MATH 103A Homework 2 - SOLUTIONS
Due January 18, 2013
Version January 23, 2013
Assigned reading: Chapters 1 and 2 of Gallian.
Recommended practice questions: Chapter 2 of Gallian, exercises
1, 2, 3, 4, 5, 6, 13, 14, 17, 23, 25, 26
Assigned questions to hand in:
(1) (Gallian Chapter 2 # 2bc) Which of the following binary operations are associative?
(Justify your answers.)
(b) division of nonzero rationals
Solution: Not associative. It’s enough to find one counterexample, so consider the
rational numbers 36, 4, 12.
9
3
36 4 12 12
4
but
1
36 4 12 36 36 3 108.
3
Since these numbers are not equal, division of nonzero rationals is not associative.
(c) function compositions of polynomials with real coefficients
Solution: Is associative. Three arbitrary elements of the set of polynomials with real
coefficients can be written as
P1 x an xn a1 x a0
P2 x bm xm b1 x b0
P3 x cr xr c1 x c0
with n, m, r
Z and ai , bj , ck
¨
R. We want to show that
P1 P2 P3 x P1 P2 P3 x.
Let’s compute the formulas for the LHS and RHS:
LHS
RHS
¨
P1 P2 cr xr c0 P1 bm cr xr c0 m b0 an bm cr xr c0 m b0 n a0
P1 P2 cr xr c0 ¨
an bm xm b0 n a0 cr xr c0 r
m
n
an bm cr x c0 b0 a0 .
These formulas are equal so function compositions of polynomials with real coefficients is associative.
Alternate solution: In fact, function composition of arbitrary real-valued functions
is associative. To see this, let f, g, h : R R. Then, for x R,
f g h x f y where
y
g h x g h x.
Thus,
f g h x f g h x.
On the other hand,
Since f
f
g h
x f
g h x f
g
g h
h x f g h x.
x for all x R, the two functions are equal.
(2) (Gallian Chapter 2 # 3bc) Which of the following binary operations are commutative?
(Justify your answers.)
(b) division of nonzero rationals
Solution: Not commutative. It’s enough to find one counterexample, so consider the
rational numbers 1, 2.
1 2 0.5
2 1 2.
These are not equal so division of nonzero rationals is not commutative.
(c) function compositions of polynomials with real coefficients
Solution: Not commutative. It’s enough to find one counterexample, so consider the
polynomials P1 x x 1 and P2 x x2 . Then
P1 P2 x P1 x2 x2 1.
P2 P1 x P2 x 1 x 12
x2 2x 1.
These functions are not equal; for example, they disagree on the image of x 2:
P1 P2 2 22 1 5
P2 P1 2 22 2 2 1 9.
Therefore, function composition of polynomials is not commutative.
(3) (Gallian Chapter 2 # 4bc) Which of the following sets are closed under the given operation? (Justify your answers.)
(b) 0, 4, 8, 12 addition mod 15
Solution: We can compute the addition table:
0 4 8 12
0 0 4 8 12
4 4 8 12 1
8 8 12 1 5
12 12 1 5 9
Since this table includes values which are not in the set 0, 4, 8, 12, this set is not
closed under the operation.
(c) 1, 4, 7, 13 multiplication mod 15
Solution: This time, we compute the multiplication table:
2
1
1 1
4 4
7 7
13 13
Since this table only includes values in
the operation.
4
4
1
13
7
the
7 13
7 13
13 7
4 1
1 4
set 1, 4, 7, 13, this set is closed under
(4) (Gallian Chapter 2 # 9) Show that 1, 2, 3 under multiplication modulo 4 is not a group
but that 1, 2, 3, 4 under multiplication modulo 5 is a group.
Solution: To prove that 1, 2, 3 under multiplication modulo 4 is not a group, we will
show that the set is not closed under the operation. It is enough to demonstrate a
counterexample to closure; namely, consider
2 2 mod 4 4 mod 4 0 1, 2, 3.
