ALGEBRA 235
HOMEWORK 3 SOLUTIONS
Question
1
.
Let
ψ
:
C
×
→
R
×
be given by
ψ
(
z
)=

z

,where

x
+
yi

=
°
x
2
+
y
2
.
(a) Show that
ψ
is a homomorphism.
(b) Calculate ker(
ψ
) and im(
ψ
).
Solution.
(a) First, observe that
ψ
is welldefned, as

z
°
=0i
±andon
ly
li
±
z
°
= 0 ±or any
z
∈
C
.W
en
e
e
dt
oc
h
e
c
kt
h
a
t
ψ
(1) = 1, but this is clear since
√
1
2
=1
. Next
,let
z
1
=
x
1
+
y
1
i
and
z
2
=
x
2
+
y
2
i
where
z
i
∈
C
×
(that is,
at least one o±
x
i
,y
i
∈
R
×
±or
i
=1
,
2). Then
ψ
(
z
1
z
2
)=
ψ
(
x
1
x
2
−
y
1
y
2
+(
x
1
y
2
+
y
1
x
2
)
i
)
=
°
(
x
1
x
2
−
y
1
y
2
)
2
+(
x
1
y
2
+
y
1
x
2
)
2
=
±
x
2
1
x
2
2
+
x
2
1
y
2
2
+
y
2
1
x
2
2
+
y
2
1
y
2
2
=
±
(
x
2
1
+
y
2
1
)(
x
2
2
+
y
2
2
)
=
ψ
(
z
1
)
ψ
(
z
2
)
.
There±ore,
ψ
is a homomorphism.
(b) By defnition, ker(
ψ
)=
{
z
∈
C
×

ψ
(
z
)=1
}
. In other words, i±
z
=
x
+
yi
∈
ker(
ψ
),

z

=1
,andhence
z
is an element o± the complex unit circle,
S
1
.N
o
t
et
h
a
t
z
∈
ker(
ψ
)i
±andon
lyi
±(
x,y
) is a real solution to
x
2
+
y
2
= 1. On the other hand, we can write
z
=
e
2
πit
±or some
t
∈
R
. There±ore,
ker(
ψ
)=
S
1
=
{
x
+
yi

x
2
+
y
2
=1
,
(
x,y
)
∈
R
2
}
=
{
e
2
πit

t
∈
R
}
.
Next, we want to compute im(
ψ
). Note that since we defne
ψ
using the positive branch o± the squareroot,

z

>
0 ±or all
z
∈
C
×
. Furthermore, since
R
>
0
⊆
C
×
and

z

=
z
±or
z
∈
R
>
0
, it ±ollows that
ψ
surjects onto
R
>
0
and im(
ψ
)=
R
>
0
. (Here
R
>
0
:=
{
x
∈
R

x>
0
}
).
°
Question
2
.
Let
φ
:
G
→
H
be a homomorphism and let
g
∈
G
have fnite order.
(a) Show that the order oF
φ
(
g
) divides the order oF
g
.
(b) IF
φ
is an isomorphism, show that

φ
(
g
)

=

g

.
Solution.
(a) Let
n
=

g

.Th
ens
in
ce
φ
is a homomorphism,
φ
(
e
G
)=
e
H
,and
φ
(
g
a
)=
φ
(
g
)
a
±or all
a
∈
N
.Thu
s
,
e
H
=
φ
(
e
G
)=
φ
(
g
n
)=
φ