Definitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations

Transcription

1 Page 1 Definitions Tuesday, May 8, :23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of is injective if by Injective, Surjective and Bijective be a function is surjective if (i.e. ) is bijective if Divides If is both injective and surjective We say divides and write, if s.t. Cartesian Product If and are sets the Cartesian product of and is Relations A relation on a set is a subset of We write Equivalence Relations if A relation on is an equivalence relation if is Reflexive If i.e. Symmetric If

2 Page 2 i.e. Transitive If i.e. If and Greatest Common Divisor, where either or A greatest common divisor of and is a positive integer s.t. and If s.t. and then We write the greatest common divisor of As a convention Equivalence Class and, if it exists, as be a set, and let ~ be an equivalence relation on If the equivalence class represented by is the set Integers Modulo The relation on given by is an equivalence relation The set of equivalence classes under is denoted as We call this set integers modulo (or integers mod n) We can check that there are We use Then Group If elements in to denote the equivalence class in is a set equipped with a binary operation that satisfies Associativity: Identity: Inverses: Then we say Abelian Group We say a group s.t. s.t. is a group under this operation is abelian, if Order of Group Element

3 Page 3 If is a group, and The order of is the smallest positive integer s.t. If is the order of, write If no such integer exists, write i.e. Symmetric Group be fixed (i.e. is the set of all permutations of ) Then is a group with operation given by function composition We call this group symmetric group of degree Cycle The element of be fixed if is denoted by Disjoint Cycles for given by and is called a cycle of length Two cycles and are disjoint if Homomorphism A function One says Isomorphism be groups is a homomorphism if "respects", or "preserves" the group operation be groups A homomorphism there is a homomorphism, and is a isomorphism if s.t. In this case, we say and are isomorphic Subgroup

4 Page 4 be a group, and let is a subgroup if (nonempty) If (closed under the operation) If (closed under inverse) If is a subgroup of, we write Regular A regular Symmetry picking up a copy of it moving it around in 3d setting it back down is a polygon with all sides and angles equal A symmetry of a regular -gon is a way of so that it exactly covers the original Dihedral Groups Cyclic Group A group is cyclic if s.t. Least Common Multiple where one of is nonzero. and If and is called -th dihedral groups A least common multiple of and is a positive integer s.t. We denote the least common multiple of and by Define Subgroups Generated by Subsets of a Group be a group and The subgroup generated by the intersection of every subgroup of is containing Finitely Generated Group A group is finitely generated if There is a finite subset of s.t.

5 Page 5 Coset If is a group,, and is called a left coset is called a right coset An element of a coset is called a representative of the coset Normal Subgroup be a group, is a normal subgroup if, In other words, If Quotient Group is closed under conjugation is normal, we write be a group, The set of left costs of is a group under the operation This group is denoted as (say " mod ") We call this group quotient group or factor group Index of a Subgroup If is a group, and The index of is the number of distinct left cosets of in Denote the index by Product of Subgroups Define Transposition Fix be a group and to be a positive integer A cycle in is a transposition Sign of Permutation (Transposition Definition) Sign of Permutation (Auxiliary Polynomial Definition)

6 Page 6 is even if is odd if is the sign of, often denoted as Alternating Group The alternative group, denoted as That is, Group Action is the kernel of contains of all even permutations in An action of on is a function, s.t. Orbit and Stabilizer Suppose a group The orbit of, denoted acts on a set, is The stabilizer of, denoted stab( ), is Centralizer be a group, and let act on itself by conjugation If This set is called the centralizer of, denoted as Center Normalizer is the set of elements in that commute with the element is the set of elements that commute with every element of be the set of subgroups of a group acts on by If This set is called the normalizer of in, denoted Note: Conjugacy Class is the set of elements in If is a group, acts on itself by conjugation: that commute with the set The orbits under this action are called conjugacy classes

7 Page 7 Denote a conjugate class represented by some element Partition A partition of is a way of writing as a sum of positive integers Example: 3 has 3 partitions: Ring A ring is a set equipped with two operations and s.t. is associative is an abelian group s.t. Distributive property: Zero-Divisor and Unit be a ring A nonzero element s.t. s.t. is called a zero-divisor if or Assume, is called a unit if Group of Unites Field A communitive ring Every nonzero element of i.e. Every nonzero element of Product Ring be rings The product ring is called a field if is a unit have a multiplicative inverse has the following ring structure For addition, it's just the product as groups For multiplication, Integral Domain A communicative ring contains no zero-divisors Subring with identity by is an integral domain (or just domain) if

8 Page 8 A subring of a ring is a additive subgroup of s.t. is closed under multiplication contains Polynomials over a ring be a commutative ring A polynomial over Degree If is a variable, and The degree of, denoted Note: If The leading term of is the sum is The leading coefficient of is, is, where Leading Term and Leading Coefficient Polynomial ring Then is a commutative ring with is a polynomial over ordinary addition and multiplication of polynomials Ideal be a subset of ring Define is a left ideal of if, and let is an additive subgroup of Right ideal is defined similarly is an ideal if is both a left and right ideal Principal Ideal is a commutative ring, and let Quotient Ring be a ring is called the principal ideal generated by If is an ideal the quotient group is a ring with multiplication

9 Page 9 Conversely, if Then is an ideal is an additive subgroup is a ring with multiplication defined above Ideal Generated by Subset be a commutative ring If is a subset of the ideal generated by is If is finite we write as Maximal Ideal An ideal in a ring is maximal if Prime Ideal An ideal, and the only ideals containing are and be a commutative ring Euclidean Domain be a domain is prime if or A norm on is a function s.t. is called a Euclidean domain if is equipped with a norm s.t. either with, s.t., and or Principal Ideal Domain A domain in which every ideal is principal is called a principal ideal domain

