Introduction to Abstract Algebra
Homework 4
October 4, 2010
Problem 1
Suppose the group
G
acts on the set
X
and
x
∈
X
.
(a) If
y
is in the kernel, then
∀
x
∈
X
:
y
·
x
=
x
. By deﬁnition of a group action, the identity
1 is in the kernel. Assume
y
is in the kernel, then
∀
x
∈
X
:
x
= 1
·
x
= (
y

1
y
)
·
x
=
y

1
·
(
y
·
x
)
=
y

1
·
x,
so
y

1
is also in the kernel, meaning we have closure under inverses. Now assume that
y
and
z
are elements in the kernel, then
(
yz
)
·
x
=
y
·
(
z
·
x
)
=
y
·
x
=
x,
so (
xy
) is also in the kernel, giving us closure. Thus, the kernel of the action is a subgroup.
(b) First, the identity 1 is in
G
x
, because 1
·
x
=
x
by the deﬁnition of an action. Assume
y
∈
G
x
, then
1
·
x
= (
y

1
y
)
·
x
=
y

1
·
(
y
·
x
)
=
y

1
·
x,
so also
y

1
∈
G
x
, meaning we have closure under inverses. Assume
y,z
∈
G
x
, then
(
yz
)
·
x
=
y
·
(
z
·
x
)
=
y
·
x
=
x,
so also
xy
∈
G
x
meaning we have closure, thus
G
x
is a subgroup.
Problem 2
Clearly, since
H
is a subgroup in itself, all we really want to show in both cases is that
H
is a
subset of
N
G
(
H
) and
C
G
(
H
) respectively.
(a) To show that
H
is a subgroup of
N
G
(
H
), we want to show that all elements of
H
are also
in
N
G
(
H
). Pick some
h
∈
H
, and consider two arbitrary
h
1
,h
2
∈
H
, we shall see that
hHh

1
=
H
, because
hh
1
h

1
=
hh
2
h

1
⇐
h
1
=
h
2
:
hh
1
h

1
=
hh
2
h

1
⇔
h

1
hh
1
h

1
=
h

1
hh
2
h

1
⇔
h
1
h

1
=
h
2
h

1
⇔
h
1
h

1
h
=
h
2
h

1
h
⇔
h
1
=
h
2
.
1/4