Homework #5 Solutions
p 83, #16.
In order to find a chain
a
1
≤
a
2
≤ · · · ≤
a
n
of subgroups of
Z
240
with
n
as large as possible, we start at the top with
a
n
= 1 so that
a
n
=
Z
240
. In general, given
a
i
we will choose
a
i

1
to be the largest proper subgroup
of
a
i
. We will make repeated use of the fundamental theorem of cyclic groups which tells
us that a cyclic group of order
m
has a unique subgroup of order
d
for any
d

m
.
The largest proper subgroup of
Z
240
has size 120 and is
2 . Since

2

= 120, the largest
proper subgroup of
2
has size 60 and is
4 . Since

4

= 60, the largest proper subgroup of
4
has size 30 and is
8 . Since

8

= 30, the largest proper subgroup of
8
has order 15
and is
16 . Since

16

= 15, the largest possible subgroup of
16
has order 5 and is
48.
Finally, since

48

= 5 is prime, the only proper subgroup of
48
is
0 . Therefore, we have
produced the maximal chain
0
≤
48
≤
16
≤
8
≤
4
≤
2
≤
1
which has length 7. Notice that the chain
0
≤
120
≤
60
≤
30
≤
15
≤
5
≤
1
also has length 7, but is produced in the opposite way, i.e. by starting with
0
and at each
stage choosing
a
i
+1
as the smallest subgroup containing
a
i
.
p 83, # 20.
Let
x
∈
G
. Since
x
35
=
e
, we know that

x

= 1
,
5
,
7 or 35. Since

G

= 35,
if
G
contains an element
x
of order 35, then
G
=
x
as desired. On the other hand, if
G
contains an element
x
of order 5 and and element
y
of order 7, then, since
G
is abelian
(
xy
)
35
=
x
35
y
35
=
ee
=
e
so that the order
k
of
xy
divides 35. That is,

xy

= 5
,
7 or 35. If

xy

= 5 then
e
= (
xy
)
5
=
x
5
y
5
=
ey
5
=
y
5
which means that 7 =

y

divides 5, a contradiction. Likewise, we have a similar problem if

xy

= 7. It follows that

xy

= 35, and as above that
G
is cyclic.
So, what we need to do is show that
G
must
have an element of order 5 and an element of
order 7. We argue by contradiction. If
G
has no elements of order 5 then every nonidentity
element of
G
has order 7. That is, there are 34 elements in
G
or order 7. However, by the
corollary to Theorem 4.4, the number of elements in
G
of order 7 is divisible by
φ
(7) = 6,
and 34 is
not
divisible by 6. Likewise, if
G
had no element of order 7 then
G
would contain
34 elements of order 5, and this number would have to be divisible by
φ
(5) = 4, which is
also impossible. It follows that
G
must have at least one element of order 5 and at least one
of order 7. As we pointed out above, this forces
G
to be cyclic.
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 Fall '10
 RyanDaileda
 β, Subgroup, τ, Cyclic group

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