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Ma
5
a
HOMEWORK
6
SOLUTION
FALL 2010
The exercises are taken from the text,
Abstract Algebra
(third edi
tion) by Dummit and Foote.
Page 138,
11
. Deﬁne a homomorphism
φ
:
N
S
p
−→
Aut
(
P
) via
g
7−→
(
x
7−→
gxg

1
). Then
φ
(
gh
)(
x
) =
ghx
(
gh
)

1
=
g
(
hxh

1
)
g

1
=
φ
(
g
)(
φ
(
h
)(
x
)) =
(
φ
(
g
)
◦
φ
(
h
))(
x
), and
gxg

1
∈
P
, so
φ
is welldeﬁned. Clearly,
ker
(
φ
) =
C
S
p
(
P
), and

N
S
p
(
P
)
C
S
p
(
P
)
 ≤
p
−
1 =
Aut
(
P
). Since the conjugacy classes
of
P
all consist of (
p
−
1)
p
−
cycles
, there are
(
p

1)!
p

1
in total, so
[
S
p
:
N
S
p
(
P
)] =
(
p

1)!
p

1
.
Therefore,

N
S
p
(
P
)

=
p
!
(
p

2)!
=
p
(
p
−
1).
Take any
g
∈
C
S
p
(
P
), and
x
= (
x
1
x
2
...x
p
) a
p
cycle, then
gxg

1
=
(
g
(
x
1
)
g
(
x
2
)
...g
(
x
p
)) = (
x
1
...x
p
).
Obvious

C
S
p
(
P
)

=

P

=
p
.
Therefore

N
S
p
(
P
)
C
S
p
(
P
)

=
Aut
(
P
), and
φ
is an isomorphism.
Page 138,
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 Fall '10
 DinakarRamakrishnan
 Algebra

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