MAT401 HW5 Solutio ns
CHAPT ER 16
22. Prove that Z[x] is not a principle idea l domain.
Consider the ideal <x, 2> = cfw_xf(x) + 2 g(x) | f(x), g(x) Z[x]
Claim: <x, 2> ca nnot be ge nerate d by a single polynomia l p(x).
Proof: Assume <x, 2 > <p(x)>, p(x)
PMath 442 Homework 1
Due Wednesday, September 21
1. Find all automorphisms of Q( 3 5). [An automorphism of a ring R is a
ring isomorphism : R R.]
Solution:
Let F = Q( 3 5). Then F isomorphic to the eld K =
is
3
3
Q[x]/ (x 5), with x corresponding to 5. We
Q1.
A hot-air balloon rises vertically from ground level as a rope attached to the base of the balloon is
released at the rate of 6 ft/sec (see the figure). The pulley that releases the rope is L = 40 feet from a
platform where passengers board the balloo
Algebra 3 (2003-04) Assignment 8 Solutions
Instructor: Dr. Eyal Goren
1) Let p < q < r be primes. Prove that a group of order pqr is not simple.
Proof. Let G be a group of order pqr and suppose G is simple. Then np > 1, nq > 1, nr > 1.
Moreover np |qr, np
Math 415/515 : Spring 2006 Homework 10, March 31 24.14 Let G be a noncylic group of order 21. How many Sylow 3-subgroups does G have? We must have n7 = 1, so if n3 = 1, G would be (isomorphic to) the internal direct product of its Sylow subgroups, and so
Debra Griffin
Math 210A
HW #9 The Sylow Theorems #1 16
Textbook Problems:
5.23 (i) Prove that if d is a positive divisor of 24, then S4 has a subgroup of order d.
Proof:
We have the following subgroups of S4:
1|24
cfw_(1) S4
2|24
cfw_(1), (12) S4
3|24
cfw
Abstract Algebra I
Fall 2009
Solutions 7
Problem 1. Let G be a simple group of order 168=7 23 3. Determine the number of
elements of G of order 7.
We have n7 = 1 mod 7 and n7 | 24. Hence n7 is either 1 or 8. If n7 = 1 the Sylow 7subgoup is unique and ther
Algebra 3 (2003-04) Assignment 2 Solutions
Instructor: Dr. Eyal Goren
1) Find the center of the following groups:
(1) The dihedral group D2n (n 3).
(2) The group GLn (F), where F is a eld.
Answer. 1) The group D2n is generated by the elements x, y that sa
Math 621, Fall 2010
Miami University
Take-Home Exam 1.
Due Friday, October 8th, 1:00 p.m.
Directions: Please work on your own. Do not talk to each other, to other
students, or to other faculty about these problems, even if this merely
involves giving hint
Chapter 3
Euclidean Constructions
The idea of constructions comes from a need to create certain objects in our proofs. A
construction is, in some sense, a physical substantiation of the abstract.
In Greek times, geometric constructions of gures and length
MODULE 6
Constructions with Compass and Straightedge
A thing constructed can only be loved after it is constructed: but a thing created is loved
before it exists. Gilbert Keith Chesterton
1. Constructible Numbers
Denition 6.1. A length is constructible if
Notes on Classical Groups
Peter J. Cameron
School of Mathematical Sciences
Queen Mary and Westeld College
London E1 4NS
U.K.
[email protected]
These notes are the content of an M.Sc. course I gave at Queen Mary and
Westeld College, London, in JanuaryM
A nite subgroup of the multiplicative group of a eld is cyclic
1 Lemma. Let G be a group. If G contains an element of order n, and m is a divisor of n, then G
contains an element of order m.
Proof. Let g G have order n, and suppose that n = km. Then g k h
Math 415/515 : Spring 2006 Homework 9, March 24 22.4 How many elements of the cyclic group GF(81) are generators? GF(81) Z/80Z. A cyclic group of order 80 has (80) = 23 (2 - 1) (5 - 1) = 32 generators, where is Euler's function. 22.16 Suppose that and bel
Math 415/515 : Spring 2006 Homework 9, March 24 22.4 How many elements of the cyclic group GF(81) are generators? GF(81) Z/80Z. A cyclic group of order 80 has (80) = 23 (2 - 1) (5 - 1) = 32 generators, where is Euler's function. 22.16 Suppose that and bel
THE IMPOSSIBILITY OF TRISECTING AN ANGLE WITH STRAIGHTEDGE AND
COMPASS: AN APPROACH USING RATIONAL TRIGONOMETRY
David G. Poole
Trent University
Peterborough, Ontario
Canada K9J 7B8
A classical problem in geometry is to trisect an angle using only a straig
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t z hp r u uwt z s r w rtwu w wz v rws h zku r ~ s
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prove that 30 is a constructible angle
MojoMaj
Best Answer - Chosen by Voters
Using a compass and a straight edge, you can construct an equilateral triangle. With the straight edge, draw a line segment. Open the
compass to the length of the line segment.
MATH 6/71051 Homework #11 Selected Solutions 3.4 #7: If G is a finite group and H whose terms is H. G, then there is a composition series of G one of
Proof. Method 1: Since G is finite, G has a composition series. Thus by the Schreier Refinement Theorem,
ALGEBRA HW 4
CLAY SHONKWILER
3.2.19 Prove that if N is a normal subgroup of the finite group G and (|N |, |G : N |) = 1 then N is the unique subgroup of G of order |N |. Proof. Let H G such that |H| = |N |. By Proposition 13, |N H| = |N |H| |N |2 |N | = =
Solutions: Abstract Algebra, 2nd edn, Dummit & Foote
Section. 1.7: Group Actions
1/44. Let be a field. Show that the multiplicative group of nonzero elements of , denoted
, acts on the set
by
that is, by left multiplication.
Soln. Let
. We have
by the de
Algebra Homework 5
Due Wednesday, February 27
1
(a) Assume F has characteristic p = 0, and let a F . Show that f =
X p a either splits or is irreducible in F [X ].
(b) Let F C, and let = e2i/p F , where p is a prime. Let a F .
Show that f = X p a either s
MATH 4000/6000 Exam 2 Solutions, 10/28/08
1. Find the fourth roots of a = 2 3 6i. (You may write the answer using trig functions.)
|a| = (2 3)2 + 62 = 48 = 4 3, a = 4 3( 1 23 i) = 4 3(cos 43 + i sin 43 )
2
4
By DeMoivres Theorem, a fourth root of a is b =
Math 482, Abstract Algebra 2, Spring 2004
Problem Set 8, due 5/14
20.2 Show that Q( 2, 3) = Q( 2 + 3).
Because Q( 2 + 3) is the intersection all elds containing Q and 2 + 3, in
of
order to
is
demonstrate that Q( 2 + 3) Q( 2, 3) it enough to show that Q