10 Page 10 Propositions Wednesday, April 4, :18 PM Proposition 1: Well-ordering of Every nonempty set of has a unique minimum element s.t. Proposition 2: The Division Algorithm, where Then s.t., and Proposition 3: Uniqueness of Greatest Common Divisor, where either or Suppose (1) (2) Then Suppose s.t. and both divide and If s.t. and and Proposition 4: Lemma for Euclidean Algorithm, where Choose s.t., and If exists exists and Proposition 5: Proposition 6: Existence of GCD If exists Proposition 7: Bézout's Identity If s.t. Proposition 8: Equivalence Classes Partition the Set be a set with equivalence relationship ~ If and are either equal or disjoint Proposition 9: Addition and Multiplication in, and let If, and in Then, and Corollary 10: Integers Modulo

11 Page 11 For, is a group under the operation We will denote this operation by So Proposition 11: is a group with operation given by multiplication Proposition 12: Properties of Group be a group has the following properties The identity of Each is unique has a unique inverse The Generalized Associative Law Proposition 13: Cancellation Law If If If If be a group, and let Corollary 14: Cancellation Law and Identity be a group, and let Proposition 15: Order of Symmetric Group Proposition 16: Isomorphism Preserves Commutativity be an isomorphism is abelian if and only if Then is abelian Proposition 16: Injective Homomorphism Preserves Order be an injective homomorphism Proposition 17: The Subgroup Criterion A subset of a group is a subgroup iff and Proposition 18: Isomorphism of Cyclic Group be a cyclic group

12 Page 12 If If Proposition 19: Order of Theorem 20: Subgroup of Cyclic Group is Cyclic be a cyclic group Then every subgroup of is cyclic More precisely, if either or, where is the smallest positive integer s.t. Theorem 20: Subgroup of Finite Cyclic Group is Determined by Order be a finite cyclic group of order For all positive integers dividing, subgroup of order Proposition 21: Construction of If If Proposition 22: Properties of Coset be a group and The relation on given by iff is an equivalence relation In particular, left/right cosets are either equal or disjoint Proposition 23 be a subgroup of a group iff Proposition 24: Quotient Group If is a group, and the set of left costs of, denoted as (say " mod ") is a group under the operation We call this group quotient group or factor group Theorem 25: Lagrange's Theorem If is finite group, and In particular, Corollary 26: Group of Prime Order is Cyclic If is a group, and is prime is cyclic, hence,

13 Page 13 Corollary 27: If is a finite group, and Corollary 28: The Fundamental Theorem of Cyclic Groups If is a finite cyclic group there is a bijection Proposition 29: Order of Product of Subgroups Proposition 30: Permutable Subgroups If iff Corollary 31: Product of Subgroup and Normal Subgroup If, and either or is normal in Theorem 32: The First Isomorphism Theorem If is a homomorphism induces an isomorphism Corollary 33: Order of Kernel and Image Theorem 34: The Second Isomorphism Theorem, and assume Theorem 35: The Third Isomorphism Theorem be a group, and, where Proposition 36: Criterion for Defining Homomorphism on Quotient be groups, and A homomorphism If and only if given by induces a homomorphism Theorem 37: The Correspondence Theorem be a group, and let there is a bijection

14 Page 14 Proposition 38: Transposition Decomposition of Permutation Every Proposition 39: is a group homomorphism If can be written as a product of transposition is transposition is a Group Homomorphism Proposition 40: Sign of Transposition Corollary 41: Equivalence of Two Definitions of Sign is well-defined, and Corollary 42: Surjectivity of If is surjective Proposition 43: Subgroup of Index 2 is Normal If is a group,, and Proposition 44: Conjugate Cycle If Then s.t. are -cycles in Theorem 45: Have No Subgroup of Order 6 have no subgroup of order 6 Proposition 46: Stabilizer is a Subgroup If acts on, and stab Proposition 47: Orbits Equivalence act on a set The relation iff s.t. is an equivalence relation on Proposition 48: Orbit-Stabilizer Theorem If acts on, and Proposition 49: Permutation Representation of Group Action be a group acting on a finite set Then each determines a by Proposition 49: Induced Homomorphism of Group Action The map Theorem 50: Cayley's Theorem is a homomorphism

15 Page 15 Every finite group is isomorphic to a subgroup of the symmetric group Theorem 51: The Class Equation be a finite group be representatives of the conjugacy classes of Corollary 52: Center of -Group is Non-Trivial If is a prime, and is a group of order Corollary 53: Group of Order Prime Squared is Abelian If is a prime, and is a group of order is abelian. In fact, either or Theorem 54: Cauchy's Theorem If is a finite group, and is a prime divisor of of order Lemma 55: Recognizing Direct Products be a group with normal subgroups The map given by is an isomorphism if and only if and Lemma 56: Coprime Decomposition of Finite Abelian Group be a finite abelian group of order, where If and the map given by is an isomorphism Moreover, if and are nontrivial Corollary 57: -Group Decomposition of Finite Abelian Group be a finite abelian group, and be a prime divisor of Choose s.t. and Then, where,, and Lemma 58: Prime Decomposition of Abelian -Group If is an abelian group of order, where is a prime Then has maximal order among all the elements of, where Theorem 59: Fundamental Theorem of Finite Abelian Groups Every finite abelian group is a product of cyclic groups Corollary 60: Number of Finite Abelian Groups of Order

16 Page 16 If, where are distinct primes Then the number of finite abelian groups of order n is Proposition 61: Properties of Ring be a ring The multiplicative identity 1 is unique Proposition 62: Criterion for Trivial Ring A ring is trivial (i.e. have only one element) iff Proposition 63: One-Sided Zero Divisor and Unit be a ring, and s.t., and s.t. s.t. Proposition 64: Units and Zero-Divisors of Every nonzero element in is either a unit or a zero-divisor Proposition 65: Criterion for Product Ring to be Domain If and are rings is a domain iff one of the or is a domain, and the other is trivial Proposition 66: Finite Domain is a Field A finite domain be a domain is a domain is a field Proposition 67: Polynomial Rings over a Domain If Proposition 68: Ideal Containing 1 is the Whole Ring is an ideal Proposition 69: Quotient Ring

