# Automorphism Groups Definition. An automorphism of a group G is an isomorphism G G. The set of automorphisms of G is denoted Aut G.

Size: px
Start display at page:

Download "Automorphism Groups Definition. An automorphism of a group G is an isomorphism G G. The set of automorphisms of G is denoted Aut G."

Transcription

1 Automorphism Groups Definition. An automorphism of a group G is an isomorphism G G. The set of automorphisms of G is denoted Aut G. Example. The identity map id : G G is an automorphism. Example. There are two automorphisms of Z: the identity map and the map µ : Z Z given by µ(x) = x. For Z is cyclic, and an isomorphism Z Z must carry a generator to a generator. Since the only generators of Z are 1 and 1, the only automorphisms are the maps sending 1 1 and 1 1. The inverse map which appeared in the last example is a special case of the following result. Lemma. Let G be an abelian group. The map µ : G G given by µ(x) = x is an automorphism. Proof. µ is a homomorphism, since µ(x+y) = (x+y) = x y = µ(x)+µ(y). Clearly, µ µ(x) = x, so µ is its own inverse. Since µ is an invertible homomorphism, it s an isomorphism. Remark. Note that if G is not abelian, µ(xy) = y 1 x 1 x 1 y 1 = µ(x)µ(y). Lemma. Let G be a group, and let g G. The map i g : G G given by i g (x) = gxg 1 is an automorphism of G. (It is called conjugation by g, or the inner automorphism corresponding to g.) Proof. i g is a homomorphism, since i g (xy) = gxyg 1 = gxg 1 gyg 1 = i g (x)i g (y). The inner automorphism i g 1(x) = g 1 xg clearly inverts i g. Since i g is an invertible homomorphism, it s an automorphism. Notation. The set of inner automorphisms of G is denoted InnG. Remark. If G is abelian, then i g (x) = gxg 1 = gg 1 x = x = id(x). That is, in an abelian group the inner automorphisms are trivial. More generally, i g = id if and only if g Z(G). Proposition. Aut G is a group under function composition. 1

2 Proof. The composite of homomorphisms is a homomorphism, and the composite of bijections is a bijection. Therefore, the composite of isomorphisms is an isomorphism, and in particular, the composite of automorphisms is an automorphism. Hence, composition is a well-defined binary operation on Aut G. Composition of functions is always associative. The identity map is an automorphism of G. Finally, an isomorphism has an inverse which is an isomorphism, so the inverse of an automorphism of G exists and is an automorphism of G. Example. From an earlier example, AutZ has order 2. Since there is only one group of order 2, AutZ Z 2. Lemma. InnG AutG. Proof. First, I need to show that InnG is a subgroup. Since id = i 1 InnG, InnG. Now suppose i g,i h InnG. Then (i h ) 1 = i h 1, so i g (i h ) 1 (x) = i g (i h 1(x)) = g(i h 1(x))g 1 = gh 1 xhg 1 = i gh 1(x). Therefore, i g (i h ) 1 = i gh 1 InnG, so InnG < AutG. For normality, suppose g G and φ AutG. Then φi g φ 1 (x) = φ ( gφ 1 (x)g 1) = φ(g)xφ(g 1 ) = φ(g)xφ(g) 1 = i φ(g) (x). Since φi g φ 1 = i φ(g) InnG, it follows that InnG is normal. Proposition. The map φ : G AutG given by φ(g) = i g is a homomorphism onto the subgroup of inner automorphisms of G. Proof. Obviously, φ maps onto InnG. I must verify that it is a homomorphism. φ(g)φ(h)(x) = i g i h (x) = i g ( hxh 1 ) = ghxh 1 g 1 = (gh)x(gh) 1 = i gh (x) = φ(gh)(x). Corollary. G/Z(G) Inn G. Proof. The preceding proposition gives a surjective map φ : G InnG. I only need to verify that ker φ = Z(G). First, suppose g Z(G). Then φ(g)(x) = i g (x) = gxg 1 = gg 1 x = x = id(x). Since φ(g) = id, g ker φ. Conversely, suppose g ker φ. Then φ(g) = id, so i g = id. Applying both sides to x G, i g (x) = x, gxg 1 = x, or gx = xg. Since x was arbitrary, g commutes with everything, so g Z(G). Hence, ker φ = Z(G) as claimed. Finally, G/Z(G) Inn G by the First Isomorphism Theorem. Example. If G is abelian, G = Z(G), so InnG = {1}, as noted earlier. 2

