Math 103A Homework Solutions
HW2
Jacek Nowacki
January 28, 2007
Chapter 2
Problem 8.
Show that the set
{
5
,
15
,
25
,
35
}
is a group under multiplication
modulo 40.
What is the identity element of this group?
Can you see any
relationship between this group and
U
(8)
?
Proof.
In order to make our calculations simpler we will notice that: 25 =

15
mod
(40) and 35 =

5
mod
(40). This will give us the following cayley
table
5
15

15

5
5

15

5
5
15
15

5

15
15
5

15
5
15

15

5

5
15
5

5

15
Notice the ”symmetry” in the above table.
We see that the set is closed under multiplication.
Also, it is quite clear
that

15, or 25 is the identity element, and each element is its own inverse.
Associativity follows from the properties of integers.
Hence, the set is a
group.
The group is closely related to
U
(8) =
{
1
,
3
,
5
,
7
}
via
25
· {
1
,
3
,
5
,
7
}
=
{
25
,
35
,
5
,
15
}
mod
(40).
Problem 12.
For any integer
n >
2, show that there are at least two ele
ments in
U
(
n
) that satisfy
x
2
= 1.
Proof.
We all know the solution of the above equaiton over the integers,
namely
x
= 1
,

1. Let’s see if we can translate it to
U
(
n
). First, we note
that since
n >
2, 1 =

1
mod
(
n
). Indeed, we will get two distinct elements
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
as desired.
Next, we see that 1
∈
U
(
n
) as
gcd
(1
, n
) = 1.
To handle

1, we notice
that

1 =
n

1
mod
(
n
).
So, all we have to do now is show that the
gcd
(
n, n

1) = 1. But, (1)
·
n
+ (

1)
·
(
n

1) = 1. Hence,
gcd
(
n, n

1) = 1
(we’re using the fact that 1 is the smallest possitive integer).
Problem 14.
Let
G
be a group with the following property: Whenever
a
,
b
, and
c
belong to
G
and
ab
=
ca
, then
b
=
c
. Prove that
G
is Abelian.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 DR.RAHULA
 Abelian group, Abelian

Click to edit the document details