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.
1