17 Page 17 be a ring If is an ideal the quotient group is a ring with multiplication Conversely, if Then is an ideal If is an additive subgroup is a ring with multiplication defined above Theorem 70: The First Isomorphism Theorem for Rings is a ring homomorphism there is an induced isomorphism, given by Proposition 71: Criterion for Maximal Ideal If is a commutative ring, and is an ideal Then is maximal is a field Proposition 72: Prime Ideas of The prime ideals of are ideals of the form, where is prime or Proposition 73: Criterion for Prime Ideal be a commutative ring, an ideal is prime is a domain In particular, is a domain 0 ideal is prime Corollary 74: Maximal Ideal is Prime If is a commutative ring, and is maximal is prime Proposition 75: Euclidean Domain is a Principal Ideal Domain Every ideal in a Euclidean domain is principal More precisely, if is an ideal, where is an element of with minimum norm Theorem 76: Polynomial Division Then be a field is a Euclidean domain More specifically, if where s.t. and

18 Page 18 Notations, Divides, Equivalence Relations Wednesday, January 24, :46 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of Definition For is injective if by Injective, Surjective and Bijective Example 1 Example 2 be a function is surjective if (i.e. ) is bijective if is injective Suppose Therefore is not surjective is both injective and surjective is injective Because the image of be given by does not contain any odd integers

19 Page 19 is injective is surjective Divides Definition If Examples Therefore is bijective Because is surjective is both injective and surjective We say divides and write, if s.t. Cartesian Product Relations, since, since, since Equivalence Relations If and are sets the Cartesian product of and is A relation on a set is a subset of We write if Equivalence Relations A relation on is an equivalence relation if is Reflexive If i.e. Symmetric If i.e. Transitive If i.e. If and

20 Page 20 Example 1 R be a relation on set A such that Then R is an equivalence relation Reflexive If Symmetric If Thus Transitive Example 2 by definition If then and Thus Reflexive be a positive integer It follows that Symmetric Suppose Choose Then Thus, Transitive Suppose Then Choose Then Thus, s.t., since, and so and s.t., and we have, and so is an equivalence relation

21 Page 21 Induction, Well-Ordering of Friday, January 26, :05 AM Induction Base case Induction step Proposition 1: Well-Ordering of Statement Every nonempty subset of has a unique minimum element That is, Proof (Existence) Assume s.t. We argue by induction on Base case is finite When Inductive step When Choose Assume Choose By induction, this is clear has a minimum value: call it Case 1: is a minimum value of Case 2: is a minimum value of is infinite Then i.e. is finite

22 Page 22 So we can choose a minimum element of If If : call it In either case,, so is a minimum element of This proves existence Proof (Uniqueness) Suppose and are both minimum elements of Thus,, and This proves uniqueness

23 Page 23 Division Algorithm, Greatest Common Divisor Monday, January 29, 2018 Statement, where Then s.t., and Proof (Existence) is not empty Then i.e. Thus, Choose Since contains a unique minimum element: call it s.t. We still need to show that, we know So we just need to show that If Then This is impossible, since Thus,, and it is less than Therefore we've proven the existence of Proof (Uniqueness) Suppose We must show that Suppose, where s.t., where and Without loss of generality, assume Then 9:47 AM Proposition 2: The Division Algorithm is the minimum element of and Thus,, but. This is impossible, thus

24 Page 24 We have Therefore we've proven the uniqueness of Note we can prove the following stronger statement If Proof (Existence) Assume and then Choose s.t., and Then This proves existence Proof (Uniqueness) Assume Suppose Then Since Therefore, and, where s.t., where, where, where, our previous result implies and Greatest Common Divisor, where either or and If s.t. and then and A greatest common divisor of and is a positive integer s.t. We write the greatest common divisor of As a convention Statement Proof, where either or Suppose (1) (2) Then s.t. and both divide and If s.t. and and Combining properties (1) and (2), we have and, if it exists, as Proposition 3: Uniqueness of Greatest Common Divisor and

25 Page 25 Statement Choose s.t. and By substitution, we get If. This is impossible since and are both positive Therefore Suppose and Proposition 4: Lemma for Euclidean Algorithm Proof, where Choose s.t., and If exists exists and Set Statement and Choose Then s.t. And we already know If s.t. and s.t. and, since Choose s.t. and Thus Since Proposition 5: Proof and We can conclude that By Proposition 3, If If This is true by our convention Certainly, and

26 Page 26 If s.t. and then Therefore

27 Page 27 Euclidean Algorithm, Bézout's Identity Wednesday, January 31, :56 AM Proposition 6: Existence of GCD Statement Proof Input Output (0) (1) If exists By Proposition 5, we may assume that Choose s.t., where We argue by induction on Base case Suppose We have and If s.t. and Therefore Inductive hypothesis exists, and equals If s.t., and Then Inductive step If Suppose exists Choose s.t., where By inductive hypothesis, By Proposition 4, The Euclidean Algorithm Algorithm (2) Note with, output Else, proceed to step (1) exists exists, and equals Since, we can find s.t., where If, output Otherwise, repeat step (1) with and playing the roles of and The algorithm terminates

28 Page 28 The algorithm terminates Since the remainder decreases at each application of step (1) By Proposition 4, the output will be Example: use the Euclidean Algorithm to compute Take Here Thus, Statement Note, so the algorithm terminates Proposition 7: Bézout's Identity Proof If s.t. If If If need not to be unique We can take In fact, any pair of or works Without loss of generality, assume Then We can take and Without loss of generality, assume Choose s.t. We argue by induction on Base case When So we can take Inductive step Suppose Choose By induction, s.t. Thus, by Proposition 4 s.t.

