Math 541
Solutions to HW #10
1. Prove that if
G
is a simple group such that 61
≤
G
≤
70, then
G
is a cyclic group.
[Feel free to use the Sylow theorems.]
Referencing the lemmas from homework 9, we can create the following table:
Order
Simple/Not Simple
Reason
61
Simple
Lemma P
62
Not Simple
Lemma PQ: 62 = 2
*
31
63
Not Simple
Sylow’s Third Theorem:
n
7

63 implies
n
7
= 1, 3, 7, 9, 21, or 63. Of
these, only 1
≡
1 (mod 7). Thus, by the corollary to Sylow’s Third
Theorem, there is a normal subroup of order 7.
64
Not Simple
Lemma Z: 64 = 2
6
65
Not Simple
Lemma PQ: 65 = 5
*
13
66
Not Simple
Sylow’s Third Theorem:
n
11

66 implies
n
11
= 1, 2, 3, 6, 11, 22,
33, or 66. Of these, only 1
≡
1 (mod 1)1. Thus, by the corollary to
Sylow’s Third Theorem, there is a normal subgroup of order 11.
67
Simple
Lemma P
68
Not Simple
Sylow’s Third Theorem:
n
17

68 implies
n
17
= 1, 2, 4, 17, 34, or
68. Of these, only 1
≡
1 (mod 17). Thus, by the corollary to Sylow’s
Third Theorem, there is a normal subgroup of order 17.
69
Not Simple
Lemma PQ: 69 = 3
*
23
70
Not Simple
Sylow’s Third Theorem:
n
7

70 implies
n
7
= 1, 2, 5, 7, 10, 14, 35, or
70. Of these, only 1
≡
1 (mod 7). Thus, by the corollary to Sylow’s
Third Theorem, there is a normal subgroup of order 7.
Notice that the only simple groups are those that have prime order. Since every group of prime order
is cyclic, the proof is complete.
2. Let
A
and
B
be normal subgroups of a group
G
such that
A
∩
B
=
{
e
}
. Prove that
ab
=
ba
for all
a
∈
A
and
b
∈
B
.
[Hint: Prove that
aba

1
b

1
∈
A
∩
B
.]
•
Note ﬁrst that for all
a
∈
A
and
b
∈
B
, we have
aba

1
∈
B
and
ba

1
b

1
∈
A
. Correspondingly,
(
aba

1
)
b

1
∈
Bb

1
=
B
and
a
(
ba

1
b

1
)
∈
aA
=
A
. Thus,
aba

1
b

1
∈
A
∩
B
. Since
A
∩
B
=
{
e
}
,
we know that
aba

1
b

1
=
e
, and therefore that
ab
=
ba
. This completes the proof.
3. Let
G
be a group of size 15. Let
a
be an element of order 3 and
b
be an element of order 5. Set
A
=
h
a
i
and
B
=
h
b
i
.
(a) Prove that
AB
=
G
.
•
Since
A
and
B
have prime orders 3 and 5, each nonidentity element of
A
and
B
also has
an order of 3 and 5, respectively. Therefore
A
∩
B
=
{
e
}
, for otherwise we could conclude

A

= 5 or

B

= 3, both of which would be contradictions.
One way to accomplish our task is to determine if the set
AB
=
{
ab

a
∈
A
and
b
∈
B
}
has
size 15. We can do this using properties of cosets. Let
A
=
{
e,a
1
,a
2
}
. Then
B
=
eB
6
=
a
1
B
and
B
=
eB
6
=
a
2
B
, since
a
1
,a
2
/
∈
B
. Since diﬀerent cosets are disjoint, the only case in
1