3 Example. Let G = S 3. Z(S 3 ) = {id}, so S 3 /Z(S 3 ) = 6. Thus, there are 6 inner automorphisms of S 3 : different elements of S 3 give rise to distinct inner automorphisms. You can verify that i (1 3) i (1 2) (1 2) = (2 3) and i (1 2) i (1 3) (1 2) = (1 3). That is, i (1 3) i (1 2) i (1 2) i (1 3). It follows that InnS 3 is a nonabelian group of order 6. Therefore, InnS 3 S 3. Proposition. Let G = a. (a) If φ : G G is an automorphism, then φ(a) is a generator of G. (b) If b is a generator of G, there is a unique automorphism φ : G G such that φ(a) = b. Proof. (a) Let φ : G G be an automorphism, and let g G. Since φ 1 (g) G = a, it follows that φ 1 (g) = a n for some n Z. Then g = φ(a n ) = φ(a) n. Since every element of G can be expressed as a power of φ(a), φ(a) generates G. (b) Suppose b generates G. Define φ : G G by φ(a n ) = b n for all n Z. I want to cite an earlier result that says a homomorphism out of a cyclic group is determined by sending a generator somewhere. If G is infinite cyclic, I may send the generator wherever I please, and so φ is a well defined homomorphism. If G is cyclic of order n, then I must be careful to map the generator a to an element that is killed by n. But b is a generator, so it has order n as Again, the result applies to show that φ is a well defined homomorphism. In both cases, the earlier result says that the map φ is unique. Next, observe that φ is invertible. In fact, the map ψ(b n ) = a n clearly inverts φ, and it is well-defined by the same argument which showed that φ was well-defined. Since φ is an invertible homomorphism from G onto G, it is an automorphism of G. This result gives us a way of computing AutZ n : Simply fix a generator and count the number of places where it could go. Definition. Let n 1 be an integer. The Euler phi-function φ(n) is the number of elements in {1,...,n} which are relatively prime to n. Example. φ(12) = 4. φ(25) = 20. Corollary. AutZ n = φ(n). n Proof. By an earlier result, the order of m Z n is (m,n). A generator of Z n must have order n, and this evidently occurs exactly when (m,n) = 1. Now 1 Z n always generates, and the Proposition I just proved implies there is exactly one automorphism of Z n for each generator (i.e. for each possible target for 1 under an automorphism). How do you compute φ(n)? Next on the agenda is a formula for φ(n) in terms of the prime factors of n. Lemma. If p is prime, then φ(p) = p 1. Proof. The numbers {1,...,p 1} are relatively prime to p. 3

4 Lemma. Let p be prime, and let n 1. Then φ(p n ) = p n p n 1. Proof. The numbers in {1,...,p n } which are not relatively prime to p n are exactly the numbers divisible by p. These are p 1,p 2,...,p p n 1. There are p n 1 numbers which are not relatively prime to p n, so there are p n p n 1 which are. Example. φ(9) = 9 3 = 6. Therefore, AutZ 9 = 6. Theorem. If m,n > 0 and (m,n) = 1, then φ(mn) = φ(m)φ(n). Proof. Write down the integers from 1 to mn: 1 m+1 2m+1... (n 1)m+1 2 m+2 2m+2... (n 1)m+2 3 m+3 2m+3... (n 1)m m 2m 3m... mn I m going to find the numbers which are relatively prime to mn. First, I only need to look in rows whose row numbers are relatively prime to m. For suppose row i has (m,i) = k > 1. Since k m and k i, it follows that k am+i (a general element of the i-th row). Since k m mn, k (am+i,mn), so (am+i,mn) 1. Therefore, look at rows i for which (i,m) = 1. Note that there are φ(m) such rows. First, observe am+i bm+i mod n. Assume without loss of generality that a > b. Then (am+i) (bm+i) = (a b)m. Now if n (a b)m, then n (a b), since (m,n) = 1. However, a b < n, so this is impossible. It follows that the elements of a row are distinct mod n. However, each row has n elements, so mod n each row reduces to {0,1,...,n 1}. Hence, exactly φ(n) elements in each row are relatively prime to n. The elements relatively prime to mn are therefore the φ(n) elements in the φ(m) rows whose row numbers are relatively prime to m. Hence, φ(mn) = φ(m)φ(n). Corollary. Let n = p e 1 1 pe m be the prime factorization of n. Then φ(n) = n (1 1p1 ) ( 1 1 p m ( Proof. If m = 1, the result says φ(p n ) = p n 1 1 ), which follows from an earlier result. p Let m > 1, and assume the result is true when n is divisible by fewer than m primes. Suppose that n = p e 1 1 pe m is the prime factorization of n. Then φ(n) = φ ( p e ) 1 1 pe m 1 m 1 φ(p e m m ) = ( m 1 i=1 This establishes the result by induction. p e i i )( m 1 i=1 4 ). ) (1 ) ( (p 1pi m 1 1 )) p m = n m i=1 (1 1pi ).

