MATH 5a; Solutions to Homework Set 1
October 2011
1. (1) For each A P , A A, so is relexive.
Suppose P Q and Q P . Then for B Q, B A P , and A B Q. Thus
B B , so as Q is a partition, B = B . Thus B = A P , so Q = P . That is is
antisymmetric.
Suppose P Q
MATH 5a: Solutions to Homework Set 3
October 2013
1. Its easy to see that GL2 (Z/2Z) is a set consisting of the following 6 elements:
I=
10
01
,A =
01
10
,B =
11
01
,C =
10
11
,D =
11
10
,E =
01
11
.
Its easy to check that 1 I , A (12), B (13), C (23), D
MATH 5a: Solutions to Homework Set 4
October 2013
1. As x and y commute, by a previous HW problem, for each integer k , (xy )k = xk y k . Let k be the
least common multiple of m and n. Thus k = na = mb for some integers a and b, so
(xy )k = xk y k = xna y
MATH 5b; Homework Set 1
Due: Wednesday, January 11, 2012 at noon
Read the rst four sections of Chapter 7 of the Text; that is read pages 222 to 256.
1. Problem 26 on page 232.
2. Problem 27 on page 232. In addition prove that p is a discrete valuation on
MATH 5b; Homework Set 2
Due: Wednesday, January 18, 2012 at noon
Read Chapter 8 in the Text; that is read pages 270 to 292.
Do Problem 33 on page 259 of Dummit and Foote. This is as much a problem in
topology as in algebra. To minimize eort spent on topol
MATH 5b; Homework Set 3
Due: Wednesday, January 25, 2012 at noon
Read the rst ve sections of Chapter 9 in the Text; that is read pages 295 to 314.
1. Problem 12 on page 279. Also prove there exists d Z with dd 1 mod (N ).
2. Problem 3 on page 282.
3. Prob
MATH 5b; Homework Set 4
Due: Wednesday, February 1, 2012 at noon
Read the rst three sections of Chapter 10; that is read pages 337 to 355.
1. Problem 4, page 298.
2. Problem 6, page 298.
3. Problem 13, page 298.
4. Problem 4, page 301.
1
MATH 5b; Homework Set 5
Due: Wednesday, February 15, 2012 at noon
Read sections 1 and 2 of Chapter 11; that is read pages 408 through 420. You can
skip the material on tensor products on pages 420-422.
1. Problem 8, page 344.
2. Problem 11, page 423.
3. P
MATH 5b; Homework Set 6
Due: Wednesday, February 22, 2012 at noon
Read section 1 of Chapter 12; that is read pages 456 through 468. I will be giving
easier proofs of weaker theorems; that is I will prove theorems for modules over EDs
rather than PIDs, alt
MATH 5a; Solutions to Homework Set 2
October 2013
1.) Problem 24: Let a and b be commuting elements in a group A and k an integer. Claim abn = bn a
for all positive integers n. This holds for n = 1 as a and b commute. Assume it holds for k n. Then
abn+1 =
MATH 5a; Solutions to Homework Set 1
October 2013
1. For each A P , A A, so is reexive.
Suppose P Q and Q P . Then for each B Q, B A P , and A B Q. Thus B B , so as
Q is a partition, B = B . Thus B = A P , so Q = P . That is is antisymmetric.
Suppose P Q
MATH 5a; Solutions to Homework Set 2
October 2011
1. First observe that for all positive integers n, and all x G, (xn ) = (x)n . Prove
by induction on n. For n = 1, this is trivial. Assume for n; then
(xn+1 ) = (xn x) = (xn )(x) = (x)n (x) = (x)n+1
by the
MATH 5a; Solutions to Homework Set 3
October 2011
1. Let = (a1 , . . . , a )(b1 , . . . , b ) (z1 , . . . , z ). Notice
(a) ai j = ai+j , bi j = bi+j , etc. with the indices read modulo cycle length.
Proof. Prove by induction on j . The remark holds if j
MATH 5a; Solutions to Homework Set 4
October 2011
1. As x and y commute, by Problem 3 on HW2, for each integer k , (xy )k = xk y k . Let
k be the least common multiple of m and n. Thus k = na = mb for some integers a and
b, so
(xy )k = xk y k = xna y mb =
MATH 5a; Solutions to Homework Set 5
November, 2011
1. If h, g G(r) then as G is abelian, (gh)r = g r hr = 1 1 = 1 by a Remark 6 from
Chapter 5 in class. Hence gh G(r). Similarly (g 1 )r = (g r )1 = 1, so g 1 G(r).
Thus G(r) G.
