 5 and the right product contains 3k factors of 5, therefore 2 + 3j = 3k, a contradiction. 0

11 Chapter 7: Cardinalities of Sets Topics: 7. Finite Sets 7.2 Denumerable and Countable Sets 7.3 Uncountable Sets Notes: This chapter does not contain a chapter review. Definition: A is a finite set if and only if there is a bijection f : A {, 2,..., n} for some n. Notation: A = n. Counting the elements of A gives rise to f. Definition: A is an infinite set if and only if it is not a finite set. Definition: A is a countably infinite set (denumerable set) if and only if there there is a bijection f : A N. Notation: A = ℵ 0. Note that A = ℵ 0 if and only if it is possible to list the elements of A sequentially. Examples of countably infinite sets: N, Z, kn, N N (one way to do this is to organize the pairs (a, b) by the size of a + b). Theorem: P (N) ℵ 0. Proof: Let S, S 2, S 3,... be any infinite sequence of distinct subsets of N. Then the set X = {n N : n S n } is not any of these sets. So you cannot list all the subsets of N sequentially. Theorem: If a, a 2, a 3,... is an infinite sequence, and if A = {a i : i N} is infinite, then A is countably infinite. Proof: The list of distinct elements in A is a n, a n2, a n3,..., where n < n 2 < n 3 <. Corollary: Q is countably infinite. Proof: List the elements of N N sequentially as (a, b ), (a 2, b 2 ), (a 3, b 3 ),.... Then the sequence a b, a 2 b 2, a 3 b 3,...

12 contains all positive rational numbers, therefore Q + = { a i b i : i N} is countable, therefore Q is contable. Theorem: If A is countably infinite and B A is infinite, then B is countably infinite. Proof: If the elements of A are a, a 2, a 3,... then the elements of B are a n, a n2, a n3,... where n < n 2 < n 3 <. Theorem: c ℵ 0. Proof: If suffices to show that R contains a subset X that is not countably infinite. Let X be the set of real numbers of the form 0.a a 2 a 3, where a i {0, } for each i. There is a one-to-one corerspondence between X and P (N). Since P (N) is not countably infinite, neither is X. Theorem: If A = ℵ 0 and B = ℵ 0 then A B = ℵ 0. Proof: The elements of A B can be listed a, b, a 2, b 2,.... Corollary: The set of irrational numbers is not countably infinite. Proof: Since k 2 is irrational for all k N, there are infinitely many irrational numbers. Let I be the set of irrational numbers. If I = ℵ 0 then since R = I Q, c = ℵ 0, a contradiction. Therefore I ℵ 0. Theorem: If A = ℵ 0 and B = ℵ 0 then A B = ℵ 0. Proof: Q Q = {(i, j ), (i 2, j 2 ), (i 3, j 3 ),... } A B = {(a i, b j ), (a i2, b j2 ), (a i3, b j3 ),... }. Equivalence class on sets: A B if and only if there is a bijection f : A B. Definition: A = B if and only if A B. We say that A and B have the same cardinality iff A = B. Definition: A B if and only if there is an injective function f : A B. So for example N < P (N) and N < c. Some facts about cardinality: () A B and B C implies A C. Reason: composition of injective functions. 2