5 Example. AutZ 11 = 10. In fact, there are two groups of order 10: Z 10 and D 5, the group of symmetries of the regular pentagon. I ll digress a little here and prove part of this claim: namely, that an abelian group of order 10 is isomorphic to Z 10. As in most extended proofs of this sort, you should try to get a feel for the kinds of techniques involved. Each classification problem of this kind presents its own difficulties, so there is not question of memorizing some kind of general method: there isn t any! Suppose then that G is abelian. I claim G is cyclic. Suppose not. Then every element of G has order 2 or order 5. I claim that there is an element of order 5 and an element of order 2. First, suppose every element besides 0 has order 2. Consider distinct elements a and b, a,b 0. Look at the subgroup a,b. I ll show that a,b = {0,a,b,a+b}. Since 2a = 2b = 0, it is easy to see by checking cases that this set is closed. However, a subset of a finite group closed under the operation is a subgroup. Now I have a contradiction, since this putative subgroup has order 4, which does not divide 10. It follows that there must be an element of order 5. On the other hand, could G contain only elements of order 5? Let a have order 5, and let b be an element of order 5 which is not in a. Since a b divides a = 5, it must be either 1 or 5. If it is 5, then a = b, which is impossible (since b / a ). Therefore, a b = 1, and so a b = {0}. This accounts for = 9 elements of G. The remaining element must generate a subgroup of order 5, which (by the preceding argument) intersect a and b in exactly {0}. I ve now accounted for at least 13 elements in G. This contradiction shows that G must can t contain only elements of order 5. The preceding arguments show that G must contain an element a of order 2 and an element b of order 5. I will now show that G is the internal direct product of a and b. Since a b must divide both 2 and 5, it can only be 1. Therefore, a b = {0}. Since G is abelian, a and b are automatically normal. Finally, I claim that G = a + b. To see this, I need only show that the right side has order 10. This will be true if the following elements are distinct: S = {ma+nb 0 m 1, 0 n 4}. Suppose then that pa+qb = ra+sb, where 0 p,r 1 and 0 q,s 4. Then (p r)a = (q s)b a b = {0}. Therefore, 2 p r and 5 q s, which (given the ranges for these parameters) force p = r and q = s. It follows that the elements of S are distinct, so Obviously, this forces G = a + b. Therefore, G a b, and G Z = S a + b. The group of automorphisms AutG is an important group which is contructed from a given group. In case G = Z n, there s another one, which turns out to be related to AutZ n. Definition. U(n) is the set of numbers in {1,...,n} relatively prime to n. Example. U(9) = {1,2,4,5,7,8}. 5

6 Lemma. U(n) is a group under multiplication mod n. Proof. If x and y are relatively prime to n, then (x,n) = 1 and (y,n) = 1. Therefore, there are numbers a, b, c, d such that 1 = ax+bn, 1 = cy +dn. Multiply the two equations: 1 = ac(xy)+(axd+bcy +bdn)n. It follows that (xy,n) = 1, so the product of two elements of U(n) is again an element of U(n). Multiplication of integers mod n is associative, and since 1 is relatively prime to n, it will serve as the identity. Finally, suppose (x,n) = 1. Write ax + bn = 1. Reducing the equation mod n, I get ax = 1 mod n. Therefore, x has a multiplicative inverse mod n, namely a (possibly reduced mod n to lie in U(n)). This shows that U(n) is a group under multiplication mod n. Clearly, U(n) = φ(n). But AutZ n = φ(n). The answer to the obvious question is: Yes. Theorem. AutZ n U(n). Proof. Define φ : U(n) AutZ n by φ(a) = τ a, where τ a is the unique automorphism of Z n which maps 1 to a. I showed earlier that this does indeed give rise to a unique automorphism, and that every automorphism of Z n arises in this way. It follows that φ is a well-defined one-to-one correspondence. I need only show that φ is a homomorphism. Let a,b U(n). Then φ(ab) = τ ab, where τ ab is the map sending 1 to ab. I must show that this is the same as φ(a)φ(b) = τ a τ b. To do this, consider the effect of the two maps on m Z n. τ ab (m) = mab, since τ ab sends 1 to ab and τ ab is a homomorphism. On the other hand, τ a τ b (m) = τ a (mb) = mba, since τ b sends 1 to b and τ a sends 1 to a, and since they re both homomorphisms. Therefore, the maps are equal, and φ is a homomorphism hence, an isomorphism. Example. AutZ 10 U(10) = {1,3,7,9} There are only two groups of order 4: Z 4 and Z 2 Z 2. Notice that 3 2 = 9 mod 10. Since 3 does not have order 2, it follows that AutZ 10 Z 2 Z 2. Therefore, AutZ 10 Z 4. c 2012 by Bruce Ikenaga 6