29 Page 29 So we can take Recall when we computed 's now find s.t. and Example: Express as where, we had Start with the second to last equation, and "back-fill" Therefore

30 Page 30 Equivalence Class, Friday, February 2, :06 AM, Group Homework 1 (a): Injective Function Has a Left Inverse and be two nonempty sets Prove that Since Define be a injective function has a left inverse is injective, Choose If If Check that If Choose Define Define Thus, Therefore in the following way s.t. is a left inverse Example of The Euclidean Algorithm Use the Euclidean Algorithm to find Find s.t. So we can take

31 Page 31 Equivalence Class be a set, and let ~ be an equivalence relation on If the equivalence class represented by is the set Proposition 8: Equivalence Classes Partition the Set Statement Proof be a set with equivalence relationship ~ If and are either equal or disjoint Suppose It suffices to show that if Suppose Suppose Integers Modulo (Symmetry) (Transitivity) (Symmetry) (Transitivity) (Symmetry) (Transitivity) (Symmetry) (Transitivity) The relation on given by is an equivalence relation The set of equivalence classes under is denoted as We call this set integers modulo (or integers mod n) We can check that there are We use Then Group Definition If elements in to denote the equivalence class in is a set equipped with a binary operation

32 Page 32 that satisfies Associativity: Identity: Inverses: Then we say s.t. s.t. is a group under this operation are groups with operation If (Similarly for ) + is certainly associative in all 4 sets 0 is the identity in each case If the inverse of is

33 Page 33 Examples of Groups, Well-definedness, Monday, February 5, :55 AM Examples of Groups Is Are Is a group under multiplication? No, because there is no inverses for 2 groups under multiplication? No, because 0 still has no multiplicative inverse and similarly Then are groups with operation given by multiplication We argue this for ; the same proof works for and Multiplication is an operation on If Associativity This is clear Identity Inverses is the identity a group with operation given by subtraction? No, because subtraction is not associative General Linear Group If Then, is a group under matrix multiplication Matrix multiplication is an operation on Associativity This is clear Identity (R), since

34 Page 34 The identity matrix is the identity Inverses Note If Abelian Group We say a group Statement, its inverse is When, the operation in is not commutative is abelian, if Proposition 9: Addition and Multiplication in Proof:, and let If, and in Then, and Choose s.t. and Then Thus, Proof: So, Choose s.t. and Then Thus, So, Well-definedness Example Say we want to "define" a map Note that But is not a function in

35 Page 35 So we say that How to check well-definedness is not well defined To check that a purported function One needs to check that Corollary 10: Addition Group of Statement Proof be fixed is a group under the operation We will denote this operation by So Well-definedness By proposition 9, the operation Associative is well-defined, is well-defined Associativity is inherited from the associativity of addition for Identity The identity is Inverses,, the inverse of is

36 Page 36 Wednesday, February 7, 2018, Properties of Group 9:56 AM is Not a Group Under Multiplication be fixed Proposition 9 implies that there is a well-defined function Check group property Identity: Definition This operation is associative is a reasonable candidate for an identity, but there is no inverse Example in Proposition 11: Define Statement Proof By HW 2 #2, s.t. is a group with operation given by multiplication Closure: If as well Associativity: Clear, from associativity of multiplication of integers Identity: List of Groups Inverses: Built in HW 2 #2 Set Operation, Matrix multiplication

37 Page 37, Proposition 12: Properties of Group be a group has the following properties The identity of is unique In other word If s.t. and Then 1 Proof Each has a unique inverse In other word If and s.t. and Then Proof, and suppose are both inverses of Then Since the inverse is unique, The Generalized Associative Law i.e. If is independent of how it is bracketed First show the result is true for Assume for any any bracketing of a product of elements can be reduced to an expression of the form Then any bracketing of the product 2 sub-products, say must break into where each sub-product is bracketed in some fashion Apply the induction assumption to each of these two sub-products Reduce the result to the form to complete the induction By the generalized associative law Notation

38 Page 38 We will apply the Generalized Associative Law without mentioning it In particular, if is a group and, we will write If If Proposition 13: Cancellation Law Statement Proof Warning If If be a group, and let This holds in abelian groups, but not in general Corollary 14: Cancellation Law and Identity be a group, and let

39 Page 39 Order, Definition of Friday, February 9, :07 AM Order Definition Note If is a group, and The order of is the smallest positive integer s.t. If is the order of, write If no such integer exists, write i.e. Example 1 The order of the identity is Example 2 Therefore, In Example 3, every nonzero element has infinite order The identity 0 has order of 1 In and, the elements of finite order are In Example 4, there are lots more Elements of order in are called roots of unity is the fourth root of unity i.e.,,, What are the orders of the elements in? Elements Order Note 1 is the identity

40 Page 40 Example 5 6 In general, if the " power" of is Note that all the orders are divisors of 6 (Lagrange Theorem) What are the orders of the elements in? Note: Definition Elements Order Note 1 is the identity be fixed, so Symmetric Group (Section 1.3) Proof (i.e. is the set of all permutations of ) Then is a group with operation given by function composition We call this group symmetric group of degree Function composition is an operation on The composition of bijective functions is still bijective Therefore, function composition is an operation on Associativity Suppose Thus Identity The identity map is the identity Inverses Bijective functions all have inverse functions

41 Page 41 Properties of Monday, February 12, 2018, Properties of Cycles 9:53 AM Proposition 15: Order of Symmetric Group Statement Proof Cycle First, we prove that If and are sets of order Then there are injective functions from to We argue by induction on When For Suppose, this is clear is injective there are possibilities for restricts to an injective function There are Thus, there are Now, take Note Definition Example such functions, by induction injective functions Since injection between finite sets of the same order is bijective We can conclude that The sets must be finite Counterexample: The element of be fixed if is denoted by for given by is not bijective and is called a cycle of length

