Mathematics 113: Homework # 4
Curtis Tekell
Dr. Poirier
October 4, 2010
Exercise 1.
Let
G
be a group acting on a set
X
and let
K
be the kernel of this action and
G
x
be the stabalizer of
x
∈
X
of the action. Then
K,G
x
≤
G
.
Proof.
Suppose
a,b
∈
K
. Then
b

1
x
= (
b

2
b
)
x
=
b

2
(
bx
) =
b

2
(1) = 1, and thus
ab

1
x
=
a
(
b

1
) =
a
1 = 1, and
ab

1
∈
K
, so
K
≤
G
. Now suppose
a,b
∈
G
x
. Then
b

1
x
=
b

1
(
bx
) =
(
b

1
b
)
x
=
x
, and thus
ab

1
x
=
ax
=
x
, and
ab

1
∈
G
x
, so
G
x
≤
G
.
Exercise 2.
Let
H
≤
G
.
(a)
H
≤
N
(
H
)
, this is not true if
H
±
G
.
(b)
H
≤
C
(
H
)
if and only if
H
is abelian.
Proof.
(a) Let
h
∈
H
. Then clearly for any
k
∈
H
,
khk

1
∈
H
, as
H
≤
G
, so
k
∈
N
(
H
), and
thus
H
≤
N
(
H
). We now exhibit a example to show that this does not hold if
H
±
G
.
Let
G
=
GL
2
(
R
), and let
H
=
±
A
=
²
1 1
0 1
³
,B
=
²
1 0
1 1
³´
. Then one can check that
ABA

1
=
²
2 0
1 0
³
/
∈
H
, thus
A /
∈
N
(
H
), and
H
±
N
(
H
).
(b) If
H
is abelian, clearly
H
≤
C
(
H
). Conversely, if
H
is abelian, all of its elements commute
with each other, so
H
≤
C
(
H
).
1