### Frank Moore Algebra 901 Notes Professor: Tom Marley Direct Products of Groups:

Frank Moore Algebra 901 Notes Professor: Tom Marley Direct Products of Groups: Definition: The external direct product is defined to be the following: Let H 1,..., H n be groups. H 1 H 2 H n := {(h 1,...,

### Solution Outlines for Chapter 6

Solution Outlines for Chapter 6 # 1: Find an isomorphism from the group of integers under addition to the group of even integers under addition. Let φ : Z 2Z be defined by x x + x 2x. Then φ(x + y) 2(x

### The Class Equation X = Gx. x X/G

The Class Equation 9-9-2012 If X is a G-set, X is partitioned by the G-orbits. 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.

### Theorems and Definitions in Group Theory

Theorems and Definitions in Group Theory Shunan Zhao Contents 1 Basics of a group 3 1.1 Basic Properties of Groups.......................... 3 1.2 Properties of Inverses............................. 3

### Math 103 HW 9 Solutions to Selected Problems

Math 103 HW 9 Solutions to Selected Problems 4. Show that U(8) is not isomorphic to U(10). Solution: Unfortunately, the two groups have the same order: the elements are U(n) are just the coprime elements

### Solutions to odd-numbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3

Solutions to odd-numbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3 3. (a) Yes; (b) No; (c) No; (d) No; (e) Yes; (f) Yes; (g) Yes; (h) No; (i) Yes. Comments: (a) is the additive group

### Isomorphisms. 0 a 1, 1 a 3, 2 a 9, 3 a 7

Isomorphisms Consider the following Cayley tables for the groups Z 4, U(), R (= the group of symmetries of a nonsquare rhombus, consisting of four elements: the two rotations about the center, R 8, and

### SUMMARY ALGEBRA I LOUIS-PHILIPPE THIBAULT

SUMMARY ALGEBRA I LOUIS-PHILIPPE 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.

### Solutions to Assignment 4

1. Let G be a finite, abelian group written additively. Let x = g G g, and let G 2 be the subgroup of G defined by G 2 = {g G 2g = 0}. (a) Show that x = g G 2 g. (b) Show that x = 0 if G 2 = 2. If G 2

### CLASSIFICATION OF GROUPS OF ORDER 60 Alfonso Gracia Saz

CLASSIFICATION OF GROUPS OF ORDER 60 Alfonso Gracia Saz Remark: This is a long problem, and there are many ways to attack various of the steps. I am not claiming this is the best way to proceed, nor the

### Elements 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

### Group Actions Definition. Let G be a group, and let X be a set. A left action of G on X is a function θ : G X X satisfying:

Group Actions 8-26-202 Definition. Let G be a group, and let X be a set. A left action of G on X is a function θ : G X X satisfying: (a) θ(g,θ(g 2,x)) = θ(g g 2,x) for all g,g 2 G and x X. (b) θ(,x) =

### A while back, we saw that hai with hai = 42 and Z 42 had basically the same structure, and said the groups were isomorphic, in some sense the same.

CHAPTER 6 Isomorphisms A while back, we saw that hai with hai = 42 and Z 42 had basically the same structure, and said the groups were isomorphic, in some sense the same. Definition (Group Isomorphism).

### Math 3140 Fall 2012 Assignment #3

Math 3140 Fall 2012 Assignment #3 Due Fri., Sept. 21. Remember to cite your sources, including the people you talk to. My solutions will repeatedly use the following proposition from class: Proposition

### 120A 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

### The 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

### φ(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 =

### x 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.

### Algebra-I, Fall Solutions to Midterm #1

Algebra-I, 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

### Solutions of exercise sheet 4

D-MATH 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

### Lecture 4.1: Homomorphisms and isomorphisms

Lecture 4.: Homomorphisms and isomorphisms Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4, Modern Algebra M. Macauley (Clemson) Lecture

### Exercises MAT2200 spring 2013 Ark 4 Homomorphisms and factor groups

Exercises MAT2200 spring 2013 Ark 4 Homomorphisms and factor groups This Ark concerns the weeks No. (Mar ) and No. (Mar ). Plans until Eastern vacations: In the book the group theory included in the curriculum

### Chapter 5 Groups of permutations (bijections) Basic notation and ideas We study the most general type of groups - groups of permutations

