LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS


 Harry Miles
 1 years ago
 Views:
Transcription
1 LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS Recall Lagrange s theorem says that for any finite group G, if H G, then H divides G. In these lectures we will be interested in establishing certain partial converses to this result. It should be noted that a true converse is not possible: for instance, the alternating group A 4 has order 12 but no subgroup of 6 elements. However, one can still deduce weaker statements about the existence of subgroups of a group of a given order. As an example we state the following fundamental theorem. Theorem (Cauchy s theorem). Let p be a prime and suppose G is a finite group for which p G. Then G contains an element of order p and, consequently, a cyclic subgroup of order p. Thus, Cauchy s theorem is a partial converse to Lagrange s theorem, in which one restricts one s attention to prime order subgroups. We will prove this result, and far reaching generalisations known as the Sylow theorems, below. The key tool in the proofs will be the theory of group actions, is reviewed presently. Group actions. Definition. Let X be a set and G be a group. An action of G on X is a map : G X X such that i) ex = x for all x X; ii) (g 1 g 2 )x = g 1 (g 2 x) for all x X and g 1, g 2 G. In this situation, we say X is a Gset. Definition. On any Gset X one may define an equivalence relation x 1 x 2 if and only if there exists some g G such that gx 1 = x 2. The equivalence classes of this relation are called the orbits in X under G. If x X, the equivalence class containing x is called the orbit of x and is denoted by orb G (x) = Gx := {gx : g G}. Definition. If X is a Gset and x X, then one may define the stabilizer stab G (x) = G x of x in G to be the subgroup stab G (x) = {g G : gx = x}. Theorem (OrbitStabilizer theorem). Let X be a Gset and x X. Then orb G (x) = (G : stab G (x)). Thus, if G is finite, then orb G (x) divides G. Definition. For any Gset X we define the set of fixed points to be X G := {x X : gx = x for all g G}; that is, the union of all the 1element orbits of X. Suppose X is a finite Gset and let x 1,..., x r X be a set of representatives for the orbits of the action of G on X so that r X = orb G (x i ). i=1 1
2 2 LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS Thus, if X G = s, then continuing with the above setup r X = X G + orb G (x i ). (0.1) Henceforth, p denotes a prime. i=s+1 Theorem. Let G be a group of order p n and X a finite Gset. Then X X G mod p. By the OrbitStabilizer theorem, p divides orb G (x i ) for each i = s + 1,..., r. Cauchy s theorem. Theorem (Cauchy s theorem). Suppose G is a finite group for which p G. Then G contains an element of order p and, consequently, a cyclic subgroup of order p. Consider the set X := { (g 1,..., g p ) G p : g 1... g p = e }. Any arbitrary choice of (g 1,..., g p 1 ) G p 1 determines a unique element of X: indeed, if we define g p := (g 1... g p 1 ) 1, then (g 1,..., g p ) X. Thus, X = G p 1 and since p G, it follows that p X. Let σ S p by the cycle (1, 2,..., p) and define an action of the group σ S p on X by permuting the coordinates. By the theorem, X X σ mod p and so p X σ. Since (g 1,..., g p ) X σ if and only if g 1 = = g p, and X σ > 1, it follows that there exists some a G such that a e and a p = e, as required. As an application of Cauchy s theorem we prove a simple classification theorem. Corollary. Let p be a prime and G be a finite group. The following are equivalent: i) Every element of G has order given by a power of p. ii) G is a power of p. If either of these (equivalent) conditions hold, we say G is a pgroup. Definition. We say H G is a psubgroup of G if H is a pgroup. Proof (of the Corollary). If G = p n for some n N {0}, then G is clearly a pgroup by Lagrange s theorem. For the converse, suppose that G is a finite group and G p n for all n N {0}. Then there exists a prime q p such that q G. By Cauchy s theorem there exists an element of order q, so G is not a pgroup. The First Sylow theorem. In this section we prove a farreaching generalisation of Cauchy s theorem. Theorem (First Sylow theorem). Let G be a finite group and suppose p n G but p n+1 G for some prime p and n N. 1) G contains a subgroup of order p i for each 1 i n. 2) Every H G of order p i is a normal subgroup of a subgroup of order p i+1 for 1 i n 1.