13 (2) A = B and B = C implies A = C. Reason: composition of bijective functions. (3) A = A and B = B implies A B = A B. Reason: it s easy to construct a bijection. (3) A B and B A implies A = B. This is called the Schroeder- Bernstein Theorem. Proof of Schroeder-Bernstein Theorem: Let f : A B and g : B A be injective. Let A be the elements in A not in the range of g. Then set A 2 = g(f(a )), A 3 = g(f(a 2 ), A 4 = g(f(a 3 )), etc. Define h : A B by { f(a) a A i for some i h(a) = g (a) a A i for all i. h is injective: Suppose h(a ) = h(a 2 ). Case : f(a ) = f(a 2 ). Then a = a 2. Case 2: f(a ) = g (a 2 ). Then g(f(a )) = a 2. So we have a A i for some i, which places a 2 A i+, which is not possible. So this case does not occur. Case 3: g (a ) = f(a 2 ). Not possible either. Case 4: g (a ) = g (a 2 ). Then a = a 2. h is surjective: Let b B be given. Then g(b) A. Case : g(b) A i for all i. Then h(g(b)) = g (g(b)) = b. Case 2: g(b) A i for some i 2. Then g(b) = g(f(a)) for some a A i, therefore b = f(a). Theorem: [0, ) = P (N). Proof: P (N) is equinumerous with binary sequences, which can be mapped injectively into [0, ) using a decimal expansions. Conversely, each x [0, ) has a decimal expansion, and we can uniquely encode 0.b b 2 b 3 by a string of 0s and s (first 0 digits encodes b, second 0 digits encodes b 2, etc), hence [0, ) can be injectively mapped into binary sequences, hence into P (N). Notation: R = c. Theorem: c = [0, ). 3

14 Proof: [0, ) can be bijectively mapped into (0, ] via x x, and (0, ] can be bijectively mapped into [0, ) via x. So we have x [0, ) [0, ). We also have (0, ) R via x ln x. To complete the proof, we just need to show [0, ) (0, ). We use the following lemma. Lemma: If X is infinite and x 0 X then X X {x 0 }. Proof: Choose a sequence of elements x, x 2, x 3,... distinct from x 0. Define f : X X {x 0 } via { x x x i for all x f(x) = x i+ x = x i. One can check that f is a bijection. Corollary: P (N) = c. Cantor s Theorem: For all sets A, A < P (A). Proof: The injection that establishes A P (A) is a {a}. Now suppose A = P (A). Then we have P (A) = {f(a) : a A} for some bijection f : A P (A). Let X = {a A : a f(a)} P (A). If X = f(b) then b X b f(b), so X f(b), a contradiction. Hence A P (A). It is not known whether or not there exists a set X such that ℵ 0 < X < c. Continuum Hypothesis: No such X exists. Generalized Continuum Hypothesis: For all infinite sets A, there is no set X satisfying A < X < P (X). Exercise: R R R Solution: It suffices to prove B B B, where B is the set of binary sequences. A bijection is (a, b) (a, b, a 2, b 2,... ). 4

15 Chapter 8: Proofs from Real Analysis Topics: 8. Sequences 8.2 Limit Theorems for Sequences 8.3 Monotone Sequences and Subsequences 8.4 Cauchy Sequences Notes: There is no chapter review for this chapter. Also, we will skip the material on Cauchy Sequences since it is not put to any use. Trichotomy Law: for all a, b R, either a > b or a = b or a < b. These are mutually exclusive conditions. The absolute value of a rational number: { r r 0 r = r r < 0. Distance between two rational numbers: r s. Triangle Inequalities for Rational Numbers: r + s r + s and r s r s. Proof: r + s = (r + s)θ = rθ + sθ r + s where θ {, }. To prove the second inequality, for any u and v we have Setting u = r s, v = s, we obtain Reversing r and s we obtain u + v v u. r s r s. s r s r = r s. So r s r s. 5

16 Limit of a sequence Let x n R for all n, and let x R. Then we define lim x n = x n if and only if for all ɛ R + there exists N N such that x N x < ɛ, x N+ x < ɛ, x N+2 x < ɛ,.... In other words, x n x < ɛ for all n beyond a certain point. x n x. Notation: Example: lim n = 0. Let ɛ > 0 be given. We want to find N such that n n N implies < ɛ. We can use any natural number N. n ɛ Example: If a R and 0 < a < then lim n a n = 0. Proof: Let ɛ > ( 0 be) given. We wish to find N so that n > N implies a n < ɛ, n or equivalently a >. Write = + b where b > 0. We proved earlier ɛ a ( ) n that a = ( + b) n + nb. We wish to require + nb >. We just ɛ need any natural N satisfying N ɛ b = ɛ a. Example: a = 99 and ɛ =. Our formula yields N(an, ɛ) = 980, a gross overestimate according to the graph below: 0.30 a[n] = n

