Answer key No.1 to selected homework problems: Math. 110B
7.1
2.
A permutation
f
either fxes one number 1
≤
j
≤
3 or two numbers or
moves all. There are 3 elements fxing only one number (those interchang
ing the two numbers diﬀerent From
j
). ±or such
f
,
f
◦
f
=
I
. There is
only one which fxes two numbers; so, it fxes actually all three numbers
1
,
2
,
3, that is,
I
.T
a
k
e
f
which moves all. Then
f
(1) =
i
and
i
±
= 1. IF
f
(
i
) = 1, then
f
interchanges 1 and
i
; so,
f
fxes the rest
j
with
j
±
=1
and
j
±
=
i
. This is impossible because
f
moves all. Thus we conclude that
f
(
i
)=
j
and
j
±
=
i
and
j
±
= 1. In other words,
f
=
±
1
ij
1
²
. The choice oF
i
and
j
are either (
i,j
)=(2
,
3) or (
)=(3
,
2); so, there are two element
with
f
◦
f
◦
f
=
I
(which can be verifed by the above shape oF
f
). In
conclusion, there are 3 elements
g
∈
S
3
with
g
◦
g
=
I
, and there are two
elements
f
with
f
◦
f
◦
f
=
I
; the rest is
I
itselF.
2.
(a)

Z
18

= 18;
(b)

D
4

= 8, Four rotations (including the identity), one ²ip aFter a
rotation (see Example in page 164167).
(c) 4!: number oF 4–permutations;
(d) 5!: number oF 5–permutations;
(e) By Theorem 2.11, nonunits oF
Z
n
are zerodivisors, which are mul
tiples oF proper divisors oF
n
= 0 (here “proper divisor
d

n
” means
1
<d
). When
n
= 18, proper divisors are: 2
,
3
,
6
,
9
,
18. Remove all
multiples oF the above numbers out oF integers 1 to 18 = 0, remaining
ones are 1
,
5
,
7
,
11
,
13
,
17; so,

U
18

=6
.
4.
(a) Yes. Consider
π
:
Z
10
→
Z
5
given by
π
(
x
x
mod 5. Check this
is a homomorphism oF rings. Then
π
:
{
2
,
4
,
6
,
8
}→
U
5
is onetoone
and onto. The product is just a product oF
U
5
. Since we know that
U
5
is a group; so,
{
2
,
4
,
6
,
8
}
is also a group.
(b) Yes, because 2
x
+
y
=2
x
2
y
. Consider a Function
f
:
Q
→
G
given by
f
(
x
)=2
x
. Then
f
is onetoone and onto. Moreover
f
(
x
+
y
f
(
x
)
*
f
(
y
). Since
Q
with addition + is a group, (
G,
*
) is a group
(the structure oF (
G,
*
) is identical with (
Q
,
+) via
f
).
(c) No. 6
∈
G
does not have inverse in
G
.
(d) Yes. Check associative law:
(
a
*
b
)
*
c
=(
a
+
b

ab
)
*
c
a
+
b

ab
+
c
)

(
a
+
b

ab
)
c
a
+
b
+
c

bc
)

a
(
b
+
c

bc
a
*
(
b
+
c

bc
a
*
(
b
*
c
)
.
The identity exists:
a
*
b
=
b
⇒
a
+
b

ab
=
b
⇒
a
(1

b
)=0
b
±
=1
=
⇒
a
=0
.
Thus
a
= 0 is the identity. Inverse exists:
a
*
b
⇒
a
+
b

ab
⇒
b
=

a
1

a
∈
G,
1
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View Full Documentbecause
a
±
= 1. Actually defning
f
:
Q
→
Q
by
f
(
x
)=1

x
, then
f
(
a
*
b

(
a
+
b

ab
)=(1

a
)(1

b
)=
f
(
a
)
f
(
b
).
f
(0) = 1.
This shows the group structure oF
G
and (
R
×
,
×
) are identical by
onetoone onto Function
f
. Here
R
×
is the number line 0 removed
(the unit group oF the ring
R
).
(e) Yes. Consider
f
:
R
→
R
f
(
x
x
+ 1. Then
f
(
a
*
b
f
(
a
+
b
+
ab
a
+
b
+
ab
+1 = (
a
+1)(
b
+1) =
f
(
a
)
f
(
b
) and
f
(

1) = 0.
Thus the group structure oF (
G,
*
) and the group structure oF (
R
×
,
×
)
are identical via the onetoone onto Function
f
.
(F) Yes. By the condition
cd
= 0, the complex numbers in
G
are either
real or purely imaginary (that is,
r
√

1 For real numbers
r
). Since
c
+
d
±
=0
,0
∈
C
is removed. Since product oF real or purely imaginary
numbers stay in real or purely imaginary,
G
is a group with identity
1.
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 Winter '08
 HIDA
 Math, Algebra, Subgroup, NH, Cyclic group

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