42 Page 42 Notice: Disjoint Cycles Definition Example Fact Algorithm Two cycles and are disjoint if Every element of are disjoint can be written as a product of disjoint cycles = {(1), (1 2), (1 3), (1 4), (2 3), (2 4), (3 4), (1 2 3), (1 2 4), (1 3 2), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3), ( ), ( ), ( ), ( ), ( ), ( ), (1 2)(3 4), (1 4)(2 3), (1 3)(2 4)} Note: We write the identity of Cycle Decomposition for Permutations as Step Begin the new cycle: If close the cycle with a right parenthesis return to step 1 If write next to a in this cycle: (1 So write Example If close the cycle with a right parenthesis So continue the cycle as: return to step 1 ( If write next to in this cycle: and repeat this step until the cycle closes Naturally this process stops when all the numbers from σ = ( )(2 1 3)

43 Page 43 have appeared in some cycle. (3)( )(6 9) Remove all cycles of length 1 σ = ( )(2 1 3) ( )(6 9) Example Take to be the following Reminder Start with 1,, so write 12 after 1. Keep going until you cycle back to 1 Start with the smallest number which hasn't yet appeared, and repeat. Repeat this step until Product of Cycles Example Read from right to left Write What is? Thus Similarly maps 1 to 1 maps 1 to 2 maps 2 to 3 Thus we close the cycle We won't write down Thus have all appeared. as a product of disjoint cycles, since it is the identity Note:, but it make sense to think of for Commutativity of In particular Therefore is not abelian is not abelian for

44 Page 44 Homomorphism, Isomorphism Wednesday, February 14, :39 AM Homomorphism Definition A function One says Trivial Examples Example 1 be groups be a group is a homomorphism if "respects", or "preserves" the group operation The identity map given by is a homomorphism The map given by is a homomorphism Then Example 2 This only works if we send every element of to 1 If, and is given by Thus This is impossible since The map Example 3 be given by is a homomorphism be a group, and let is a homomorphism This homomorphism is called conjugation by Is Example 4 Only when Is be fixed be fixed a homomorphism? a homomorphism?

45 Page 45 Only when When For Example 5 Example 6 Only constant mapping to identity is a homomorphism But 1 is not the identity (0 is) But this is not always true For instance, when, for be fixed is a homomorphism The previous examples is a special case of the following: Define be a group, and is a homomorphism Proof: homomorphism Say Since Thus Proof: abelian is a homomorphism is abelian First, suppose abelian We argue by induction on If Suppose homomorphism, this is obvious is abelian

46 Page 46 Isomorphism Definition Fact Example Now suppose Then So is a homomorphism, by the above argument Now, take and to obtain the result be groups A homomorphism there is a homomorphism, and is a isomorphism if s.t. In this case, we say and are isomorphic is an isomorphism Proof: isomorphism This is clear Proof: bijective homomorphism We need to show that is a bijective homomorphism bijective homomorphism is a homomorphism Choose s.t. and Then Define Then Moreover, Observation The inverse of If isomorphism is a group under multiplication where is a homomorphism is an isomorphism is are isomorphic groups

47 Page 47 Homomorphism, Isomorphism, Subgroup Friday, February 16, :05 AM Proposition 16: Isomorphism Preserves Commutativity Statement Proof be an isomorphism is abelian if and only if Choose Then Statement Then Suppose is abelian s.t. is abelian Apply the same argument with Proposition 16: Injective Homomorphism Preserves Order Proof be an injective homomorphism By Cancellation Law, When (This last equality follows from an induction argument) Therefore, Now, apply this same argument with So we can conclude that When If The above argument shows This is impossible Thus, replaced by Groups with Same Order is Not Necessarily Isomorphic

48 Page 48 Example 1: are groups, and, is it the case that? No In fact, any homomorphism If Obviously Assume otherwise By induction, be a homomorphism is not surjective is not surjective By assumption,, since otherwise Example 2: Thus is not surjective, but Because is abelian, but is not Also in, but have no element of order 6 Orders of Elements in If, where are disjoint cycles If Subgroup Definition Note is a -cycle Example 1 Example 2 be a group, and let is a subgroup if (nonempty) If (closed under the operation) If (closed under inverse) If is a subgroup of, we write Subgroups of a group are also groups If is a group and

49 Page 49 If Example 3 be a group, and let Then is called the cyclic subgroup generated by, since If

50 Page 50, Subgroup Criterion, Special Subgroups Monday, February 19, :58 AM Regular A regular is a polygon with all sides and angles equal Symmetry Definition Examples Definition Note A symmetry of a regular -gon is a way of picking up a copy of it moving it around in 3d setting it back down so that it exactly covers the original Rotations Reflection Dihedral Groups (Section 1.2) Example: (proof on page 24) There are rotations and reflections Symmetries of -gons are determined by the permutations of the vertices they induce Rotations is called -th dihedral groups

51 Page 51 Reflections Example: Rotations Reflections

52 Page 52 Fact Statement In general is isomorphic to a subgroup of Every finite group is isomorphic to a subgroup of a symmetric group Proposition 17: The Subgroup Criterion A subset of a group is a subgroup iff and Recall the original definition Proof Proof A subset of a group is a subgroup iff This is Clear Closed under multiplication Thus, Closed under inversion So Thus, then

53 Page 53 Examples of Subgroups Example 1 Example 2 Definition Fix Claim Proof, since is called the special linear group Example 3 Definition If Claim Proof If, so Also so is a group, since is closed under multiplication is closed under inversion is called the center or

