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Abstract Algebra Solutions
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· August 2015
DOI: 10.13140/RG.2.1.2799.6004
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Evan Oman
MATH 5371
Joe Gallian
September 16, 2013
Homework 1
Gallian 2.6
: In each case, perform the indicated operation:
a)
In
C
*
, (7 + 5
i
) (

3 + 2
i
)
(7 + 5
i
) (

3 + 2
i
) =

21 + 14
i

15
i
+ 10
i
2
=

21

i
+ 10 (

1)
=

(
i
+ 31)
∈
C
*
b)
In
GL
(2
, Z
13
), det
7
4
1
5
First recall the definition of the determinant of a 2
×
2 matrix:
det
a
b
c
d
=
a
·
d

b
·
c
So in our case we would have:
det
7
4
1
5
= 7
·
5

4
·
1
= 31
≡
13
5
∈
Z
13
c)
In
GL
(2
,
R
),
6
3
8
2

1
Recall the definition of the inverse of a 2
×
2 matrix:
A

1
=
a
b
c
d

1
=
1
det
A
d

b

c
a
So if we let
B
=
6
3
8
2
, we find that
det[
B
] = 6
·
2

8
·
3
= 12

24
=

12
∈
R
1
Therefore
B

1
=

1
12
2

3

8
6
=

1
6
1
4
2
3

1
2
∈
GL
(2
,
R
)
d)
In
GL
(2
, Z
13
),
6
3
8
2

1
Here we use the same formula but since the entries are elements of
Z
13
, we must adjust
the values according to this new modulus. So from part
c)
: det[
B
] =

12
≡
13
1.
So our inverse becomes:
B

1
=
1
1
2

3

8
6
≡
13
2
10
5
6
∈
GL
(2
, Z
13
)
Gallian 2.20
: For any integer
n >
2, show that there are at least 2 elements in
U
(
n
) that
satisfy
x
2
= 1.
Let
n
∈
Z
such that
n >
2. Then we know that that
U
(
n
), the group of integers modulo
n
under multiplication, contains only the elements which are relatively prime to
n
. So we can
write
U
(
n
) =
{
x
mod
n

gcd(
n, x
) = 1
}
So clearly 1
∈
U
(
n
) because the gcd(
n,
1) = 1(which is true for any number) and because
1 =
e
under multiplication which must be an element of
U
(
n
) by
U
(
n
) being a group.
Then since 1 is the identity it follows that (1)
2
= 1 giving one element satisfying
x
2
= 1.
The next element we will consider is
n

1. By Lemma 0.1 we know that
n
and (
n

1) are
relatively prime so
n

1
∈
U
(
n
).
Then squaring (
n

1)
2
yields:
(
n

1)
2
=
n
2

2
n
+ 1
=
n
2

2
n
+ 1
≡
n
1
Therefore we have found two elements that satisfy
x
2
= 1, guaranteeing at least 2 elements
∀
U
(
n
).
Lemma 0.1
Let
n
∈
Z
such that
n >
2. Assume by contradiction that
n, n

1 are not relatively prime.
2
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Then
∃
a
∈
N
\{
1
}
such that
a

n
and
a

(
n

1) which is to say that
∃
l, m
∈
N
{
1
}
such that
m
·
a
=
n
and
l
·
a
= (
n

1).
Rearranging we have
ma
=
la
+ 1
⇒
a
=
1
m

l
.
However since
a, l, m
∈
N
\{
1
}
, we arrive at a contradiction.
Therefore we conclude that
n, n

1 must be relatively prime for
n >
2.
Gallian 2.34
: Prove that in a group, (
ab
)
2
=
a
2
b
2
if and only if
ab
=
ba
.
(
⇒
)
Assume (
ab
)
2
=
a
2
b
2
.
Then
(
ab
)
2
=
a
2
b
2
⇒
abab
=
a
2
b
2
⇒
a

1
abab
=
a

1
a
2
b
2
⇒
bab
=
ab
2
⇒
babb

1
=
ab
2
b

1
⇒
ba
=
ab
Therefore (
ab
)
2
=
a
2
b
2
⇒
ab
=
ba
.
(
⇐
)
Assume
ab
=
ba
.
Then
ab
=
ba
⇒
a
(
ab
) =
a
(
ba
)
⇒
a
2
b
=
aba
⇒
(
a
2
b
)
b
= (
aba
)
b
⇒
a
2
b
2
= (
ab
)
2
Therefore
ab
=
ba
⇒
(
ab
)
2
=
a
2
b
2
.
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