3 LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS 3 Note part 1) with i = 1 recovers Cauchy s theorem. The proof of the First Sylow theorem will require the notion of a normaliser of a group. Let G be a group and S denote the set of all subgroups of G. We can define an action G S S of G on S by conjugation: (g, H) ghg 1. Recall, for any H G it follows that stab G (H) := {g G : ghg 1 = H} is a subgroup of G and, furthermore, it follows by definition that H stab G (H). In fact, clearly stab G (H) is the largest subgroup of G for which H is a normal subgroup. Definition. If G is a group and H G, then we define the normaliser of H to be the subgroup N[H] := stab G (H), where the stabilizer is as defined above. Lemma. If G is a finite group and H G, then g N[H] if and only if ghg 1 H for all h H. Proof. If g N[H], then by definition ghg 1 = H and so ghg 1 H for all h H. Conversely, suppose ghg 1 H for all h H so that the conjugation map ι g : H H given by ι g (h) := ghg 1 is welldefined and clearly injective. Since H is a finite set, it must also be surjective so that ghg 1 = H, as required. Lemma. If H is a psubgroup of a finite group G, then (N[H] : H) (G : H) mod p. In particular, if p (G : H), then N[H] H. Let L denote the set of left cosets of H and let H act of L by left translation: H L L, (h, xh) hxh. Note that L = (G : H) by definition. By the previous lemma, the set L H of fixed points of the action satisfies L H = {xh L : hxh = xh for all h H} = {xh L : xh N[H]} and so L H = (N[H] : H) Since H is a psubgroup, it has order given by a power of p and so L L H mod p. For the final part of the lemma, if p (G : H), then p (N[H] : H) and so (N[H] : H) > 1. We are now in a position to prove the First Sylow theorem. Proof (of First Sylow theorem). The key ideas of the proof of part 1) are: Proceed by induction on i. The case i = 1 is Cauchy s theorem. Suppose, by way of induction hypothesis that H G has order p i for some 1 i n 1. Since p (G : H), it follows that p (N[H] : H) so that p N[H]/H. By Cauchy s theorem, there exists K N[H]/H with K = p. If π : N[H] N[H]/H is the canonical homomorphism, then π 1 K G satisfies H π 1 K. Since π 1 K/H = K by the first isomorphism theorem, it follows that π 1 K = p i+1. The key ideas of the proof of part 2) are: Arguing as before, H π 1 K N[H] where π 1 K = p i+1. Since H N[H], it follows that H π 1 K.
4 4 LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS We consider some simple consequences of the First Sylow Theorem. Theorem. i) Every group of primepower order is solvable. ii) No group of order p r for some r 2 is simple. The Second and Third Sylow theorems. Definition. A Sylow psubgroup of a finite group G is a maximal psubgroup. Thus, by the First Sylow theorem if p n G but p n+1 G, then the Sylow psubgroups are precisely the subgroups of G of order p n. The remaining Sylow theorems give insight into the structure of the Sylow p subgroups. Lemma. If G is a finite group and P a Sylow psubgroup then for any g G the conjugation gp g 1 is also a Sylow psubgroup. Corollary. Suppose G is a finite group and p, q are distict prime divisors of G. If G has only one Sylow psubgroup, then this group is normal and so G is not simple. Proof. If G has only one Sylow psubgroup, then this group is a nontrivial proper subgroup of G which is fixed under conjugation, and therefore normal. The converse of the lemma also holds. Theorem (Second Sylow theorem). If P 1, P 2 are Sylow psubgroups of a finite group G, then they are conjugate. That is, there exists some g G such that P 2 = gp 1 g 1. Let L denote the left cosets of P 1 and let P 2 act of L by left translation. Note that L P2 L mod p and p L = (G : P 1 ) so that L P2 0. If xp 1 L P2, then P 1 = x 1 P 2 x. Theorem (Third Sylow theorem). If G is a finite group and p G, then the number of Sylow psubgroups is congruent to 1 modulo p and divides G. Let S denote the set of all Sylow psubgroups and fix some P S. Then P acts on S by conjugation: (x, Q) xqx 1 for all (x, Q) P S. If Q S P, then P, Q N[Q] are Sylow psubgroups of N[Q]. By Sylow II, it follows P, Q are conjugate in N[Q]. By the definition of N[Q], it follows that P = Q, so that S P = {P }. S S P = 1 mod p. Let G act on S by conjugation. By Sylow II there is only one orbit and so, by the OrbitStabilizer theorem, S = (G : stab G (P )) for any P S. In particular, S G. We conclude with some examples which demonstrate the utility of the Sylow theorems. Example. No group of order 20 is simple. Indeed, by Sylow III the number of Sylow 5subgroups is congruent to 1 modulo 5 and divides 20. Thus there is precisely 1 Sylow 5subgroup which must be normal.
