Homework 7 Solutions
1. Let

G

=
p
n
for
n
≥
0.
(
⇒
) Suppose
G
is simple. Since
G
is simple,
n
6
= 0. In addition, we know that
G
has normal
subgroups of order
p
k
for
k
≤
n
. Since
G
is simple,
n
must equal 1. Thus

G

=
p
.
(
⇐
) Suppose

G

=
p
. By Lagrange’s theorem, since
p
is prime, the only subgroups of
G
are
{
e
}
and
G
itself. Thus
G
has no nontrivial, proper normal subgroups.
2.
(a)
C
G
(
x
)
∩
H
=
H
∩ {
g
∈
G

gxg

1
=
x
}
=
{
g
∈
H

gxg

1
=
x
}
=
C
H
(
x
)
(b) [
G
:
H
] = 2
⇒
H
C
G
. By the second isomorphism theorem we know
C
G
/
(
H
∩
C
G
)
∼
=
HC
G
/H
and
HC
G
≤
G
. In addition, since [
G
:
H
] = 2 then
HC
G
=
H
or
G
.
By part (a) we have
C
G
/
(
H
∩
C
G
) =
C
G
/C
H
∼
=
HC
G
/H
. Since

x
H

=

H

/

C
H

and

C
G
/C
H

=

HC
G
/H

we have

x
H

=

HC
G

/

C
G

.
Since
HC
G
=
H
or
G
and

G

= 2

H

we have

x
H

=
1
2

G

/

C
G

or

G

/

C
G

and thus

x
H

=
1
2

x
G

or

x
G

.
3.
(a) Assume there exists
H
≤
G
such that [
G
:
H
] =
n >
1 and
G
is simple. The generalized
Cayley’s theorem implies there exists a homomorphism
ϕ
:
G
→
S
n
with
kerϕ
≤
H
.
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 Spring '11
 Johnson
 Algebra, Group Theory, Addition, Normal subgroup, Sn, BWoC, kerϕ

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