Math 113 Homework 3 – Solutions
1.
Fraleigh, section 6, exercise 32 (as always, carefully justify your answers):
Mark each of the following true or false.
a. Every cyclic group is abelian.
True: any cyclic group has a generator
a
, and
a
m
a
n
=
a
m
+
n
=
a
n
a
m
(see also Theorem 6.1 in Fraleigh).
b. Every abelian group is cyclic.
False: the Klein 4-group is abelian but not cyclic.
c.
(
Q
,
+)
is a cyclic group.
False.
Indeed, the subgroup
(
a
)
generated by an element
a
∈
Q
consists of all integer multiples of
a
. When
a
= 0, we get
(
a
)
=
{
0
}
; when
a
negationslash
= 0, the
subgroup
(
a
)
does not contain
a
2
∈
Q
. So in all cases
(
a
)
is a proper subgroup of
Q
, hence
Q
is
not cyclic (it is not generated by one of its elements).
d.
Every element of every cyclic group generates the group.
False, for example
2
∈
Z
4
does not generate
Z
4
. Another example: for
n
≥
2 the element
0 does not generate
Z
n
.
e. There is at least one abelian group of every finite order
>
0
.
True: for any integer
n>
0, (
Z
n
,
+) is an abelian group of order
n
.
f. Every group of order
≤
4
is cyclic.
False: the Klein 4-group is not cyclic.
g.
All generators of
Z
20
are prime numbers.
False:
9 is a generator of
Z
20
, but 9
isn’t prime. Since we have defined
Z
20
to be a set of congruence classes, the statement doesn’t
even quite make sense anyway, because being prime is a property of an integer, while being
a generator of
Z
20
is a property of a congruence class. (For example,
7 =
27
∈
Z
20
, however
the integer 7 is prime while 27 is not). In general you should be careful to distinguish between
integers and their equivalence classes mod
n
whenever it is not clear from the context.
h. If
G
and
G
′
are groups, then
G
∩
G
′
is a group.
False.
G
and
G
′
might not intersect
at all (if their elements had completely different names!); and even if they do intersect,
G
∩
G
′
may not have a well-defined binary operation on it. Namely, the group operations on
G
and
G
′
might not agree on
G
∩
G
′
: for instance how do you intersect (
R
+
,
·
) with (
Z
,
+)? So it doesn’t
even make sense to ask whether
G
∩
G
′
is a group.
i. If
H
and
K
are subgroups of a group
G
, then
H
∩
K
is a group.
True, by section 5
exercise 54 (cf. homework 2).
j. Every cyclic group of order
>
2
has at least two distinct generators.
True. Any
finite cyclic group of order
n >
2 is isomorphic to
Z
n
, where both 1 and
−
1 are generators
(distinct since we assumed that
n>
2; for most
n
there are other generators too). Any infinite
cyclic group is isomorphic to
Z
, which also has two distinct generators, namely 1 and
−
1.
2. Fraleigh, section 6, exercises 44 and 56(a).
# 44: Let
G
be a cyclic group with generator
a
, and let
G
′
be a group isomorphic
to
G
. If
φ
:
G
→
G
′
is an isomorphism, show that, for every
x
∈
G
,
φ
(
x
)
is completely
determined by the value
φ
(
a
)
.
That is, if
φ
:
G
→
G
′
and
ψ
:
G
→
G
′
are two
isomorphisms such that
φ
(
a
) =
ψ
(
a
)
, then
φ
(
x
) =
ψ
(
x
)
for all
x
∈
G
.

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