Chapter 5 Groups of permutations (bijections) Basic notation and ideas We study the most general type of groups - groups of permutations (bijections). Definition A bijection from a set A to itself is also

### Teddy Einstein Math 4320

Teddy Einstein Math 4320 HW4 Solutions Problem 1: 2.92 An automorphism of a group G is an isomorphism G G. i. Prove that Aut G is a group under composition. Proof. Let f, g Aut G. Then f g is a bijective

### Cosets and Normal Subgroups

Cosets and Normal Subgroups (Last Updated: November 3, 2017) These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from

### M3P10: 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

### Homework 10 M 373K by Mark Lindberg (mal4549)

Homework 10 M 373K by Mark Lindberg (mal4549) 1. Artin, Chapter 11, Exercise 1.1. Prove that 7 + 3 2 and 3 + 5 are algebraic numbers. To do this, we must provide a polynomial with integer coefficients

### Lecture 7 Cyclic groups and subgroups

Lecture 7 Cyclic groups and subgroups Review Types of groups we know Numbers: Z, Q, R, C, Q, R, C Matrices: (M n (F ), +), GL n (F ), where F = Q, R, or C. Modular groups: Z/nZ and (Z/nZ) Dihedral groups:

### Groups 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

### Two subgroups and semi-direct products

Two subgroups and semi-direct 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

### MAT1100HF ALGEBRA: ASSIGNMENT II. Contents 1. Problem Problem Problem Problem Problem Problem

MAT1100HF ALEBRA: ASSINMENT II J.A. MRACEK 998055704 DEPARTMENT OF MATHEMATICS UNIVERSITY OF TORONTO Contents 1. Problem 1 1 2. Problem 2 2 3. Problem 3 2 4. Problem 4 3 5. Problem 5 3 6. Problem 6 3 7.

### Introduction to Groups

Introduction to Groups Hong-Jian Lai August 2000 1. Basic Concepts and Facts (1.1) A semigroup is an ordered pair (G, ) where G is a nonempty set and is a binary operation on G satisfying: (G1) a (b c)

### Basic Definitions: Group, subgroup, order of a group, order of an element, Abelian, center, centralizer, identity, inverse, closed.

Math 546 Review Exam 2 NOTE: An (*) at the end of a line indicates that you will not be asked for the proof of that specific item on the exam But you should still understand the idea and be able to apply

### PROBLEMS 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.

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

### book 2005/1/23 20:41 page 132 #146

book 2005/1/23 20:41 page 132 #146 132 2. BASIC THEORY OF GROUPS Definition 2.6.16. Let a and b be elements of a group G. We say that b is conjugate to a if there is a g G such that b = gag 1. You are

### School of Mathematics and Statistics. MT5824 Topics in Groups. Problem Sheet I: Revision and Re-Activation

MRQ 2009 School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet I: Revision and Re-Activation 1. Let H and K be subgroups of a group G. Define HK = {hk h H, k K }. (a) Show that HK

### MA441: 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

### Lecture 3. Theorem 1: D 6

Lecture 3 This week we have a longer section on homomorphisms and isomorphisms and start formally working with subgroups even though we have been using them in Chapter 1. First, let s finish what was claimed

### Homomorphisms. The kernel of the homomorphism ϕ:g G, denoted Ker(ϕ), is the set of elements in G that are mapped to the identity in G.

10. Homomorphisms 1 Homomorphisms Isomorphisms are important in the study of groups because, being bijections, they ensure that the domain and codomain groups are of the same order, and being operation-preserving,

### Normal Subgroups and Factor Groups

Normal Subgroups and Factor Groups Subject: Mathematics Course Developer: Harshdeep Singh Department/ College: Assistant Professor, Department of Mathematics, Sri Venkateswara College, University of Delhi

### Extra exercises for algebra

Extra exercises for algebra These are extra exercises for the course algebra. They are meant for those students who tend to have already solved all the exercises at the beginning of the exercise session

### MATH 420 FINAL EXAM J. Beachy, 5/7/97

MATH 420 FINAL EXAM J. Beachy, 5/7/97 1. (a) For positive integers a and b, define gcd(a, b). (b) Compute gcd(1776, 1492). (c) Show that if a, b, c are positive integers, then gcd(a, bc) = 1 if and only

### Definition : Let G be a group on X a set. A group action (or just action) of G on X is a mapping: G X X

1 Frank Moore Algebra 901 Notes Professor: Tom Marley Group Actions on Sets Definition : Let G be a group on X a set. A group action (or just action) of G on X is a mapping: G X X Such the following properties

