MATH 405/505 ABSTARCT ALGEBRA HOMEWORK
MAIN IDEAS OF SOLUTIONS
2.65
Prove that
U
(
I
9
) is isomrophic to
I
6
and that
U
(
I
15
) is isomorphic to
I
4
×
I
2
.
Solution:
For the first part note
U
(
I
9
) contains 6 elements. Now exhibit an element of
order 6. For the second part note that
U
(
I
15
) contains 8 elements. Now exhibit an element
a
of order 4 and an element
b
of order 2 so that
h
a
i ∩ h
b
i
= 1 and
h
a
ih
b
i
=
U
(
I
15
).
2.67
Prove that (i)
Aut
(
V
)
∼
=
S
3
, (ii)
Aut
(
S
3
)
∼
=
S
3
, and (iii)
Aut
(
Z
)
∼
=
I
2
.
Solution:
(i) Note that
V
∼
=
I
2
×
I
2
, so
V
∼
=
{
0 = (0
,
0)
, a
= (1
,
0)
, b
= (0
,
1)
, c
= (1
,
1)
}
.
Every automorphism of
V
fixes 0 and permutes
a
,
b
,
c
.
This shows that
Aut
(
V
) is a
subgroup of
S
X
, where
X
=
{
a, b, c
}
. Now check that every permutation of
X
does also
define an automorphism.
(ii) Note that
S
3
contains 3 elements of order 2,
x
= (12),
y
= (13), and
z
= (23). An
element of
Aut
(
S
3
) permutes these 3 elements, so
Aut
(
S
3
) is a subgroup of
S
X
, where
X
=
{
x, y, z
}
. Now check that every permutation of
X
does also define an automorphism.
(iii) Every automophism of
Z
send a generator to a generator. The result follows from the
fact that the only generators of
Z
are
±
1.
2.68
If
G
is a finite group such that
Aut
(
G
) is the trivial group, then
G
is trivial or
G
∼
=
I
2
.
Solution:
Let
γ
g
be the inner automorphism that is conjugation by
g
, so
γ
g
(
g
0
) =
gg
0
g

1
for all
g
0
∈
G
. Since
Aut
(
G
) is trivial, i.e. only containes the identity,
γ
g
is the identity.
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 Winter '11
 PhanThuongCang
 Normal subgroup, Group isomorphism, Conjugacy, G. Since, ABSTARCT ALGEBRA HOMEWORK

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