5 LECTURES 1113: CAUCHY S THEOREM AND THE SYLOW THEOREMS 5 Example. No group of order 30 is simple. By Sylow III the number of Sylow 5subgroups is congruent to 1 modulo 5 and divides 30, so there are either 1 or 6 such subgroups. Similarly, the number of Sylow 3subgroups is congruent to 1 modulo 3 and divides 30, so there are either 1 or 10 such subgroups. If in either case there is only 1 subgroup, then we are done so we may suppose there are 6 Sylow 5subgroup and 10 Sylow 3subgroups. Each nonidentity element of a given Sylow 5subgroup is a generator and thus has order 5 and is distinct from all the nonidentity elements of the other Sylow 5subgroups. Hence G contains precisely 24 = 4 6 elements of order 5. Similarly, G contains 20 = 2 10 elements of order 3. But in this case, the number of elements of G is 45, a contradiction. Jonathan Hickman, Department of mathematics, University of Chicago, 5734 S. University Avenue, Eckhart hall Room 414, Chicago, Illinois, address:
Stab(t) = {h G h t = t} = {h G h (g s) = g s} = {h G (g 1 hg) s = s} = g{k G k s = s} g 1 = g Stab(s)g 1.
1. Group Theory II In this section we consider groups operating on sets. This is not particularly new. For example, the permutation group S n acts on the subset N n = {1, 2,...,n} of N. Also the group
More informationDefinitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations
Page 1 Definitions Tuesday, May 8, 2018 12:23 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
More informationGROUP ACTIONS EMMANUEL KOWALSKI
GROUP ACTIONS EMMANUEL KOWALSKI Definition 1. Let G be a group and T a set. An action of G on T is a map a: G T T, that we denote a(g, t) = g t, such that (1) For all t T, we have e G t = t. (2) For all
More informationFINITE GROUP THEORY: SOLUTIONS FALL MORNING 5. Stab G (l) =.
FINITE GROUP THEORY: SOLUTIONS TONY FENG These are hints/solutions/commentary on the problems. They are not a model for what to actually write on the quals. 1. 2010 FALL MORNING 5 (i) Note that G acts
More information3. G. Groups, as men, will be known by their actions.  Guillermo Moreno
3.1. The denition. 3. G Groups, as men, will be known by their actions.  Guillermo Moreno D 3.1. An action of a group G on a set X is a function from : G X! X such that the following hold for all g, h
More informationAlgebraI, Fall Solutions to Midterm #1
AlgebraI, Fall 2018. Solutions to Midterm #1 1. Let G be a group, H, K subgroups of G and a, b G. (a) (6 pts) Suppose that ah = bk. Prove that H = K. Solution: (a) Multiplying both sides by b 1 on the
More informationGROUP ACTIONS RYAN C. SPIELER
GROUP ACTIONS RYAN C. SPIELER Abstract. In this paper, we examine group actions. Groups, the simplest objects in Algebra, are sets with a single operation. We will begin by defining them more carefully
More informationGroups and Symmetries
Groups and Symmetries Definition: Symmetry A symmetry of a shape is a rigid motion that takes vertices to vertices, edges to edges. Note: A rigid motion preserves angles and distances. Definition: Group
More informationChapter 25 Finite Simple Groups. Chapter 25 Finite Simple Groups
Historical Background Definition A group is simple if it has no nontrivial proper normal subgroup. The definition was proposed by Galois; he showed that A n is simple for n 5 in 1831. It is an important
More informationLecture 5.6: The Sylow theorems
Lecture 5.6: The Sylow theorems Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson) Lecture 5.6:
More informationSelected exercises from Abstract Algebra by Dummit and Foote (3rd edition).
Selected exercises from Abstract Algebra by Dummit Foote (3rd edition). Bryan Félix Abril 12, 2017 Section 4.1 Exercise 1. Let G act on the set A. Prove that if a, b A b = ga for some g G, then G b = gg
More informationSF2729 GROUPS AND RINGS LECTURE NOTES
SF2729 GROUPS AND RINGS LECTURE NOTES 20110301 MATS BOIJ 6. THE SIXTH LECTURE  GROUP ACTIONS In the sixth lecture we study what happens when groups acts on sets. 1 Recall that we have already when looking
More informationDISCRETE MATH (A LITTLE) & BASIC GROUP THEORY  PART 3/3. Contents
DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY  PART 3/3 T.K.SUBRAHMONIAN MOOTHATHU Contents 1. Cayley s Theorem 1 2. The permutation group S n 2 3. Center of a group, and centralizers 4 4. Group actions
More informationPseudo Sylow numbers
Pseudo Sylow numbers Benjamin Sambale May 16, 2018 Abstract One part of Sylow s famous theorem in group theory states that the number of Sylow p subgroups of a finite group is always congruent to 1 modulo
More informationCourse 311: Abstract Algebra Academic year
Course 311: Abstract Algebra Academic year 200708 D. R. Wilkins Copyright c David R. Wilkins 1997 2007 Contents 1 Topics in Group Theory 1 1.1 Groups............................... 1 1.2 Examples of Groups.......................