### MODEL ANSWERS TO HWK #4. ϕ(ab) = [ab] = [a][b]

MODEL ANSWERS TO HWK #4 1. (i) Yes. Given a and b Z, ϕ(ab) = [ab] = [a][b] = ϕ(a)ϕ(b). This map is clearly surjective but not injective. Indeed the kernel is easily seen to be nz. (ii) No. Suppose that

### MAT534 Fall 2013 Practice Midterm I The actual midterm will consist of five problems.

MAT534 Fall 2013 Practice Midterm I The actual midterm will consist of five problems. Problem 1 Find all homomorphisms a) Z 6 Z 6 ; b) Z 6 Z 18 ; c) Z 18 Z 6 ; d) Z 12 Z 15 ; e) Z 6 Z 25 Proof. a)ψ(1)

### CONJUGATION IN A GROUP

CONJUGATION IN A GROUP KEITH CONRAD 1. Introduction A reflection across one line in the plane is, geometrically, just like a reflection across any other line. That is, while reflections across two different

### ENTRY GROUP THEORY. [ENTRY GROUP THEORY] Authors: started Mark Lezama: October 2003 Literature: Algebra by Michael Artin, Mathworld.

ENTRY GROUP THEORY [ENTRY GROUP THEORY] Authors: started Mark Lezama: October 2003 Literature: Algebra by Michael Artin, Mathworld Group theory [Group theory] is studies algebraic objects called groups.

### Math 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

### Algebra 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

### Your Name MATH 435, EXAM #1

MATH 435, EXAM #1 Your Name You have 50 minutes to do this exam. No calculators! No notes! For proofs/justifications, please use complete sentences and make sure to explain any steps which are questionable.

### and this makes M into an R-module by (1.2). 2

1. Modules Definition 1.1. Let R be a commutative ring. A module over R is set M together with a binary operation, denoted +, which makes M into an abelian group, with 0 as the identity element, together

### S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 2 BASIC PROPERTIES

S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 2 BASIC PROPERTIES 1 Some Definitions For your convenience, we recall some of the definitions: A group G is called simple if it has

### Abstract Algebra II Groups ( )

Abstract Algebra II Groups ( ) Melchior Grützmann / melchiorgfreehostingcom/algebra October 15, 2012 Outline Group homomorphisms Free groups, free products, and presentations Free products ( ) Definition

### Chapter 5. Modular arithmetic. 5.1 The modular ring

Chapter 5 Modular arithmetic 5.1 The modular ring Definition 5.1. Suppose n N and x, y Z. Then we say that x, y are equivalent modulo n, and we write x y mod n if n x y. It is evident that equivalence

### MA441: Algebraic Structures I. Lecture 14

MA441: Algebraic Structures I Lecture 14 22 October 2003 1 Review from Lecture 13: We looked at how the dihedral group D 4 can be viewed as 1. the symmetries of a square, 2. a permutation group, and 3.

### 5 Group theory. 5.1 Binary operations

5 Group theory This section is an introduction to abstract algebra. This is a very useful and important subject for those of you who will continue to study pure mathematics. 5.1 Binary operations 5.1.1

### MAT301H1F Groups and Symmetry: Problem Set 2 Solutions October 20, 2017

MAT301H1F Groups and Symmetry: Problem Set 2 Solutions October 20, 2017 Questions From the Textbook: for odd-numbered questions, see the back of the book. Chapter 5: #8 Solution: (a) (135) = (15)(13) is

### The Chinese Remainder Theorem

Chapter 5 The Chinese Remainder Theorem 5.1 Coprime moduli Theorem 5.1. Suppose m, n N, and gcd(m, n) = 1. Given any remainders r mod m and s mod n we can find N such that N r mod m and N s mod n. Moreover,

### 2MA105 Algebraic Structures I

2MA105 Algebraic Structures I Per-Anders Svensson http://homepage.lnu.se/staff/psvmsi/2ma105.html Lecture 7 Cosets once again Factor Groups Some Properties of Factor Groups Homomorphisms November 28, 2011

### 1.5 Applications Of The Sylow Theorems

14 CHAPTER1. GROUP THEORY 8. The Sylow theorems are about subgroups whose order is a power of a prime p. Here is a result about subgroups of index p. Let H be a subgroup of the finite group G, and assume

### Homework #11 Solutions

Homework #11 Solutions p 166, #18 We start by counting the elements in D m and D n, respectively, of order 2. If x D m and x 2 then either x is a flip or x is a rotation of order 2. The subgroup of rotations

### Assigment 1. 1 a b. 0 1 c A B = (A B) (B A). 3. In each case, determine whether G is a group with the given operation.