54 Page 54 Properties of Cyclic Group, Order of Wednesday, February 21, :56 AM Cyclic Group Definition Note Example 1: Example 2: A group is cyclic if s.t. A finite group of order is cyclic iff s.t. If Note: If is cyclic is cyclic is a -cycle Every element in have order 1,2, or 3 So is not cyclic cannot be cyclic Proposition 18: Isomorphism of Cyclic Group If be a cyclic group Choose Define a map s.t. Well-definedness We need to check that given by is well-defined. That is we must show that if in, suppose in Choose Thus, Homomorphism Thus, Surjectivity s,t, is well-defined is a homomorphism Surjectivity is clear by definition

55 Page 55 If Injectivity If Thus Choose Define a map If then Thus, is injective s.t. Homomorphism Surjectivity given by is a homomorphism Surjectivity is clear Injectivity Suppose Then Without loss of generality, assume Then Since i.e. Thus is injective Least Common Multiple Definition where one of is nonzero. A least common multiple of and is a positive integer s.t. and If and We denote the least common multiple of and by Define Uniqueness Similar to the proof of uniqueness of greatest common divisor

56 Page 56 and Suppose Choose s.t. and Choose s.t. Thus and Statement Proposition 19: Order of Proof So assume By HW3 #1, Thus, is a common multiple of and

57 Page 57

58 Page 58 Subgroups of Cyclic Groups, Friday, February 23, :07 AM Theorem 20: Subgroup of Cyclic Group is Cyclic Statement Proof be a cyclic group Then every subgroup of is cyclic More precisely, if either or, where is the smallest positive integer s.t. Assume Statement Proof Choose s.t. Thus, contains some positive power of, and so By the Well-Ordering Principle, Therefore, for some Choose s.t. If, which is impossible since The minimality of Therefore Therefore forces be a finite cyclic group of order contains a minimum element Theorem 20: Subgroup of Finite Cyclic Group is Determined by Order For all positive integers dividing, subgroup of order Existence

59 Page 59 Uniqueness Suppose and Then,, where is the smallest positive integer s.t. Statement Proof Thus So Since i.e., we have Lemma: Intersection of Subgroups is Again a Subgroup Definition Since Then Subgroups Generated by Subsets of a Group (Section 2.4) be a group and The subgroup generated by the intersection of every subgroup of is containing Example If If

60 Page 60, Finitely Generated Group Monday, February 26, :01 AM Proposition 21: Construction of Statement Proof If Example Note Definition Note: When, we get Denote the right hand side by If Then Therefore, since (take ) Because, and is the smallest subgroup of containing Because every subgroup of containing (i.e. ) must contain every finite product of elements of Therefore If is a group, and When is abelian and we have Finitely Generated Group A group Example 1 Example 2 Example 3 is finitely generated if There is a finite subset of s.t. Cyclic groups are finitely generated Finite groups are finitely generated and their inverses. If are finitely generated is also finitely generated

61 Page 61 For instance, is finitely generated by In particular, products of cyclic groups are finitely generated Every finitely generated abelian group is a product of cyclic groups (This is called the Fundamental Theorem of Finite Abelian Groups) Example 4 Every finitely generated subgroup of is cyclic. It follows that is not finitely generated, since is not cyclic Without loss of generality, assume Applying the Well-Ordering Principle We can choose a minimum element be fixed Set Choose s.t, The minimality of This shows forces

62 Page 62 Coset, Normal Subgroup Wednesday, February 28, :59 AM Coset If is a group,, and is called a left coset is called a right coset An element of a coset is called a representative of the coset Proposition 22: Properties of Coset be a group and For ( ) Choose s.t. (since ) Therefore Choose Therefore Reflexive If So Symmetric s.t. The relation on given by iff is an equivalence relation If, and i.e. Thus So Transitive Suppose This means for some, which means and and Choose s.t., and Then So In particular, left/right cosets are either equal or disjoint Suppose Suppose, and

63 Page 63 So This implies that To get, exchange the roles of and Therefore Example 1 If Thus be a group, By closure under the operation, Therefore Example 2, and unique subgroup of of order 2 Left cosets of Note in G,, and has 3 distinct cosets If is a finite group, and, and Normal Subgroup Definition This is called the Lagrange's Theorem be a group, is a normal subgroup if, In other words, If Example 1 is closed under conjugation is normal, we write If is abelian, every subgroup of is normal Suppose

64 Page 64 Then Example 2, Suppose Then Therefore Example 3 Note In Example 4 in and is normal, conjugation amounts to changing basis is change of basis matrix Statement If Thus Therefore Suppose is normal be a homomorphism, since Proposition 23: Criteria for a Subgroup to be Normal Proof be a subgroup of a group Therefore

65 Page 65 Proof Suppose We must show that Choose s.t. Then Therefore

66 Page 66 Quotient Group, Index, Lagrange's Theorem Monday, March 5, :41 AM Proposition 24: Quotient Group Statement Proof be a group, The set of left costs of is a group under the operation This group is denoted as (say " mod ") We call this group quotient group or factor group Check, given by is well-defined Suppose, and Note Therefore So the operation is well-defined Identity Inverse Since Associativity

67 Page 67 If Example 1 Since there is a surjective homomorphism This shows that, if Example 2 Definition Note be a subgroup of given by with Then since is abelian Since is cyclic, is also cyclic So we can write There is isomorphism If is a group and is the kernel of a homomorphism from, where is the trivial group of order 1 Intuition: The bigger the subgroup, the smaller the quotient Index of a Subgroup If is a group, and The index of is the number of distinct left cosets of in Denote the index by If Example Statement Theorem 25: Lagrange's Theorem Notice Proof If is finite group, and In particular, If in the setting of Lagrange's Theorem, to some other group

68 Page 68, and Cosets partition be the representatives of the distinct cosets of in (In other words: if ) By proposition 22, left costs are either equal or disjoint So, Cosets have the same size there is a function given by is certainly surjective is also injective since if Thus, Therefore

