Sample Exam 1 Solutions
1. (a) (Z, #) is a group. Clearly it is closed under # since the integers are closed under addition. x# 2 = x and 2#x = x for all x Z so 2 is the identity. x#(y #z ) = x#(y + z +2) = x + y + z +4 and (x#y )#z = (x + y +2)#z = x + y
MATH 373: Sample Exam 1
Monday, February 15, 2010
Answer all questions in the answer booklet. Please do these in order. You are expected to justify your answers in a manner that an average MATH 373 student should be able to follow. This includes sentence
Quotient Group Examples
1. The basic example: Z/nZ. Let Z = (Z, +) and nZ = (nZ, +). You should verify nZ Z. Since Z is abelian, we get nZ Z for free. A typical coset is a + nZ for a Z. Z/nZ = cfw_a + nZ : a Z = cfw_i + nZ : 0 i < n = Zn (the latter notat
Homework 11
Due: NEVER
1. Describe the elements of Q( 3 5). Justify carefully using an appropriate theorem. 2. Show Q( 2, 3) = Q( 2 + 3). 3. Find the splitting eld for x3 1 over Q. Express your answer in the form Q(a). 4. Find the splitting eld for x4 +
Homework 10
Due: Monday, April 19, 2010
Problems after the double bar will not be collected but it is recommended that you do them. My assumption is that you will do them. 1. Let F be a nite eld. Show that F [x] contains irreducible polynomials of arbitra
Homework 9
Due: Monday, April 12, 2010
Problems after the double bar will not be collected but it is recommended that you do them. My assumption is that you will do them. 1. Show that there is an innite number of solutions to x2 = 1 in the ring of quatern
Homework 8
Due: Friday, April 2, 2010
1. Answer True or False and justify your answer in a sentence or two. (a) If G is a nite group and p is prime then G has exactly one Sylow p-subgroup. (b) If G is nite abelian group and p is prime then G has exactly o
Homework 7
Due: Wednesday 3/24/10
Problems after the double bar will not be collected but it is recommended that you do them. My assumption is that you will do them. 1. Prove that a nite p-group G is simple if and only if |G| = p. 2. (a) If H is a subgrou
Homework 6
Due: Monday 3/15/10
1. Let G = cfw_1, 7, 43, 49, 51, 57, 93, 99, 101, 107, 143, 149, 151, 157, 193, 199 under multiplication modulo 200. Express G as a direct product of cyclic groups. 2. Suppose G is an Abelian group of order 9. What is the ma
Homework 5
Due: Friday 2/26/10
Problems after the double bar will not be collected but it is recommended that you do them. My assumption is that you will do them. 1. The commutator subgroup G of a group G is the subgroup generated by the set cfw_x1 y 1 xy
Homework 4
Due: Wednesday 2/10/10
Problems after the double bar will not be collected but it is recommended that you do them. My assumption is that you will do them. 1. Let G be an abelian group. Dene H as H = cfw_x2 | x G. Prove that H is a subgroup of G
Homework 3
Due: Wednesday 2/3/10
Problems after the double bar will not be collected but it is recommended that you do them. My assumption is that you will do them. 1. Classify each of the following as a group or not. For each which is a group give the id
Homework 2
Due: Wednesday 1/27/10
Problems after the double bar will not be collected but it is recommended that you do them. My assumption is that you will do them. 1. (Problem 2.25) (a) If is an r-cycle, show that r = (1) (b) If is an r-cycle, show that
Problem Set 1
1. Prove case 2 of the Division Algorithm 2. (1.51*) If is a root of unity, prove that there is a positive integer d with d = 1 such that if k = 1 then d|k 3. (1.56*) Let a and b be integers and let sa + tb = 1 for s, t Z. Prove that a and b
Proof from 3/15/10 Theorem 0.1. If G acts on X and x X then |O(x)| = [G : Gx ]. Proof. Let G/Gx denote the family of cosets of Gx in G (not necessarily a quotient group. If y O(x) then y = gx for some g G. Let : O(x) G/Gx such that (y ) = gGx . (Our goal
Homework 11 Solutions 1. Since 3 5 is the root of x3 5 which is irreducible over Q we know Q( 3 5) Q[x]/(x3 5) = and the elements in the latter have the form ax2 + + c + (x3 5) and so the corresponding bx elements in the former have the form a( 3 5)2 + b(
Homework 10 Solutions
1. BWOC, suppose the highest degree irreducible polynomial in F [x] is n. Since F is nite, there are only a nite number of irreducible polynomials of each degree, and we dont have irreducible polynomials above degree n, thus our assu
Homework 8 Solutions
1. (a) False. e.g. S3 has 9 Sylow 2-subgroups. (b) True. G abelian implies all subgroups are normal, which implies that each Sylow subgroup is unique. (c) True. Sylows Theorem part 1. (d) True. If y = gx then Gy = gGx g 1 . (e) False.
