Math 4124
Monday, February 7
February 7, Ungraded Homework
Exercise 2.3.12 on page 60
Prove that the following groups are
not
cyclic.
(a)
Z
/
2
Z
×
Z
/
2
Z
(b)
Z
/
2
Z
×
Z
(c)
Z
×
Z
(a)

Z
/
2
Z
×
Z
/
2
Z

=
4, so if the group was cyclic, it would have an element of order 4.
This is not the case because all nonidentity elements have order 2.
(b) Since
Z
/
2
Z
×
Z
is infinite, if it is also cyclic, it must be isomorphic to
Z
. However the
given group has an element of order 2, namely
(
¯
1
,
0
)
, whereas all nonidentity elements
of
Z
have infinite order.
(c) Suppose
Z
×
Z
is cyclic. Then it would have a generator
(
a
,
b
)
, where
a
,
b
∈
Z
. This
would mean that every element of the group could be written as
(
na
,
nb
)
for some
n
∈
Z
.
Thus there would be
p
,
q
∈
Z
such that
(
pa
,
pb
) = (
1
,
0
)
and
(
qa
,
qb
) = (
0
,
1
)
. The we
have
pa
=
1 and
qb
=
1, so
p
,
q
=
0, and then
pb
=
0 and
qa
=
0 yields
a
=
b
=
0 and
we have a contradiction.
Exercise 2.3.15 on page 60
Prove that
Q
×
Q
is not cyclic.
Z
×
Z
is a subgroup of
Q
×
Q
. From the previous problem
Z
×
Z
is not cyclic, which proves
the result because subgroups of cyclic groups are cyclic.
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 Spring '08
 Staff
 Math, Algebra, Normal subgroup, Subgroup, Cyclic group, Homomorphism

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