Math 4124
Wednesday, February 23
February 23, Ungraded Homework
Exercise 3.3.4 on page 101
Let
C
be a normal subgroup of the group
A
and let
D
be a
normal subgroup of the group
B
.
Prove that
(
C
×
D
)
(
A
×
B
)
and
(
A
×
B
)
/
(
C
×
D
)
∼
=
(
A
/
C
)
×
(
B
/
C
)
.
Define
θ
:
A
×
B
→
(
A
/
C
)
×
(
B
/
D
)
by
θ
(
a
,
b
) = (
aC
,
bD
)
. Let
a
,
a
1
∈
A
and
b
,
b
1
∈
B
. Then
θ
(
(
a
,
b
)(
a
1
,
b
1
)
)
=
θ
(
aa
1
,
bb
1
) = (
aa
1
C
,
bb
1
D
) = (
aCa
1
C
,
bDb
1
D
)
= (
aC
,
bD
)(
a
1
C
,
b
1
D
) =
θ
(
a
,
b
)
θ
(
a
1
,
b
1
)
Thus
θ
is a homomorphism.
θ
is onto because given
(
aC
,
bD
)
∈
(
A
/
C
)
×
(
B
/
D
)
,
θ
(
a
,
b
) = (
aC
,
bD
)
.
ker
θ
=
{
(
c
,
d
)

θ
(
c
,
d
) =
e
}
, which is
{
(
c
,
d
)

(
cC
,
dD
) = (
C
,
D
)
}
. Now
cC
=
C
if and only
if
c
∈
C
, and
dD
=
D
if and only if
d
∈
D
. We deduce that ker
θ
=
C
×
D
. The result now
follows from the fundamental homomorphism theorem (Theorem 16 on page 97).
Prove that
C
×
/
{
1
,

1
}
∼
=
C
×
.
Let
G
=
C
×
. Of course,
G
is the nonzero complex numbers under multiplication. Define
θ
:
C
→
C
by
θ
(
x
) =
x
2
for
x
∈
G
. Then
θ
is a mapping, because if
x
6
=
0, then
x
2
6
=
0
and so certainly
θ
maps
G
to
G
.
Next
θ
is a homomorphism, because if
x
,
y
∈
G
, then
θ
(
xy
) = (
xy
)
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 Spring '08
 Staff
 Math, Algebra, Normal subgroup, Homomorphism, kernel, Group homomorphism, Ker θ

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