Math 3124
Thursday, October 27
Seventh Homework Solutions
1.
S
5
×
S
4
has elements of order 15, for example ((1 2 3 4 5), (1 2 3)). On the other
there is no element of order 15 in
S
6
(the only way to have a permutation of order 15
is if there is a 15 cycle, or a product of a 5 cycle and a 3 cycle (which are disjoint),
which would require at least
S
8
). Since the elements of
Z
4
have order 1, 2 or 4 and
o
(
x
,
y
) = [
o
(
x
)
,
o
(
y
)]
, we see that
S
6
×
Z
4
can have an element of order 15 only if
S
6
has an element of order 15. Therefore
S
6
×
Z
4
has no element of order 15 and it
follows that
S
5
×
S
4
±
S
6
×
Z
4
.
2. Problem 21.6(a) on page 109. Show that
α
:
Z
3
→
Z
6
by
α
([
a
]
3
) = [
a
]
6
is not well
deﬁned.
[
0
]
3
= [
3
]
3
, so for
α
to be well deﬁned, we must have
α
[
0
]
3
=
α
[
3
]
3
, that is
[
0
]
6
= [
3
]
6
.
But
[
0
]
6
6
= [
3
]
6
and it follows that
α
is not welldeﬁned.
3. Problem 21.27 on page 110. Prove that if
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 Fall '08
 PARRY
 Math, Algebra, Ker, Coset, Conjugacy class, Ker θ

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