69 Page 69 Lagrange's Theorem, Product of Subgroups Wednesday, March 7, :56 AM Corollary 26: Group of Prime Order is Cyclic Statement Proof Order If is a group, and is prime is cyclic Hence, If By Lagrange's Theorem, Thus, It follows that if Therefore i.e. is cyclic Groups of Small Order Property 2 Cyclic 3 Cyclic 4 Cyclic or 5 Cyclic 6 Cyclic or Corollary 27: Statement Proof If is a finite group, and By Lagrange's Theorem, Since Thus,, we have for some integer Corollary 28: The Fundamental Theorem of Cyclic Groups Statement Proof If is a finite cyclic group there is a bijection

70 Page 70 Define Divisor of the unique subgroup with order Subgroup Product of Subgroups Statement be a group and of Proposition 29: Order of Product of Subgroups Proof Notice that is the union of left cosets of In the proof of Lagrange's Theorem, we know that B, the number of distinct cosets of the form is Note:, But Statement Note is not always a subgroup is not a divisor of By Lagrange's Theorem, If is not a subgroup of Proposition 30: Permutable Subgroups

71 Page 71 Proof Proof is not equivalent to It implies that every product is of the form and conversely This is true because Set So, Then, since We must show that for some Choose Then Example Therefore, Thus Statement Therefore is not a subgroup of Corollary 31: Product of Subgroup and Normal Subgroup Proof If, and either or is normal in Without loss of generality, assume Therefore

72 Page 72 The First & Second Isomorphism Theorems Friday, March 9, :06 AM Theorem 32: The First Isomorphism Theorem Statement Intuition If is a homomorphism induces an isomorphism This is an analogue of the Rank-Nullity Theorem in Linear Algebra Given vector space and a linear transformation Proof is well-defined and injective Thus is surjective Choose Then is well-defined and injective s.t. is a homomorphism

73 Page 73 Statement Example Corollary 33: Order of Kernel and Image Proof be coprimes Then any homomorphism i.e. be such a homomorphism is trivial Note Thus, Statement, so The same proof tells us that If are finite groups such that All homomorphism between them are trivial Theorem 34: The Second Isomorphism Theorem If Intuition, and

74 Page 74 Note by Corollary 31 Proof We have homomorphisms given by is certainly surjective Compute Thus, The First Isomorphism Theorem gives an isomorphism

75 Page 75 The Third & Fourth Isomorphism Theorem Monday, March 12, :57 AM Theorem 35: The Third Isomorphism Theorem Statement Note Intuition be a group, and, where Proof Then Then and Define a homomorphism is well-defined Suppose given by

76 Page 76 Then Since i.e. is surjective If Compute, we have By First Isomorphism Theorem Example Then the Third Isomorphism Theorem tells us that The map given by is well-defined and surjective Proposition 36: Criterion for Defining Homomorphism on Quotient Statement Proof Proof be groups, and A homomorphism If and only if Thus, i.e. And Therefore given by induces a homomorphism since homomorphisms preserve identities, by definition of certainly meets the Subgroup Criteria Suppose is well-defined, we must check that (since )

77 Page 77 Statement is a homomorphism Theorem 37: The Correspondence Theorem Proof be a group, and let there is a bijection Define If Thus, This also shows that If is a subgroup of and containing are the identity maps

78 Page 78 Transposition, Sign of Permutation Wednesday, March 14, :56 AM Transposition Fix to be a positive integer A cycle in is a transposition Proposition 38: Transposition Decomposition of Permutation Statement Every Example Proof Note Fix Intuition We may assume that By induction on, we claim Base case: Inductive step: is generated by Sign of Permutation Definition can be written as a product of transposition is a cycle The numbers of transposition used to write some is not well-defined, but it is always either even or odd (Transposition Definition) Then is a group homomorphism is the alternating group of degree

79 Page 79 Sign of Permutation Auxiliary Polynomial (Auxiliary Polynomial Definition) Example Then is always either or Definition and Statement is the sign of, often denoted as is even if is odd if Proposition 39: Example Proof Fix is a group homomorphism is a Group Homomorphism, since

80 Page 80 Suppose has "reversed factor" (i.e. factors, where ) Therefore

81 Page 81 Homework 6 Friday, March 16, :51 AM Homework 6 Question 1 Statement Proof Suppose Define a map Check Compute is a homomorphism Therefore and Prove surjectivity Choose Then s.t.

82 Page 82 Therefore is surjective By the First Isomorphism theorem, there is an isomorphism Note Statement Given two homomorphism Then their direct product is also a homomorphism Homework 6 Question 2 Proof Proof is abelian Suppose So Therefore Suppose Choose given by is abelian is cyclic is the trivial group is cyclic is cyclic and s.t. for some Choose and s.t So, Then Homework 6 Question 4 Statement is cyclic of order Choose a generator Then is a generator of, and Proof: If is a cyclic group and is also cyclic

83 Page 83 If Choose Therefore Proof s.t. By Lagrange's Theorem

84 Page 84 Sign of Permutation, Monday, March 19, :50 AM Recall Proposition 40: Sign of Transposition Statement If Example Proof Suppose Suppose Say Then Thus So Suppose is transposition is a factor of denote the following permutation

85 Page 85 Statement Without loss of generality, assume For We know is well-defined, and is a homomorphism, so Corollary 41: Equivalence of Two Definitions of Sign Proof Say where is a transposition If Statement Proof is odd cannot be written as a product of an even number of transpositions So for with odd, and vice verse This shows is well-defined, and Corollary 42: Surjectivity of If is surjective Since has only 2 elements, is surjective Alternating Group Definition Order of The alternative group, denoted as That is, is the kernel of contains of all even permutations in By the First Isomorphism Theorem