1. Show that the set G = multiplication. Assigment 1 1 a b 0 1 c a, b, c R 0 0 1 is a group under matrix 2. Let U be a set and G = {A A U}. Show that G ia an abelian group under the operation defined by

### Section 15 Factor-group computation and simple groups

Section 15 Factor-group computation and simple groups Instructor: Yifan Yang Fall 2006 Outline Factor-group computation Simple groups The problem Problem Given a factor group G/H, find an isomorphic group

### Math 210A: Algebra, Homework 6

Math 210A: Algebra, Homework 6 Ian Coley November 13, 2013 Problem 1 For every two nonzero integers n and m construct an exact sequence For which n and m is the sequence split? 0 Z/nZ Z/mnZ Z/mZ 0 Let

### Section 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

### its image and kernel. A subgroup of a group G is a non-empty 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

### MATH 436 Notes: Cyclic groups and Invariant Subgroups.

MATH 436 Notes: Cyclic groups and Invariant Subgroups. Jonathan Pakianathan September 30, 2003 1 Cyclic Groups Now that we have enough basic tools, let us go back and study the structure of cyclic groups.

### SPRING BREAK PRACTICE PROBLEMS - WORKED SOLUTIONS

Math 330 - Abstract Algebra I Spring 2009 SPRING BREAK PRACTICE PROBLEMS - WORKED SOLUTIONS (1) Suppose that G is a group, H G is a subgroup and K G is a normal subgroup. Prove that H K H. Solution: We

### Algebra I: Final 2015 June 24, 2015

1 Algebra I: Final 2015 June 24, 2015 ID#: Quote the following when necessary. A. Subgroup H of a group G: Name: 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

### CHAPTER 9. Normal Subgroups and Factor Groups. Normal Subgroups

Normal Subgroups CHAPTER 9 Normal Subgroups and Factor Groups If H apple G, we have seen situations where ah 6= Ha 8 a 2 G. Definition (Normal Subgroup). A subgroup H of a group G is a normal subgroup

### MATH 4107 (Prof. Heil) PRACTICE PROBLEMS WITH SOLUTIONS Spring 2018

MATH 4107 (Prof. Heil) PRACTICE PROBLEMS WITH SOLUTIONS Spring 2018 Here are a few practice problems on groups. You should first work through these WITHOUT LOOKING at the solutions! After you write your

### Homework #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

### Physics 251 Solution Set 1 Spring 2017

Physics 5 Solution Set Spring 07. Consider the set R consisting of pairs of real numbers. For (x,y) R, define scalar multiplication by: c(x,y) (cx,cy) for any real number c, and define vector addition

### ABSTRACT ALGEBRA 1, LECTURE NOTES 5: HOMOMORPHISMS, ISOMORPHISMS, SUBGROUPS, QUOTIENT ( FACTOR ) GROUPS. ANDREW SALCH

ABSTRACT ALGEBRA 1, LECTURE NOTES 5: HOMOMORPHISMS, ISOMORPHISMS, SUBGROUPS, QUOTIENT ( FACTOR ) GROUPS. ANDREW SALCH 1. Homomorphisms and isomorphisms between groups. Definition 1.1. Let G, H be groups.

### D-MATH Algebra I HS 2013 Prof. Brent Doran. Solution 3. Modular arithmetic, quotients, product groups

D-MATH Algebra I HS 2013 Prof. Brent Doran Solution 3 Modular arithmetic, quotients, product groups 1. Show that the functions f = 1/x, g = (x 1)/x generate a group of functions, the law of composition

### MA441: Algebraic Structures I. Lecture 15

MA441: Algebraic Structures I Lecture 15 27 October 2003 1 Correction for Lecture 14: I should have used multiplication on the right for Cayley s theorem. Theorem 6.1: Cayley s Theorem Every group is isomorphic

### MODEL ANSWERS TO THE FIFTH HOMEWORK

MODEL ANSWERS TO THE FIFTH HOMEWORK 1. Chapter 3, Section 5: 1 (a) Yes. Given a and b Z, φ(ab) = [ab] = [a][b] = φ(a)φ(b). This map is clearly surjective but not injective. Indeed the kernel is easily

### Discrete Mathematics with Applications MATH236

Discrete Mathematics with Applications MATH236 Dr. Hung P. Tong-Viet School of Mathematics, Statistics and Computer Science University of KwaZulu-Natal Pietermaritzburg Campus Semester 1, 2013 Tong-Viet

### IDEAL CLASSES AND RELATIVE INTEGERS

IDEAL CLASSES AND RELATIVE INTEGERS KEITH CONRAD The ring of integers of a number field is free as a Z-module. It is a module not just over Z, but also over any intermediate ring of integers. That is,

### Lecture Note of Week 2

Lecture Note of Week 2 2. Homomorphisms and Subgroups (2.1) Let G and H be groups. A map f : G H is a homomorphism if for all x, y G, f(xy) = f(x)f(y). f is an isomorphism if it is bijective. If f : G

### Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.