If g G(m) G(k ) then g m =
MATH 5a; Solutions to Homework Set 6
November, 2011
1. Let G = A1 An and H = (A1 /B1 ) (An /Bn ). Dene : G H by
: (a1 , . . . , an ) (B1 a1 , . . . , Bn an )
Observe that is a group homomorphism:
(a1 , . . . , an )(c1 , . . . , cn ) = (a1 c1 , . . . , an
MATH 5a; Solutions to Homework Set 7
November, 2011
1. Let B be a block of A and B = cfw_Bg : g G. By Lemma 4B, G is representated
as a group of permutions on the set P of all subsets of A and by construction, B is an
orbit of G on P . Further
G(B ) = cfw
MATH 5a; Solutions to Homework Set 8
November, 2011
1. Let G be S4 on A = cfw_1, 2, 3, 4. Then |G| = 24 = 3 8, so a Sylow 2-group of G is
of order 8 and the number k of Sylow 2-groups divides 3, so k = 1 or 3. Dene
X = cfw_1, (1, 2)(3, 4), (1, 3)(2, 4), (
CALIFORNIA INSTITUTE OF TECHNOLOGY
Department of Mathematics
Solutions to Math 5a Midterm, October 2011
1. We rst prove:
Lemma 1. If : X Y is a group homomorphism and X is nite then X Y and
|X| divides |X |.
Proof. By Lemma 3A.3, X Y . As X is the image o
MATH 5b; Homework Set 7
Due: Wednesday, February 29, 2012 at noon
Read section 2 of Chapter 5; that is read pages 158 to 165. Also I will continue to
give a slightly dierent treatment of the material in section 1 of Chapter 12, so you may
wish to read mor
MATH 5b; Homework Set 8
Due: Wednesday, March 7, 2012 at noon
Read section 2 of Chapter 12.
1. Let A be a nite dimensional F -algebra, a A a unit of nite order m in the group
of units of A, k a positive integer, and p(x) the minimal polynomial of a. Prove
MATH 5a; Homework Set 1
Due: Wednesday, October 5, 2011 at noon
Read Chapter 0 and the rst section of Chapter 1 of the text. That is read pages 1-21
of Dummit and Foote.
In Problem 2, you may use the fact that addition and multiplication of integers modul
M a 5a
HOMEWORK ASSIGNMENT 4
D. RAMAKRISHNAN 278 SLOAN
FALL 2010
4348
INSTRUCTOR:
It is advisable to do all the problems. But the only ones which need to be turned in are the ve which are underlined and in bold, like 14. Give details; part of what one mus
M a 5a HOMEWORK ASSIGNMENT 4 SOLUTION FALL 2010
The exercises are taken from the text, Abstract Algebra (third edition) by Dummitt and Foote. Page 65,16. (a)Suppose H itself is not maximal, then there is a subgroup H1 of G such that H H1 . Similarly, if H
M a 5a
HOMEWORK ASSIGNMENT 4
D. RAMAKRISHNAN 278 SLOAN
FALL 2010
4348
INSTRUCTOR:
It is advisable to do all the problems. But the only ones which need to be turned in are the ve which are underlined and in bold, like 14. Give details; part of what one mus
M a 5a HW ASSIGNMENT 3 FALL 2010 SOLUTIONS
The exercises are taken from the text, Abstract Algebra (third edition) by Dummitt and Foote. Page 45, 23. Lets call the set of 3 pairs of opposite faces of the cube: S = cfw_a1 , a2 , cfw_b1 , b2 , cfw_c1 , c2 ,
M a 5a
HOMEWORK ASSIGNMENT 3
D. RAMAKRISHNAN 278 SLOAN
FALL 2010
4348
INSTRUCTOR:
It is advisable to do all the problems. But the only ones which need to be turned in are the ve which are underlined and in bold, like 14. Give details; part of what one mus
M a 5a
HOMEWORK ASSIGNMENT 2 SOLUTIONS FALL 2010
The exercises are taken from the text, Abstract Algebra (third edition) by Dummitt and Foote. Page 22,25. Take a, b G. Then ab G, and (ab)(ba) = ab2 a = 1 = (ab)(ab). Multiplying both sides of (ab)(ba) = (a
M a 5a
HOMEWORK ASSIGNMENT 2
D. RAMAKRISHNAN 278 SLOAN
FALL 2010
4348
INSTRUCTOR:
It is advisable to do all the problems. But the only ones which need to be turned in are the ve which are underlined and in bold, like 14. Give details; part of what one mus
M a 5a HW ASSIGNMENT 1 FALL 09 SOLUTIONS
The exercises are taken from the text, Abstract Algebra (third edition) by Dummitt and Foote. We have followed the books notation here for Z+ as the set of all positive integers. (This is not standard, and many use