Chapter 2
2.3
Since
m
is not a prime, it can be factored as the product of two integers
a
and
b
,
m
=
a
·
b
with
1
< a, b < m
. It is clear that both
a
and
b
are in the set
{
1
,
2
,
· · ·
, m

1
}
. It follows
from the definition of modulo
m
multiplication that
a
¡
b
= 0
.
Since
0
is not an element in the set
{
1
,
2
,
· · ·
, m

1
}
, the set is not closed under the modulo
m
multiplication and hence can not be a group.
2.5
It follows from Problem 2.3 that, if
m
is not a prime, the set
{
1
,
2
,
· · ·
, m

1
}
can not be a
group under the modulo
m
multiplication. Consequently, the set
{
0
,
1
,
2
,
· · ·
, m

1
}
can not
be a field under the modulo
m
addition and multiplication.
2.7
First we note that the set of sums of unit element contains the zero element
0
. For any
1
≤
‘ < λ
,
‘
X
i
=1
1 +
λ

‘
X
i
=1
1 =
λ
X
i
=1
1 = 0
.
Hence every sum has an inverse with respect to the addition operation of the field GF
(
q
)
. Since
the sums are elements in GF
(
q
)
, they must satisfy the associative and commutative laws with
respect to the addition operation of GF
(
q
)
. Therefore, the sums form a commutative group
under the addition of GF
(
q
)
.
Next we note that the sums contain the unit element 1 of GF
(
q
)
. For each nonzero sum
‘
X
i
=1
1
with
1
≤
‘ < λ
, we want to show it has a multiplicative inverse with respect to the multipli
cation operation of GF
(
q
)
. Since
λ
is prime,
‘
and
λ
are relatively prime and there exist two
1