More informationCONSEQUENCES OF THE SYLOW THEOREMS
CONSEQUENCES OF THE SYLOW THEOREMS KEITH CONRAD For a group theorist, Sylow s Theorem is such a basic tool, and so fundamental, that it is used almost without thinking, like breathing. Geoff Robinson 1.
More informationits image and kernel. A subgroup of a group G is a nonempty subset K of G such that k 1 k 1
10 Chapter 1 Groups 1.1 Isomorphism theorems Throughout the chapter, we ll be studying the category of groups. Let G, H be groups. Recall that a homomorphism f : G H means a function such that f(g 1 g
More informationExercises on chapter 1
Exercises on chapter 1 1. Let G be a group and H and K be subgroups. Let HK = {hk h H, k K}. (i) Prove that HK is a subgroup of G if and only if HK = KH. (ii) If either H or K is a normal subgroup of G
More informationA SIMPLE PROOF OF BURNSIDE S CRITERION FOR ALL GROUPS OF ORDER n TO BE CYCLIC
A SIMPLE PROOF OF BURNSIDE S CRITERION FOR ALL GROUPS OF ORDER n TO BE CYCLIC SIDDHI PATHAK Abstract. This note gives a simple proof of a famous theorem of Burnside, namely, all groups of order n are cyclic
More informationGroups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems
Group Theory Groups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems Groups Definition : A nonempty set ( G,*)
More informationMath 429/581 (Advanced) Group Theory. Summary of Definitions, Examples, and Theorems by Stefan Gille
Math 429/581 (Advanced) Group Theory Summary of Definitions, Examples, and Theorems by Stefan Gille 1 2 0. Group Operations 0.1. Definition. Let G be a group and X a set. A (left) operation of G on X is
More informationSupplementary Notes: Simple Groups and Composition Series
18.704 Supplementary Notes: Simple Groups and Composition Series Genevieve Hanlon and Rachel Lee February 2325, 2005 Simple Groups Definition: A simple group is a group with no proper normal subgroup.
More informationMA441: Algebraic Structures I. Lecture 26
MA441: Algebraic Structures I Lecture 26 10 December 2003 1 (page 179) Example 13: A 4 has no subgroup of order 6. BWOC, suppose H < A 4 has order 6. Then H A 4, since it has index 2. Thus A 4 /H has order
More informationSchool of Mathematics and Statistics. MT5824 Topics in Groups. Problem Sheet I: Revision and ReActivation
MRQ 2009 School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet I: Revision and ReActivation 1. Let H and K be subgroups of a group G. Define HK = {hk h H, k K }. (a) Show that HK
More informationS11MTH 3175 Group Theory (Prof.Todorov) Quiz 6 (PracticeSolutions) Name: 1. Let G and H be two groups and G H the external direct product of G and H.
Some of the problems are very easy, some are harder. 1. Let G and H be two groups and G H the external direct product of G and H. (a) Prove that the map f : G H H G defined as f(g, h) = (h, g) is a group
More informationTHE SYLOW THEOREMS AND THEIR APPLICATIONS
THE SYLOW THEOREMS AND THEIR APPLICATIONS AMIN IDELHAJ Abstract. This paper begins with an introduction into the concept of group actions, along with the associated notions of orbits and stabilizers, culminating
More informationPROBLEMS FROM GROUP THEORY
PROBLEMS FROM GROUP THEORY Page 1 of 12 In the problems below, G, H, K, and N generally denote groups. We use p to stand for a positive prime integer. Aut( G ) denotes the group of automorphisms of G.
More informationMath 451, 01, Exam #2 Answer Key
Math 451, 01, Exam #2 Answer Key 1. (25 points): If the statement is always true, circle True and prove it. If the statement is never true, circle False and prove that it can never be true. If the statement
More informationφ(xy) = (xy) n = x n y n = φ(x)φ(y)
Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =
More informationHomework 2 /Solutions
MTH 912 Group Theory 1 F18 Homework 2 /Solutions #1. Let G be a Frobenius group with complement H and kernel K. Then K is a subgroup of G if and only if each coset of H in G contains at most one element
More informationLECTURE 22: COUNTABLE AND UNCOUNTABLE SETS
LECTURE 22: COUNTABLE AND UNCOUNTABLE SETS 1. Introduction To end the course we will investigate various notions of size associated to subsets of R. The simplest example is that of cardinality  a very
More informationElements of solution for Homework 5
Elements of solution for Homework 5 General remarks How to use the First Isomorphism Theorem A standard way to prove statements of the form G/H is isomorphic to Γ is to construct a homomorphism ϕ : G Γ
More informationOn Maximal Subgroups of a Group with Unique Order
Available online at wwwscholarsresearchlibrarycom European Journal of Applied Engineering and Scientific Research, 208, 6():23 On Maximal Subgroups of a Group with Unique Order ISSN: 2278004 M Bello
More informationSchool of Mathematics and Statistics. MT5824 Topics in Groups. Handout 0: Course Information
MRQ 2009 School of Mathematics and Statistics MT5824 Topics in Groups Handout 0: Course Information Lecturer: Martyn Quick, Room 326. Prerequisite: MT4003 Groups. Lectures: Mon (odd), Wed & Fri 10am, Lecture
More informationSUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT
SUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT Contents 1. Group Theory 1 1.1. Basic Notions 1 1.2. Isomorphism Theorems 2 1.3. Jordan Holder Theorem 2 1.4. Symmetric Group 3 1.5. Group action on Sets 3 1.6.
