MATH 731, FALL 2008
MIDTERM sample solutions
A. Let G be a group and : G G a group homomorphism with 2 = . Show that G is
an internal semidirect product of A = ker and B = im .
Proof. Kernels are normal, so A G . Images may not be, which is why we only wi
1. A nonzero ideal in a principal ideal domain is maximal if and only if it is prime.
Proof: Let R be a P10 and let I be a preper ideal in R. Then R has an identity 1 75 0. Thus if I is
a maximal ideal, then R/I is a eld (by Theorem III-2.20: For a commut
G: Let R be a commutative ring. Let N denote the set of nilpotent elements in R. As
explained in class, N is an ideal of R. Prove that the quotient ring R/N has no nonzero
nilpotent elements.
SOLUTION: Consider the quotient ring R/N. Everr element of R/N
2. Prove that the set of zerodivisors of R is a union of prime ideals. (Hint: Note that the
set of nonzerodivisors is a mnltiplieatively closed set of R.)
Solution: Given an I be an ideal of R and a multiplicativcly closed set S such that
I 0 S = [3, ther
Algebra I Homework Three
Name:
Instruction: In the following questions, you should work out the solutions in a clear and concise manner. Three
questions will be randomly selected and checked for correctness; they count 50% grades of this homework set. The
NILPOTEHT AND SOLVABLE GROUPS
Yu
August 12, 2011
1 Exercise
1. (a) A4 is not the direct product of its Sylow subgroups,but A4 does have
the property:mn = 12 and (m, n) = 1 imply there is a subgroup of order m.
(b) S3 has subgroups of order 1, 2, 3, and 6
Algebra I Homework Five
Name:
Instruction: In the following questions, you should work out the solutions in a clear and concise manner.
Three questions will be randomly selected and checked for correctness; they count 50% grades of this
homework set. The
Solution for Homework 6 part 1, problem 1, 2, 4 and 5
Recording and typing by 9622534, Bin Yeh
April 15, 2009
1. Show that no group is the union of its two proper subgroups.
Suppose G = N H where N , H are proper subgroups of G.
If one of N , H contains t
Solutions to Assignment 2
1. Let H and K be finite subgroups of a group G.
(a) Recall that HK = cfw_hk : h H, k K, and show
|HK| =
|H|K|
.
|H K|
(b) For x G, the set HxK = cfw_hxk : h H, k K is called a double coset of H and K.
Show that G is a disjoint u
MATH 731, FALL 2008
FINAL EXAM sample solutions
A. Let G be a group and let N, K be normal subgroups with G = N K .
(1) Prove that K is Abelian if and only if K Z (G) .
Proof. If K Z (G), then certainly K is Abelian. Suppose conversely that K is
Abelian a
MATH 731, FALL 2008
HOMEWORK SET 1 sample solutions
A. Let A, B be arbitrary sets and let f : A B be a function. Prove that the following
statements are equivalent.
(1) f is onto.
(2) f is right invertible, that is, there exists g : B A with f g = idB .
(
MATH 731, FALL 2008
HOMEWORK SET 2 sample solutions
A. Let , be a bilinear form on V and let y V . Show that the function v v , y is an
element of the dual space V .
Proof. Dene : V F by (v ) = v , y . We are asked to show is linear. If
v, v V , then (v +
MATH 731, FALL 2008
HOMEWORK SET 3 sample solutions
A. (i) Let V be the vector space of continuous complex-valued functions with domain [0, 1] .
1
Show V does not become an inner product space when we dene f , g = 0 f (x)g (x) dx
for f, g V .
1
1
1
Proof.
MATH 731, FALL 2008
HOMEWORK SET 4 sample solutions
It will be useful to make the following observation. If U is a unitary matrix, then U v, U w =
v, w for any v, w Cn . To see this, note that U v, U w = v, U U w = v, w . If we
put v = w , we get U v = v
MATH 731, FALL 2008
HOMEWORK SET 5 sample solutions
Note: AEij () is obtained from A by adding times column i of A to column j .
A. Let F be a eld and let G be a subgroup of GLn (F ) .
(1) Show that any scalar matrix in G is in Z (G) .
Proof. If A is any
MATH 731, FALL 2008
HOMEWORK SET 6 sample solutions
A. Suppose that G is a group with |G| = pn for some prime number p and nonnegative
integer n . Show that G is polycyclic.
Proof. The key fact we need, a result of the Class Equation, is that any group of
MATH 731, FALL 2008
HOMEWORK SET 9 sample solutions
A. Find the minimal
2+ 3 over Q (and verify it is the minimal polynomial).
polynomial of
Show that Q[ 2, 3] = Q[ 2 + 3] .
Answer & Proof. Let = 2 + 3. Then 2 = 5 + 2 6, so (2 5)2 = 24. This
means is a
(b) Note that for any (an) E S, (0,1)(r,n) : (mt) : (r,n)(0,1) and so (0,1) is the identity of
S. It follows that the element (0,1) + A is the identitr of S/ A.
Dene f:R1>Sbyf(r)=(r,0)+A. Then
f(r1+r2):(r1+cr'20)IA:(rh)+A+(r20)+A:f(r1)if(r2),
and
f(T1T2:(