Answer key No.3 to selected homework problems: Math. 110B
8.3
1.
Count the number of 4–subsets of the set
{
1
,
2
,
3
,
4
,
5
,
6
}
:
C
(6
,
4) = (
6
4
) =
15. The 4–cycle (1234) generates a cyclic subgroup
{
I,
(1234)
,
(12)(34)
,
(4321)
}
.
Another 4–cycle (1324) (made of the same four numbers) generates a
different cyclic subgroup
{
I,
(1324)
,
(13)(24)
,
(4231)
}
.
Thus one subset of four elements gives rise to at least two cyclic subgroups
generated by a 4–cycle.
There are 30 of such subgroups.
For each four
cycle, the product of the 4–cycle and the unique disjoint trasnposition gives
an element of order 4. Such element generates again 30 cyclic subgroups;
so, there are at least 60 subgroups of order 4.
3.
Note

A
4

= 12. By the third Sylow theorem, the number
x
of 2–Sylow sub
group is odd and is a factor of 12; so, it has to be either 1 or 3. Since order 4
elements in
S
4
are only 4–cycles, which are not in
A
4
; so,
A
4
does not have
order 4 elements. Thus a Sylow 2–subgroup is made of products of two
transpositions: (12) and (34), which is
{
(12)(34)
,
(13)(24)
,
(14)(23)
,
(1)
}
.
The number
y
of Sylow 3subgroup is a factor of 12 and
y
≡
1
mod 3.
Thus they are either 1 or 4. However one of such
(123)
is not normal;
so, there are four of them:
(123) ,
(234) ,
(134)
and
(124) .
5.
(a) Use the third Sylow theorem (Theorem 8.17).
6.
(a) Let
G
be a group of order 115. 115 = 5
·
23 for primes 5 and 23. The
number
x
of 5–Sylow subgroup is a factor of 115 and
x
≡
1
mod 5. The
only possibility is
x
= 1, because factors of 115 are 1
,
5
,
23 and 115 and only
1 satisfies the congruence
x
≡
1
mod 5. Thus the Sylow 5–subgroup
H
is normal (Corollary 8.16). By the same reason, there is only one Sylow
23–subgroup
K
.
Then
H
∼
=
Z
5
and
K
∼
=
Z
23
by Theorem 7.28.
Since
K
∩
H
is a subgroup of
K
and
H
, it has order which is a common divisor
of 5 and 23. Thus
K
∩
H
=
{
e
}
. If
kh
=
k h
ofr
k, k
∈
K
and
h, h
∈
H
,
then
K
k

1
k
=
h h

1
∈
H
; so,
k

1
k
=
h h

1
∈
K
∩
H
=
{
e
}
. We
find
h h

1
=
e
⇒
h
=
h
and similarly,
k
=
k
. Thus all elements of the
form
hk
are distinct; so,

HK

=

H

K

= 5
·
23 =

G

. Thus
G
=
HK
.
Then by Exercise 25 in Section 8.1, we find that
G
∼
=
H
×
K
∼
=
Z
5
×
Z
23
.
8.
Let
G
be a finite
p
–group. If a prime
q

G

, then there exists an element
of order
q
by Cauchy’s theorem. Thus
q
has to be the prime
p
, because
every element of
G
has
p
–power order (see the definition of
p
–group in
page 255 of the text).
Thus
p
is the only prime factor of

G

; in other
words,

G

=
p
n
for some
n >
0.
9.
Pick
x
∈
G
. Since
G/N
is a
p
–group [
x
]
p
k
= [
e
], where [
x
] =
xN
=
Nx
.
Thus
x
p
k
∈
N
. Since
N
is a
p
–group, (
x
p
k
)
p
=
e
; thus,
x
p
k
+
=
e
by the
exponent law because
p
k
·
p
=
p
k
+
. This shows that every element of
G
has a
p
–power order; so,
G
is a
p
–group.
11.
No.
Take an inner automorphism by
f
:
S
3
→
S
3
given by
f
(
x
) =
(12)
x
(12)

1
. Check yourself that
f
brings the Sylow 2–subsgroup
(23)
onto another different Sylow 2–subsgroup
(13) .
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 Winter '08
 HIDA
 Algebra, Group Theory, Sets, Subgroup, Coset, Index of a subgroup, Sylow, Sylow psubgroup

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