e
=
e
k
= (
a
k
*
a

k
) =
a
k
(
a

1
)
k
=
a
k
But we know that
k
cannot be the order of
a
. Hence

a

1

=
n
.
Next, see infinite order case. Let
a
has infinite order and
a

1
dose not, then
we can say that

a

1

=
n
. Moveover finite inverse of
a

1
that is (
a

1
)

1
has
same number of order. But this cannot happen. Thus
a

1
has infinite order.
Page67:6
Let
x
belong to a group. If
x
2
6
=
e
and
x
6
=
e
, prove that
x
4
6
=
e
and
x
5
6
=
e
.
What can we say about the order of
x
?
Obviously,
x
6
=
e
because
x
n
=
e
for all
x
∈
Z
. Then we can determine
that
x
6
=
e
=
x
4
·
x
2
=
x
2
if
x
4
= 2. Also
x
6
=
x
5
·
x
=
x
=
e
if
x
5
=
e
. Those
cases are not true so that
x
4
6
=
e
and
x
5
6
=
e
. Further we can say
x
3
=
e
and
x
6
=
e
. That’s,
x
has order of 3 either 6.
Page67:10
Prove that an Abelian group with two elements of order 2 must have a sub
group of order 4.
Let
G
be an Abelian group with distinct elements
a, b
such that
a
2
=
b
2
=
e
.
Then the set of
H
=
{
e, a, b, ab
}
has order 4 and it is the subgroup of
G
by
Finite subgroup test.
Page67:12
Suppose that
H
is a proper subgroup of
Z
under addition and
H
contains
18
,
30, and 40. Determine
H
.
As it stated,
H
is closed under addition,
H
must be linear combination of
18
,
30, and 40. We know that
gcd
(18
,
30
,
40) = 2 so that 2 = 18
r
+ 30
s
+ 40
t
for some integers
r, s,
and
t
. This means that 2
∈
H
but
H
6
=
Z
. That’s
H
= 2
Z
.
Page67:15
Let
G
be a group. Show that
Z
(
G
) =
T
a
∈
G
C
(
a
)
.
Suppose
x
∈
Z
(
G
). Then
x
commutes with every
a
∈
G
and
x
∈
C
(
a
) for all
a
∈
G
. This means that
x
∈
T
a
∈
G
C
(
a
) and
Z
(
G
)
⊆
T
a
∈
G
C
(
a
).
Conversely, suppose
x
∈
T
a
∈
G
C
(
a
), this implies that
x
∈
C
(
a
) =
{
y
∈
G
:
ay
=
ya
}
for all
a
∈
G
and
x
commutes with all
a
∈
G
, i.e.,
x
∈
Z
(
g
).
Page67:18
If
a
and
b
are distinct group elements, prove that either
a
2
6
=
b
2
, or
a
3
6
=
b
3
.
This problem requires us to prove that one
a
2
6
=
b
2
and if not, prove an
2