Because the abelian groups
Z
and
Z
n
are of such importance, it is a good idea to completely
characterize all subgroups of these abelian groups. As the following two theorems show, the sub
groups in the above examples are the
only
subgroups of these groups.
Theorem 4.7
If
G
is a subgroup of
Z
, then there exists a unique nonnegative integer
m
such that
G
=
m
Z
.
Proof.
Actually, we have already proven this.
One only needs to observe that a subset
G
is a
subgroup if and only if it is an ideal (as defined in
§
1.2), and then apply Theorem 1.4.
2
Theorem 4.8
If
G
is a subgroup of
Z
n
, then there exists a unique positive integer
m
dividing
n
such that
G
=
m
Z
n
.
Proof.
Let
G
be a subgroup of
Z
n
.
Define
G
0
:=
{
a
∈
Z
: [
a
]
∈
G
}
.
It is easy to see that
G
=
{
[
a
] :
a
∈
G
0
}
.
First, we claim that
G
0
is a subgroup of
Z
. Suppose that
a, b
∈
G
0
. This means that [
a
]
∈
G
and [
b
]
∈
G
, which implies that [
a
+
b
] = [
a
] + [
b
]
∈
G
, and hence
a
+
b
∈
G
0
. Similarly, if [
a
]
∈
G
,
then [

a
] =

[
a
]
∈
G
, and hence

a
∈
G
0
.
By the previous theorem, it follows that
G
0
is of the form
m
Z
for some nonnegative integer
m
.
Moreover, note that
n
∈
G
0
, since [
n
] = [0] is the identity element of
Z
n
, and hence belongs to
G
.
Therefore,
m

n
.
So we have
G
=
{
[
a
] :
a
∈
m
Z
}
=
m
Z
n
.
From the observations in Example 4.17, the uniqueness of
m
is clear.
2
Of course, not all abelian groups have such a simple subgroup structure.
Example 4.18
Consider the group
G
=
Z
2
×
Z
2
. For any nonzero
α
∈
G
,
α
+
α
= 0
G
. From this,
it is easy to see that the set
H
=
{
0
G
, α
}
is a subgroup of
G
. However, for any integer
m
,
mG
=
G
if
m
is odd, and
mG
=
{
0
G
}
if
m
is even. Thus, the subgroup
H
is not of the form
mG
for any
m
.
2
Example 4.19
Consider the group
Z
*
n
discussed in Example 4.9. The subgroup (
Z
*
n
)
2
plays an
important role in some situations. Integers
a
such that [
a
]
∈
(
Z
*
n
)
2
are called
quadratic residues
modulo
n
.
2
Example 4.20
Consider again the group
Z
*
n
, for
n
= 15, discussed in Example 4.10. As discussed
there, we have
Z
*
15
=
{
[
±
1]
,
[
±
2]
,
[
±
4]
,
[
±
7]
}
. Therefore, the elements of (
Z
*
15
)
2
are
[1]
2
= [1]
,
[2]
2
= [4]
,
[4]
2
= [16] = [1]
,
[7]
2
= [49] = [4];
thus, (
Z
*
15
)
2
has order 2, consisting as it does of the two distinct elements [1] and [4].
Going further, one sees that (
Z
*
15
)
4
=
{
[1]
}
. Thus,
α
4
= [1] for all
α
∈
Z
*
15
.
By direct calculation, one can determine that (
Z
*
15
)
3
=
Z
*
15
; that is, cubing simply permutes
Z
*
15
.
For any integer
m
, write
m
= 4
q
+
r
, where 0
≤
r <
4.
Then for any
α
∈
Z
*
15
, we have
α
m
=
α
4
q
+
r
=
α
4
q
α
r
=
α
r
. Thus, (
Z
*
15
)
m
is either
Z
*
15
, (
Z
*
15
)
2
, or
{
[1]
}
.
However, there are certainly other subgroups of
Z
*
15
— for example, the subgroup
{
[
±
1]
}
.
2
23