To prove that 1, 2, 3, 4 is a group under multiplication modulo 5, we need to prove that
closure, associativity, existence of identity, and existence of inverses all hold. For all these
properties, it will be convenient to refer to the multiplication table:
1 2 3 4
1 1 2 3 4
2 2 4 1 3
3 3 1 4 2
4 4 3 2 1
For closure, observe that each entry in the table is in the set 1, 2, 3, 4. For associativity,
we appeal to the fact that multiplication modulo n (for any positive n) is associative.
The identity element is 1 (as evidenced by the top row and left-most column of the table).
The inverses are:
11
21
1,
31
3,
41
2,
4.
(5) (Gallian Chapter 2 # 10) Show that group GL 2, R of Example 9 is non-Abelian by
exhibiting a pair of matrices A and B in GL 2, R such that AB BA.
¢
Solution: Consider A
ª
1 2
0 1
¢
and B
ª
1 1
. Each of these matrices has nonzero
1 0
determinant:
det A 1 0 1
det B 0 1 1
so each of these matrices is in GL 2, R. Then
¢
AB
1 2
0 1
and
¢
BA 1 1
1 0
ª¢
ª¢
1 1
1 0
1 2
0 1
ª
¢
ª
¢
3 1
1 0
ª
ª
1 3
.
1 2
(6) (Gallian Chapter 2 # 12) Give an example of group elements a and b with the property
that a1 ba b.
3
Solution: Consider the group GL 2, R and pick a A, b B from previous question.
A1 Then
a1 ba ¢
1 0
2 1
ª¢
1
det A
1 1
1 0
ª¢
¢
1 2
0 1
1
0
2
ª
¢
ª
1
¢
1
1
ª
1 0
.
2 1
1
1
ª¢
1 2
0 1
ª
¢
1
1
ª
3
,
3
which is not equal to b.
(7) (Gallian Chapter 2 # 22) Let G be a group with the property that for any x, y, z in the
group, xy zx implies y z. Prove that G is Abelian.
Solution: Let G be a group with the stated property. To prove that G is abelian, we need
to prove that for any a, b G, ab ba. Let a, b G. Consider x a1 . Note that x
exists and is an element of G by the existence and definition of inverses. Then,
xab a1 ab
a1 ab
Assoc.
Inverses
eb
Identity
b
and
bax baa1 b aa1 be b.
In particular, if we write y ab and z ba then we have proved that xy
assumption on G, this implies that y z. In other words, that ab ba.
Inverses
Assoc.
Identity
b
zx. By
(8) (Gallian Chapter 2 # 27) For any elements a and b from a group and any integer n, prove
that a1 ban a1 bn a.
Solution: For non-negative values of n, we proceed by induction.
Base case n 0. By definition, x0 e for any x G. Therefore, the formula we
want to show because e a1 ea. To prove it,
a1 ea
a1 ea
Assoc.
Identity
a1 a
Inverse
e.
Induction step. We assume that
a1 ban
a1 bn a.
Now, compute
a1 ban1
I.H.
Assoc.
a1 ban a1 ba a1 bn a a1 ba a1 bn aa1 ba
Inverse
a1 bn eba
Identity
a1 bn ba a1 bn1 a,
as required.
For negative values of n, we recall the definition (p. 51) that xm x1 m (for any
x G). In particular, this implies that xm xm xmm x0 e. Applying this to our
context,
n 0
e a1 ban a1 ban
a1 ban a1 bn a.
Multiplying both sides (on the right) by a1 bn a, we get an equation with
Since
LHS
Identity
e a1 bn a a1 bn a
4
and
RHS
Assoc.
a1 ban a1 bn a a1 bn a a1 ban a1 bn aa1 bn a
a1 ban a1 bn bn a
a1 ban a1 a
Assoc.
Assoc.
a1 ban a1 bn bn a
a1 ban .
Since LHS = RHS, we have the desired result: a1 bn a a1 ban .
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