86 Page 86 We have an isomorphism By Lagrange's Theorem, Note We showed earlier that, if cycle is even when is odd, and vise versa Thus, is odd Examples trivial group {(1), (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3), (1 2) (3 4), (1 3)(2 4), (1 4)(2 3)} Subgroups of Order 1 Subgroup 2 {(1), (1 2)(3 4)} {(1), (1 3)(2 4)} {(1), (1 4)(2 3)} 3 {(1), (1 2 3), (1 3 2)} {(1), (1 2 4), (1 4 2)} {(1), (1 3 4), (1 4 3)} {(1), (2 3 4), (2 4 3)} 4 {(1), (12)(34), (13)(24), (14)(23)} 6 None 12 Converse of Lagrange's Theorem has no subgroup of order 6 This shows that the converse of Lagrange's Theorem is false If, there is not necessarily a subgroup of with order But the converse does hold for finite cyclic groups Cauchy's Theorem If is a prime, and contains a subgroup of order Sylow's Theorem If, where is prime and Then contains a subgroup of order

87 Page 87 Subgroups of Wednesday, March 21, 2018, Group Action, Orbit, Stabilizer 9:57 AM Proposition 43: Subgroup of Index 2 is Normal Statement Proof If is a group,, and If If Therefore So If Corollary (See HW8 #2) Statement Proof If Then Choose be the smallest prime dividing By HW 7 #1, Proposition 44: Conjugate Cycle Statement s.t. s.t. are -cycles in Theorem 45: Have No Subgroup of Order 6 Proof have no subgroup of order 6 By way of contradiction, suppose Then and thus, and Since contains eight 3-cycles, must contain some 3-cycle Write So far, we have Also, since Thus, is closed under inverses,, which makes a contradiction

88 Page 88 Group Action Definition Examples Therefore have no subgroup of order 6 An action of on is a function, s.t. Set Group Action Group Group Set of cosets of Set of all subgroups of group If, and If, since Group Group Group Group Proof: Conjugation on subgroup is a group action Orbit and Stabilizer Suppose a group The orbit of, denoted acts on a set, is The stabilizer of, denoted stab( ), is Proposition 46: Stabilizer is a Subgroup Statement Proof If acts on, and stab stab( ) Centralizer, because be a group, and let act on itself by conjugation If This set is called the centralizer of, denoted as

89 Page 89 Center is the set of elements in that commute with the element Intersections of subgroups are subgroup Normalizer is the set of elements that commute with every element of be the set of subgroups of a group acts on by If This set is called the normalizer of in, denoted Note: is the set of elements in that commute with the set

90 Page 90 Orbit, Stabilizer, Cayley's Theorem Friday, March 23, :07 AM Proposition 47: Orbits Equivalence Statement Proof Note act on a set The relation s.t. is an equivalence relation Reflexive Statement Symmetric Suppose Transitive Suppose Choose s.t. and Then The equivalence classes are the orbits of the group action Thus, the orbits partition Proposition 48: Orbit-Stabilizer Theorem Proof If acts on, and Define a function So is injective is surjective This is clear Therefore

91 Page 91 Proposition 49: Permutation Representation of Group Action Statement Proof Statement be a group acting on a finite set Then each determines a by The map, given by is bijection Injectivity: Surjectivity: So each determines a permutation where Proposition 49: Induced Homomorphism of Group Action Proof The map Suppose Then Write Statement Therefore for some for some Theorem 50: Cayley's Theorem Proof is a homomorphism Every finite group is isomorphic to a subgroup of the symmetric group, where act on itself by left multiplication Then this action determines a homomorphism is injective Thus Example Klein 4 group

92 Page Label the group elements with 1, 2, 3, 4 Therefore since since since

93 Page 93 Conjugacy Class, The Class Equation Monday, April 2, :57 AM Conjugacy Class Definition Example 1 If is a group, acts on itself by conjugation: The orbits under this action are called conjugacy classes Denote a conjugate class represented by some element If, and Since Example 2 The converse is also true: If If If For instance More generally is a -cycle by be a product of disjoint cycles Statement Then Theorem 51: The Class Equation be a finite group be representatives of the conjugacy classes of not contained in the center that are Recall: Proof is the disjoint union of its disjoint conjugate classes

94 Page 94 Corollary 52: Center of -Group is Non-Trivial Statement Note Proof If is a prime, and is a group of order for prime is called a -group representatives of the conjugate classes of By Lagrange's Theorem, Combing previous two results, not contained in Corollary 53: Group of Order Prime Squared is Abelian Statement Proof If is a prime, and is a group of order is abelian. In fact, either or By Corollary 52 and Lagrange's Theorem, Suppose or By Corollary 26, By HW6 #2, In this case Therefore Suppose is abelian is cyclic is impossible

95 Page 95 If If We have So is abelian is cyclic clearly is not cyclic, we need to show that Set Since any non-identity element of or is a generator For instance, if This is impossible, so for some By HW6 #1, there exists an isomorphism Similarly for Therefore

96 Page 96 Cauchy's Theorem, Recognizing Direct Products Wednesday, April 4, :48 AM Theorem 54: Cauchy's Theorem Statement Proof If is a finite group, and is a prime divisor of of order Write We argue by strong induction on When, this is trivial, since any non-identity element of has order Suppose, and of order If is abelian If If By the Fundamental Theorem of Cyclic Groups, Set contains a (unique) subgroup of order By the Lagrange's Theorem, Since or If If Since is cyclic, contains a (unique) subgroup of order It follows that contains a subgroup of order, so, by induction, s.t. So we only need to prove If is not abelian If Therefore Now, take is a group homomorphism, the usual surjection Since is cyclic, contains a (unique) subgroup of order It follows that By the Lagrange's Theorem, Since contains a subgroup of order or If for some