17 Notation: When (a n ) is a sequence which converges to a limit a we will denote by N(a n, ɛ) a number such that n N(a n, ɛ) = a n a < ɛ. We can summarize our last two examples by saying N( n, ɛ) = ɛ and N(a n, ɛ) = ɛ. a Example of a sequence that does not have a limit: x n = ( ) n. For any x R, every other value of x n x is. Properties of Convergent Sequences Lemma: A convergent sequence is bounded from above and from below. Proof: Suppose a n a. Then for n N, a n a <, which implies a < a n < a +. Therefore for all n we have min(a,..., a N, a ) a n max(a,..., a N, a + ). Sum rule: a k a and b k b implies a k + b k a + b. Details: a k + b k a b a k a + b k b 0. Say that n > N(a n, ɛ) = a n a < ɛ and n > N(b n, ɛ) = b n b < ɛ. Then we can use N(a n + b n, ɛ) = max ( N ( a n, ɛ 2 ), N ( b n, ɛ )). 2 Product rule: a k a and b k b implies a k b k ab. Details: a k b k ab = a k b k ab k + ab k ab a k a b k + a b k b a k a B + A b k b 0 where A and B are positive upper bounds for ( a n ) and ( b n ). We can use N(a n b n, ɛ) = max ( N ( a n, ɛ 2B ), N ( b n, ɛ 2A )). 7

18 Reciprocal rule: a k a and a k, a 0 implies a k. Details: a a k = a a k a a a k a k a 0, assuming a a (/2) a 2 k a. We can use 2 ( ) ( ( N, ɛ = max N a n, a ) )), N (a n, a 2 a n 2 2 ɛ. Quotient rule: a k a and b k b and b k, b 0 implies a k b k a. Details: b this follows from product rule and reciprocal rule. Assuming A is a positive upper bound for ( a k ) and B is a positive upper bound for ( b k ), we can use ( ) ( an ( ɛ ) ( )) ɛ N, ɛ = max N a n,, N,. b n 2B b n 2A Example: Let x n = n 2 5n+. To prove lim n x n = 0 using an ɛ-n argument, apply limit properties to x n = n 2 5 n + n 2. We have N(, ɛ) =, and an upper bound for ( ) is, hence using the n ɛ n product rule we have N( ( n, ɛ) = N 2 n, ɛ ) = 2 2 ɛ. Note that for n 6, 5 n + n 2 > 6, hence x n 0 < 6 n 2. So a solution is N(x n, ɛ) = max(6, 3ɛ ). Example: Let a n = (2 + n )2. To prove that lim n (2 + n )2 = 4 using an ɛ-n argument, observe that for all n. We have n ( 2n Exercise: Compute N )., ɛ n 3 3n 2 3 N(a n, ɛ) = N((2 + /n)(2 + /n), ɛ) = N(2 + n, ɛ 6 ) = N( n, ɛ 6 ) = 6 ɛ. 8

19 Exercise: Assume that (a n ) converges to a and that ( a n ) has positive upper bound A. Prove that for k N, N(a k n, ɛ) = N(a n, ɛ 2 k A k ) (hint: use induction). Using this, find N so that n > N = (2 + n ) < 0. Theorem: Suppose that (x n ) converges to a and (x n ) converges to b. Then a = b. Proof: If a > b then beyond a certain point x n a < a b and x 2 n b < a b which implies that beyond a certain point a b = a b = a x n +x n b a x n + x n b < a b 2 + a b 2 a contradiction. Therefore a b. Similarly, b a, therefore a = b. 2, = a b, Theorem: If a n b n c n for all n and lim n a n = lim n c n = L, then lim n b n = L. Solution: Let ɛ > 0 be given. Beyond a certain point a n > L ɛ and c n < L ɛ, hence L ɛ < a n b n c n < L + ɛ, hence b n L < ɛ. Bounded Monotonic Sequences Definition. Example: Let s n = n 2. The sequence (s n ) is bounded and monotonic increasing. Reason: k s 2 k = + i=0 (2 i + ) + 2 (2 i + 2) (2 i+ ) + k 2 i=0 2 i 2 k = + 22i 2. Hence the subsequence (s 2 k) is bounded above + 2, hence the entire sequence is bounded above by this number since the sequence is increasing. The sequence is plotted below through s

