Math330
Solutions to Midterm 1
Fall 2008
1. The set
{
1
,
2
,
3
}
is not a group under multiplication modulo 4 because it is not closed.
In particular, 2
×
2 = 4 so 2
·
4
2 = 0
.
Therefore,
{
1
,
2
,
3
}
cannot be a group. One could also
note that 1 is the identity and 2 does not have an inverse (since 2
·
4
1 = 2, 2
·
4
2 = 0 and
2
·
4
3 = 2).
2. The group
U
(13) =
{
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
,
11
,
12
}
is a group under multiplication
modulo 13.
1. Given that
U
(13) is a cyclic group of order 12, we can determine how many gener
ators the group has by taking
φ
(12) = 4. (Corresponding to 1
,
5
,
7
,
11; the numbers
relatively prime to 12.) The cyclic group
U
(13) has 4 generators.
2. Given that the number 2 is a generator of
U
(13), we know that

2

=

U
(13)

= 12
.
3. To calculate the order of the element 2
8
in
U
(13) we use Theorem 4.2 on page 76.
Thus,

2
8

= 12
/
gcd(12
,
8) = 12
/
4 = 3
.
4. The number of generators of