More information1 Chapter 6  Exercise 1.8.cf
1 CHAPTER 6  EXERCISE 1.8.CF 1 1 Chapter 6  Exercise 1.8.cf Determine 1 The Class Equation of the dihedral group D 5. Note first that D 5 = 10 = 5 2. Hence every conjugacy class will have order 1, 2
More informationAlgebra SEP Solutions
Algebra SEP Solutions 17 July 2017 1. (January 2017 problem 1) For example: (a) G = Z/4Z, N = Z/2Z. More generally, G = Z/p n Z, N = Z/pZ, p any prime number, n 2. Also G = Z, N = nz for any n 2, since
More informationMORE ON THE SYLOW THEOREMS
MORE ON THE SYLOW THEOREMS 1. Introduction Several alternative proofs of the Sylow theorems are collected here. Section 2 has a proof of Sylow I by Sylow, Section 3 has a proof of Sylow I by Frobenius,
More informationGroup Theory (Math 113), Summer 2014
Group Theory (Math 113), Summer 2014 George Melvin University of California, Berkeley (July 8, 2014 corrected version) Abstract These are notes for the first half of the upper division course Abstract
More informationAlgebra. Travis Dirle. December 4, 2016
Abstract Algebra 2 Algebra Travis Dirle December 4, 2016 2 Contents 1 Groups 1 1.1 Semigroups, Monoids and Groups................ 1 1.2 Homomorphisms and Subgroups................. 2 1.3 Cyclic Groups...........................
More informationGroup Theory
Group Theory 2014 2015 Solutions to the exam of 4 November 2014 13 November 2014 Question 1 (a) For every number n in the set {1, 2,..., 2013} there is exactly one transposition (n n + 1) in σ, so σ is
More informationMath 210A: Algebra, Homework 5
Math 210A: Algebra, Homework 5 Ian Coley November 5, 2013 Problem 1. Prove that two elements σ and τ in S n are conjugate if and only if type σ = type τ. Suppose first that σ and τ are cycles. Suppose
More informationSolutions of exercise sheet 4
DMATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 4 The content of the marked exercises (*) should be known for the exam. 1. Prove the following two properties of groups: 1. Every
More informationThe Gordon game. EWU Digital Commons. Eastern Washington University. Anthony Frenk Eastern Washington University
Eastern Washington University EWU Digital Commons EWU Masters Thesis Collection Student Research and Creative Works 2013 The Gordon game Anthony Frenk Eastern Washington University Follow this and additional
More informationHall subgroups and the pronormality
1 1 Sobolev Institute of Mathematics, Novosibirsk, Russia revin@math.nsc.ru Novosibirsk, November 14, 2013 Definition A subgroup H of a group G is said to be pronormal if H and H g are conjugate in H,
More informationMTH Abstract Algebra II S17. Review for the Final Exam. Part I
MTH 4111 Abstract Algebra II S17 Review for the Final Exam Part I You will be allowed to use the textbook (Hungerford) and a printout of my online lecture notes during the exam. Nevertheless, I recommend
More informationM3P10: GROUP THEORY LECTURES BY DR. JOHN BRITNELL; NOTES BY ALEKSANDER HORAWA
M3P10: GROUP THEORY LECTURES BY DR. JOHN BRITNELL; NOTES BY ALEKSANDER HORAWA These are notes from the course M3P10: Group Theory taught by Dr. John Britnell, in Fall 2015 at Imperial College London. They
More informationMath 250A, Fall 2004 Problems due October 5, 2004 The problems this week were from Lang s Algebra, Chapter I.
