Ma
5
a
HOMEWORK ASSIGNMENT
4
SOLUTION
FALL 2010
The exercises are taken from the text,
Abstract Algebra
(third edi
tion) by Dummitt and Foote.
Page 65,
16
. (a)Suppose
H
itself is not maximal, then there is a sub
group
H
1
of
G
such that
H
(
H
1
. Similarly, if
H
1
is not maximal, there
is
H
2
in
G
such that
H
(
H
1
(
H
2
. We get a sequence
{
H
1
, H
2
, ...
}
,
and for each
i
, there is some
g
i
∈
H
i
such that
g
i
/
∈
H
i

1
. Note that
g
i
is distinct. Since
G
is finite,
{
g
i
}
must also be finite. Therefore, there
is a maximal subgroup
H
m
in the sequence, and
H
(
H
m
.
(b)Let
A
=
{
1
, r, r
2
, r
3
}
, then

A

= 4. If there is a subgroup
H
of
D
8
containing
A
, then

H
  
D
4

= 8, and 4 =

A
  
H

.
But then

H

= 4 or 8, and
H
=
D
8
or
H
=
A
. Hence
A
must be maximal.
(c)Suppose
H
=
< x
p
>
.
Since
p
divides
n
,

H

=
n
p
< n
=

G

.
Therefore
H
is a proper subgroup of
G
. Suppose there is a subgroup
H
0
=
< x
q
>
⊆
G
such that
H
(
H
0
. If
p
divides
q
, then
q
=
pk
for
some
k
, and
x
q
= (
x
p
)
k
∈
H
. Therefore
H
0
⊆
H
. Contradiction. For
q
not divisible by
p
, since
p
is prime, (
p, q
) = 1. Now there exist
a
,
b
such that
pa
+
qb
= 1. Now (
x
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 Fall '10
 DinakarRamakrishnan
 Algebra, Group Theory, Normal subgroup, Cyclic group, Group isomorphism, proper subgroup

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