Homework 7 Solutions
1. Let |G| = pn for n 0. () Suppose G is simple. Since G is simple, n = 0. In addition, we know that G has normal subgroups of order pk for k n. Since G is simple, n must equal 1. Thus |G| = p. () Suppose |G| = p. By Lagranges theorem
Homework 6 Solutions
1. Table of Elements and their Orders:
element order
1 1
7 4
43 4
49 2
51 2
57 4
93 4
99 2
101 2
107 4
143 4
149 2
151 2
157 4
193 4
199 2
Since |G| = 16 and G only has elements of 4 or 2 (or 1) we know either G Z4 Z4 or = G Z4 Z2 Z2
Homework 5 Solutions
1. (a) Note: for an arbitrary commutator from G, x1 y 1 xy , with x, y G and for g G, gx1 y 1 xyg 1 = (gx1 g 1 )(gy 1 g 1 )(gxg 1 )(gyg 1 ) = (gxg 1 )1 (gyg 1 )1 (gxg 1 )(gyg 1 ) Thus the conjugate of a commutator is also a commutator
Homework 4 Solutions 1. e2 = e thus e H and H = . Now let x, y H , hence a, b G such that x = a2 , y = b2 . xy 1 = a2 (b2 )1 = (ab1 )2 since G is abelian. Thus xy 1 H and H G by the one step subgroup test. 2. (1) = (12)(12) thus (1) An and An = . Let a, b
Homework 3 Solutions 1. (a) Not a group. Neither associative nor does it have an identity. Since a e = a we would need e = 0, but 0 a = a (for a = 0). Since no identity it has no inverses. (b) Is a group. It is non-abelian and the identity is: I= 10 01
(c
Homework 2 Solutions
1. Suppose Sn is an r-cycle. Let = (x1 x2 . . . xr ) where xi = xj for i = j . (a) only permutes the elements x1 through xr and leaves any other elements from cfw_1, . . . n xed. Thus in order to show that r = (1) it suces to show tha
Homework 1 Solutions
1. Case 2 of the Division Algorithm follows almost identically to Case 1. The big change is replacing q + 1 in the existence proof with q 1 since a is now negative. 2. Problem 1.51: Let I = cfw_k Z : k = 1 Z. We claim that I meets the
Final Exam Solutions
1. (a) Yes, x2 x 3 + x2 x + 3 = x2 x + 3 6 + x2 x + 3 = 6 + x2 x + 3 (b) True. Eisensteins criterion with p = 3 works. (c) It neither has an identity nor closure under addition. (d) [S5 : H ] = 120/10 = 12 (e) False. Consider = (12) S
Math 21-373 Algebraic Structures Syllabus, Spring 2010
Time and Place: MWF 1:30-2:20, DH 1117 Instructor: Dr. Greggo M. Johnson Oce: Wean Hall 6211 (temporary) Phone: 412-268-6471 E-mail: [email protected] Class webpage: http:/www.math.cmu.edu/greggo Oc
MATH 373: Final Exam
Monday, May 3, 2010
Answer all questions in the answer booklet. Please do these in order. You are expected to justify your answers in a manner that an average MATH 373 student should be able to follow. This includes sentences for cla
Exam 3 Solutions 1. Since 0 A and is a homomorphism, (0) = 0 (A) thus (A) = . Let (a), (b) (A) and hence a, b A. (a) (b) = (a b) because is a homomorphism. a b A because A is an ideal, hence (a) (b) (A) and (A) is an additive subgroup. Let s S and (a) (A)