Math 250A, Fall 2004 Problems due October 5, 2004 The problems this week were from Lang s Algebra, Chapter I. 24. We basically know already that groups of order p 2 are abelian. Indeed, pgroups have nontrivial
More informationDefinition List Modern Algebra, Fall 2011 Anders O.F. Hendrickson
Definition List Modern Algebra, Fall 2011 Anders O.F. Hendrickson On almost every Friday of the semester, we will have a brief quiz to make sure you have memorized the definitions encountered in our studies.
More informationThe Class Equation X = Gx. x X/G
The Class Equation 992012 If X is a Gset, X is partitioned by the Gorbits. So if X is finite, X = x X/G ( x X/G means you should take one representative x from each orbit, and sum over the set of representatives.
More informationWritten Homework # 2 Solution
Math 516 Fall 2006 Radford Written Homework # 2 Solution 10/09/06 Let G be a nonempty set with binary operation. For nonempty subsets S, T G we define the product of the sets S and T by If S = {s} is
More informationA characterisation of psoluble groups
A characterisation of psoluble groups Paul Flavell The School of Mathematics and Statistics The University of Birmingham Birmingham B15 2TT United Kingdom email: p.j.flavell@bham.ac.uk If p is a prime
More information120A LECTURE OUTLINES
120A LECTURE OUTLINES RUI WANG CONTENTS 1. Lecture 1. Introduction 1 2 1.1. An algebraic object to study 2 1.2. Group 2 1.3. Isomorphic binary operations 2 2. Lecture 2. Introduction 2 3 2.1. The multiplication
More informationKevin James. pgroups, Nilpotent groups and Solvable groups
pgroups, Nilpotent groups and Solvable groups Definition A maximal subgroup of a group G is a proper subgroup M G such that there are no subgroups H with M < H < G. Definition A maximal subgroup of a
More informationAlgebra I: Final 2012 June 22, 2012
1 Algebra I: Final 2012 June 22, 2012 Quote the following when necessary. A. Subgroup H of a group G: H G = H G, xy H and x 1 H for all x, y H. B. Order of an Element: Let g be an element of a group G.
More informationExamples: The (left or right) cosets of the subgroup H = 11 in U(30) = {1, 7, 11, 13, 17, 19, 23, 29} are
Cosets Let H be a subset of the group G. (Usually, H is chosen to be a subgroup of G.) If a G, then we denote by ah the subset {ah h H}, the left coset of H containing a. Similarly, Ha = {ha h H} is the
More informationHigher Algebra Lecture Notes
Higher Algebra Lecture Notes October 2010 Gerald Höhn Department of Mathematics Kansas State University 138 Cardwell Hall Manhattan, KS 665062602 USA gerald@math.ksu.edu This are the notes for my lecture
More informationConverse to Lagrange s Theorem Groups
Converse to Lagrange s Theorem Groups Blain A Patterson Youngstown State University May 10, 2013 History In 1771 an Italian mathematician named Joseph Lagrange proved a theorem that put constraints on
More informationAlgebra Exam Fall Alexander J. Wertheim Last Updated: October 26, Groups Problem Problem Problem 3...
Algebra Exam Fall 2006 Alexander J. Wertheim Last Updated: October 26, 2017 Contents 1 Groups 2 1.1 Problem 1..................................... 2 1.2 Problem 2..................................... 2
More informationA conjugacy criterion for Hall subgroups in finite groups
MSC2010 20D20, 20E34 A conjugacy criterion for Hall subgroups in finite groups E.P. Vdovin, D.O. Revin arxiv:1004.1245v1 [math.gr] 8 Apr 2010 October 31, 2018 Abstract A finite group G is said to satisfy
More informationTwo subgroups and semidirect products
Two subgroups and semidirect products 1 First remarks Throughout, we shall keep the following notation: G is a group, written multiplicatively, and H and K are two subgroups of G. We define the subset
More informationSchool of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet IV: Composition series and the Jordan Hölder Theorem (Solutions)
CMRD 2010 School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet IV: Composition series and the Jordan Hölder Theorem (Solutions) 1. Let G be a group and N be a normal subgroup of G.
