2.3
J.A.Beachy
1
Review Problems
from A Study Guide for Beginner’s by J.A.Beachy,
a supplement to Abstract Algebra by Beachy / Blair
1. For the function f : Z16 → Z16 defined by f ([x]16 ) = [x2 ]16 , for all [x]16 ∈ Z16 ,
describe the equivalence relation ∼f on Z16 that is determined by f .
Solution: We begin by computing the image of f . We have [02 ]16 = [0]16 , [(±1)2 ]16 =
[1]16 , [(±2)2 ]16 = [4]16 , [(±3)2 ]16 = [9]16 , [(±4)2 ]16 = [16]16 = [0]16 , [(±5)2 ]16 =
[25]16 = [9]16 , [(±6)2 ]16 = [36]16 = [4]16 , [(±7)2 ]16 = [49]16 = [1]16 , and [82 ]16 =
[64]16 = [0]16 . Thus the image of f is the set {[0]16 , [1]16 , [4]16 , [9]16 }. The equivalence
classes that correspond to these elements are (in order)
{[0]16 , [4]16 , [8]16 , [12]16 }, {[1]16 , [7]16 , [9]16 , [15]16 },
{[2]16 , [6]16 , [10]16 , [14]16 }, and {[3]16 , [5]16 , [11]16 , [13]16 }.
2. Define f : Z7 → Z7 by f ([x]7 ) = [x3 + 3x − 5]7 , for all [x]7 ∈ Z7 . Is f a one-to-one
correspondence?
Solution: We need to calculate the values of the function. We have f ([0]7 ) = [−5]7 =
[2]7 , f ([1]7 ) = [13 + 3 · 1 − 5]7 = [−1]7 = [6]7 , f ([−1]7 ) = [(−1)3 + 3 · (−1) − 5]7 =
[−9]7 = [5]7 , f ([2]7 ) = [23 + 3 · 2 − 5]7 = [9]7 = [2]7 . Since f ([2]7 ) = f ([0]7 ), the
function is not a one-to-one correspondence.
Just out of curiosity we will find the remaining values. We have f ([−2]7 ) = [(−2)3 +
3 · (−2) − 5]7 = [−19]7 = [2]7 , f ([3]7 ) = [33 + 3 · 3 − 5]7 = [31]7 = [3]7 , and f ([−3]7 ) =
[(−3)3 + 3 · (−3) − 5]7 = [−41]7 = [1]7 .
3. On the set Q of rational numbers, define x ∼ y if x − y is an integer. Show that ∼ is
an equivalence relation.
Solution: If x ∈ Q, then x ∼ x since x − x = 0 is an integer. If x, y ∈ Q and x ∼ y,
then x − y ∈ Z, so y − x = −(x − y) ∈ Z, and therefore y ∼ x. If x, y, z ∈ Q with
x ∼ y and y ∼ z, then x − y and y − z are integers, so the sum x − z = (x − y) + (y − z)
is an integer, showing that x ∼ z.
4. In S10 , let α = (1, 3, 5, 7, 9), β = (1, 2, 6), and γ = (1, 2, 5, 3). For σ = αβγ, write σ
as a product of disjoint cycles, and use this to find its order and its inverse. Is σ even
or odd?
Solution: We have σ = (1, 6, 3, 2, 7, 9), so σ has order 6, and
σ −1 = (1, 9, 7, 2, 3, 6). Since σ has length 6, it can be written as a product of 5
transpositions, so it is an odd permutation.
×
×
−1
5. Define the function φ : Z×
17 → Z17 by φ(x) = x , for all x ∈ Z17 . Is φ one to one? Is
φ onto? If possible, find the inverse function φ−1 .
−1
−1 −1 = x, so φ = φ−1 ,
Solution: For all x ∈ Z×
17 we have φ(φ(x)) = φ(x ) = (x )
which also shows that φ is one-to-one and onto.
2.3
J.A.Beachy
2
6. (a) Let α be a fixed element of Sn . Show that φα : Sn → Sn defined by φα (σ) = ασα−1 ,
for all σ ∈ Sn , is a one-to-one and onto function.
Solution: If φα (σ) = φα (τ ), for σ, τ ∈ Sn , then ασα−1 = ατ α−1 . We can multiply
on the left by α−1 and on the right by α, to get σ = τ , so φα is one-to-one. Finally,
given τ ∈ Sn , we have φα (σ) = τ for σ = α−1 τ α, and so φα is onto.
Alternate solution: To show that φα is one-to-one and onto you could also show that
it has an inverse function. A short computation shows that (φα )−1 = φα−1 .
(b) In S3 , let α = (1, 2). Compute φα .
Solution: Since (1, 2) is its own inverse, direct computations show that
φα ((1)) = (1), φα ((1, 2)) = (1, 2), φα ((1, 3)) = (2, 3), φα ((2, 3)) = (1, 3),
φα ((1, 2, 3)) = (1, 3, 2), and φα ((1, 3, 2)) = (1, 2, 3).
7. Let S be the set of all n × n matrices with real entries. For A, B ∈ S, define A ∼ B
if there exists an invertible matrix P such that B = P AP −1 . Prove that ∼ is an
equivalence relation.
Comment: When multiplying matrices, be careful not to mix up the sides, because
matrices don’t necessarily satisfy the commutative law. If they did, similarity would
be the same as equality. Review your linear algebra textbook if you need to.
Solution: Let A be any n × n matrix. Since A = IAI −1 for the identity matrix I, it
follows that A ∼ A. (You could use P = A instead of P = I.)
Let A, B be n × n matrices with A ∼ B. Then there exists an invertible matrix P
with B = P AP −1 . To solve for A, multiply on the left by P −1 and on the right by P .
So we get P −1 BP = A. To put this in the right form, note that P = (P −1 )−1 . Then
we have P −1 B(P −1 )−1 = A and so B ∼ A.
Suppose that A, B, C are n × n matrices with A ∼ B and B ∼ C. Then there exist
invertible matrices P and Q with B = P AP −1 and C = QBQ−1 . (Don’t make the
mistake of assuming that you can use P in both places.) Substituting for B in the
second equation, C = Q(P AP −1 )Q−1 = (QP )A(QP )−1 , and thus A ∼ C.
We have verified that the reflexive, symmetric, and transitive properties hold, so ∼ is
an equivalence relation.
8. Show that Sn contains n(n − 1)/2 transpositions.
Solution: In constructing a transposition (a, b), we have n choices for a and n − 1
choices for b. But since in cycle notation we have (b, a) = (a, b), we have really only
constructed n(n − 1)/2 different transpositions.