Homework 8 Solutions
1.
(a) False. e.g.
S
3
has 9 Sylow 2subgroups.
(b) True.
G
abelian implies all subgroups are normal, which implies that each Sylow sub
group is unique.
(c) True. Sylow’s Theorem part 1.
(d) True. If
y
=
gx
then
G
y
=
gG
x
g

1
.
(e) False. e.g.
G
=
S
3
and
H
=
(12) .
N
G
(
H
) =
H
G
.
(f) True. Definition of normalizer.
(g) False. If the Sylow
p
subgroup isn’t normal then this doesn’t hold.
(h) True. The size of the Sylow
p
subgroup is
p
n
where
n
is the largest power of
p
appearing
in the order of the group. Since

G

=

H

,
n
is the same for both of them.
(i) False. 8
≡
1 (mod 5).
2. Let
G
=
S
3
×
S
3
. Subgroups of size 4 are the Sylow 2subgroups. Let
P
=
(12)
×
(12) ,
Q
=
(13)
×
(13)
and
R
=
(12)
×
(13) .
P
,
Q
and
R
are all Sylow 2subgroups.
P
∩
Q
=
{
((1)
,
(1))
}
=
{
e
}
but
P
∩
R
=
{
e,
((12)
,
(1))
}
.
3.
(a) 300 = 2
2
·
3
·
5
2
.
n
5
≡
1 (mod 5) and
n
5

12. Thus
n
5
= 1 or 6. If
n
5
= 1 then the Sylow
5subgroup is normal and thus the group isn’t simple. So we need to deal with the case
of
n
5
= 6
Suppose
n
5
= 6. This implies [
G
:
N
(
P
)] = 6 where
P
is one of the Sylow 5subgroups
(it doesn’t matter which one). Generalized Cayley’s theorem implies that there exists
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 Spring '11
 Johnson
 Algebra, Group Theory, Normal subgroup, Prime number, Subgroup, Sylow

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