Proof.
Exercise.
2
Theorem 4.26
If
G
is a finite abelian group of order
n
, and
m
is an integer relatively prime to
n
, then
mG
=
G
.
Proof.
Consider the
m
multiplication map on
G
.
We claim that the kernel of this map is
{
0
G
}
. Indeed,
ma
= 0
G
, implies ord(
a
) divides
m
, and
since ord(
a
) also divides
n
and gcd(
m, n
) = 1, we must have ord(
a
) = 1, i.e.,
a
= 0
G
. That proves
the claim.
Thus, the
m
multiplication map is injective, and because
G
is finite, it must be surjective as
well.
2
Theorem 4.27
If
G
is an abelian group of prime order, then
G
is cyclic.
Proof.
Let

G

=
p
. Let
a
∈
G
with
a
6
= 0
G
. Since ord(
a
)

p
, we have ord(
a
) = 1 or ord(
a
) =
p
.
Since
a
6
= 0
G
, we must have ord(
a
)
6
= 1, and so ord(
a
) =
p
, which implies
a
generates
G
.
2
Theorem 4.28
Suppose that
a
is an element of an abelian group, and for some prime
p
and
e
≥
1
,
we have
p
e
a
= 0
G
and
p
e

1
a
6
= 0
G
. Then
a
has order
p
e
.
Proof.
If
m
is the order of
a
, then since
p
e
a
= 0
G
, we have
m

p
e
. So
m
=
p
f
for some 0
≤
f
≤
e
.
If
f < e
, then
p
e

1
a
= 0
G
, contradicting the assumption that
p
e

1
a
6
= 0
G
.
2
Theorem 4.29
Suppose
G
is an abelian group with
a
1
, a
2
∈
G
such that the
a
1
is of finite order
n
1
and
a
2
is of finite order
n
2
, and
d
= gcd(
n
1
, n
2
)
.
Then
ord(
a
1
+
a
2
)

n
1
n
2
/d
.
Moreover,
ord(
a
1
+
a
2
) =
n
1
n
2
if and only if
d
= 1
.
Proof.
Since (
n
1
n
2
/d
)(
a
1
+
a
2
) = (
n
2
/d
)(
n
1
a
1
) + (
n
1
/d
)(
n
2
a
1
) = 0
G
+ 0
G
= 0
G
, the order of
a
1
+
a
2
must divide
n
1
n
2
/d
. On the one hand, if
d >
1, then clearly
a
1
+
a
2
cannot have order
n
1
n
2
. On the other hand, if
d
= 1, then
m
(
a
1
+
a
2
) = 0
G
implies
ma
1
=

ma
2
; since

ma
2
has
order dividing
n
2
, so does
ma
1
; also,
ma
1
has order dividing
n
1
, and so we conclude that the order
of
ma
1