Math 4124
Wednesday, January 26
First Homework Solutions
1.
1.1.9(b) on page 22
Let
G
=
{
a
+
b
√
2
∈
R

a
,
b
∈
Q
}
. Prove that the nonzero
elements of
G
form a group under multiplication.
If
x
,
y
∈
G
then we may write
x
=
a
+
b
√
2 and
y
=
c
+
d
√
2 where
a
,
b
,
c
,
d
∈
Q
, and
then
xy
= (
ac
+
2
bd
)+(
ad
+
bc
)
√
2. Also if
x
,
y
±
=
0, then
xy
±
=
0. It follows that
multiplication is a binary operation on
G
\
0. Also multiplication is associative and
the identity is 1. Finally we need to check for inverses. If
a
+
b
√
2
∈
G
\
0, then
the inverse will be 1
/
(
a
+
b
√
2
)
; the only problem we might have is that this is not
obviously in
G
\
0. However by multiplying top and bottom by
a

b
√
2, this is
a
a
2

2
b
2

b
√
2
a
2

2
b
2
.
Since
a
2

2
b
2
is a nonzero rational number, it is now clear that the inverse is in
G
\
0
(if
a
2

2
b
2
=
0, then
√
2
∈
Q
) and the result follows.
2.
1.1.14 on page 22
Find the orders of the following elements of the multiplicative
group
(
Z
/
36
Z
)
×
:
1,

1,
5,
13,

13,
17.
Answer: 1,2,6,3,6,2. I won’t give explanations for all the answers here, just for
5. We
have modulo 36, 5
1
=
5, 5
2
=
25, 5
3
=
17, 5
4
=
13, 5
5
=
29, 5
6
=
1. Therefore the
least positive power of
5 which is the identity is 6, consequently the order of
5 is 6.
3.
1.1.25 on page 22
Prove that if
G
is a group and
x
2
=
1 for all
x
∈
G
, then
G
is
abelian.
Let
x
,
y
∈
G
. Then
x
2
=
y
2
= (
xy
)
2
=
1. Therefore
x
2
y
2
=
1
= (
xy
)
2
=
xyxy
.
Multiplying by
x

1
on the left and
y

1
on the right, we obtain
xy
=
yx
as required.
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View Full DocumentMath 4124
Wednesday, February 1
Second Homework Solutions
1.
2.1.10(a) on page 48
Prove that if
H
and
K
are subgroups of the group
G
, then so is
H
∩
K
.
Let
e
be the identity of
G
. Then
e
∈
H
,
K
because
H
and
K
are subgroups of
G
,
consequently
e
∈
H
∩
K
. Next let
x
,
y
∈
H
∩
K
. Then
xy
∈
H
because
H
≤
G
and
xy
∈
K
because
K
≤
G
. Therefore
xy
∈
H
∩
K
. Finally
x

1
∈
H
and
K
because
H
and
K
are subgroups of
G
, and we deduce that
x

1
∈
H
∩
K
. It now follows that
H
∩
K
is
a subgroup of
G
as required.
2.
1.2.3 on page 27
Use the standard generators and relations to show that every ele
ment of
D
2
n
which is not a power of
r
has order 2. Deduce that
D
2
n
is generated by
the two elements
s
and
sr
, both of which have order 2.
Each element of
D
2
n
can be written uniquely in the form
r
i
or
sr
i
, where 0
≤
i
≤
n

1
(see page 25). The elements
r
i
are of course powers of
r
, so we need to prove that
sr
i
has order 2. Since
sr
i
±
=
e
, it will be sufﬁcient to show that
(
sr
i
)
2
=
e
; we shall show
that this is true for all
i
. We have
(
sr
i
)
2
=
sr
i
sr
i
. Now the relation
rs
=
sr

1
applied
i
times shows that
r
i
s
=
sr

i
, consequently
(
sr
i
)
2
=
ssr

i
r
i
=
s
2
r

i
+
i
=
ee
=
e
, which
is what is required.
Finally we need to show that
D
2
n
is generated by the two elements
s
and
sr
, both
of which have order two. That they both have order two follows from the previous
paragraph. Also every element of
D
2
n
can be written as a product of
r
,
s
,
r

1
,
s

1
,
and since
r
=
s

1
(
sr
)
, we see that every element can be written as a product of
s
,
(
sr
)
,
s

1
,
(
sr
)

1
. This establishes that
D
2
n
is generated by
s
and
sr
.
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 Spring '08
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 Algebra, Multiplication, Cyclic group

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