20 Example: Let s n = n. The sequence (s k) is unbounded. Reason: We have s 2 k = + k i=0 2 i i k i+ i=0 2 i 2 i+ = + k 2. Since (s 2 k) is unbounded, (s n ) is unbounded and does not converge. The sequence is plotted below through s Theorem: Every increasing and bounded sequence in R converges to a limit. Proof: Proved in Math

21 Therefore ( n k= ) converges to a limit, which we call k 2 k= shown using methods of complex analysis (Math 5) that k= It can be k 2 = π2 k 2 6 Remark: More generally, every subset of R that has an upper bound has a least upper bound in R. Hence R forms a complete ordered field. This is not true of the rational numbers. For example, the decimal expansion of 2 yields an increasing sequence of rational numbers bounded above by 2 but does not have a rational least upper bound. In fact, the least upper bound is 2, which is irrational. Bolzano-Weierstrass Theorem: Every bounded sequence in R has a convergent subsequence. Proof: If suffices to find a monotonic subsequence. If (x i ) has a a strictly increasing subsequence, we re done. Now suppose (x i ) does not have a strictly increasing subsequence. Then for any k the sequence (x k, x k+,... ) must have a maximum element, otherwise it yields a strictly increasing subsequence of (x i ). Let x k be the maximum element of (x, x 2,..., ), let x k2 be the maximum element of (x k +, x k +2,... ), let x k3 be the maximum element of (x k2 +, x k2 +2,... ). We have x k x k2 x k3, therefore (x ki ) is a decreasing subsequence of (x i ). 2

22 Chapter 9: Proofs from Group Theory Topics: 9. Binary Operations and Algebraic Structures 9.2 Groups Subgroups and Cyclic Groups Notes: Examples of groups: Z 7 under addition, Z 7 under multiplication. Give the operation tables. Group in general: a set and a binary operation on the set. The binary operation must be associative, there must be an identity element, and every element should have an inverse. Some examples of sets with binary opertions are given in Exercises 9., pp Some examples of groups are given in Exercises 9.2, pp Z n is always a group under addition. Z 6 is not a group under multiplication. Reason: identity is [] but [2] does not exist. Z p is always a group under multiplication. Reason: For a p, p a, so there is a solution to px + ay =. Therefore [a][y] = [], and moreover [y] Z p since [y] [0]. Groups also arise as automorphisms of a geometric figure. Example: automorphisms of a square. The elements of the group are functions from {, 2, 3, 4} to {, 2, 3, 4} that preserve the edges 2, 23, 34, 4. The operation table for D 4 shows that the group is not Abelian. For example, σ = (, 2, 3, 4) D 4 and τ = (, 2)(3, 4) D 4, yet στ = (3, )(2, 4) and τσ = ()(2, 4)(3). A subgroup of a group is simply a subset that forms its own group using the same group operation. Cyclic subgroup generated by g: g = {g n : n Z}. Cyclic subgroups are always abelian. (, 2, 3, 4) is an abelian subgroup of S 4. D 4 is a non-abelian subgroup of S 4. 22

23 Some examples of subgroups are given in Exercises 9.3, pp We can derive theorems about numbers by thinking in terms of groups. For example, given any g in a finite group G, the list g, g 2, g 3,... must eventually repeat itself, so that g i = g j where i < j, which implies g j i = e. The least positive value of k such that g k = e is called the order of g. It is a theorem that o(g) o(g) when G is finite. To prove this, form the function σ : G G via σ(a) = ga. Then σ is a permutation. If we write it in cycle structure, we will see that all the cycles have length equal to o(g). Since all the elements of G are present in the cycles, O(G) = k o(g) where k is the number of cycles of σ. Example: Let G = Z 7 and g = [2]. Then σ = ([], [2], [4])([3], [6], [5]). We can see that o(g) = 3 divides o(g) = 6. There are some order problems in Exercises