More informationSemiregular automorphisms of vertextransitive cubic graphs
Semiregular automorphisms of vertextransitive cubic graphs Peter Cameron a,1 John Sheehan b Pablo Spiga a a School of Mathematical Sciences, Queen Mary, University of London, Mile End Road, London E1
More informationINVERSE LIMITS AND PROFINITE GROUPS
INVERSE LIMITS AND PROFINITE GROUPS BRIAN OSSERMAN We discuss the inverse limit construction, and consider the special case of inverse limits of finite groups, which should best be considered as topological
More informationFundamental Theorem of Finite Abelian Groups
Monica Agana Boise State University September 1, 2015 Theorem (Fundamental Theorem of Finite Abelian Groups) Every finite Abelian group is a direct product of cyclic groups of primepower order. The number
More informationModern Algebra Homework 9b Chapter 9 Read Complete 9.21, 9.22, 9.23 Proofs
Modern Algebra Homework 9b Chapter 9 Read 9.19.3 Complete 9.21, 9.22, 9.23 Proofs Megan Bryant November 20, 2013 First Sylow Theorem If G is a group and p n is the highest power of p dividing G, then
More informationPh.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018
Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018 Do 6 problems with at least 2 in each section. Group theory problems: (1) Suppose G is a group. The
More informationMath 120: Homework 6 Solutions
Math 120: Homewor 6 Solutions November 18, 2018 Problem 4.4 # 2. Prove that if G is an abelian group of order pq, where p and q are distinct primes then G is cyclic. Solution. By Cauchy s theorem, G has
More informationbut no smaller power is equal to one. polynomial is defined to be
13. Radical and Cyclic Extensions The main purpose of this section is to look at the Galois groups of x n a. The first case to consider is a = 1. Definition 13.1. Let K be a field. An element ω K is said
More informationModern Algebra I. Circle the correct answer; no explanation is required. Each problem in this section counts 5 points.
1 2 3 style total Math 415 Please print your name: Answer Key 1 True/false Circle the correct answer; no explanation is required. Each problem in this section counts 5 points. 1. Every group of order 6
More informationMath 4400, Spring 08, Sample problems Final Exam.
Math 4400, Spring 08, Sample problems Final Exam. 1. Groups (1) (a) Let a be an element of a group G. Define the notions of exponent of a and period of a. (b) Suppose a has a finite period. Prove that
More informationAnswers to Final Exam
Answers to Final Exam MA441: Algebraic Structures I 20 December 2003 1) Definitions (20 points) 1. Given a subgroup H G, define the quotient group G/H. (Describe the set and the group operation.) The quotient
More informationHOMEWORK Graduate Abstract Algebra I May 2, 2004
Math 5331 Sec 121 Spring 2004, UT Arlington HOMEWORK Graduate Abstract Algebra I May 2, 2004 The required text is Algebra, by Thomas W. Hungerford, Graduate Texts in Mathematics, Vol 73, Springer. (it
More information5 Structure of 2transitive groups
Structure of 2transitive groups 25 5 Structure of 2transitive groups Theorem 5.1 (Burnside) Let G be a 2transitive permutation group on a set Ω. Then G possesses a unique minimal normal subgroup N and
More informationAbstract Algebra Study Sheet
Abstract Algebra Study Sheet This study sheet should serve as a guide to which sections of Artin will be most relevant to the final exam. When you study, you may find it productive to prioritize the definitions,
More informationSection 10: Counting the Elements of a Finite Group
Section 10: Counting the Elements of a Finite Group Let G be a group and H a subgroup. Because the right cosets are the family of equivalence classes with respect to an equivalence relation on G, it follows
More informationDMATH Algebra II FS18 Prof. Marc Burger. Solution 21. Solvability by Radicals
DMATH Algebra II FS18 Prof. Marc Burger Solution 21 Solvability by Radicals 1. Let be (N, ) and (H, ) be two groups and ϕ : H Aut(N) a group homomorphism. Write ϕ h := ϕ(h) Aut(N) for each h H. Define
More informationProblem 1.1. Classify all groups of order 385 up to isomorphism.
Math 504: Modern Algebra, Fall Quarter 2017 Jarod Alper Midterm Solutions Problem 1.1. Classify all groups of order 385 up to isomorphism. Solution: Let G be a group of order 385. Factor 385 as 385 = 5
More informationMATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:0010:00 PM
MATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:0010:00 PM Basic Questions 1. Compute the factor group Z 3 Z 9 / (1, 6). The subgroup generated by (1, 6) is
More informationTC10 / 3. Finite fields S. Xambó
TC10 / 3. Finite fields S. Xambó The ring Construction of finite fields The Frobenius automorphism Splitting field of a polynomial Structure of the multiplicative group of a finite field Structure of the
More informationSelected Solutions to Math 4107, Set 4
Selected Solutions to Math 4107, Set 4 November 9, 2005 Page 90. 1. There are 3 different classes: {e}, {(1 2), (1 3), (2 3)}, {(1 2 3), (1 3 2)}. We have that c e = 1, c (1 2) = 3, and c (1 2 3) = 2.
More informationx 2 = xn xn = x 2 N = N = 0
Potpourri. Spring 2010 Problem 2 Let G be a finite group with commutator subgroup G. Let N be the subgroup of G generated by the set {x 2 : x G}. Then N is a normal subgroup of G and N contains G. Proof.
