Solutions Chapters 1–5
Section 1.1
1. Under multiplication, the positive integers form a monoid but not a group, and the
positive even integers form a semigroup but not a monoid.
2. With

a

denoting the order of
a
, we have

0

=1
,

1

=6
,

2

=3
,

3

=2
,

4

,
and

5

=6.
3. There is a subgroup of order 6
/d
for each divisor
d
of 6. We have
Z
6
itself (
d
= 1),
{
0
}
(
d
=6)
,
{
0
,
2
,
4
}
(
d
= 2), and
{
0
,
3
}
(
d
= 3).
4.
S
forms a group under addition. The inverse operation is subtraction, and the zero
matrix is the additive identity.
5.
S
∗
does not form a group under multiplication, since a nonzero matrix whose deter
minant is 0 does not have a multiplicative inverse.
6. If
d
is the smallest positive integer in
H
, then
H
consists of all multiples of
d
. For if
x
∈
H
we have
x
=
qd
+
r
where 0
≤
r<d
. But then
r
=
x
−
∈
H
,so
r
must be
0.
7. Consider the rationals with addition mod 1, in other words identify rational numbers
that di±er by an integer. Thus, for example, 1
/
3=4
/
3=7
/
3, etc. The group is
in²nite, but every element generates a ²nite subgroup. For example, the subgroup
generated by 1
/
3is
{
1
/
3
,
2
/
3
,
0
}
.
8. (
ab
)
mn
=(
a
m
)
n
(
b
n
)
m
= 1, so the order of
ab
divides
mn
.Thu
s

ab

=
m
1
n
1
where
m
1
divides
m
and
n
1
divides
n
. Consequently,
a
m
1
n
1
b
m
1
n
1
= 1
(1)
If
m
=
m
1
m
2
, raise both sides of (1) to the power
m
2
to get
b
mn
1
= 1. The order of
b
,
namely
n
, must divide
mn
1
, and since
m
and
n
are relatively prime,
n
must divide
n
1
. But
n
1
divides
n
, hence
n
=
n
1
. Similarly, if
n
=
n
1
n
2
we raise both sides of (1)
to the power
n
2
and conclude as above that
m
=
m
1
. But then

ab

=
m
1
n
1
=
mn
,
as asserted.
If
c
belongs to both
h
a
i
and
h
b
i
then since
c
is a power of
a
and also a power of
b
,we
have
c
m
=
c
n
= 1. But then the order of
c
divides both
m
and
n
, and since
m
and
n
are
relatively prime,
c
has order 1, i.e.,
c
=1.
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9. Let

a

=
m
,

b

=
n
.I
f[
m,n
] is the least common multiple, and (
) the greatest
common divisor, of
m
and
n
, then [
]=
mn/
(
). Examine the prime factoriza
tions of
m
and
n
:
m
=(
p
t
1
1
···
p
t
i
i
)(
p
t
i
+1
i
+1
p
t
j
j
)=
rr
0
n
p
u
1
1
p
u
i
i
)(
p
u
i
+1
i
+1
p
u
j
j
s
0
s
where
t
k
≤
u
k
for 1
≤
k
≤
i
, and
t
k
≥
u
k
for
i
+1
≤
k
≤
j
.
Now
a
r
has order
m/r
and
b
s
has order
n/s
, with
m/r
(=
r
0
) and
n/s
(=
s
0
) relatively
prime. By Problem 8,
a
r
b
s
has order
mn/rs
=
mn/
(
)=[
]. Thus given
elements of orders
m
and
n
, we can construct another element whose order is the
least common multiple of
m
and
n
. Since the least common multiple of
m
,
n
and
q
is [[
]
,q
], we can inductively Fnd an element whose order is the least common
multiple of the orders of all elements of
G
.
10. Choose an element
a
that belongs to
H
but not
K
, and an element
b
that belongs to
K
but not
H
, where H and K are subgroups whose union is
G
. Then
ab
must belong
to either
H
or
K
,say
ab
=
h
∈
H
. But then
b
=
a
−
1
h
∈
H
, a contradiction. If
ab
=
k
∈
K
, then
a
=
kb
−
1
∈
K
, again a contradiction. To prove the last statement,
note that if
H
∪
K
is a subgroup, the Frst result with
G
replaced by
H
∪
K
implies
that
H
=
H
∪
K
or
K
=
H
∪
K
, in other words,
K
⊆
H
or
H
⊆
K
.
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 Spring '11
 sdd
 Multiplication, Integers, Prime number, Rational number, Greatest common divisor

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