MATH 6/71051
Exam I Solutions
1. Let A be an abelian group and let : A A be a homomorphism. Show that
H = cfw_a A | (a) = a2 is a subgroup of A.
Proof. Since is a homomorphism, (1) = 1 = 12 , and so 1 H and H is nonempty. If
a, b H , then (a) = a2 and (b
Abstract Algebra I
Fall 2013
Solutions 3
Problem 1. Let N be a normal subgroup of a group G. Given |G : N | = n, show that
an N for all a G.
The order of the quotient group G/N is n. By Lagranges theorem applied to the quotient
group, we have (aN )n = N f
Abstract Algebra I
Fall 2013
Solutions 2
Problem 1. Are the following groups isomorphic:
(a) R and Q ;
No as Q is countable while R is not.
(b) additive groups Q and Z;
No. Let : Z Q be an isomorphism. Denote (1) Q by a. Since (0) = 0 and
is injective, a
Abstract Algebra I
Fall 2013
Solutions 1
Problem 1. How many elements of order 3 are there in S6 ?
The order of a cycle is its length and the order of a product of cycles is the least common
multiple of the orders of the factors. Hence, an element in S6 h
QUALIFYING EXAM IN ALGEBRA
January 2013
1. There are 18 problems on the exam. Work and turn in 10 problems, in the following
categories.
I.
Linear Algebra
1 problem
II.
Group Theory
3 problems
III.
Ring Theory
2 problems
IV.
Field Theory
3 problems
An
Abstract Algebra I
Fall 2013
Homework 4
Problem 1. Show that An is generated by 3-cycles. Use the fact that every permutation in
An is a product of an even number of transpositions and rewrite every (ij )(kl) as a product
of 3-cycles.
Pick any element in
Abstract Algebra I
Fall 2013
Solutions 5
Problem 1. Let the G be a group of order 203 and let H
order 7. Show that G is abelian.
G be its normal subgroup of
Since H is a normal subgroup, G acts on H by conjugation. The kernel of this action is
CG (H ), so
Abstract Algebra I
Fall 2013
Solutions 10
Let R be a ring with 1 = 0.
Problem 1. Let R be integral domain. Show that if two principal ideals a and b are
equal, we have a = ub for some unit u R.
We have a b , so a = ub for some u R. Similarly, b q , and he
Abstract Algebra I
Fall 2013
Solutions 9
Problem 1. Let I be an ideal of a ring R. Show that the correspondence A A/I is an
inclusion preserving bijection between the set of subrings A of R that contain I and the set
of subrings of R/I . Furthermore, show
Abstract Algebra I
Fall 2013
Solutions 8
Problem 1. Let G be a nite group and p be prime. Let N
that P N/N Sylp (G/N ) and P N Sylp (N ).
G and P Sylp (G). Show
Since N G, P N is a subgroup of G and P N/N is a subgroup of G/N . Let |G| = pa m
and |N | = p
Abstract Algebra I
Fall 2013
Homework 6
Denition 0.1. A subgroup H of a group G is called characteristic in G, denoted H char G,
if every automorphism of G maps H to itself, that is, (H ) = H for all Aut(G).
Problem 1.
(a) Show that characteristic subgrou
MATH 6/71051
Homework #12
Solutions
7.1 #7: The center, Z (R) = cfw_z R | zr = rz r R, of a ring R with 1 is a subring
of R containing the identity. The center of a division ring is a eld.
Proof. We know that 1R commutes with every element of R, so 1R Z (
MATH 6/71051
Homework #11
Selected Solutions
#I. Let H and K be subgroups of G with K
solvable.
G. If H and K are solvable, then HK is
Proof. Since K G, we have that HK is a subgroup of G. By the Second Isomorphism
Theorem, HK/K H/H K , and since H is sol
MATH 6/71051
Exam II Solutions
1. Let G be a nite group. Show that if G has a normal subgroup N of order 3 that is not
contained in the center of G, then G has a subgroup of index 2.
Proof. Let 1 = x N , so that N = x , and therefore x is not in the cente
MATH 6/71051
Homework #1
Selected Solutions
Lemma: Let G be a group and let g G. If g m = 1 for some positive integer m, then
|g | = n < and n | m.
Proof. It follows from the denition of order that |g | = n < (and, in fact, that |g | m).
By the Division A
MATH 6/71051
Homework #3
Selected Solutions
Lemma 1: If G is an abelian group and a Z, then the map a : G G dened by
a (x) = xa for all x G is a homomorphism.
Proof. By 1.1 Exercise 24, (xy )n = xn y n for all x, y G and all n Z since G is abelian.
Hence
MATH 6/71051
Homework #4
Selected Solutions
2.4 #17: If G = g1 , g2 , . . . , gn is a nontrivial nitely generated group, then G has a
maximal subgroup.
Proof. Let S be the set of all proper subgroups of G, partially ordered by inclusion. Note
that since G
MATH 6/71051
Homework #5
Solutions
3.1 #36: If G/Z (G) is cyclic, then G is abelian.
Proof. Let G/Z (G) = xZ (G) . Then every element of G/Z (G), that is, every coset of Z (G),
is of the form (xZ (G)a = xa Z (G) for some integer a. If g is any element of
MATH 6/71051
Homework #6
Selected Solutions
4.3 #19: If H G and K is a conjugacy class of G contained in H , then K is a union
of k conjugacy classes of equal size in H , where k = |G : HCG (x)| for x K.
In particular, a conjugacy class of Sn consisting o
MATH 6/71051
Homework #7
Selected Solutions
Lemma 1. If n
3, then the center of Sn is trivial.
Proof. Let n 3 and 1 = Sn . Then (a) = a for some a cfw_1, 2, . . . , n. Let (a) = b.
Since n
3, there is a c cfw_1, 2, . . . , n with c = a and c = b. Let = (a
MATH 6/71051
Homework #8
Selected Solutions
#I. Let G be a nite group and p a prime. If N is a normal subgroup of G and P is a
Sylow p-subgroup of G, then
(a) P N/N is a Sylow p-subgroup of G/N , and
(b) P N is a Sylow p-subgroup of N .
Proof. For an inte
MATH 6/71051
Homework #9
Selected Solutions
5.2 #5: If G is a nite abelian group of type (n1 , n2 , . . . , nt ), then G contains an element
of order m if and only if m | n1 . Moreover, the exponent of G is n1 .
Proof. Let G = Zn1 Zn2 Znt , with Zni a mul
MATH 6/71051
Homework #10
Solutions
Reference: The main theorem on equivalent conditions for nilpotence in a nite group will
be cited as NGT (or Nilpotent Group Theorem). This theorem was proved in class on
November 7 & 8 (see page 106 of the notes) and i
Abstract Algebra I
Fall 2013
Homework 11
Problem 1. Prove that R[x]/ x2 + 1 is a eld isomorphic to the eld of complex numbers.
Dene a map : R[x]/ x2 + 1 by sending bx + a + x2 + 1 to a + bi. This map is clearly
well-dened, 1-1 and onto. The addition is al