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66
(vii)
If
G
and
H
are abelian groups, then
G
×
H
is an abelian group.
Solution.
True.
(viii)
If
G
and
H
are cyclic groups, then
G
×
H
is a cyclic group.
Solution.
False.
(ix)
If every subgroup of a group
G
is a normal subgroup, then
G
is
abelian.
Solution.
False.
(x)
If
G
is a group, then
{
1
}
C
G
and
G
/
{
1
}
∼
=
G
.
Solution.
True.
2.96
Prove that
U
(
I
9
)
∼
=
I
6
and
U
(
I
15
)
∼
=
I
4
×
I
2
.
Solution.
The 6 elements in
U
(
I
9
)
are 1, 2, 4, 5,7, 8, and they form a cyclic
group with 2 as a generator.
The 8 elements of
U
(
I
15
)
are 1, 2, 4, 7, 8, 11, 13, 14. Now 2 has order 4:
h
2
i=
1
,
2
,
4
,
8
,
and 11 has order 2. The intersection of these two subgroups is
{
1
}
and
they generate all of
U
(
I
1
5
)
(if
S
=h
2
,
11
i
, then

S
≥
5; by Lagrange
’
s
theorem,

S

is a divisor of 8; hence,

S
=
8 and
S
=
U
(
I
1
5
)
.
2.97
(i)
If
H
and
K
are groups, prove, without using the
f
rst isomorphism
theorem, that
H
∗
={
(
h
,
1
)
:
h
∈
H
}
and
K
∗
={
(
1
,
k
)
:
k
∈
K
}
are normal subgroups of
H
×
K
with
H
∼
=
H
∗
and
K
∼
=
K
∗
.
Solution.
First we show that
H
∗
is a subgroup of
H
×
K
.I
t
contains
(
1
,
1
)
, which is the identity of
H
×
K
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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