Homework #4
Solutions
Due: September 21, 2011
1. Two results concerning groups are (1):
If
H
and
K
are subgroups of
G
, then
H
\
K
is a
subgroup
(Theorem 0.4.4), and (2):
Every subgroup of a cyclic group is cyclic
(Theorem
0.4.5). For each of the following subgroups
H
and
K
of a cyclic group
G
= [
a
], identify
H
\
K
by giving a generator. That is, write
H
\
K
= [
b
] for some
b
2
G
.
(a)
G
=
Z
,
H
= 6
Z
,
K
= 45
Z
.
I
Solution.
H
\
K
=
f
m
2
Z
: 6
j
m
and 45
j
m
g
. Thus
H
\
K
is the set of all
common multiples of 6 and 45. Since
H
\
K
is a subgroup of
Z
it follows, from
the Theorem proven in class, (or see the proof of Theorem 0.4.5, Page 25) that
H
\
K
=
b
Z
where
b
is the smallest positive integer in
H
\
K
. That is,
b
is the
least common multiple of 6 and 45, which is 90. Thus,
H
\
K
= 90
Z
.
J
(b)
G
= [
a
] with
o
(
a
) = 20,
H
= [
a
14
],
K
= [
a
15
].
I
Solution.
Since
o
(
a
14
) = 20
=
gcd(20
;
14) = 10 (Theorem 0.3.4), it follows
that the order of the cyclic group
H
is 10. According to Theorem 0.4.6, there is
a unique subgroup of
G
of order 10, namely [
a
20
=
10
] = [
a
2
]. Thus,
H
= [
a
2
]. In
other words,
a
2
is also a generator of
H
along with
a
14
. (In fact, (
a
14
)
3
=
a
2
.)
Similarly,
o
(
a
15
) = 20
=
gcd(15
;
20) = 20
=
5 = 4. Thus,