More informationLagrange s Theorem. Philippe B. Laval. Current Semester KSU. Philippe B. Laval (KSU) Lagrange s Theorem Current Semester 1 / 10
Lagrange s Theorem Philippe B. Laval KSU Current Semester Philippe B. Laval (KSU) Lagrange s Theorem Current Semester 1 / 10 Introduction In this chapter, we develop new tools which will allow us to extend
More informationLandau s Theorem for πblocks of πseparable groups
Landau s Theorem for πblocks of πseparable groups Benjamin Sambale October 13, 2018 Abstract Slattery has generalized Brauer s theory of pblocks of finite groups to πblocks of πseparable groups where
More informationNotas de Aula Grupos Profinitos. Martino Garonzi. Universidade de Brasília. Primeiro semestre 2018
Notas de Aula Grupos Profinitos Martino Garonzi Universidade de Brasília Primeiro semestre 2018 1 Le risposte uccidono le domande. 2 Contents 1 Topology 4 2 Profinite spaces 6 3 Topological groups 10 4
More informationFinite Groups with ssembedded Subgroups
International Journal of Algebra, Vol. 11, 2017, no. 2, 93101 HIKARI Ltd, www.mhikari.com https://doi.org/10.12988/ija.2017.7311 Finite Groups with ssembedded Subgroups Xinjian Zhang School of Mathematical
More informationAbstract Algebra: Supplementary Lecture Notes
Abstract Algebra: Supplementary Lecture Notes JOHN A. BEACHY Northern Illinois University 1995 Revised, 1999, 2006 ii To accompany Abstract Algebra, Third Edition by John A. Beachy and William D. Blair
More informationMA441: Algebraic Structures I. Lecture 18
MA441: Algebraic Structures I Lecture 18 5 November 2003 1 Review from Lecture 17: Theorem 6.5: Aut(Z/nZ) U(n) For every positive integer n, Aut(Z/nZ) is isomorphic to U(n). The proof used the map T :
More informationThe Outer Automorphism of S 6
Meena Jagadeesan 1 Karthik Karnik 2 Mentor: Akhil Mathew 1 Phillips Exeter Academy 2 Massachusetts Academy of Math and Science PRIMES Conference, May 2016 What is a Group? A group G is a set of elements
More informationHomework #5 Solutions
Homework #5 Solutions p 83, #16. In order to find a chain a 1 a 2 a n of subgroups of Z 240 with n as large as possible, we start at the top with a n = 1 so that a n = Z 240. In general, given a i we will
More informationMATH HL OPTION  REVISION SETS, RELATIONS AND GROUPS Compiled by: Christos Nikolaidis
MATH HL OPTION  REVISION SETS, RELATIONS AND GROUPS Compiled by: Christos Nikolaidis PART B: GROUPS GROUPS 1. ab The binary operation a * b is defined by a * b = a+ b +. (a) Prove that * is associative.
More informationGROUPS AS GRAPHS. W. B. Vasantha Kandasamy Florentin Smarandache
GROUPS AS GRAPHS W. B. Vasantha Kandasamy Florentin Smarandache 009 GROUPS AS GRAPHS W. B. Vasantha Kandasamy email: vasanthakandasamy@gmail.com web: http://mat.iitm.ac.in/~wbv www.vasantha.in Florentin
More informationSome practice problems for midterm 2
Some practice problems for midterm 2 Kiumars Kaveh November 14, 2011 Problem: Let Z = {a G ax = xa, x G} be the center of a group G. Prove that Z is a normal subgroup of G. Solution: First we prove Z is
More informationChapter 9: Group actions
Chapter 9: Group actions Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Summer I 2014 M. Macauley (Clemson) Chapter 9: Group actions
More informationSUMMARY OF GROUPS AND RINGS GROUPS AND RINGS III Week 1 Lecture 1 Tuesday 3 March.
SUMMARY OF GROUPS AND RINGS GROUPS AND RINGS III 2009 Week 1 Lecture 1 Tuesday 3 March. 1. Introduction (Background from Algebra II) 1.1. Groups and Subgroups. Definition 1.1. A binary operation on a set
More information1 Finite abelian groups
Last revised: May 16, 2014 A.Miller M542 www.math.wisc.edu/ miller/ Each Problem is due one week from the date it is assigned. Do not hand them in early. Please put them on the desk in front of the room
More informationMath 121 Homework 5: Notes on Selected Problems
Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements
More informationCHAPTER III NORMAL SERIES
CHAPTER III NORMAL SERIES 1. Normal Series A group is called simple if it has no nontrivial, proper, normal subgroups. The only abelian simple groups are cyclic groups of prime order, but some authors
More information