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Math 417 Exam I: July 2, 2009; Solutions 1. If a and b are relatively prime and each of them divides an integer n, prove that their product ab also divides n. Here are two proofs (of course, either one suces for full credit). By hypothesis, n = ak = b .
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Homework I: June 19, 2009
1.9. Find a formula for 1 + is correct.
n j =1
j !j ; use induction to prove that your formula
A list of the sums for n = 1, 2, 3, 4, 5 is 2, 6, 24, 120, 720. These are factorials; n better, they are 2!, 3!, 4!, 5!, 6!. If we w
MATH 417 HOMEWORK 4 SOLUTIONS
(1) Prove: There is no surjective group homomorphism from D4 to Z4 .
Solution: This was one of the in-class problems. First of all, if there
was such a surjective homomorphism, then its kernel would be a normal
subgroup of or
MATH 417 QUIZ 2 9/30/14
Name:
(1) Find the cosets of S 1 = cfw_ei : [0, 2) in C . Give a geometric picture.
Solution: The cosets are cfw_rS 1 : r R>0 and consist of all circles about
the origin of xed radius r, not including the origin.
(2) Find a group
MATH 417 WEEK 7: GROUP ACTIONS ON SETS
(1) Draw the hexagon with 2 black and 4 white vertices. Draw all possible such hexagons.
Now divide them up into orbits under the action of (1) D6 (rotations and ips) and
(2) C6 (rotations).
(2) Prove: The group G ac
MATH 417 FINAL EXAM 12/12/2014
Name:
Special instructions: Write your name legibly at the top of each page!
Cellphones and other communication devices are to remain on the desk at the front
of the room for the duration of the exam.
All answers need a p
MATH 417 MIDTERM 1 SOLUTIONS, FALL 2014
(1) (5 points each) True/False questions.
(a) Let G be a group and H be a non-trivial proper subgroup of G. The action of G on G/H by
left multiplication is transitive.
Solution: True, by denition: gi H = gi H, so a
MATH 417 SOLUTIONS TO EXAM III 12/10/13
All answers need a proof! If you use a theorem, lemma or denition then state it completely
rst.
(1) (15 points) Prove or disprove, stating your reasoning carefully, stating any theorems
you are using: The polynomial
MATH 417 SOLUTIONS TO EXAM II FALL 2014
(1) (a) (2 points) T/F: Let I Q[x] be the ideal generated by two polynomials f (x), g(x).
Then there exists a polynomial h(x) Q(x) such that I = (h(x).
Solutions: True. Q[x] is a PID and h(x) = gcd(f, g).
(b) (2 poi
MATH 417 QUIZ 1 SOLUTIONS
(1) Let i = (i, i + 1) be the element of S3 which interchanges i and i + 1, with i = 1, 2.
Compute
1 2 1 (1) = 3
1 2 1 (2) = 2
1 2 1 (3) = 1
2 1 2 (1) = 3
2 1 2 (2) = 2
What can you say about these two products?
2 1 2 (3) = 1
The
MATH 417 EXAM II 10/31/13
Name:
All answers need a proof!
(1) (10 points) Prove or disprove the following statement: The map dened as
:
RZ
x x (The integer part of x)
is a group homomorphism.
1
2
MATH 417 EXAM II 10/31/13
(2) (10 points) Let G = GLn (R) b
MATH 417 EXAM I 9/26/13
Name:
All answers need a proof!
(1) The group D4 acts on the set of vertices cfw_1, 2, 3, 4 of the square by reections and
rotations. Here is a rotation by /2 and = 1 is a reection about the x-axis.
4
3
2
2
1
3
1
4
The convention i
ALGEBRA
ABSTRACT
AND
CONCRETE
E DITION 2.6
F REDERICK M. G OODMAN
SemiSimple Press
Iowa City, IA
Last revised on May 1, 2015.
Algebra: abstract and concrete / Frederick M. Goodman ed. 2.6
ISBN 978-0-9799142-1-8
c 2014, 2006, 2003, 1998 by Frederick M. Goo
ALGEBRA
ABSTRACT
AND
CONCRETE
E DITION 2.6
F REDERICK M. G OODMAN
SemiSimple Press
Iowa City, IA
Last revised on May 1, 2015.
Algebra: abstract and concrete / Frederick M. Goodman ed. 2.6
ISBN 978-0-9799142-1-8
c 2014, 2006, 2003, 1998 by Frederick M. Goo
SYLLABUS: MATH 417 FALL 2014
PROFESSOR RINAT KEDEM
1. Vital data:
(1) Course website: We will use learn.illinois.edu.
If you are not registered for the course yet, send me an email with your netid,
and I will manually give you access to the course web pa
SOLUTIONS TO SOME OF THE PROBLEMS FROM THURSDAYS
LECTURE
the ideal generated by (x 1) and (x 2) in Q[x] is a principal ideal. Can you nd
the polynomial that generates this ideal?
Solution: Their g.c.d. is 1, so the ideal is generated by 1, it is Q[x].
W
T403 Homework Hints and Solutions II
2.1.1 The symmetries of a non square rhombus is isomorphic to that of a (non-square) rectangle. It has
four elements and is abelian. Denoting the 180 rotation by and the reection across one of the
diagonals by the elem