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
6
=

1
mod
(
n
). Indeed, we will get two distinct elements
1