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Ma
5
a
HOMEWORK
5
SOLUTION
FALL 2010
The exercises are taken from the text,
Abstract Algebra
(third edi
tion) by Dummitt and Foote.
Page 106,
11
. We will ﬁrst show that if
H
is a normal subgroup of
a solvable group
G
. Then
H
is also solvable. Let 1 =
G
0
▹
G
1
▹
· · ·
▹
G
n
▹
G
be a normal series with
G
i
+1
/G
i
abelian. Consider
H
i
=
H
∩
G
i
, then each
H
i
is clearly normal in
H
i
+1
, and we have a
normal series 1 =
H
0
▹
H
1
▹
· · ·
▹
H
n
=
H
. Also, take
g, h
∈
H
i
+1
=
G
i
+1
∩
H
, then since
G
i
+1
/G
i
is abelian,
ghg

1
h

1
∈
G
i
∩
H
=
H
i
.
Therefore
H
i
+1
/H
i
is abelian, and
H
is solvable. Now
H
has a derived
series
H
=
H
(0)
◃
H
(1)
◃
· · ·
◃
H
(
n
)
= 1. We claim that
H
(
k
)
is normal
in
G
for any 0
6
k
6
n
. Since
H
is trivial, the base case holds. Assume
H
(
k

1)
is normal. Take
hkh

1
k

1
∈
H
(
k
)
with
h
,
k
∈
H
(
k

1)
. For any
g
∈
G
,
ghkh

1
k

1
g

1
=
ghg

1
gkg

1
(
ghg

1
)

1
(
gkg

1
)

1
. Since
H
(
k

1)
is normal,
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This note was uploaded on 03/03/2011 for the course MATH 5A taught by Professor Dinakarramakrishnan during the Fall '10 term at Caltech.
 Fall '10
 DinakarRamakrishnan
 Algebra

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