Math 3124
Thursday, October 13
Fifth Homework Solutions
1. Problem 14.33 on page 81. Prove or give a counterexample. If a group has a subgroup
of order
n
, then it has an element of order
n
.
S
3
is a subgroup of
S
3
of order 6. But the elements of
S
3
have order 1, 2 or 3, so we
have a counterexample.
2. Problem 15.2 on page 84. Prove that 24
,

36
,
54
=
6 .
24
∈
6 because 24
=
4
*
6,

36
∈
6 because

36
=

6
*
6, and 54
∈
6 because
54
=
9
*
6. Therefore 24
,

36
,
54
⊆
6 .
Conversely 6
=
24
+
2
*
(

36
) +
54, so 6
∈
24
,

36
,
54
and we deduce that
6
⊆
24
,

36
,
54 . The required equality follows.
3. Problem 15.21 on page 84.
(a) List the elements of
S
3
×
Z
2
.
(b) List the elements of the cyclic subgroup
((
1 2
)
,
[
1
])
of
S
3
×
Z
2
.
(c) List the elements of the cyclic subgroup
((
1 2 3
)
,
[
1
])
of
S
3
×
Z
2
.
(a)
((1), [0])
((2 3), [0])
((3 1), [0])
((1 2), [0])
((1 2 3), [0])
((1 3 2), [0])
((1), [1])
((2 3), [1])
((3 1), [1])
((1 2), [1])
((1 2 3), [1])
((1 3 2), [1])
(b) ((1), [0]), ((1 2), [1])
(c) ((1), [0]), ((1 2 3), [1]), ((1 3 2), [0]), ((1), [1]), ((1 2 3), [0]), ((1 3 2), [1])
4. Problem 16.2 on page 87. Determine the right cosets of
[
3
]
in
Z
12
.
{
[0], [3], [6], [9]
} {
[1], [4], [7], [10]
} {
[2], [5], [8], [11]
}
5. Problem 16.17 on page 88. Compute the right cosets of
((
1 2
)
,
[
1
])
in
S
3
×
Z
2
.
Let
H
=
((
1 2
)
,
[
1
])
.
Then
H
=
{
((
1
)
,
[
0
])
,
((
1 2
)
,
[
1
])
}
, a subgroup of order 2.
Since the order of
S
3
×
Z
2
is 6
*
2
=
12, we expect 12/2 = 6 cosets. The cosets are
H
((
1
)
,
[
0
])
=
{
((
1
)
,
[
0
])
,
((
1 2
)
,
[
1
])
}
H
((
2 3
)
,
[
0
])
=
{
((
2 3
)
,
[
0
])
,
((
1 2 3
)
,
[
1
])
}
H
((
3 1
)
,
[
0
])
=
{
((
3 1
)
,
[
0
])
,
((
1 3 2
)
,
[
1
])
}
H
((
1 2
)
,
[
0
])
=
{
((
1 2
)
,
[
0
])
,
((
1
)
,
[
1
])
}
H
((
1 2 3
)
,
[
0
])
=
{
((
1 2 3
)
,
[
0
])
,
((
2 3
)
,
[
1
])
}
H
((
1 3 2
)
,
[
0
])
=
{
((
1 3 2
)
,
[
0
])
,
((
1 3
)
,
[
1
])
}