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.