Glossary 1 Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.23 Abelian Group. A group G, (or just G for short) is

### Modern Algebra (MA 521) Synopsis of lectures July-Nov 2015 semester, IIT Guwahati

Modern Algebra (MA 521) Synopsis of lectures July-Nov 2015 semester, IIT Guwahati Shyamashree Upadhyay Contents 1 Lecture 1 4 1.1 Properties of Integers....................... 4 1.2 Sets, relations and

### ALGEBRA QUALIFYING EXAM PROBLEMS

ALGEBRA QUALIFYING EXAM PROBLEMS Kent State University Department of Mathematical Sciences Compiled and Maintained by Donald L. White Version: August 29, 2017 CONTENTS LINEAR ALGEBRA AND MODULES General

### Quiz 2 Practice Problems

Quiz 2 Practice Problems Math 332, Spring 2010 Isomorphisms and Automorphisms 1. Let C be the group of complex numbers under the operation of addition, and define a function ϕ: C C by ϕ(a + bi) = a bi.

### FINITE 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

### Inverses and Elementary Matrices

Inverses and Elementary Matrices 1-12-2013 Matrix inversion gives a method for solving some systems of equations Suppose a 11 x 1 +a 12 x 2 + +a 1n x n = b 1 a 21 x 1 +a 22 x 2 + +a 2n x n = b 2 a n1 x

### Math 546, Exam 2 Information.

Math 546, Exam 2 Information. 10/21/09, LC 303B, 10:10-11:00. Exam 2 will be based on: Sections 3.2, 3.3, 3.4, 3.5; The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/546fa09/546.html)

### a = mq + r where 0 r m 1.

8. Euler ϕ-function We have already seen that Z m, the set of equivalence classes of the integers modulo m, is naturally a ring. Now we will start to derive some interesting consequences in number theory.

### A. (Groups of order 8.) (a) Which of the five groups G (as specified in the question) have the following property: G has a normal subgroup N such that

MATH 402A - Solutions for the suggested problems. A. (Groups of order 8. (a Which of the five groups G (as specified in the question have the following property: G has a normal subgroup N such that N =

### ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008

ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved completely

### = 1 2x. x 2 a ) 0 (mod p n ), (x 2 + 2a + a2. x a ) 2

8. p-adic numbers 8.1. Motivation: Solving x 2 a (mod p n ). Take an odd prime p, and ( an) integer a coprime to p. Then, as we know, x 2 a (mod p) has a solution x Z iff = 1. In this case we can suppose

### Normal Subgroups and Quotient Groups

Normal Subgroups and Quotient Groups 3-20-2014 A subgroup H < G is normal if ghg 1 H for all g G. Notation: H G. Every subgroup of an abelian group is normal. Every subgroup of index 2 is normal. If H

### Quizzes for Math 401

Quizzes for Math 401 QUIZ 1. a) Let a,b be integers such that λa+µb = 1 for some inetegrs λ,µ. Prove that gcd(a,b) = 1. b) Use Euclid s algorithm to compute gcd(803, 154) and find integers λ,µ such that

### SUPPLEMENT ON THE SYMMETRIC GROUP

SUPPLEMENT ON THE SYMMETRIC GROUP RUSS WOODROOFE I presented a couple of aspects of the theory of the symmetric group S n differently than what is in Herstein. These notes will sketch this material. You

### A Readable Introduction to Real Mathematics

Solutions to selected problems in the book A Readable Introduction to Real Mathematics D. Rosenthal, D. Rosenthal, P. Rosenthal Chapter 7: The Euclidean Algorithm and Applications 1. Find the greatest

### ABSTRACT ALGEBRA: REVIEW PROBLEMS ON GROUPS AND GALOIS THEORY

ABSTRACT ALGEBRA: REVIEW PROBLEMS ON GROUPS AND GALOIS THEORY John A. Beachy Northern Illinois University 2000 ii J.A.Beachy This is a supplement to Abstract Algebra, Second Edition by John A. Beachy and

### Math 121 Homework 3 Solutions

Math 121 Homework 3 Solutions Problem 13.4 #6. Let K 1 and K 2 be finite extensions of F in the field K, and assume that both are splitting fields over F. (a) Prove that their composite K 1 K 2 is a splitting