Solutions to problem Set 7
Math 322, Fall 2013
Problem 2. Answer: n = 8. To prove this, we need to
(a) exhibit three non-isomorphic abelian groups of order 8, and
(b) show that for every n = 1, 2, . . . , 7, there are at most two non-isomorphic abelian
gr
Practice problems for Math 322 final exam
December 2016
(1) Let f : G H be a surjective homomorphism. Denote the centres of G and H by
Z(G) and Z(H), respectively.
(a) Show that if g Z(G) then f (g) Z(H).
(b) Show that the converse to part (a) fails. That
Mathematics 322. Midterm 2 solutions. November 2016
Problem 1. Let G1 and G2 be groups and f : G1 G2 be a homomorphism.
(a) If H is a subgroup of G1 , show that f (H) = cfw_f (h) | h H is a subgroup of G2 .
(b) If f is onto and H is a normal subgroup of G
Problem Set 2. Solutions.
Math 322, Fall 2016
(1) Explain your reasoning in each part.
(a) Let R be a relation on the set Z of integers given by (x, y) R if and only if
xy > 0. Is R an equivalence relation?
(b) Let S be a relation on Z given by (x, y) S i
Problem Set 3. Due in class Tuesday, October 18.
Math 322, Fall 2016
(1) Give an example of a set X and a relation R on X such that R is symmetric and
transitive but not refllexive.
Solution : We can take the relation from last week, with X = Z and (x, y)
Mathematics 322. Solutions to Midterm 1
October 2016. Instructor: Z. Reichstein.
Problem 1. Let G be a group. Consider the relation R on the set G defined as follows:
(a, b) R if b = an for some integer n. Is this relation reflexive? symmetric? transitive
Mathematics 322. Solutions to practice problems for Midterm 1
October 2016. Instructor: Z. Reichstein.
(1) Show by induction on n that
1
1
1
n
+
+ +
=
.
12 23
n (n + 1)
n+1
1
1
Solution: When n = 1, the right hand side is
= and the left hand side is
12
2
Problem Set 5. Solutions.
Math 322, Fall 2016
(1) Let G be a group and a, b G. Recall that a commutator [a, b] is defined as
aba1 b1 . Show that the inverse of a commutator is again a commutator. That
is, [a, b]1 = [c, d] for some c, d G. Find c and d (as
1
Exercise set 1
1. Induction on n.
Base case, n = 1. We check that 1 2 =
1(2)(3)
.
3
Induction step, n n + 1 : Suppose that the result is true for some n 1, let
us prove that it is true for n + 1.
n+1
n
k=1
k=1
k(k + 1) = k(k + 1) + (n + 1)(n + 2)
n(n +
Solutions to Problem Set 4.
Math 322, Fall 2016. Instructor: Reichstein
(1) Let G be a cyclic group of order pqr, where p, q and r are distinct primes. How
many subgroups does G have (counting the trivial subgroup cfw_e and G itself).
Solution: Let G be a
The University of British Columbia. Mathematics 322
Final Examination - Monday, December 10, 2012, 3:30-6pm. Instructor: Reichstein
First
Last Name
Signature
Student Number
Every problem is worth 5 points.
Rules governing examinations
Each examination ca
The University of British Columbia. Mathematics 322
Final Examination - Monday, December 10, 2012, 3:30-6pm. Instructor: Reichstein
Problem 1: Let f1 : A B and f2 : B C be homomorphisms of groups.
(a) Show that the composition f : A C dened by
f (a) = f2
Mathematics 322. Midterm 1 solutions.
Problem 1 (4 marks). Find a number x between 0 and 100 such that
121212434343 x (mod 101) .
Explain your answer.
Solution: The key observation here is that 102 1 (mod 101) and thus 104 1
(mod 101), 106 1 (mod 101), 10
Mathematics 322. Review problems for Midterm 2
November 5, 2013. Instructor: Z. Reichstein.
Problem 1. Let G be a group and D = cfw_(g, g ) | g G be the diagonal in G G.
(a) Show that D is a subgroup of G G.
(b) Show that D is a normal subgroup of G G if
Mathematics 322. Midterm 2 solutions
November 2013. Instructor: Z. Reichstein.
Problem 1 (4 marks). Let G be a group and H , K be subgroups of G. Recall that the
product set HK is dened as
HK := cfw_hk | h H, k K .
(a) Show that if H is normal in G then H
Solutions to problem Set 6
Math 322, Fall 2013
Problem 8. Let G be a non-abelian group of order 39 = 3 13. Let Z (G) be the centre
of G.
Claim 1: |Z (G)| = 1.
Indeed, by Lagranges theorem, |Z (G)| divides 39, i.e., is 1, 3, 13 or 39. If |Z (G)| = 39,
then
Solutions to problem Set 5
Math 322, Fall 2013
pp. 156160. Problem 52. Since 16, 17, 18 and 19 do not divide | A5 | = 60, we only
need to show that A5 does not contain a subgroup of order 15 or 20. I will show that A5
does not have a subgroup of order 15
Solutions to Problem Set 2
Math 322, Fall 2013
Problem 58. We need to check that our relation is reexive, symmetric and transitive.
Reexive: x x = 0 Z
Symmetric: x y Z = y x = (x y ) Z
Transitive: x y Z and y z Z = x z = (z y ) + (y z ) Z
The equivalence
Solutions to problem Set 3
Math 322, Fall 2013
15
.
gcd(15, k )
In part (a), where k = 3.6.9 or 12, gcd(15, k ) = 3 and thus the order of ak is 5.
Similarly, in part (b), where k = 5 or 10, gcd(15, k ) = 5 and the order of ak is 3.
Finally, in part (a), k
Solutions to problem Set 4
Math 322, Fall 2013
pp. 138142.
Problem 2. Let f : Z Z be an automorphism. Then
f (n) = f (1 + 1 + + 1) = f (1) + f (1) + + f (1) = nf (1) .
n times
n times
Since f is onto, f (1) divides every integer. Hence, f (1) = 1 of f (1)
Solutions to Problem Set 1
Math. 322, Fall 2013
Problem 4. We are looking for an integer s such that 11s 1 mod 7 or equivalently,
4s 1 mod 7. We only need to check s = 0, 1, . . . , 6. By trial and error, s = 2 works,
1 11 2 = 21, so (s, t) = (3, 2) is a
Solutions to Problem Set 6.
Math 322, Fall 2016
(1) Suppose a p-group G acts on a finite set X. If this action has n fixed points in X,
show that n |X| (mod p).
(Recall that x X is called a fixed point if g x = x for every g G.)
Solution: Decompose X into